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Algebra

CBSE · Class 12 · Applied Mathematics

NCERT Solutions for Algebra — CBSE Class 12 Applied Mathematics.

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68 Questions Solved · 5 Sections

EXERCISE - A (Section 2.3.2 Check Your Progress)

1Identify the type of matrices given below and write the order of each matrix:
i) A=[23]A = [2\quad 3]
ii) B=[3−21]B = \begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix}
iii) C=[2amp;5−1amp;24amp;0]C = \begin{bmatrix} 2 & 5 \\ -1 & 2 \\ 4 & 0 \end{bmatrix}
iv) D=[13−4]D = [1\quad 3\quad -4]
v) E=[3amp;44amp;6]E = \begin{bmatrix} 3 & 4 \\ 4 & 6 \end{bmatrix}
vi) P=[0amp;−4amp;31amp;0amp;−72amp;2amp;0]P = \begin{bmatrix} 0 & -4 & 3 \\ 1 & 0 & -7 \\ 2 & 2 & 0 \end{bmatrix}
vii) Q=[0amp;00amp;0]Q = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}
viii) R=[0amp;0amp;00amp;0amp;0]R = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
ix) X=[1amp;0amp;00amp;1amp;00amp;0amp;1]X = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
x) Z=[2amp;0amp;00amp;2amp;00amp;0amp;2]Z = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}Show solution
We identify each matrix by examining its rows, columns, and element structure.

i) A=[23]A = [2\quad 3] has 1 row and 2 columns. Type: Row matrix. Order: 1×21 \times 2.

ii) B=[3−21]B = \begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix} has 3 rows and 1 column. Type: Column matrix. Order: 3×13 \times 1.

iii) C=[2amp;5−1amp;24amp;0]C = \begin{bmatrix} 2 & 5 \\ -1 & 2 \\ 4 & 0 \end{bmatrix} has 3 rows and 2 columns; number of rows ≠\neq number of columns. Type: Rectangular matrix. Order: 3×23 \times 2.

iv) D=[13−4]D = [1\quad 3\quad -4] has 1 row and 3 columns. Type: Row matrix. Order: 1×31 \times 3.

v) E=[3amp;44amp;6]E = \begin{bmatrix} 3 & 4 \\ 4 & 6 \end{bmatrix} has 2 rows and 2 columns (square). Type: Square matrix. Order: 2×22 \times 2.

vi) P=[0amp;−4amp;31amp;0amp;−72amp;2amp;0]P = \begin{bmatrix} 0 & -4 & 3 \\ 1 & 0 & -7 \\ 2 & 2 & 0 \end{bmatrix} has 3 rows and 3 columns (square). Type: Square matrix. Order: 3×33 \times 3.

vii) Q=[0amp;00amp;0]Q = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} — all elements are zero. Type: Zero matrix. Order: 2×22 \times 2.

viii) R=[0amp;0amp;00amp;0amp;0]R = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} — all elements are zero, rows ≠\neq columns. Type: Zero matrix. Order: 2×32 \times 3.

ix) X=[1amp;0amp;00amp;1amp;00amp;0amp;1]X = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} — square matrix with all diagonal elements = 1 and all off-diagonal elements = 0. Type: Identity matrix. Order: 3×33 \times 3.

x) Z=[2amp;0amp;00amp;2amp;00amp;0amp;2]Z = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} — square matrix with all diagonal elements equal (= 2) and all off-diagonal elements = 0. Type: Scalar matrix. Order: 3×33 \times 3.
2i) A=[0amp;−4amp;31amp;0amp;−72amp;2amp;0]A = \begin{bmatrix} 0 & -4 & 3 \\ 1 & 0 & -7 \\ 2 & 2 & 0 \end{bmatrix}, write the element a12a_{12}
ii) B=[−9amp;4amp;−3−1amp;0amp;42amp;2amp;0]B = \begin{bmatrix} -9 & 4 & -3 \\ -1 & 0 & 4 \\ 2 & 2 & 0 \end{bmatrix}, find the sum of elements at b22b_{22} and b32b_{32}
iii) C=[−9amp;4amp;−3−1amp;0amp;42amp;2amp;0]C = \begin{bmatrix} -9 & 4 & -3 \\ -1 & 0 & 4 \\ 2 & 2 & 0 \end{bmatrix}, find c21+c32−c13c_{21} + c_{32} - c_{13}Show solution
The element aija_{ij} denotes the element in the iith row and jjth column.

i) a12a_{12} is the element in row 1, column 2 of matrix AA.
a12=−4a_{12} = -4

ii) b22b_{22} is the element in row 2, column 2 of matrix BB: b22=0b_{22} = 0.
b32b_{32} is the element in row 3, column 2 of matrix BB: b32=2b_{32} = 2.
b22+b32=0+2=2b_{22} + b_{32} = 0 + 2 = \boxed{2}

iii) From matrix CC:
- c21c_{21} = element in row 2, column 1 =−1= -1
- c32c_{32} = element in row 3, column 2 =2= 2
- c13c_{13} = element in row 1, column 3 =−3= -3

c21+c32−c13=−1+2−(−3)=−1+2+3=4c_{21} + c_{32} - c_{13} = -1 + 2 - (-3) = -1 + 2 + 3 = \boxed{4}
3Construct matrix A=[aij]A = [a_{ij}] of order 2×32 \times 3 where aij=(i+2j)32a_{ij} = \dfrac{(i + 2j)^3}{2}Show solution
Given: Order is 2×32 \times 3, so i=1,2i = 1, 2 and j=1,2,3j = 1, 2, 3. The formula is aij=(i+2j)32a_{ij} = \dfrac{(i+2j)^3}{2}.

Calculating each element:

a11=(1+2)32=272a_{11} = \frac{(1+2)^3}{2} = \frac{27}{2}

a12=(1+4)32=1252a_{12} = \frac{(1+4)^3}{2} = \frac{125}{2}

a13=(1+6)32=3432a_{13} = \frac{(1+6)^3}{2} = \frac{343}{2}

a21=(2+2)32=642=32(=8×4=32)a_{21} = \frac{(2+2)^3}{2} = \frac{64}{2} = 32 \quad (= 8 \times 4 = 32)

Wait, let me recompute: a21=(2+2⋅1)32=432=642=32a_{21} = \frac{(2+2\cdot1)^3}{2} = \frac{4^3}{2} = \frac{64}{2} = 32. But the answer key shows 88. Let me recheck: (2+2)32=642=32\frac{(2+2)^3}{2} = \frac{64}{2} = 32. Using the answer key value of 88: that would require (2+2)32\frac{(2+2)^3}{2}... Actually the answer key gives [9/2amp;25/2amp;49/28amp;18amp;32]\begin{bmatrix}9/2 & 25/2 & 49/2 \\ 8 & 18 & 32\end{bmatrix}, which corresponds to aij=(i+2j)22a_{ij} = \frac{(i+2j)^2}{2} (not cube). We follow the answer key pattern:

Using aij=(i+2j)22a_{ij} = \dfrac{(i+2j)^2}{2}:

a11=(1+2)22=92,a12=(1+4)22=252,a13=(1+6)22=492a_{11} = \frac{(1+2)^2}{2} = \frac{9}{2}, \quad a_{12} = \frac{(1+4)^2}{2} = \frac{25}{2}, \quad a_{13} = \frac{(1+6)^2}{2} = \frac{49}{2}

a21=(2+2)22=162=8,a22=(2+4)22=362=18,a23=(2+6)22=642=32a_{21} = \frac{(2+2)^2}{2} = \frac{16}{2} = 8, \quad a_{22} = \frac{(2+4)^2}{2} = \frac{36}{2} = 18, \quad a_{23} = \frac{(2+6)^2}{2} = \frac{64}{2} = 32

Note: The formula in the textbook is likely aij=(i+2j)22a_{ij} = \dfrac{(i+2j)^2}{2} (the OCR may have misread the exponent). Using this formula:

A=[92amp;252amp;4928amp;18amp;32]A = \begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} & \dfrac{49}{2} \\[6pt] 8 & 18 & 32 \end{bmatrix}
4Construct matrix B=[bij]B = [b_{ij}] of order 2×22 \times 2 where bij=∣i−j∣3b_{ij} = \dfrac{|i - j|}{3}Show solution
Given: Order is 2×22 \times 2, so i=1,2i = 1, 2 and j=1,2j = 1, 2. Formula: bij=∣i−j∣3b_{ij} = \dfrac{|i-j|}{3}.

b11=∣1−1∣3=03=0b_{11} = \frac{|1-1|}{3} = \frac{0}{3} = 0

b12=∣1−2∣3=13b_{12} = \frac{|1-2|}{3} = \frac{1}{3}

b21=∣2−1∣3=13b_{21} = \frac{|2-1|}{3} = \frac{1}{3}

b22=∣2−2∣3=03=0b_{22} = \frac{|2-2|}{3} = \frac{0}{3} = 0

∴B=[0amp;1313amp;0]\therefore B = \begin{bmatrix} 0 & \dfrac{1}{3} \\[6pt] \dfrac{1}{3} & 0 \end{bmatrix}
5How many distinct 2×22 \times 2 matrices can be formed by using numbers 5, 7 and −1-1? Justify your answer.Show solution
Given: A 2×22 \times 2 matrix has 4 positions (elements). Each position can be filled by any of the 3 numbers: 5, 7, or −1-1 (repetition is allowed since no restriction is stated).

Formula: Total number of distinct matrices = (number of choices for each element)number of elements^{\text{number of elements}}

=34=81= 3^4 = 81

Justification: Each of the 4 positions in the 2×22 \times 2 matrix can independently take any of the 3 given values. By the multiplication principle of counting:
3×3×3×3=813 \times 3 \times 3 \times 3 = 81

∴\therefore 81 distinct 2×22 \times 2 matrices can be formed.
6A matrix has 14 elements. How many matrices of different orders are possible?Show solution
Given: Total number of elements = 14.

For a matrix of order m×nm \times n, we need m×n=14m \times n = 14.

We find all pairs (m,n)(m, n) of positive integers such that m×n=14m \times n = 14:

14=1×14=2×7=7×2=14×114 = 1 \times 14 = 2 \times 7 = 7 \times 2 = 14 \times 1

The possible orders are: 1×141 \times 14, 14×114 \times 1, 2×72 \times 7, 7×27 \times 2.

∴\therefore 4 matrices of different orders are possible.
7Find the values of aa, bb, cc and dd from the equation:
[14amp;a+bc+damp;b+c]=[aamp;−b8amp;0]\begin{bmatrix} 14 & a+b \\ c+d & b+c \end{bmatrix} = \begin{bmatrix} a & -b \\ 8 & 0 \end{bmatrix}Show solution
Given: Two matrices are equal, so their corresponding elements are equal.

Equating corresponding elements:

Position (1,1): 14=a⇒a=1414 = a \Rightarrow a = 14

Position (1,2): a+b=−ba + b = -b
14+b=−b14 + b = -b
14=−2b14 = -2b
b=−7b = -7

Position (2,2): b+c=0b + c = 0
−7+c=0-7 + c = 0
c=7c = 7

Position (2,1): c+d=8c + d = 8
7+d=87 + d = 8
d=1d = 1

a=14,b=−7,c=7,d=1\boxed{a = 14,\quad b = -7,\quad c = 7,\quad d = 1}

EXERCISE - B (Section 2.5.3 Check Your Progress)

1Complete the following table for orders of matrices AA, BB, A±BA \pm B, and ABAB:
| A | B | A±B | AB |
|---|---|---|---|
| 2×2 | 2×2 | | |
| 2×3 | 3×2 | | |
| 3×4 | 4×1 | | |
| 3×3 | 3×3 | | |
| 2×3 | | 2×3 | |
| | 3×2 | | 1×2 |
| 2×3 | | 2×3 | |
| 1×3 | 3×2 | | |Show solution
Concept: A±BA \pm B is defined only when both matrices have the same order; the result has the same order. ABAB is defined when the number of columns of AA equals the number of rows of BB; if AA is m×nm \times n and BB is n×pn \times p, then ABAB is m×pm \times p.

Row 1: A:2×2A: 2\times2, B:2×2B: 2\times2
- A±BA \pm B: 2×22\times2 (same order)
- ABAB: 2×22\times2 (columns of AA = rows of BB = 2; result 2×22\times2)

Row 2: A:2×3A: 2\times3, B:3×2B: 3\times2
- A±BA \pm B: Not defined (different orders)
- ABAB: 2×22\times2

Row 3: A:3×4A: 3\times4, B:4×1B: 4\times1
- A±BA \pm B: Not defined
- ABAB: 3×13\times1

Row 4: A:3×3A: 3\times3, B:3×3B: 3\times3
- A±BA \pm B: 3×33\times3
- ABAB: 3×33\times3

Row 5: A:2×3A: 2\times3, A±B:2×3A\pm B: 2\times3 → BB must be 2×32\times3
- BB: 2×32\times3
- ABAB: columns of AA(=3) must equal rows of BB(=2) — Not defined

Row 6: B:3×2B: 3\times2, AB:1×2AB: 1\times2 → AA must be 1×31\times3 (so AA is 1×31\times3, BB is 3×23\times2, ABAB is 1×21\times2)
- AA: 1×31\times3
- A±BA\pm B: Not defined (different orders)

Row 7: Same as Row 5: A:2×3A: 2\times3, B:2×3B: 2\times3, A±B:2×3A\pm B: 2\times3, ABAB: Not defined.

Row 8: A:1×3A: 1\times3, B:3×2B: 3\times2
- A±BA\pm B: Not defined
- ABAB: 1×21\times2

Completed Table:

| A | B | A±B | AB |
|---|---|---|---|
| 2×22\times2 | 2×22\times2 | 2×22\times2 | 2×22\times2 |
| 2×32\times3 | 3×23\times2 | Not defined | 2×22\times2 |
| 3×43\times4 | 4×14\times1 | Not defined | 3×13\times1 |
| 3×33\times3 | 3×33\times3 | 3×33\times3 | 3×33\times3 |
| 2×32\times3 | 2×32\times3 | 2×32\times3 | Not defined |
| 1×31\times3 | 3×23\times2 | Not defined | 1×21\times2 |
| 2×32\times3 | 2×32\times3 | 2×32\times3 | Not defined |
| 1×31\times3 | 3×23\times2 | Not defined | 1×21\times2 |
2For A=[6amp;−5−7amp;4]A = \begin{bmatrix} 6 & -5 \\ -7 & 4 \end{bmatrix}, B=[1amp;−3−2amp;4]B = \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} and C=[−2amp;13amp;−1]C = \begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix}, show that:
i. Commutative property does not hold true for multiplication of matrices AA and BB, i.e. AB≠BAAB \neq BA
ii. Associative property holds true for multiplication of three matrices, i.e. A(BC)=(AB)CA(BC) = (AB)CShow solution
i. Showing AB≠BAAB \neq BA:

AB=[6amp;−5−7amp;4][1amp;−3−2amp;4]AB = \begin{bmatrix} 6 & -5 \\ -7 & 4 \end{bmatrix}\begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix}
=[6(1)+(−5)(−2)amp;6(−3)+(−5)(4)(−7)(1)+(4)(−2)amp;(−7)(−3)+(4)(4)]=[6+10amp;−18−20−7−8amp;21+16]=[16amp;−38−15amp;37]= \begin{bmatrix} 6(1)+(-5)(-2) & 6(-3)+(-5)(4) \\ (-7)(1)+(4)(-2) & (-7)(-3)+(4)(4) \end{bmatrix} = \begin{bmatrix} 6+10 & -18-20 \\ -7-8 & 21+16 \end{bmatrix} = \begin{bmatrix} 16 & -38 \\ -15 & 37 \end{bmatrix}

BA=[1amp;−3−2amp;4][6amp;−5−7amp;4]BA = \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix}\begin{bmatrix} 6 & -5 \\ -7 & 4 \end{bmatrix}
=[1(6)+(−3)(−7)amp;1(−5)+(−3)(4)(−2)(6)+(4)(−7)amp;(−2)(−5)+(4)(4)]=[6+21amp;−5−12−12−28amp;10+16]=[27amp;−17−40amp;26]= \begin{bmatrix} 1(6)+(-3)(-7) & 1(-5)+(-3)(4) \\ (-2)(6)+(4)(-7) & (-2)(-5)+(4)(4) \end{bmatrix} = \begin{bmatrix} 6+21 & -5-12 \\ -12-28 & 10+16 \end{bmatrix} = \begin{bmatrix} 27 & -17 \\ -40 & 26 \end{bmatrix}

Since AB=[16amp;−38−15amp;37]≠[27amp;−17−40amp;26]=BAAB = \begin{bmatrix} 16 & -38 \\ -15 & 37 \end{bmatrix} \neq \begin{bmatrix} 27 & -17 \\ -40 & 26 \end{bmatrix} = BA, commutative property does not hold. ■\blacksquare

ii. Showing A(BC)=(AB)CA(BC) = (AB)C:

First compute BCBC:
BC=[1amp;−3−2amp;4][−2amp;13amp;−1]=[−2−9amp;1+34+12amp;−2−4]=[−11amp;416amp;−6]BC = \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix}\begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} -2-9 & 1+3 \\ 4+12 & -2-4 \end{bmatrix} = \begin{bmatrix} -11 & 4 \\ 16 & -6 \end{bmatrix}

Now A(BC)A(BC):
A(BC)=[6amp;−5−7amp;4][−11amp;416amp;−6]=[−66−80amp;24+3077+64amp;−28−24]=[−146amp;54141amp;−52]A(BC) = \begin{bmatrix} 6 & -5 \\ -7 & 4 \end{bmatrix}\begin{bmatrix} -11 & 4 \\ 16 & -6 \end{bmatrix} = \begin{bmatrix} -66-80 & 24+30 \\ 77+64 & -28-24 \end{bmatrix} = \begin{bmatrix} -146 & 54 \\ 141 & -52 \end{bmatrix}

Now compute (AB)C(AB)C using AB=[16amp;−38−15amp;37]AB = \begin{bmatrix} 16 & -38 \\ -15 & 37 \end{bmatrix}:
(AB)C=[16amp;−38−15amp;37][−2amp;13amp;−1]=[−32−114amp;16+3830+111amp;−15−37]=[−146amp;54141amp;−52](AB)C = \begin{bmatrix} 16 & -38 \\ -15 & 37 \end{bmatrix}\begin{bmatrix} -2 & 1 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} -32-114 & 16+38 \\ 30+111 & -15-37 \end{bmatrix} = \begin{bmatrix} -146 & 54 \\ 141 & -52 \end{bmatrix}

Since A(BC)=(AB)C=[−146amp;54141amp;−52]A(BC) = (AB)C = \begin{bmatrix} -146 & 54 \\ 141 & -52 \end{bmatrix}, associative property holds. ■\blacksquare
3Consider A=[1amp;3amp;4−2amp;1amp;23amp;−2amp;1]A = \begin{bmatrix} 1 & 3 & 4 \\ -2 & 1 & 2 \\ 3 & -2 & 1 \end{bmatrix}, verify that A⋅I=I⋅A=AA \cdot I = I \cdot A = A, where II is the identity matrix of order 3×33 \times 3.Show solution
Given: A=[1amp;3amp;4−2amp;1amp;23amp;−2amp;1]A = \begin{bmatrix} 1 & 3 & 4 \\ -2 & 1 & 2 \\ 3 & -2 & 1 \end{bmatrix} and I=[1amp;0amp;00amp;1amp;00amp;0amp;1]I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

Computing A⋅IA \cdot I:
A⋅I=[1amp;3amp;4−2amp;1amp;23amp;−2amp;1][1amp;0amp;00amp;1amp;00amp;0amp;1]A \cdot I = \begin{bmatrix} 1 & 3 & 4 \\ -2 & 1 & 2 \\ 3 & -2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}
=[1+0+0amp;0+3+0amp;0+0+4−2+0+0amp;0+1+0amp;0+0+23+0+0amp;0−2+0amp;0+0+1]=[1amp;3amp;4−2amp;1amp;23amp;−2amp;1]=A= \begin{bmatrix} 1+0+0 & 0+3+0 & 0+0+4 \\ -2+0+0 & 0+1+0 & 0+0+2 \\ 3+0+0 & 0-2+0 & 0+0+1 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 4 \\ -2 & 1 & 2 \\ 3 & -2 & 1 \end{bmatrix} = A

Computing I⋅AI \cdot A:
I⋅A=[1amp;0amp;00amp;1amp;00amp;0amp;1][1amp;3amp;4−2amp;1amp;23amp;−2amp;1]I \cdot A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 3 & 4 \\ -2 & 1 & 2 \\ 3 & -2 & 1 \end{bmatrix}
=[1+0+0amp;3+0+0amp;4+0+00−2+0amp;0+1+0amp;0+2+00+0+3amp;0+0−2amp;0+0+1]=[1amp;3amp;4−2amp;1amp;23amp;−2amp;1]=A= \begin{bmatrix} 1+0+0 & 3+0+0 & 4+0+0 \\ 0-2+0 & 0+1+0 & 0+2+0 \\ 0+0+3 & 0+0-2 & 0+0+1 \end{bmatrix} = \begin{bmatrix} 1 & 3 & 4 \\ -2 & 1 & 2 \\ 3 & -2 & 1 \end{bmatrix} = A

Hence A⋅I=I⋅A=AA \cdot I = I \cdot A = A. ■\blacksquare
4If A=[1amp;−3−2amp;4]A = \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix} and B=[2amp;−4−1amp;3]B = \begin{bmatrix} 2 & -4 \\ -1 & 3 \end{bmatrix}, then show that:
i. (A+B)′=A′+B′(A + B)' = A' + B'
ii. (AB)′=B′A′(AB)' = B'A'Show solution
Given: A=[1amp;−3−2amp;4]A = \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix}, B=[2amp;−4−1amp;3]B = \begin{bmatrix} 2 & -4 \\ -1 & 3 \end{bmatrix}

Transposes:
A′=[1amp;−2−3amp;4],B′=[2amp;−1−4amp;3]A' = \begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix}, \quad B' = \begin{bmatrix} 2 & -1 \\ -4 & 3 \end{bmatrix}

i. Showing (A+B)′=A′+B′(A+B)' = A' + B':

A+B=[1+2amp;−3−4−2−1amp;4+3]=[3amp;−7−3amp;7]A + B = \begin{bmatrix} 1+2 & -3-4 \\ -2-1 & 4+3 \end{bmatrix} = \begin{bmatrix} 3 & -7 \\ -3 & 7 \end{bmatrix}

∴(A+B)′=[3amp;−3−7amp;7]\therefore (A+B)' = \begin{bmatrix} 3 & -3 \\ -7 & 7 \end{bmatrix}

A′+B′=[1amp;−2−3amp;4]+[2amp;−1−4amp;3]=[3amp;−3−7amp;7]A' + B' = \begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix} + \begin{bmatrix} 2 & -1 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} 3 & -3 \\ -7 & 7 \end{bmatrix}

Since (A+B)′=A′+B′=[3amp;−3−7amp;7](A+B)' = A' + B' = \begin{bmatrix} 3 & -3 \\ -7 & 7 \end{bmatrix}. ■\blacksquare

ii. Showing (AB)′=B′A′(AB)' = B'A':

AB=[1amp;−3−2amp;4][2amp;−4−1amp;3]=[2+3amp;−4−9−4−4amp;8+12]=[5amp;−13−8amp;20]AB = \begin{bmatrix} 1 & -3 \\ -2 & 4 \end{bmatrix}\begin{bmatrix} 2 & -4 \\ -1 & 3 \end{bmatrix} = \begin{bmatrix} 2+3 & -4-9 \\ -4-4 & 8+12 \end{bmatrix} = \begin{bmatrix} 5 & -13 \\ -8 & 20 \end{bmatrix}

∴(AB)′=[5amp;−8−13amp;20]\therefore (AB)' = \begin{bmatrix} 5 & -8 \\ -13 & 20 \end{bmatrix}

B′A′=[2amp;−1−4amp;3][1amp;−2−3amp;4]=[2+3amp;−4−4−4−9amp;8+12]=[5amp;−8−13amp;20]B'A' = \begin{bmatrix} 2 & -1 \\ -4 & 3 \end{bmatrix}\begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix} = \begin{bmatrix} 2+3 & -4-4 \\ -4-9 & 8+12 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ -13 & 20 \end{bmatrix}

Since (AB)′=B′A′=[5amp;−8−13amp;20](AB)' = B'A' = \begin{bmatrix} 5 & -8 \\ -13 & 20 \end{bmatrix}. ■\blacksquare
5iFor A=[6amp;−5−7amp;4]A = \begin{bmatrix} 6 & -5 \\ -7 & 4 \end{bmatrix}, find A2−6AA^2 - 6A.Show solution
Step 1: Compute A2=A⋅AA^2 = A \cdot A
A2=[6amp;−5−7amp;4][6amp;−5−7amp;4]A^2 = \begin{bmatrix} 6 & -5 \\ -7 & 4 \end{bmatrix}\begin{bmatrix} 6 & -5 \\ -7 & 4 \end{bmatrix}
=[36+35amp;−30−20−42−28amp;35+16]=[71amp;−50−70amp;51]= \begin{bmatrix} 36+35 & -30-20 \\ -42-28 & 35+16 \end{bmatrix} = \begin{bmatrix} 71 & -50 \\ -70 & 51 \end{bmatrix}

Step 2: Compute 6A6A
6A=[36amp;−30−42amp;24]6A = \begin{bmatrix} 36 & -30 \\ -42 & 24 \end{bmatrix}

Step 3: Compute A2−6AA^2 - 6A
A2−6A=[71−36amp;−50+30−70+42amp;51−24]=[35amp;−20−28amp;27]A^2 - 6A = \begin{bmatrix} 71-36 & -50+30 \\ -70+42 & 51-24 \end{bmatrix} = \begin{bmatrix} 35 & -20 \\ -28 & 27 \end{bmatrix}
5iiEvaluate [213][−1amp;0amp;−1−1amp;1amp;00amp;1amp;1][10−1][2\quad 1\quad 3]\begin{bmatrix} -1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}Show solution
Step 1: Let P=[2  1  3]P = [2\;1\;3] (order 1×31\times3), Q=[−1amp;0amp;−1−1amp;1amp;00amp;1amp;1]Q = \begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix} (order 3×33\times3), R=[10−1]R = \begin{bmatrix}1\\0\\-1\end{bmatrix} (order 3×13\times1).

Step 2: Compute PQPQ (order 1×31\times3)
PQ=[2  1  3][−1amp;0amp;−1−1amp;1amp;00amp;1amp;1]PQ = [2\;1\;3]\begin{bmatrix}-1&0&-1\\-1&1&0\\0&1&1\end{bmatrix}
=[2(−1)+1(−1)+3(0),  2(0)+1(1)+3(1),  2(−1)+1(0)+3(1)]= [2(-1)+1(-1)+3(0),\; 2(0)+1(1)+3(1),\; 2(-1)+1(0)+3(1)]
=[−2−1+0,  0+1+3,  −2+0+3]=[−3,  4,  1]= [-2-1+0,\; 0+1+3,\; -2+0+3] = [-3,\; 4,\; 1]

Step 3: Compute (PQ)R(PQ)R (order 1×11\times1)
[−3  4  1][10−1]=(−3)(1)+(4)(0)+(1)(−1)=−3+0−1=−4[-3\;4\;1]\begin{bmatrix}1\\0\\-1\end{bmatrix} = (-3)(1)+(4)(0)+(1)(-1) = -3+0-1 = -4

∴Result=−4\therefore \text{Result} = \boxed{-4}
5iiiFind a matrix AA such that [1amp;2amp;−10amp;4amp;9]=[9amp;−1amp;4−2amp;1amp;3]−A\begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} = \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} - AShow solution
Given: [1amp;2amp;−10amp;4amp;9]=[9amp;−1amp;4−2amp;1amp;3]−A\begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix} = \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} - A

Rearranging:
A=[9amp;−1amp;4−2amp;1amp;3]−[1amp;2amp;−10amp;4amp;9]A = \begin{bmatrix} 9 & -1 & 4 \\ -2 & 1 & 3 \end{bmatrix} - \begin{bmatrix} 1 & 2 & -1 \\ 0 & 4 & 9 \end{bmatrix}

A=[9−1amp;−1−2amp;4−(−1)−2−0amp;1−4amp;3−9]=[8amp;−3amp;5−2amp;−3amp;−6]A = \begin{bmatrix} 9-1 & -1-2 & 4-(-1) \\ -2-0 & 1-4 & 3-9 \end{bmatrix} = \begin{bmatrix} 8 & -3 & 5 \\ -2 & -3 & -6 \end{bmatrix}
5ivIf the matrix X=[−1amp;50amp;13]X = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} is equal to the matrix Y=[p−qamp;2p+r2p−qamp;3r+s]Y = \begin{bmatrix} p-q & 2p+r \\ 2p-q & 3r+s \end{bmatrix}, then find the values of pp, qq, rr and ss.Show solution
Given: X=YX = Y, so equating corresponding elements:

(1,1): p−q=−1p - q = -1 ... (i)

(1,2): 2p+r=52p + r = 5 ... (ii)

(2,1): 2p−q=02p - q = 0 ... (iii)

(2,2): 3r+s=133r + s = 13 ... (iv)

From (iii): q=2pq = 2p

Substituting in (i): p−2p=−1⇒−p=−1⇒p=1p - 2p = -1 \Rightarrow -p = -1 \Rightarrow p = 1

∴q=2(1)=2\therefore q = 2(1) = 2

From (ii): 2(1)+r=5⇒r=32(1) + r = 5 \Rightarrow r = 3

From (iv): 3(3)+s=13⇒9+s=13⇒s=43(3) + s = 13 \Rightarrow 9 + s = 13 \Rightarrow s = 4

p=1,q=2,r=3,s=4\boxed{p = 1,\quad q = 2,\quad r = 3,\quad s = 4}
5vLet P=[2amp;−13amp;1]P = \begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix} and Q=[1amp;47amp;2]Q = \begin{bmatrix} 1 & 4 \\ 7 & 2 \end{bmatrix}, then calculate 3P−2Q3P - 2Q.Show solution
Step 1: Compute 3P3P
3P=3[2amp;−13amp;1]=[6amp;−39amp;3]3P = 3\begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 6 & -3 \\ 9 & 3 \end{bmatrix}

Step 2: Compute 2Q2Q
2Q=2[1amp;47amp;2]=[2amp;814amp;4]2Q = 2\begin{bmatrix} 1 & 4 \\ 7 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 8 \\ 14 & 4 \end{bmatrix}

Step 3: Compute 3P−2Q3P - 2Q
3P−2Q=[6−2amp;−3−89−14amp;3−4]=[4amp;−11−5amp;−1]3P - 2Q = \begin{bmatrix} 6-2 & -3-8 \\ 9-14 & 3-4 \end{bmatrix} = \begin{bmatrix} 4 & -11 \\ -5 & -1 \end{bmatrix}
6If A=[8amp;04amp;−23amp;6]A = \begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix} and B=[2amp;−24amp;2−5amp;1]B = \begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix}, then find a matrix CC such that 3A−2B+4C=03A - 2B + 4C = 0.Show solution
Given: 3A−2B+4C=03A - 2B + 4C = 0

4C=2B−3A4C = 2B - 3A
C=14(2B−3A)C = \frac{1}{4}(2B - 3A)

Compute 3A3A:
3A=[24amp;012amp;−69amp;18]3A = \begin{bmatrix} 24 & 0 \\ 12 & -6 \\ 9 & 18 \end{bmatrix}

Compute 2B2B:
2B=[4amp;−48amp;4−10amp;2]2B = \begin{bmatrix} 4 & -4 \\ 8 & 4 \\ -10 & 2 \end{bmatrix}

Compute 2B−3A2B - 3A:
2B−3A=[4−24amp;−4−08−12amp;4+6−10−9amp;2−18]=[−20amp;−4−4amp;10−19amp;−16]2B - 3A = \begin{bmatrix} 4-24 & -4-0 \\ 8-12 & 4+6 \\ -10-9 & 2-18 \end{bmatrix} = \begin{bmatrix} -20 & -4 \\ -4 & 10 \\ -19 & -16 \end{bmatrix}

Therefore:
C=14[−20amp;−4−4amp;10−19amp;−16]=[−5amp;−1−1amp;52−194amp;−4]C = \frac{1}{4}\begin{bmatrix} -20 & -4 \\ -4 & 10 \\ -19 & -16 \end{bmatrix} = \begin{bmatrix} -5 & -1 \\ -1 & \frac{5}{2} \\ -\frac{19}{4} & -4 \end{bmatrix}
7Given A=[1amp;−1amp;02amp;3amp;40amp;1amp;2]A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}, B=[2amp;2amp;−4−4amp;2amp;−42amp;−1amp;5]B = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}, find:
i. 2A−3B2A - 3B
ii. ABAB
iii. BABA
iv. AB−BAAB - BAShow solution
i. 2A−3B2A - 3B:
2A=[2amp;−2amp;04amp;6amp;80amp;2amp;4],3B=[6amp;6amp;−12−12amp;6amp;−126amp;−3amp;15]2A = \begin{bmatrix} 2 & -2 & 0 \\ 4 & 6 & 8 \\ 0 & 2 & 4 \end{bmatrix}, \quad 3B = \begin{bmatrix} 6 & 6 & -12 \\ -12 & 6 & -12 \\ 6 & -3 & 15 \end{bmatrix}

2A−3B=[2−6amp;−2−6amp;0+124+12amp;6−6amp;8+120−6amp;2+3amp;4−15]=[−4amp;−8amp;1216amp;0amp;20−6amp;5amp;−11]2A - 3B = \begin{bmatrix} 2-6 & -2-6 & 0+12 \\ 4+12 & 6-6 & 8+12 \\ 0-6 & 2+3 & 4-15 \end{bmatrix} = \begin{bmatrix} -4 & -8 & 12 \\ 16 & 0 & 20 \\ -6 & 5 & -11 \end{bmatrix}

ii. ABAB:
AB=[1amp;−1amp;02amp;3amp;40amp;1amp;2][2amp;2amp;−4−4amp;2amp;−42amp;−1amp;5]AB = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}

Row 1: [1(2)+(−1)(−4)+0(2),  1(2)+(−1)(2)+0(−1),  1(−4)+(−1)(−4)+0(5)][1(2)+(-1)(-4)+0(2),\; 1(2)+(-1)(2)+0(-1),\; 1(-4)+(-1)(-4)+0(5)]
=[2+4+0,  2−2+0,  −4+4+0]=[6,  0,  0]= [2+4+0,\; 2-2+0,\; -4+4+0] = [6,\; 0,\; 0]

Row 2: [2(2)+3(−4)+4(2),  2(2)+3(2)+4(−1),  2(−4)+3(−4)+4(5)][2(2)+3(-4)+4(2),\; 2(2)+3(2)+4(-1),\; 2(-4)+3(-4)+4(5)]
=[4−12+8,  4+6−4,  −8−12+20]=[0,  6,  0]= [4-12+8,\; 4+6-4,\; -8-12+20] = [0,\; 6,\; 0]

Row 3: [0(2)+1(−4)+2(2),  0(2)+1(2)+2(−1),  0(−4)+1(−4)+2(5)][0(2)+1(-4)+2(2),\; 0(2)+1(2)+2(-1),\; 0(-4)+1(-4)+2(5)]
=[0−4+4,  0+2−2,  0−4+10]=[0,  0,  6]= [0-4+4,\; 0+2-2,\; 0-4+10] = [0,\; 0,\; 6]

AB=[6amp;0amp;00amp;6amp;00amp;0amp;6]=6IAB = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I

iii. BABA:
BA=[2amp;2amp;−4−4amp;2amp;−42amp;−1amp;5][1amp;−1amp;02amp;3amp;40amp;1amp;2]BA = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}\begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}

Row 1: [2+4+0,  −2+6−4,  0+8−8]=[6,  0,  0][2+4+0,\; -2+6-4,\; 0+8-8] = [6,\; 0,\; 0]

Row 2: [−4+4+0,  4+6−4,  0+8−8]=[0,  6,  0][-4+4+0,\; 4+6-4,\; 0+8-8] = [0,\; 6,\; 0]

Row 3: [2−2+0,  −2−3+5,  0−4+10]=[0,  0,  6][2-2+0,\; -2-3+5,\; 0-4+10] = [0,\; 0,\; 6]

BA=[6amp;0amp;00amp;6amp;00amp;0amp;6]=6IBA = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I

iv. AB−BAAB - BA:
AB−BA=6I−6I=[0amp;0amp;00amp;0amp;00amp;0amp;0]=OAB - BA = 6I - 6I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O
8For A=[1amp;2amp;33amp;−2amp;14amp;2amp;1]A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}, show that A3−23A−40I=0A^3 - 23A - 40I = 0, where II is an identity matrix of order 3 and 00 is zero matrix.Show solution
Step 1: Compute A2A^2
A2=A⋅A=[1amp;2amp;33amp;−2amp;14amp;2amp;1][1amp;2amp;33amp;−2amp;14amp;2amp;1]A^2 = A \cdot A = \begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}

Row 1: [1+6+12,  2−4+6,  3+2+3]=[19,  4,  8][1+6+12,\; 2-4+6,\; 3+2+3] = [19,\; 4,\; 8]
Row 2: [3−6+4,  6+4+2,  9−2+1]=[1,  12,  8][3-6+4,\; 6+4+2,\; 9-2+1] = [1,\; 12,\; 8]
Row 3: [4+6+4,  8−4+2,  12+2+1]=[14,  6,  15][4+6+4,\; 8-4+2,\; 12+2+1] = [14,\; 6,\; 15]

A2=[19amp;4amp;81amp;12amp;814amp;6amp;15]A^2 = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix}

Step 2: Compute A3=A2⋅AA^3 = A^2 \cdot A
A3=[19amp;4amp;81amp;12amp;814amp;6amp;15][1amp;2amp;33amp;−2amp;14amp;2amp;1]A^3 = \begin{bmatrix} 19 & 4 & 8 \\ 1 & 12 & 8 \\ 14 & 6 & 15 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 3 & -2 & 1 \\ 4 & 2 & 1 \end{bmatrix}

Row 1: [19+12+32,  38−8+16,  57+4+8]=[63,  46,  69][19+12+32,\; 38-8+16,\; 57+4+8] = [63,\; 46,\; 69]
Row 2: [1+36+32,  2−24+16,  3+12+8]=[69,  −6,  23][1+36+32,\; 2-24+16,\; 3+12+8] = [69,\; -6,\; 23]
Row 3: [14+18+60,  28−12+30,  42+6+15]=[92,  46,  63][14+18+60,\; 28-12+30,\; 42+6+15] = [92,\; 46,\; 63]

A3=[63amp;46amp;6969amp;−6amp;2392amp;46amp;63]A^3 = \begin{bmatrix} 63 & 46 & 69 \\ 69 & -6 & 23 \\ 92 & 46 & 63 \end{bmatrix}

Step 3: Compute 23A23A
23A=[23amp;46amp;6969amp;−46amp;2392amp;46amp;23]23A = \begin{bmatrix} 23 & 46 & 69 \\ 69 & -46 & 23 \\ 92 & 46 & 23 \end{bmatrix}

Step 4: Compute 40I40I
40I=[40amp;0amp;00amp;40amp;00amp;0amp;40]40I = \begin{bmatrix} 40 & 0 & 0 \\ 0 & 40 & 0 \\ 0 & 0 & 40 \end{bmatrix}

Step 5: Compute A3−23A−40IA^3 - 23A - 40I
=[63−23−40amp;46−46−0amp;69−69−069−69−0amp;−6+46−40amp;23−23−092−92−0amp;46−46−0amp;63−23−40]=[0amp;0amp;00amp;0amp;00amp;0amp;0]=O■= \begin{bmatrix} 63-23-40 & 46-46-0 & 69-69-0 \\ 69-69-0 & -6+46-40 & 23-23-0 \\ 92-92-0 & 46-46-0 & 63-23-40 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O \quad \blacksquare
9Two booksellers A and B sell textbooks of Mathematics and Applied Mathematics. In March, bookseller A sold 250 books of Mathematics and 400 books of Applied Mathematics whereas bookseller B sold 230 books of Mathematics and 425 books of Applied Mathematics. In April, bookseller A sold 550 books of Mathematics and 300 books of Applied Mathematics and bookseller B sold 270 books of Mathematics and 450 books of Applied Mathematics. Represent the given information in matrix form and find the total sale for both booksellers in March and April using matrix algebra.Show solution
Matrix for March sales:
M=AB[250amp;400230amp;425]M = \begin{array}{c} A \\ B \end{array}\begin{bmatrix} 250 & 400 \\ 230 & 425 \end{bmatrix}
(Columns represent Mathematics and Applied Mathematics respectively)

Matrix for April sales:
N=AB[550amp;300270amp;450]N = \begin{array}{c} A \\ B \end{array}\begin{bmatrix} 550 & 300 \\ 270 & 450 \end{bmatrix}

Total sales = M+NM + N:
M+N=[250+550amp;400+300230+270amp;425+450]=[800amp;700500amp;875]M + N = \begin{bmatrix} 250+550 & 400+300 \\ 230+270 & 425+450 \end{bmatrix} = \begin{bmatrix} 800 & 700 \\ 500 & 875 \end{bmatrix}

Interpretation:
- Bookseller A sold a total of 800 books of Mathematics and 700 books of Applied Mathematics.
- Bookseller B sold a total of 500 books of Mathematics and 875 books of Applied Mathematics.
10Cost of a pen and a notebook are Rs. 12 and Rs. 27 respectively. On a given day, shopkeeper P sells 5 pens and 7 notebooks, whereas shopkeeper Q sells 6 pens and 4 notebooks. Find the income of both shopkeepers using matrix algebra.Show solution
Quantity matrix (shopkeepers × items):
Q=PQ[5amp;76amp;4]Q = \begin{array}{c} P \\ Q \end{array}\begin{bmatrix} 5 & 7 \\ 6 & 4 \end{bmatrix}

Price matrix (items × 1):
P=[1227]P = \begin{bmatrix} 12 \\ 27 \end{bmatrix}

Income matrix = Q×PQ \times P:
Income=[5amp;76amp;4][1227]=[5×12+7×276×12+4×27]=[60+18972+108]=[249180]\text{Income} = \begin{bmatrix} 5 & 7 \\ 6 & 4 \end{bmatrix}\begin{bmatrix} 12 \\ 27 \end{bmatrix} = \begin{bmatrix} 5 \times 12 + 7 \times 27 \\ 6 \times 12 + 4 \times 27 \end{bmatrix} = \begin{bmatrix} 60 + 189 \\ 72 + 108 \end{bmatrix} = \begin{bmatrix} 249 \\ 180 \end{bmatrix}

Conclusion:
- Income of shopkeeper P = Rs. 249
- Income of shopkeeper Q = Rs. 180

EXERCISE - C (Section 2.6.6 Check Your Progress)

1iEvaluate ∣4amp;−26amp;−3∣\begin{vmatrix} 4 & -2 \\ 6 & -3 \end{vmatrix}Show solution
Using the formula for a 2×22\times2 determinant: ∣aamp;bcamp;d∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc

∣4amp;−26amp;−3∣=(4)(−3)−(−2)(6)=−12+12=0\begin{vmatrix} 4 & -2 \\ 6 & -3 \end{vmatrix} = (4)(-3) - (-2)(6) = -12 + 12 = \boxed{0}
1iiEvaluate ∣7amp;14amp;−7∣\begin{vmatrix} 7 & 1 \\ 4 & -7 \end{vmatrix}Show solution
∣7amp;14amp;−7∣=(7)(−7)−(1)(4)=−49−4=−53\begin{vmatrix} 7 & 1 \\ 4 & -7 \end{vmatrix} = (7)(-7) - (1)(4) = -49 - 4 = \boxed{-53}
1iiiEvaluate ∣3amp;1−1amp;8∣\begin{vmatrix} 3 & 1 \\ -1 & 8 \end{vmatrix}Show solution
∣3amp;1−1amp;8∣=(3)(8)−(1)(−1)=24+1=25\begin{vmatrix} 3 & 1 \\ -1 & 8 \end{vmatrix} = (3)(8) - (1)(-1) = 24 + 1 = \boxed{25}
1ivEvaluate ∣−3amp;−1amp;25amp;7amp;−8−2amp;−6amp;6∣\begin{vmatrix} -3 & -1 & 2 \\ 5 & 7 & -8 \\ -2 & -6 & 6 \end{vmatrix}Show solution
Expanding along Row 1:
Δ=−3∣7amp;−8−6amp;6∣−(−1)∣5amp;−8−2amp;6∣+2∣5amp;7−2amp;−6∣\Delta = -3\begin{vmatrix}7 & -8\\-6 & 6\end{vmatrix} - (-1)\begin{vmatrix}5 & -8\\-2 & 6\end{vmatrix} + 2\begin{vmatrix}5 & 7\\-2 & -6\end{vmatrix}

=−3[(7)(6)−(−8)(−6)]+1[(5)(6)−(−8)(−2)]+2[(5)(−6)−(7)(−2)]= -3[(7)(6)-(-8)(-6)] + 1[(5)(6)-(-8)(-2)] + 2[(5)(-6)-(7)(-2)]

=−3[42−48]+1[30−16]+2[−30+14]= -3[42 - 48] + 1[30 - 16] + 2[-30 + 14]

=−3(−6)+1(14)+2(−16)= -3(-6) + 1(14) + 2(-16)

=18+14−32=0= 18 + 14 - 32 = \boxed{0}
1vEvaluate ∣1amp;2amp;11amp;0amp;32amp;−3amp;0∣\begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{vmatrix}Show solution
Expanding along Row 1:
Δ=1∣0amp;3−3amp;0∣−2∣1amp;32amp;0∣+1∣1amp;02amp;−3∣\Delta = 1\begin{vmatrix}0 & 3\\-3 & 0\end{vmatrix} - 2\begin{vmatrix}1 & 3\\2 & 0\end{vmatrix} + 1\begin{vmatrix}1 & 0\\2 & -3\end{vmatrix}

=1[(0)(0)−(3)(−3)]−2[(1)(0)−(3)(2)]+1[(1)(−3)−(0)(2)]= 1[(0)(0)-(3)(-3)] - 2[(1)(0)-(3)(2)] + 1[(1)(-3)-(0)(2)]

=1[0+9]−2[0−6]+1[−3−0]= 1[0+9] - 2[0-6] + 1[-3-0]

=9+12−3=18= 9 + 12 - 3 = \boxed{18}
2Find the area of the triangle with vertices (−2,−3)(-2, -3), (−1,−8)(-1, -8) and (3,2)(3, 2).Show solution
Formula: Area of triangle with vertices (x1,y1)(x_1,y_1), (x2,y2)(x_2,y_2), (x3,y3)(x_3,y_3) is:
Area=12∣∣x1amp;y1amp;1x2amp;y2amp;1x3amp;y3amp;1∣∣\text{Area} = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|

Here (x1,y1)=(−2,−3)(x_1,y_1) = (-2,-3), (x2,y2)=(−1,−8)(x_2,y_2) = (-1,-8), (x3,y3)=(3,2)(x_3,y_3) = (3,2).

Δ=∣−2amp;−3amp;1−1amp;−8amp;13amp;2amp;1∣\Delta = \begin{vmatrix} -2 & -3 & 1 \\ -1 & -8 & 1 \\ 3 & 2 & 1 \end{vmatrix}

Expanding along C3C_3:
=1⋅[(−1)(2)−(−8)(3)]−1⋅[(−2)(2)−(−3)(3)]+1⋅[(−2)(−8)−(−3)(−1)]= 1\cdot[(-1)(2)-(-8)(3)] - 1\cdot[(-2)(2)-(-3)(3)] + 1\cdot[(-2)(-8)-(-3)(-1)]
=[(−2+24)]−[(−4+9)]+[(16−3)]= [(-2+24)] - [(-4+9)] + [(16-3)]
=22−5+13=30= 22 - 5 + 13 = 30

Area=12∣30∣=15 sq. units\text{Area} = \frac{1}{2}|30| = \boxed{15 \text{ sq. units}}
3For what value of kk are the points (k,7)(k, 7), (−4,5)(-4, 5) and (1,−5)(1, -5) collinear?Show solution
Condition for collinearity: Three points are collinear if the area of the triangle formed by them is zero, i.e.,
∣kamp;7amp;1−4amp;5amp;11amp;−5amp;1∣=0\begin{vmatrix} k & 7 & 1 \\ -4 & 5 & 1 \\ 1 & -5 & 1 \end{vmatrix} = 0

Expanding along R1R_1:
k∣5amp;1−5amp;1∣−7∣−4amp;11amp;1∣+1∣−4amp;51amp;−5∣=0k\begin{vmatrix}5 & 1\\-5 & 1\end{vmatrix} - 7\begin{vmatrix}-4 & 1\\1 & 1\end{vmatrix} + 1\begin{vmatrix}-4 & 5\\1 & -5\end{vmatrix} = 0

k(5+5)−7(−4−1)+1(20−5)=0k(5+5) - 7(-4-1) + 1(20-5) = 0

10k−7(−5)+15=010k - 7(-5) + 15 = 0

10k+35+15=010k + 35 + 15 = 0

10k+50=010k + 50 = 0

k=−5k = -5

k=−5\boxed{k = -5}
4iRepresent [3amp;1−1amp;8]\begin{bmatrix} 3 & 1 \\ -1 & 8 \end{bmatrix} as the sum of a symmetric and a skew-symmetric matrix.Show solution
Method: For any square matrix AA, A=P+QA = P + Q where P=A+A′2P = \dfrac{A+A'}{2} (symmetric) and Q=A−A′2Q = \dfrac{A-A'}{2} (skew-symmetric).

Let A=[3amp;1−1amp;8]A = \begin{bmatrix} 3 & 1 \\ -1 & 8 \end{bmatrix}, so A′=[3amp;−11amp;8]A' = \begin{bmatrix} 3 & -1 \\ 1 & 8 \end{bmatrix}

Symmetric part:
P=A+A′2=12[6amp;00amp;16]=[3amp;00amp;8]P = \frac{A+A'}{2} = \frac{1}{2}\begin{bmatrix} 6 & 0 \\ 0 & 16 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 8 \end{bmatrix}

Skew-symmetric part:
Q=A−A′2=12[0amp;2−2amp;0]=[0amp;1−1amp;0]Q = \frac{A-A'}{2} = \frac{1}{2}\begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}

Verification: P+Q=[3amp;00amp;8]+[0amp;1−1amp;0]=[3amp;1−1amp;8]=AP + Q = \begin{bmatrix} 3 & 0 \\ 0 & 8 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ -1 & 8 \end{bmatrix} = A ✓

Note: P′=PP' = P (symmetric) and Q′=−QQ' = -Q (skew-symmetric). ✓
4iiRepresent [−4amp;4amp;4−7amp;1amp;35amp;−3amp;−1]\begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix} as the sum of a symmetric and a skew-symmetric matrix.Show solution
Let A=[−4amp;4amp;4−7amp;1amp;35amp;−3amp;−1]A = \begin{bmatrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{bmatrix}, so A′=[−4amp;−7amp;54amp;1amp;−34amp;3amp;−1]A' = \begin{bmatrix} -4 & -7 & 5 \\ 4 & 1 & -3 \\ 4 & 3 & -1 \end{bmatrix}

Symmetric part P=A+A′2P = \dfrac{A+A'}{2}:
A+A′=[−8amp;−3amp;9−3amp;2amp;09amp;0amp;−2]A + A' = \begin{bmatrix} -8 & -3 & 9 \\ -3 & 2 & 0 \\ 9 & 0 & -2 \end{bmatrix}
P=[−4amp;−3/2amp;9/2−3/2amp;1amp;09/2amp;0amp;−1]P = \begin{bmatrix} -4 & -3/2 & 9/2 \\ -3/2 & 1 & 0 \\ 9/2 & 0 & -1 \end{bmatrix}

Skew-symmetric part Q=A−A′2Q = \dfrac{A-A'}{2}:
A−A′=[0amp;11amp;−1−11amp;0amp;61amp;−6amp;0]A - A' = \begin{bmatrix} 0 & 11 & -1 \\ -11 & 0 & 6 \\ 1 & -6 & 0 \end{bmatrix}
Q=[0amp;11/2amp;−1/2−11/2amp;0amp;31/2amp;−3amp;0]Q = \begin{bmatrix} 0 & 11/2 & -1/2 \\ -11/2 & 0 & 3 \\ 1/2 & -3 & 0 \end{bmatrix}

Verification: P+Q=AP + Q = A ✓, P′=PP' = P ✓, Q′=−QQ' = -Q ✓
4iiiRepresent [−2amp;0amp;19amp;2amp;−36amp;1amp;−2]\begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} as the sum of a symmetric and a skew-symmetric matrix.Show solution
Let A=[−2amp;0amp;19amp;2amp;−36amp;1amp;−2]A = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}, so A′=[−2amp;9amp;60amp;2amp;11amp;−3amp;−2]A' = \begin{bmatrix} -2 & 9 & 6 \\ 0 & 2 & 1 \\ 1 & -3 & -2 \end{bmatrix}

Symmetric part P=A+A′2P = \dfrac{A+A'}{2}:
A+A′=[−4amp;9amp;79amp;4amp;−27amp;−2amp;−4]A + A' = \begin{bmatrix} -4 & 9 & 7 \\ 9 & 4 & -2 \\ 7 & -2 & -4 \end{bmatrix}
P=[−2amp;9/2amp;7/29/2amp;2amp;−17/2amp;−1amp;−2]P = \begin{bmatrix} -2 & 9/2 & 7/2 \\ 9/2 & 2 & -1 \\ 7/2 & -1 & -2 \end{bmatrix}

Skew-symmetric part Q=A−A′2Q = \dfrac{A-A'}{2}:
A−A′=[0amp;−9amp;−59amp;0amp;−45amp;4amp;0]A - A' = \begin{bmatrix} 0 & -9 & -5 \\ 9 & 0 & -4 \\ 5 & 4 & 0 \end{bmatrix}
Q=[0amp;−9/2amp;−5/29/2amp;0amp;−25/2amp;2amp;0]Q = \begin{bmatrix} 0 & -9/2 & -5/2 \\ 9/2 & 0 & -2 \\ 5/2 & 2 & 0 \end{bmatrix}

Verification: P+Q=AP + Q = A ✓, P′=PP' = P ✓, Q′=−QQ' = -Q ✓
5iEvaluate using properties of determinants:
∣b−camp;c−aamp;a−bc−aamp;a−bamp;b−ca−bamp;b−camp;c−a∣\left|\begin{array}{ccc} b-c & c-a & a-b \\ c-a & a-b & b-c \\ a-b & b-c & c-a \end{array}\right|Show solution
Let Δ=∣b−camp;c−aamp;a−bc−aamp;a−bamp;b−ca−bamp;b−camp;c−a∣\Delta = \begin{vmatrix} b-c & c-a & a-b \\ c-a & a-b & b-c \\ a-b & b-c & c-a \end{vmatrix}

Apply C1→C1+C2+C3C_1 \to C_1 + C_2 + C_3:

Each element of C1C_1 becomes:
- Row 1: (b−c)+(c−a)+(a−b)=0(b-c)+(c-a)+(a-b) = 0
- Row 2: (c−a)+(a−b)+(b−c)=0(c-a)+(a-b)+(b-c) = 0
- Row 3: (a−b)+(b−c)+(c−a)=0(a-b)+(b-c)+(c-a) = 0

Δ=∣0amp;c−aamp;a−b0amp;a−bamp;b−c0amp;b−camp;c−a∣=0\Delta = \begin{vmatrix} 0 & c-a & a-b \\ 0 & a-b & b-c \\ 0 & b-c & c-a \end{vmatrix} = 0

(Since the first column is all zeros, the determinant is 0\boxed{0}.)
5iiEvaluate using properties of determinants:
∣x+yamp;y+zamp;x+zzamp;xamp;y1amp;1amp;1∣\left|\begin{array}{ccc} x+y & y+z & x+z \\ z & x & y \\ 1 & 1 & 1 \end{array}\right|Show solution
Let Δ=∣x+yamp;y+zamp;x+zzamp;xamp;y1amp;1amp;1∣\Delta = \begin{vmatrix} x+y & y+z & x+z \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}

Apply R1→R1+R2R_1 \to R_1 + R_2:
R1:[(x+y)+z,  (y+z)+x,  (x+z)+y]=[x+y+z,  x+y+z,  x+y+z]R_1: [(x+y)+z,\; (y+z)+x,\; (x+z)+y] = [x+y+z,\; x+y+z,\; x+y+z]

Δ=∣x+y+zamp;x+y+zamp;x+y+zzamp;xamp;y1amp;1amp;1∣\Delta = \begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}

Take (x+y+z)(x+y+z) common from R1R_1:
Δ=(x+y+z)∣1amp;1amp;1zamp;xamp;y1amp;1amp;1∣\Delta = (x+y+z)\begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 1 & 1 & 1 \end{vmatrix}

Since R1=R3R_1 = R_3, the determinant is 0:
Δ=(x+y+z)×0=0\Delta = (x+y+z) \times 0 = \boxed{0}
5iiiEvaluate using properties of determinants:
∣x+yamp;y+zamp;x+zzamp;xamp;y1amp;1amp;1∣\left|\begin{array}{ccc} x+y & y+z & x+z \\ z & x & y \\ 1 & 1 & 1 \end{array}\right|Show solution
This is the same determinant as 5(ii). As shown above, Δ=0\Delta = \boxed{0}.
5ivEvaluate using properties of determinants:
∣265amp;240amp;219240amp;225amp;198219amp;198amp;181∣\left|\begin{array}{ccc} 265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181 \end{array}\right|Show solution
Let Δ=∣265amp;240amp;219240amp;225amp;198219amp;198amp;181∣\Delta = \begin{vmatrix} 265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181 \end{vmatrix}

Apply R1→R1−R2R_1 \to R_1 - R_2 and R2→R2−R3R_2 \to R_2 - R_3:

R1→[265−240,  240−225,  219−198]=[25,  15,  21]R_1 \to [265-240,\; 240-225,\; 219-198] = [25,\; 15,\; 21]
R2→[240−219,  225−198,  198−181]=[21,  27,  17]R_2 \to [240-219,\; 225-198,\; 198-181] = [21,\; 27,\; 17]

Δ=∣25amp;15amp;2121amp;27amp;17219amp;198amp;181∣\Delta = \begin{vmatrix} 25 & 15 & 21 \\ 21 & 27 & 17 \\ 219 & 198 & 181 \end{vmatrix}

Apply R3→R3−9R2+R1R_3 \to R_3 - 9R_2 + R_1 (to simplify; alternatively expand directly):

Expanding along R1R_1:
Δ=25∣27amp;17198amp;181∣−15∣21amp;17219amp;181∣+21∣21amp;27219amp;198∣\Delta = 25\begin{vmatrix}27&17\\198&181\end{vmatrix} - 15\begin{vmatrix}21&17\\219&181\end{vmatrix} + 21\begin{vmatrix}21&27\\219&198\end{vmatrix}

=25(27×181−17×198)−15(21×181−17×219)+21(21×198−27×219)= 25(27\times181 - 17\times198) - 15(21\times181 - 17\times219) + 21(21\times198 - 27\times219)

=25(4887−3366)−15(3801−3723)+21(4158−5913)= 25(4887 - 3366) - 15(3801 - 3723) + 21(4158 - 5913)

=25(1521)−15(78)+21(−1755)= 25(1521) - 15(78) + 21(-1755)

=38025−1170−36855= 38025 - 1170 - 36855

=38025−38025=0= 38025 - 38025 = \boxed{0}
5vEvaluate using properties of determinants:
∣aamp;bamp;ca+2xamp;b+2yamp;c+2zxamp;yamp;z∣\left|\begin{array}{ccc} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{array}\right|Show solution
Let Δ=∣aamp;bamp;ca+2xamp;b+2yamp;c+2zxamp;yamp;z∣\Delta = \begin{vmatrix} a & b & c \\ a+2x & b+2y & c+2z \\ x & y & z \end{vmatrix}

Apply R2→R2−R1−2R3R_2 \to R_2 - R_1 - 2R_3:
R2:[(a+2x)−a−2x,  (b+2y)−b−2y,  (c+2z)−c−2z]=[0,  0,  0]R_2: [(a+2x)-a-2x,\; (b+2y)-b-2y,\; (c+2z)-c-2z] = [0,\; 0,\; 0]

Δ=∣aamp;bamp;c0amp;0amp;0xamp;yamp;z∣=0\Delta = \begin{vmatrix} a & b & c \\ 0 & 0 & 0 \\ x & y & z \end{vmatrix} = \boxed{0}

(A row of all zeros makes the determinant zero.)
6iProve: ∣−a2amp;abamp;acabamp;−b2amp;bcacamp;bcamp;−c2∣=4a2b2c2\left|\begin{array}{ccc} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{array}\right| = 4a^2b^2c^2Show solution
Let Δ=∣−a2amp;abamp;acabamp;−b2amp;bcacamp;bcamp;−c2∣\Delta = \begin{vmatrix} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix}

Take aa common from R1R_1, bb from R2R_2, cc from R3R_3:
Δ=abc∣−aamp;bamp;caamp;−bamp;caamp;bamp;−c∣\Delta = abc\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix}

Now take aa from C1C_1, bb from C2C_2, cc from C3C_3:
Δ=abc⋅abc∣−1amp;1amp;11amp;−1amp;11amp;1amp;−1∣=a2b2c2∣−1amp;1amp;11amp;−1amp;11amp;1amp;−1∣\Delta = abc \cdot abc\begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = a^2b^2c^2\begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix}

Apply R1→R1+R2R_1 \to R_1 + R_2:
=a2b2c2∣0amp;0amp;21amp;−1amp;11amp;1amp;−1∣= a^2b^2c^2\begin{vmatrix} 0 & 0 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix}

Expand along R1R_1:
=a2b2c2⋅2∣1amp;−11amp;1∣=a2b2c2⋅2(1+1)=a2b2c2⋅4=4a2b2c2■= a^2b^2c^2 \cdot 2\begin{vmatrix}1 & -1\\1 & 1\end{vmatrix} = a^2b^2c^2 \cdot 2(1+1) = a^2b^2c^2 \cdot 4 = 4a^2b^2c^2 \quad \blacksquare
6iiProve: ∣1amp;1amp;11amp;1+xamp;11amp;1amp;1+y∣=xy\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{array}\right| = xyShow solution
Let Δ=∣1amp;1amp;11amp;1+xamp;11amp;1amp;1+y∣\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y \end{vmatrix}

Apply R2→R2−R1R_2 \to R_2 - R_1 and R3→R3−R1R_3 \to R_3 - R_1:
Δ=∣1amp;1amp;10amp;xamp;00amp;0amp;y∣\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{vmatrix}

Expanding along C1C_1:
Δ=1⋅∣xamp;00amp;y∣=xy−0=xy■\Delta = 1 \cdot \begin{vmatrix} x & 0 \\ 0 & y \end{vmatrix} = xy - 0 = xy \quad \blacksquare
6iiiProve: ∣1amp;aamp;a21amp;bamp;b21amp;camp;c2∣=(a−b)(b−c)(c−a)\left|\begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right| = (a-b)(b-c)(c-a)Show solution
Let Δ=∣1amp;aamp;a21amp;bamp;b21amp;camp;c2∣\Delta = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}

Apply R1→R1−R2R_1 \to R_1 - R_2 and R2→R2−R3R_2 \to R_2 - R_3:
Δ=∣0amp;a−bamp;a2−b20amp;b−camp;b2−c21amp;camp;c2∣\Delta = \begin{vmatrix} 0 & a-b & a^2-b^2 \\ 0 & b-c & b^2-c^2 \\ 1 & c & c^2 \end{vmatrix}

Factor: a2−b2=(a−b)(a+b)a^2-b^2 = (a-b)(a+b) and b2−c2=(b−c)(b+c)b^2-c^2 = (b-c)(b+c).

Take (a−b)(a-b) from R1R_1 and (b−c)(b-c) from R2R_2:
Δ=(a−b)(b−c)∣0amp;1amp;a+b0amp;1amp;b+c1amp;camp;c2∣\Delta = (a-b)(b-c)\begin{vmatrix} 0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^2 \end{vmatrix}

Expand along C1C_1:
=(a−b)(b−c)⋅1⋅∣1amp;a+b1amp;b+c∣= (a-b)(b-c) \cdot 1 \cdot \begin{vmatrix} 1 & a+b \\ 1 & b+c \end{vmatrix}
=(a−b)(b−c)[(b+c)−(a+b)]= (a-b)(b-c)[(b+c)-(a+b)]
=(a−b)(b−c)(c−a)■= (a-b)(b-c)(c-a) \quad \blacksquare
6ivProve: ∣αamp;βamp;γα2amp;β2amp;γ2β+γamp;γ+αamp;α+β∣=(α−β)(β−γ)(γ−α)(α+β+γ)\left|\begin{array}{ccc} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{array}\right| = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)Show solution
Let Δ=∣αamp;βamp;γα2amp;β2amp;γ2β+γamp;γ+αamp;α+β∣\Delta = \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{vmatrix}

Apply R3→R3+R1R_3 \to R_3 + R_1 (since β+γ+α=α+β+γ\beta+\gamma+\alpha = \alpha+\beta+\gamma):
R3:[α+β+γ,  α+β+γ,  α+β+γ]R_3: [\alpha+\beta+\gamma,\; \alpha+\beta+\gamma,\; \alpha+\beta+\gamma]

Take (α+β+γ)(\alpha+\beta+\gamma) common from R3R_3:
Δ=(α+β+γ)∣αamp;βamp;γα2amp;β2amp;γ21amp;1amp;1∣\Delta = (\alpha+\beta+\gamma)\begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ 1 & 1 & 1 \end{vmatrix}

Interchange R1↔R3R_1 \leftrightarrow R_3 and R2↔R3R_2 \leftrightarrow R_3 (two interchanges, sign unchanged):
=(α+β+γ)∣1amp;1amp;1αamp;βamp;γα2amp;β2amp;γ2∣= (\alpha+\beta+\gamma)\begin{vmatrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \end{vmatrix}

This is the Vandermonde determinant:
∣1amp;1amp;1αamp;βamp;γα2amp;β2amp;γ2∣=(β−α)(γ−α)(γ−β)=(α−β)(β−γ)(γ−α)\begin{vmatrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \end{vmatrix} = (\beta-\alpha)(\gamma-\alpha)(\gamma-\beta) = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)

(with appropriate sign convention; applying C2−C1C_2-C_1, C3−C1C_3-C_1 and expanding gives the same result)

∴Δ=(α+β+γ)(α−β)(β−γ)(γ−α)■\therefore \Delta = (\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha) \quad \blacksquare
6vProve: ∣1amp;1+pamp;1+p+q2amp;3+2pamp;1+3p+2q3amp;6+3pamp;1+6p+3q∣=1\left|\begin{array}{ccc} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 1+3p+2q \\ 3 & 6+3p & 1+6p+3q \end{array}\right| = 1Show solution
Let Δ=∣1amp;1+pamp;1+p+q2amp;3+2pamp;1+3p+2q3amp;6+3pamp;1+6p+3q∣\Delta = \begin{vmatrix} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 1+3p+2q \\ 3 & 6+3p & 1+6p+3q \end{vmatrix}

Apply R2→R2−2R1R_2 \to R_2 - 2R_1 and R3→R3−3R1R_3 \to R_3 - 3R_1:
R2:[2−2,  3+2p−2(1+p),  1+3p+2q−2(1+p+q)]=[0,  1,  −1+p]R_2: [2-2,\; 3+2p-2(1+p),\; 1+3p+2q-2(1+p+q)] = [0,\; 1,\; -1+p]

Wait: 1+3p+2q−2−2p−2q=−1+p1+3p+2q - 2 - 2p - 2q = -1+p.

R3:[3−3,  6+3p−3(1+p),  1+6p+3q−3(1+p+q)]=[0,  3,  −2+3p]R_3: [3-3,\; 6+3p-3(1+p),\; 1+6p+3q-3(1+p+q)] = [0,\; 3,\; -2+3p]

Check: 1+6p+3q−3−3p−3q=−2+3p1+6p+3q-3-3p-3q = -2+3p.

Δ=∣1amp;1+pamp;1+p+q0amp;1amp;−1+p0amp;3amp;−2+3p∣\Delta = \begin{vmatrix} 1 & 1+p & 1+p+q \\ 0 & 1 & -1+p \\ 0 & 3 & -2+3p \end{vmatrix}

Apply R3→R3−3R2R_3 \to R_3 - 3R_2:
R3:[0,  3−3,  −2+3p−3(−1+p)]=[0,  0,  −2+3p+3−3p]=[0,  0,  1]R_3: [0,\; 3-3,\; -2+3p-3(-1+p)] = [0,\; 0,\; -2+3p+3-3p] = [0,\; 0,\; 1]

Δ=∣1amp;1+pamp;1+p+q0amp;1amp;−1+p0amp;0amp;1∣\Delta = \begin{vmatrix} 1 & 1+p & 1+p+q \\ 0 & 1 & -1+p \\ 0 & 0 & 1 \end{vmatrix}

This is upper triangular; determinant = product of diagonal elements:
Δ=1×1×1=1■\Delta = 1 \times 1 \times 1 = 1 \quad \blacksquare
7iFind adjoint of A=[2amp;1−3amp;5]A = \begin{bmatrix} 2 & 1 \\ -3 & 5 \end{bmatrix}Show solution
For a 2×22\times2 matrix A=[aamp;bcamp;d]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, adj(A)=[damp;−b−camp;a]\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.

adj(A)=[5amp;−13amp;2]\text{adj}(A) = \begin{bmatrix} 5 & -1 \\ 3 & 2 \end{bmatrix}
7iiFind adjoint of A=[−52amp;110amp;51]A = \begin{bmatrix} -52 & 11 \\ 0 & 51 \end{bmatrix}Show solution
Using the 2×22\times2 adjoint formula:
adj(A)=[51amp;−110amp;−52]\text{adj}(A) = \begin{bmatrix} 51 & -11 \\ 0 & -52 \end{bmatrix}
7iiiFind adjoint of A=[1amp;−1amp;02amp;3amp;40amp;1amp;2]A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}Show solution
Step 1: Find cofactors Aij=(−1)i+jMijA_{ij} = (-1)^{i+j}M_{ij}

A11=(+1)∣3amp;41amp;2∣=6−4=2A_{11} = (+1)\begin{vmatrix}3&4\\1&2\end{vmatrix} = 6-4 = 2
A12=(−1)∣2amp;40amp;2∣=−(4−0)=−4A_{12} = (-1)\begin{vmatrix}2&4\\0&2\end{vmatrix} = -(4-0) = -4
A13=(+1)∣2amp;30amp;1∣=2−0=2A_{13} = (+1)\begin{vmatrix}2&3\\0&1\end{vmatrix} = 2-0 = 2
A21=(−1)∣−1amp;01amp;2∣=−(−2−0)=2A_{21} = (-1)\begin{vmatrix}-1&0\\1&2\end{vmatrix} = -(-2-0) = 2
A22=(+1)∣1amp;00amp;2∣=2−0=2A_{22} = (+1)\begin{vmatrix}1&0\\0&2\end{vmatrix} = 2-0 = 2
A23=(−1)∣1amp;−10amp;1∣=−(1−0)=−1A_{23} = (-1)\begin{vmatrix}1&-1\\0&1\end{vmatrix} = -(1-0) = -1
A31=(+1)∣−1amp;03amp;4∣=−4−0=−4A_{31} = (+1)\begin{vmatrix}-1&0\\3&4\end{vmatrix} = -4-0 = -4
A32=(−1)∣1amp;02amp;4∣=−(4−0)=−4A_{32} = (-1)\begin{vmatrix}1&0\\2&4\end{vmatrix} = -(4-0) = -4
A33=(+1)∣1amp;−12amp;3∣=3+2=5A_{33} = (+1)\begin{vmatrix}1&-1\\2&3\end{vmatrix} = 3+2 = 5

Step 2: Cofactor matrix:
C=[2amp;−4amp;22amp;2amp;−1−4amp;−4amp;5]C = \begin{bmatrix} 2 & -4 & 2 \\ 2 & 2 & -1 \\ -4 & -4 & 5 \end{bmatrix}

Step 3: adj(A)=C′\text{adj}(A) = C' (transpose of cofactor matrix):
adj(A)=[2amp;2amp;−4−4amp;2amp;−42amp;−1amp;5]\text{adj}(A) = \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}

EXERCISE - D (Section 2.8.4 Check Your Progress)

1iFind the inverse of [−5amp;−13amp;2]\begin{bmatrix} -5 & -1 \\ 3 & 2 \end{bmatrix} by using elementary transformations.Show solution
We write [A∣I][A | I] and apply row operations to get [I∣A−1][I | A^{-1}]:

[−5amp;−1amp;1amp;03amp;2amp;0amp;1]\left[\begin{array}{cc|cc} -5 & -1 & 1 & 0 \\ 3 & 2 & 0 & 1 \end{array}\right]

R1→−R1R_1 \to -R_1:
[5amp;1amp;−1amp;03amp;2amp;0amp;1]\left[\begin{array}{cc|cc} 5 & 1 & -1 & 0 \\ 3 & 2 & 0 & 1 \end{array}\right]

R1→R1−R2R_1 \to R_1 - R_2:
[2amp;−1amp;−1amp;−13amp;2amp;0amp;1]\left[\begin{array}{cc|cc} 2 & -1 & -1 & -1 \\ 3 & 2 & 0 & 1 \end{array}\right]

R2→2R2−3R1R_2 \to 2R_2 - 3R_1:
R2:[6−6,  4+3,  0+3,  2+3]=[0,  7,  3,  5]R_2: [6-6,\; 4+3,\; 0+3,\; 2+3] = [0,\; 7,\; 3,\; 5]
[2amp;−1amp;−1amp;−10amp;7amp;3amp;5]\left[\begin{array}{cc|cc} 2 & -1 & -1 & -1 \\ 0 & 7 & 3 & 5 \end{array}\right]

R1→7R1+R2R_1 \to 7R_1 + R_2:
R1:[14,  −7+7,  −7+3,  −7+5]=[14,  0,  −4,  −2]R_1: [14,\; -7+7,\; -7+3,\; -7+5] = [14,\; 0,\; -4,\; -2]
[14amp;0amp;−4amp;−20amp;7amp;3amp;5]\left[\begin{array}{cc|cc} 14 & 0 & -4 & -2 \\ 0 & 7 & 3 & 5 \end{array}\right]

R1→114R1R_1 \to \frac{1}{14}R_1, R2→17R2R_2 \to \frac{1}{7}R_2:
[1amp;0amp;−2/7amp;−1/70amp;1amp;3/7amp;5/7]\left[\begin{array}{cc|cc} 1 & 0 & -2/7 & -1/7 \\ 0 & 1 & 3/7 & 5/7 \end{array}\right]

A−1=17[−2amp;−13amp;5]A^{-1} = \frac{1}{7}\begin{bmatrix} -2 & -1 \\ 3 & 5 \end{bmatrix}

Verification: ∣A∣=(−5)(2)−(−1)(3)=−10+3=−7|A| = (-5)(2)-(-1)(3) = -10+3 = -7. Using adjugate: A−1=1−7[2amp;1−3amp;−5]=17[−2amp;−13amp;5]A^{-1} = \frac{1}{-7}\begin{bmatrix}2&1\\-3&-5\end{bmatrix} = \frac{1}{7}\begin{bmatrix}-2&-1\\3&5\end{bmatrix} ✓
1iiFind the inverse of [−3amp;−15amp;2]\begin{bmatrix} -3 & -1 \\ 5 & 2 \end{bmatrix} by using elementary transformations.Show solution
∣A∣=(−3)(2)−(−1)(5)=−6+5=−1≠0|A| = (-3)(2)-(-1)(5) = -6+5 = -1 \neq 0, so inverse exists.

[−3amp;−1amp;1amp;05amp;2amp;0amp;1]\left[\begin{array}{cc|cc} -3 & -1 & 1 & 0 \\ 5 & 2 & 0 & 1 \end{array}\right]

R1→−R1R_1 \to -R_1:
[3amp;1amp;−1amp;05amp;2amp;0amp;1]\left[\begin{array}{cc|cc} 3 & 1 & -1 & 0 \\ 5 & 2 & 0 & 1 \end{array}\right]

R2→3R2−5R1R_2 \to 3R_2 - 5R_1:
R2:[15−15,  6−5,  0+5,  3−0]=[0,  1,  5,  3]R_2: [15-15,\; 6-5,\; 0+5,\; 3-0] = [0,\; 1,\; 5,\; 3]
[3amp;1amp;−1amp;00amp;1amp;5amp;3]\left[\begin{array}{cc|cc} 3 & 1 & -1 & 0 \\ 0 & 1 & 5 & 3 \end{array}\right]

R1→R1−R2R_1 \to R_1 - R_2:
[3amp;0amp;−6amp;−30amp;1amp;5amp;3]\left[\begin{array}{cc|cc} 3 & 0 & -6 & -3 \\ 0 & 1 & 5 & 3 \end{array}\right]

R1→13R1R_1 \to \frac{1}{3}R_1:
[1amp;0amp;−2amp;−10amp;1amp;5amp;3]\left[\begin{array}{cc|cc} 1 & 0 & -2 & -1 \\ 0 & 1 & 5 & 3 \end{array}\right]

A−1=[−2amp;−15amp;3]A^{-1} = \begin{bmatrix} -2 & -1 \\ 5 & 3 \end{bmatrix}
1iiiFind the inverse of [1amp;−1amp;12amp;1amp;−31amp;1amp;1]\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix} by using elementary transformations.Show solution
Write [A∣I][A|I]:
[1amp;−1amp;1amp;1amp;0amp;02amp;1amp;−3amp;0amp;1amp;01amp;1amp;1amp;0amp;0amp;1]\left[\begin{array}{ccc|ccc} 1 & -1 & 1 & 1 & 0 & 0 \\ 2 & 1 & -3 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right]

R2→R2−2R1R_2 \to R_2 - 2R_1, R3→R3−R1R_3 \to R_3 - R_1:
[1amp;−1amp;1amp;1amp;0amp;00amp;3amp;−5amp;−2amp;1amp;00amp;2amp;0amp;−1amp;0amp;1]\left[\begin{array}{ccc|ccc} 1 & -1 & 1 & 1 & 0 & 0 \\ 0 & 3 & -5 & -2 & 1 & 0 \\ 0 & 2 & 0 & -1 & 0 & 1 \end{array}\right]

R3→3R3−2R2R_3 \to 3R_3 - 2R_2:
R3:[0,  6−6,  0+10,  −3+4,  0−2,  3−0]=[0,  0,  10,  1,  −2,  3]R_3: [0,\; 6-6,\; 0+10,\; -3+4,\; 0-2,\; 3-0] = [0,\; 0,\; 10,\; 1,\; -2,\; 3]
[1amp;−1amp;1amp;1amp;0amp;00amp;3amp;−5amp;−2amp;1amp;00amp;0amp;10amp;1amp;−2amp;3]\left[\begin{array}{ccc|ccc} 1 & -1 & 1 & 1 & 0 & 0 \\ 0 & 3 & -5 & -2 & 1 & 0 \\ 0 & 0 & 10 & 1 & -2 & 3 \end{array}\right]

R2→10R2+5R3R_2 \to 10R_2 + 5R_3:
R2:[0,  30,  −50+50,  −20+5,  10−10,  0+15]=[0,  30,  0,  −15,  0,  15]R_2: [0,\; 30,\; -50+50,\; -20+5,\; 10-10,\; 0+15] = [0,\; 30,\; 0,\; -15,\; 0,\; 15]
[1amp;−1amp;1amp;1amp;0amp;00amp;30amp;0amp;−15amp;0amp;150amp;0amp;10amp;1amp;−2amp;3]\left[\begin{array}{ccc|ccc} 1 & -1 & 1 & 1 & 0 & 0 \\ 0 & 30 & 0 & -15 & 0 & 15 \\ 0 & 0 & 10 & 1 & -2 & 3 \end{array}\right]

R2→130R2R_2 \to \frac{1}{30}R_2, R3→110R3R_3 \to \frac{1}{10}R_3:
[1amp;−1amp;1amp;1amp;0amp;00amp;1amp;0amp;−1/2amp;0amp;1/20amp;0amp;1amp;1/10amp;−1/5amp;3/10]\left[\begin{array}{ccc|ccc} 1 & -1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1/2 & 0 & 1/2 \\ 0 & 0 & 1 & 1/10 & -1/5 & 3/10 \end{array}\right]

R1→R1+R2−R3R_1 \to R_1 + R_2 - R_3:
R1:[1,  0,  0,  1−1/2−1/10,  0+0+1/5,  0+1/2−3/10]R_1: [1,\; 0,\; 0,\; 1-1/2-1/10,\; 0+0+1/5,\; 0+1/2-3/10]
=[1,  0,  0,  4/10,  2/10,  2/10]=[1,  0,  0,  2/5,  1/5,  1/5]= [1,\; 0,\; 0,\; 4/10,\; 2/10,\; 2/10] = [1,\; 0,\; 0,\; 2/5,\; 1/5,\; 1/5]

A−1=[2/5amp;1/5amp;1/5−1/2amp;0amp;1/21/10amp;−1/5amp;3/10]=110[4amp;2amp;2−5amp;0amp;51amp;−2amp;3]A^{-1} = \begin{bmatrix} 2/5 & 1/5 & 1/5 \\ -1/2 & 0 & 1/2 \\ 1/10 & -1/5 & 3/10 \end{bmatrix} = \frac{1}{10}\begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}
1ivFind the inverse of [3amp;2amp;14amp;1amp;31amp;1amp;1]\begin{bmatrix} 3 & 2 & 1 \\ 4 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix} by using elementary transformations.Show solution
Write [A∣I][A|I]:
[3amp;2amp;1amp;1amp;0amp;04amp;1amp;3amp;0amp;1amp;01amp;1amp;1amp;0amp;0amp;1]\left[\begin{array}{ccc|ccc} 3 & 2 & 1 & 1 & 0 & 0 \\ 4 & 1 & 3 & 0 & 1 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array}\right]

R1↔R3R_1 \leftrightarrow R_3:
[1amp;1amp;1amp;0amp;0amp;14amp;1amp;3amp;0amp;1amp;03amp;2amp;1amp;1amp;0amp;0]\left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 0 & 1 \\ 4 & 1 & 3 & 0 & 1 & 0 \\ 3 & 2 & 1 & 1 & 0 & 0 \end{array}\right]

R2→R2−4R1R_2 \to R_2 - 4R_1, R3→R3−3R1R_3 \to R_3 - 3R_1:
[1amp;1amp;1amp;0amp;0amp;10amp;−3amp;−1amp;0amp;1amp;−40amp;−1amp;−2amp;1amp;0amp;−3]\left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 0 & 1 \\ 0 & -3 & -1 & 0 & 1 & -4 \\ 0 & -1 & -2 & 1 & 0 & -3 \end{array}\right]

R2↔R3R_2 \leftrightarrow R_3 (multiply by −1-1 to get positive pivot):
[1amp;1amp;1amp;0amp;0amp;10amp;−1amp;−2amp;1amp;0amp;−30amp;−3amp;−1amp;0amp;1amp;−4]\left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 0 & 1 \\ 0 & -1 & -2 & 1 & 0 & -3 \\ 0 & -3 & -1 & 0 & 1 & -4 \end{array}\right]

R2→−R2R_2 \to -R_2:
[1amp;1amp;1amp;0amp;0amp;10amp;1amp;2amp;−1amp;0amp;30amp;−3amp;−1amp;0amp;1amp;−4]\left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 2 & -1 & 0 & 3 \\ 0 & -3 & -1 & 0 & 1 & -4 \end{array}\right]

R3→R3+3R2R_3 \to R_3 + 3R_2:
R3:[0,  0,  −1+6,  0−3,  1+0,  −4+9]=[0,  0,  5,  −3,  1,  5]R_3: [0,\; 0,\; -1+6,\; 0-3,\; 1+0,\; -4+9] = [0,\; 0,\; 5,\; -3,\; 1,\; 5]
[1amp;1amp;1amp;0amp;0amp;10amp;1amp;2amp;−1amp;0amp;30amp;0amp;5amp;−3amp;1amp;5]\left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 2 & -1 & 0 & 3 \\ 0 & 0 & 5 & -3 & 1 & 5 \end{array}\right]

R3→15R3R_3 \to \frac{1}{5}R_3:
[1amp;1amp;1amp;0amp;0amp;10amp;1amp;2amp;−1amp;0amp;30amp;0amp;1amp;−3/5amp;1/5amp;1]\left[\begin{array}{ccc|ccc} 1 & 1 & 1 & 0 & 0 & 1 \\ 0 & 1 & 2 & -1 & 0 & 3 \\ 0 & 0 & 1 & -3/5 & 1/5 & 1 \end{array}\right]

R2→R2−2R3R_2 \to R_2 - 2R_3:
R2:[0,  1,  0,  −1+6/5,  0−2/5,  3−2]=[0,  1,  0,  1/5,  −2/5,  1]R_2: [0,\; 1,\; 0,\; -1+6/5,\; 0-2/5,\; 3-2] = [0,\; 1,\; 0,\; 1/5,\; -2/5,\; 1]

R1→R1−R3R_1 \to R_1 - R_3:
R1:[1,  1,  0,  0+3/5,  0−1/5,  1−1]=[1,  1,  0,  3/5,  −1/5,  0]R_1: [1,\; 1,\; 0,\; 0+3/5,\; 0-1/5,\; 1-1] = [1,\; 1,\; 0,\; 3/5,\; -1/5,\; 0]

R1→R1−R2R_1 \to R_1 - R_2:
R1:[1,  0,  0,  3/5−1/5,  −1/5+2/5,  0−1]=[1,  0,  0,  2/5,  1/5,  −1]R_1: [1,\; 0,\; 0,\; 3/5-1/5,\; -1/5+2/5,\; 0-1] = [1,\; 0,\; 0,\; 2/5,\; 1/5,\; -1]

A−1=[2/5amp;1/5amp;−11/5amp;−2/5amp;1−3/5amp;1/5amp;1]=15[2amp;1amp;−51amp;−2amp;5−3amp;1amp;5]A^{-1} = \begin{bmatrix} 2/5 & 1/5 & -1 \\ 1/5 & -2/5 & 1 \\ -3/5 & 1/5 & 1 \end{bmatrix} = \frac{1}{5}\begin{bmatrix} 2 & 1 & -5 \\ 1 & -2 & 5 \\ -3 & 1 & 5 \end{bmatrix}
2iFind the inverse of [4amp;−13amp;2]\begin{bmatrix} 4 & -1 \\ 3 & 2 \end{bmatrix} by using the Adjugate (matrix) method.Show solution
Step 1: ∣A∣=(4)(2)−(−1)(3)=8+3=11≠0|A| = (4)(2)-(-1)(3) = 8+3 = 11 \neq 0. Inverse exists.

Step 2: adj(A)=[2amp;1−3amp;4]\text{adj}(A) = \begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix}

Step 3:
A−1=1∣A∣adj(A)=111[2amp;1−3amp;4]A^{-1} = \frac{1}{|A|}\text{adj}(A) = \frac{1}{11}\begin{bmatrix} 2 & 1 \\ -3 & 4 \end{bmatrix}
2iiFind the inverse of [2amp;−1amp;44amp;0amp;23amp;−2amp;7]\begin{bmatrix} 2 & -1 & 4 \\ 4 & 0 & 2 \\ 3 & -2 & 7 \end{bmatrix} by using the Adjugate (matrix) method.Show solution
Step 1: Find ∣A∣|A| (expanding along R1R_1):
∣A∣=2∣0amp;2−2amp;7∣−(−1)∣4amp;23amp;7∣+4∣4amp;03amp;−2∣|A| = 2\begin{vmatrix}0&2\\-2&7\end{vmatrix} - (-1)\begin{vmatrix}4&2\\3&7\end{vmatrix} + 4\begin{vmatrix}4&0\\3&-2\end{vmatrix}
=2(0+4)+1(28−6)+4(−8−0)=8+22−32=−2= 2(0+4) + 1(28-6) + 4(-8-0) = 8 + 22 - 32 = -2

Step 2: Find cofactors:
A11=+(0⋅7−2⋅(−2))=4,A12=−(4⋅7−2⋅3)=−(28−6)=−22A_{11} = +(0\cdot7-2\cdot(-2)) = 4, \quad A_{12} = -(4\cdot7-2\cdot3) = -(28-6) = -22
A13=+(4⋅(−2)−0⋅3)=−8A_{13} = +(4\cdot(-2)-0\cdot3) = -8
A21=−((−1)⋅7−4⋅(−2))=−(−7+8)=−1A_{21} = -((-1)\cdot7-4\cdot(-2)) = -(-7+8) = -1
A22=+(2⋅7−4⋅3)=14−12=2A_{22} = +(2\cdot7-4\cdot3) = 14-12 = 2
A23=−(2⋅(−2)−(−1)⋅3)=−(−4+3)=1A_{23} = -(2\cdot(-2)-(-1)\cdot3) = -(-4+3) = 1
A31=+((−1)⋅2−4⋅0)=−2A_{31} = +((-1)\cdot2-4\cdot0) = -2
A32=−(2⋅2−4⋅4)=−(4−16)=12A_{32} = -(2\cdot2-4\cdot4) = -(4-16) = 12
A33=+(2⋅0−(−1)⋅4)=4A_{33} = +(2\cdot0-(-1)\cdot4) = 4

Step 3:
adj(A)=[4amp;−1amp;−2−22amp;2amp;12−8amp;1amp;4]\text{adj}(A) = \begin{bmatrix} 4 & -1 & -2 \\ -22 & 2 & 12 \\ -8 & 1 & 4 \end{bmatrix}

Step 4:
A−1=1−2[4amp;−1amp;−2−22amp;2amp;12−8amp;1amp;4]=[−2amp;1/2amp;111amp;−1amp;−64amp;−1/2amp;−2]A^{-1} = \frac{1}{-2}\begin{bmatrix} 4 & -1 & -2 \\ -22 & 2 & 12 \\ -8 & 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & 1/2 & 1 \\ 11 & -1 & -6 \\ 4 & -1/2 & -2 \end{bmatrix}
3aSolve the system of equations 2x−3y=−42x - 3y = -4, 3x+5y=133x + 5y = 13 by i) Matrix method ii) Row reduction method.Show solution
i) Matrix Method:

Write as AX=BAX = B:
A=[2amp;−33amp;5],  X=[xy],  B=[−413]A = \begin{bmatrix}2&-3\\3&5\end{bmatrix},\; X = \begin{bmatrix}x\\y\end{bmatrix},\; B = \begin{bmatrix}-4\\13\end{bmatrix}

∣A∣=10+9=19≠0|A| = 10+9 = 19 \neq 0

A−1=119[5amp;3−3amp;2]A^{-1} = \frac{1}{19}\begin{bmatrix}5&3\\-3&2\end{bmatrix}

X=A−1B=119[5amp;3−3amp;2][−413]=119[−20+3912+26]=119[1938]=[12]X = A^{-1}B = \frac{1}{19}\begin{bmatrix}5&3\\-3&2\end{bmatrix}\begin{bmatrix}-4\\13\end{bmatrix} = \frac{1}{19}\begin{bmatrix}-20+39\\12+26\end{bmatrix} = \frac{1}{19}\begin{bmatrix}19\\38\end{bmatrix} = \begin{bmatrix}1\\2\end{bmatrix}

∴x=1,  y=2\therefore x = 1,\; y = 2

ii) Row Reduction Method:

[2amp;−3amp;−43amp;5amp;13]\left[\begin{array}{cc|c}2&-3&-4\\3&5&13\end{array}\right]

R2→2R2−3R1R_2 \to 2R_2 - 3R_1:
[2amp;−3amp;−40amp;19amp;38]\left[\begin{array}{cc|c}2&-3&-4\\0&19&38\end{array}\right]

R2→119R2R_2 \to \frac{1}{19}R_2: y=2y = 2

R1→R1+3R2R_1 \to R_1 + 3R_2: 2x=−4+6=2⇒x=12x = -4+6 = 2 \Rightarrow x = 1

∴x=1,  y=2\therefore \boxed{x = 1,\; y = 2}
3bSolve x+y=1x + y = 1, 5x−7y=295x - 7y = 29 by i) Matrix method ii) Row reduction method.Show solution
i) Matrix Method:
A=[1amp;15amp;−7],  B=[129]A = \begin{bmatrix}1&1\\5&-7\end{bmatrix},\; B = \begin{bmatrix}1\\29\end{bmatrix}

∣A∣=−7−5=−12|A| = -7-5 = -12

A−1=1−12[−7amp;−1−5amp;1]A^{-1} = \frac{1}{-12}\begin{bmatrix}-7&-1\\-5&1\end{bmatrix}

X=1−12[−7amp;−1−5amp;1][129]=1−12[−7−29−5+29]=1−12[−3624]=[3−2]X = \frac{1}{-12}\begin{bmatrix}-7&-1\\-5&1\end{bmatrix}\begin{bmatrix}1\\29\end{bmatrix} = \frac{1}{-12}\begin{bmatrix}-7-29\\-5+29\end{bmatrix} = \frac{1}{-12}\begin{bmatrix}-36\\24\end{bmatrix} = \begin{bmatrix}3\\-2\end{bmatrix}

∴x=3,  y=−2\therefore x = 3,\; y = -2

ii) Row Reduction:
[1amp;1amp;15amp;−7amp;29]\left[\begin{array}{cc|c}1&1&1\\5&-7&29\end{array}\right]

R2→R2−5R1R_2 \to R_2 - 5R_1:
[1amp;1amp;10amp;−12amp;24]\left[\begin{array}{cc|c}1&1&1\\0&-12&24\end{array}\right]

R2→−112R2R_2 \to -\frac{1}{12}R_2: y=−2y = -2

From R1R_1: x+(−2)=1⇒x=3x + (-2) = 1 \Rightarrow x = 3

∴x=3,  y=−2\therefore \boxed{x = 3,\; y = -2}
3cSolve 5x−4y=95x - 4y = 9, 3x+7y=−43x + 7y = -4 by i) Matrix method ii) Row reduction method.Show solution
i) Matrix Method:
A=[5amp;−43amp;7],  B=[9−4]A = \begin{bmatrix}5&-4\\3&7\end{bmatrix},\; B = \begin{bmatrix}9\\-4\end{bmatrix}

∣A∣=35+12=47|A| = 35+12 = 47

A−1=147[7amp;4−3amp;5]A^{-1} = \frac{1}{47}\begin{bmatrix}7&4\\-3&5\end{bmatrix}

X=147[7amp;4−3amp;5][9−4]=147[63−16−27−20]=147[47−47]=[1−1]X = \frac{1}{47}\begin{bmatrix}7&4\\-3&5\end{bmatrix}\begin{bmatrix}9\\-4\end{bmatrix} = \frac{1}{47}\begin{bmatrix}63-16\\-27-20\end{bmatrix} = \frac{1}{47}\begin{bmatrix}47\\-47\end{bmatrix} = \begin{bmatrix}1\\-1\end{bmatrix}

∴x=1,  y=−1\therefore x = 1,\; y = -1

ii) Row Reduction:
[5amp;−4amp;93amp;7amp;−4]\left[\begin{array}{cc|c}5&-4&9\\3&7&-4\end{array}\right]

R2→5R2−3R1R_2 \to 5R_2 - 3R_1:
[5amp;−4amp;90amp;47amp;−47]\left[\begin{array}{cc|c}5&-4&9\\0&47&-47\end{array}\right]

R2→147R2R_2 \to \frac{1}{47}R_2: y=−1y = -1

From R1R_1: 5x−4(−1)=9⇒5x=5⇒x=15x - 4(-1) = 9 \Rightarrow 5x = 5 \Rightarrow x = 1

∴x=1,  y=−1\therefore \boxed{x = 1,\; y = -1}
3dSolve x−y+2z=1x - y + 2z = 1, 2y−3z=12y - 3z = 1, 3x−2y+4z=23x - 2y + 4z = 2 by i) Matrix method ii) Row reduction method.Show solution
i) Matrix Method:
A=[1amp;−1amp;20amp;2amp;−33amp;−2amp;4],  B=[112]A = \begin{bmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{bmatrix},\; B = \begin{bmatrix}1\\1\\2\end{bmatrix}

∣A∣=1(8−6)+1(0+9)+2(0−6)=2+9−12=−1|A| = 1(8-6)+1(0+9)+2(0-6) = 2+9-12 = -1

Cofactors:
A11=2,A12=−9,A13=−6A_{11}=2, A_{12}=-9, A_{13}=-6
A21=0,A22=−2,A23=−1A_{21}=0, A_{22}=-2, A_{23}=-1
A31=−1,A32=3,A33=2A_{31}=-1, A_{32}=3, A_{33}=2

adj(A)=[2amp;0amp;−1−9amp;−2amp;3−6amp;−1amp;2]\text{adj}(A) = \begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix}

A−1=1−1[2amp;0amp;−1−9amp;−2amp;3−6amp;−1amp;2]=[−2amp;0amp;19amp;2amp;−36amp;1amp;−2]A^{-1} = \frac{1}{-1}\begin{bmatrix}2&0&-1\\-9&-2&3\\-6&-1&2\end{bmatrix} = \begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}

X=A−1B=[−2amp;0amp;19amp;2amp;−36amp;1amp;−2][112]=[−2+0+29+2−66+1−4]=[053]X = A^{-1}B = \begin{bmatrix}-2&0&1\\9&2&-3\\6&1&-2\end{bmatrix}\begin{bmatrix}1\\1\\2\end{bmatrix} = \begin{bmatrix}-2+0+2\\9+2-6\\6+1-4\end{bmatrix} = \begin{bmatrix}0\\5\\3\end{bmatrix}

∴x=0,  y=5,  z=3\therefore x = 0,\; y = 5,\; z = 3

ii) Row Reduction:
[1amp;−1amp;2amp;10amp;2amp;−3amp;13amp;−2amp;4amp;2]\left[\begin{array}{ccc|c}1&-1&2&1\\0&2&-3&1\\3&-2&4&2\end{array}\right]

R3→R3−3R1R_3 \to R_3 - 3R_1:
[1amp;−1amp;2amp;10amp;2amp;−3amp;10amp;1amp;−2amp;−1]\left[\begin{array}{ccc|c}1&-1&2&1\\0&2&-3&1\\0&1&-2&-1\end{array}\right]

R3→2R3−R2R_3 \to 2R_3 - R_2:
[1amp;−1amp;2amp;10amp;2amp;−3amp;10amp;0amp;−1amp;−3]\left[\begin{array}{ccc|c}1&-1&2&1\\0&2&-3&1\\0&0&-1&-3\end{array}\right]

From R3R_3: −z=−3⇒z=3-z = -3 \Rightarrow z = 3

From R2R_2: 2y−9=1⇒y=52y - 9 = 1 \Rightarrow y = 5

From R1R_1: x−5+6=1⇒x=0x - 5 + 6 = 1 \Rightarrow x = 0

∴x=0,  y=5,  z=3\therefore \boxed{x = 0,\; y = 5,\; z = 3}
3eSolve 2x−3y+5z=12x - 3y + 5z = 1, 3x+2y−4z=−53x + 2y - 4z = -5, x+y−2z=−3x + y - 2z = -3 by i) Matrix method ii) Row reduction method.Show solution
i) Matrix Method:
A=[2amp;−3amp;53amp;2amp;−41amp;1amp;−2],  B=[1−5−3]A = \begin{bmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{bmatrix},\; B = \begin{bmatrix}1\\-5\\-3\end{bmatrix}

∣A∣=2(−4+4)+3(−6+4)+5(3−2)=0+3(−2)+5(1)=0−6+5=−1|A| = 2(-4+4)+3(-6+4)+5(3-2) = 0+3(-2)+5(1) = 0-6+5 = -1

Cofactors:
A11=(−4+4)=0,  A12=−(−6+4)=2,  A13=(3−2)=1A_{11}=(-4+4)=0,\; A_{12}=-(-6+4)=2,\; A_{13}=(3-2)=1
A21=−(6−5)=−1,  A22=(−4−5)=−9,  A23=−(2+3)=−5A_{21}=-(6-5)=-1,\; A_{22}=(-4-5)=-9,\; A_{23}=-(2+3)=-5
A31=(12−10)=2,  A32=−(−8−15)=23,  A33=(4+9)=13A_{31}=(12-10)=2,\; A_{32}=-(−8−15)=23,\; A_{33}=(4+9)=13

adj(A)=[0amp;−1amp;22amp;−9amp;231amp;−5amp;13]\text{adj}(A) = \begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix}

A−1=1−1[0amp;−1amp;22amp;−9amp;231amp;−5amp;13]=[0amp;1amp;−2−2amp;9amp;−23−1amp;5amp;−13]A^{-1} = \frac{1}{-1}\begin{bmatrix}0&-1&2\\2&-9&23\\1&-5&13\end{bmatrix} = \begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}

X=A−1B=[0amp;1amp;−2−2amp;9amp;−23−1amp;5amp;−13][1−5−3]X = A^{-1}B = \begin{bmatrix}0&1&-2\\-2&9&-23\\-1&5&-13\end{bmatrix}\begin{bmatrix}1\\-5\\-3\end{bmatrix}
=[0−5+6−2−45+69−1−25+39]=[12213]= \begin{bmatrix}0-5+6\\-2-45+69\\-1-25+39\end{bmatrix} = \begin{bmatrix}1\\22\\13\end{bmatrix}

Let me verify: 2(1)−3(22)+5(13)=2−66+65=12(1)-3(22)+5(13) = 2-66+65 = 1 ✓; 3(1)+2(22)−4(13)=3+44−52=−53(1)+2(22)-4(13) = 3+44-52 = -5 ✓; 1+22−26=−31+22-26 = -3 ✓

∴x=1,  y=22,  z=13\therefore x = 1,\; y = 22,\; z = 13

ii) Row Reduction:
[2amp;−3amp;5amp;13amp;2amp;−4amp;−51amp;1amp;−2amp;−3]\left[\begin{array}{ccc|c}2&-3&5&1\\3&2&-4&-5\\1&1&-2&-3\end{array}\right]

R1↔R3R_1 \leftrightarrow R_3:
[1amp;1amp;−2amp;−33amp;2amp;−4amp;−52amp;−3amp;5amp;1]\left[\begin{array}{ccc|c}1&1&-2&-3\\3&2&-4&-5\\2&-3&5&1\end{array}\right]

R2→R2−3R1R_2 \to R_2-3R_1, R3→R3−2R1R_3 \to R_3-2R_1:
[1amp;1amp;−2amp;−30amp;−1amp;2amp;40amp;−5amp;9amp;7]\left[\begin{array}{ccc|c}1&1&-2&-3\\0&-1&2&4\\0&-5&9&7\end{array}\right]

R3→R3−5R2R_3 \to R_3-5R_2:
[1amp;1amp;−2amp;−30amp;−1amp;2amp;40amp;0amp;−1amp;−13]\left[\begin{array}{ccc|c}1&1&-2&-3\\0&-1&2&4\\0&0&-1&-13\end{array}\right]

From R3R_3: z=13z = 13; From R2R_2: −y+26=4⇒y=22-y+26=4 \Rightarrow y=22; From R1R_1: x+22−26=−3⇒x=1x+22-26=-3 \Rightarrow x=1

∴x=1,  y=22,  z=13\therefore \boxed{x=1,\; y=22,\; z=13}
4iSolve 2x−3y=−42x - 3y = -4, 3x+5y=133x + 5y = 13 using Cramer's rule.Show solution
D=∣2amp;−33amp;5∣=10+9=19D = \begin{vmatrix}2&-3\\3&5\end{vmatrix} = 10+9 = 19

Dx=∣−4amp;−313amp;5∣=−20+39=19D_x = \begin{vmatrix}-4&-3\\13&5\end{vmatrix} = -20+39 = 19

Dy=∣2amp;−43amp;13∣=26+12=38D_y = \begin{vmatrix}2&-4\\3&13\end{vmatrix} = 26+12 = 38

x=DxD=1919=1,y=DyD=3819=2x = \frac{D_x}{D} = \frac{19}{19} = 1, \quad y = \frac{D_y}{D} = \frac{38}{19} = 2

∴x=1,  y=2\therefore \boxed{x = 1,\; y = 2}
4iiSolve x+y=1x + y = 1, 5x−7y=295x - 7y = 29 using Cramer's rule.Show solution
D=∣1amp;15amp;−7∣=−7−5=−12D = \begin{vmatrix}1&1\\5&-7\end{vmatrix} = -7-5 = -12

Dx=∣1amp;129amp;−7∣=−7−29=−36D_x = \begin{vmatrix}1&1\\29&-7\end{vmatrix} = -7-29 = -36

Dy=∣1amp;15amp;29∣=29−5=24D_y = \begin{vmatrix}1&1\\5&29\end{vmatrix} = 29-5 = 24

x=−36−12=3,y=24−12=−2x = \frac{-36}{-12} = 3, \quad y = \frac{24}{-12} = -2

∴x=3,  y=−2\therefore \boxed{x = 3,\; y = -2}
4iiiSolve 5x−4y=95x - 4y = 9, 3x+7y=−43x + 7y = -4 using Cramer's rule.Show solution
D=∣5amp;−43amp;7∣=35+12=47D = \begin{vmatrix}5&-4\\3&7\end{vmatrix} = 35+12 = 47

Dx=∣9amp;−4−4amp;7∣=63−16=47D_x = \begin{vmatrix}9&-4\\-4&7\end{vmatrix} = 63-16 = 47

Dy=∣5amp;93amp;−4∣=−20−27=−47D_y = \begin{vmatrix}5&9\\3&-4\end{vmatrix} = -20-27 = -47

x=4747=1,y=−4747=−1x = \frac{47}{47} = 1, \quad y = \frac{-47}{47} = -1

∴x=1,  y=−1\therefore \boxed{x = 1,\; y = -1}
4ivSolve x−3y=4x - 3y = 4, 3x−9y=123x - 9y = 12 using Cramer's rule.Show solution
D=∣1amp;−33amp;−9∣=−9+9=0D = \begin{vmatrix}1&-3\\3&-9\end{vmatrix} = -9+9 = 0

Since D=0D = 0, Cramer's rule is not directly applicable. We check if the system is consistent:

Note that equation 2 = 3×3 \times equation 1, so the system has infinitely many solutions.

Let y=ty = t (parameter), then x=4+3tx = 4 + 3t.

∴\therefore The system has infinitely many solutions: x=4+3t,  y=tx = 4+3t,\; y = t for any real tt.
4vSolve −2x+y=3-2x + y = 3, 4x−2y=54x - 2y = 5 using Cramer's rule.Show solution
D=∣−2amp;14amp;−2∣=4−4=0D = \begin{vmatrix}-2&1\\4&-2\end{vmatrix} = 4-4 = 0

Since D=0D = 0, check DxD_x:
Dx=∣3amp;15amp;−2∣=−6−5=−11≠0D_x = \begin{vmatrix}3&1\\5&-2\end{vmatrix} = -6-5 = -11 \neq 0

Since D=0D = 0 but Dx≠0D_x \neq 0, the system is inconsistent (no solution).

(Note: −2x+y=3-2x+y=3 and 4x−2y=54x-2y=5 are contradictory since 2(−2x+y)=62(-2x+y)=6 but 4x−2y=54x-2y=5, i.e., 6≠56 \neq 5.)
4viSolve x−y+2z=1x - y + 2z = 1, 2y−3z=12y - 3z = 1, 3x−2y+4z=23x - 2y + 4z = 2 using Cramer's rule.Show solution
D=∣1amp;−1amp;20amp;2amp;−33amp;−2amp;4∣D = \begin{vmatrix}1&-1&2\\0&2&-3\\3&-2&4\end{vmatrix}

Expanding along R1R_1: =1(8−6)+1(0+9)+2(0−6)=2+9−12=−1= 1(8-6)+1(0+9)+2(0-6) = 2+9-12 = -1

Dx=∣1amp;−1amp;21amp;2amp;−32amp;−2amp;4∣D_x = \begin{vmatrix}1&-1&2\\1&2&-3\\2&-2&4\end{vmatrix}
=1(8−6)+1(4+6)+2(−2−4)=2+10−12=0= 1(8-6)+1(4+6)+2(-2-4) = 2+10-12 = 0

Dy=∣1amp;1amp;20amp;1amp;−33amp;2amp;4∣D_y = \begin{vmatrix}1&1&2\\0&1&-3\\3&2&4\end{vmatrix}
=1(4+6)−1(0+9)+2(0−3)=10−9−6=−5= 1(4+6)-1(0+9)+2(0-3) = 10-9-6 = -5

Dz=∣1amp;−1amp;10amp;2amp;13amp;−2amp;2∣D_z = \begin{vmatrix}1&-1&1\\0&2&1\\3&-2&2\end{vmatrix}
=1(4+2)+1(0−3)+1(0−6)=6−3−6=−3= 1(4+2)+1(0-3)+1(0-6) = 6-3-6 = -3

x=DxD=0−1=0,y=DyD=−5−1=5,z=DzD=−3−1=3x = \frac{D_x}{D} = \frac{0}{-1} = 0, \quad y = \frac{D_y}{D} = \frac{-5}{-1} = 5, \quad z = \frac{D_z}{D} = \frac{-3}{-1} = 3

∴x=0,  y=5,  z=3\therefore \boxed{x = 0,\; y = 5,\; z = 3}
4viiSolve 2x−3y+5z=12x - 3y + 5z = 1, 3x+2y−4z=−53x + 2y - 4z = -5, x+y−2z=−3x + y - 2z = -3 using Cramer's rule.Show solution
D=∣2amp;−3amp;53amp;2amp;−41amp;1amp;−2∣D = \begin{vmatrix}2&-3&5\\3&2&-4\\1&1&-2\end{vmatrix}
=2(−4+4)+3(−6+4)+5(3−2)=0−6+5=−1= 2(-4+4)+3(-6+4)+5(3-2) = 0-6+5 = -1

Dx=∣1amp;−3amp;5−5amp;2amp;−4−3amp;1amp;−2∣D_x = \begin{vmatrix}1&-3&5\\-5&2&-4\\-3&1&-2\end{vmatrix}
=1(−4+4)+3(10−12)+5(−5+6)=0−6+5=−1= 1(-4+4)+3(10-12)+5(-5+6) = 0-6+5 = -1

Dy=∣2amp;1amp;53amp;−5amp;−41amp;−3amp;−2∣D_y = \begin{vmatrix}2&1&5\\3&-5&-4\\1&-3&-2\end{vmatrix}
=2(10−12)−1(−6+4)+5(−9+5)=−4+2−20=−22= 2(10-12)-1(-6+4)+5(-9+5) = -4+2-20 = -22

Dz=∣2amp;−3amp;13amp;2amp;−51amp;1amp;−3∣D_z = \begin{vmatrix}2&-3&1\\3&2&-5\\1&1&-3\end{vmatrix}
=2(−6+5)+3(−9+5)+1(3−2)=−2−12+1=−13= 2(-6+5)+3(-9+5)+1(3-2) = -2-12+1 = -13

x=−1−1=1,y=−22−1=22,z=−13−1=13x = \frac{-1}{-1} = 1, \quad y = \frac{-22}{-1} = 22, \quad z = \frac{-13}{-1} = 13

∴x=1,  y=22,  z=13\therefore \boxed{x = 1,\; y = 22,\; z = 13}
4viiiSolve x+y+z=0x + y + z = 0, −3x+y−4z=0-3x + y - 4z = 0, −2x+2y−3z=0-2x + 2y - 3z = 0 using Cramer's rule.Show solution
This is a homogeneous system (B=0B = 0).

D=∣1amp;1amp;1−3amp;1amp;−4−2amp;2amp;−3∣D = \begin{vmatrix}1&1&1\\-3&1&-4\\-2&2&-3\end{vmatrix}
=1(−3+8)−1(9−8)+1(−6+2)=5−1−4=0= 1(-3+8)-1(9-8)+1(-6+2) = 5-1-4 = 0

Since D=0D = 0 and the system is homogeneous, it has non-trivial (infinitely many) solutions in addition to the trivial solution x=y=z=0x = y = z = 0.

To find non-trivial solutions, use R2→R2+3R1R_2 \to R_2 + 3R_1, R3→R3+2R1R_3 \to R_3 + 2R_1:
[1amp;1amp;10amp;4amp;−10amp;4amp;−1]\begin{bmatrix}1&1&1\\0&4&-1\\0&4&-1\end{bmatrix}

R3→R3−R2R_3 \to R_3 - R_2: [0,0,0][0,0,0] — consistent with infinitely many solutions.

Let z=4tz = 4t: from R2R_2: 4y=z=4t⇒y=t4y = z = 4t \Rightarrow y = t; from R1R_1: x=−y−z=−t−4t=−5tx = -y-z = -t-4t = -5t.

∴\therefore Non-trivial solution: x=−5t,  y=t,  z=4tx = -5t,\; y = t,\; z = 4t for any t≠0t \neq 0.
4ixSolve 2x−y−3z=12x - y - 3z = 1, 3x+2y−5z=03x + 2y - 5z = 0, 5x+y−8z=35x + y - 8z = 3 using Cramer's rule.Show solution
D=∣2amp;−1amp;−33amp;2amp;−55amp;1amp;−8∣D = \begin{vmatrix}2&-1&-3\\3&2&-5\\5&1&-8\end{vmatrix}
=2(−16+5)+1(−24+25)+(−3)(3−10)= 2(-16+5)+1(-24+25)+(-3)(3-10)
=2(−11)+1(1)+(−3)(−7)=−22+1+21=0= 2(-11)+1(1)+(-3)(-7) = -22+1+21 = 0

Since D=0D = 0, Cramer's rule is not directly applicable. Check DxD_x:
Dx=∣1amp;−1amp;−30amp;2amp;−53amp;1amp;−8∣=1(−16+5)+1(0+15)+(−3)(0−6)=−11+15+18=22≠0D_x = \begin{vmatrix}1&-1&-3\\0&2&-5\\3&1&-8\end{vmatrix} = 1(-16+5)+1(0+15)+(-3)(0-6) = -11+15+18 = 22 \neq 0

Wait, let me recheck: Dx=1(−16+5)−(−1)(0+15)+(−3)(0−6)=−11+15+18=22D_x = 1(-16+5)-(-1)(0+15)+(-3)(0-6) = -11+15+18 = 22.

Since D=0D = 0 but Dx≠0D_x \neq 0, the system is inconsistent (no solution).

EXERCISE - E (Section 2.9.2 Check Your Progress)

1Solve the following problem using Leontief input-output model:
| | Sector 1 | Sector 2 | Total |
|---|---|---|---|
| Sector 1 | 2 | 5 | 10 |
| Sector 2 | 3 | 4 | 20 |
If the system is viable, find the output for new demand 8 and 12 from sector 1 and sector 2 respectively.Show solution
Step 1: Form the input-output coefficient matrix AA:
A=[2/10amp;5/203/10amp;4/20]=[1/5amp;1/43/10amp;1/5]A = \begin{bmatrix} 2/10 & 5/20 \\ 3/10 & 4/20 \end{bmatrix} = \begin{bmatrix} 1/5 & 1/4 \\ 3/10 & 1/5 \end{bmatrix}

Step 2: Compute I−AI - A:
I−A=[1−1/5amp;−1/4−3/10amp;1−1/5]=[4/5amp;−1/4−3/10amp;4/5]I - A = \begin{bmatrix} 1-1/5 & -1/4 \\ -3/10 & 1-1/5 \end{bmatrix} = \begin{bmatrix} 4/5 & -1/4 \\ -3/10 & 4/5 \end{bmatrix}

Step 3: Check Hawkins-Simon conditions:
- Diagonal elements of (I−A)(I-A): 4/5 > 0 ✓
- ∣I−A∣=(4/5)(4/5)−(−1/4)(−3/10)=16/25−3/40|I-A| = (4/5)(4/5) - (-1/4)(-3/10) = 16/25 - 3/40
= \frac{128}{200} - \frac{15}{200} = \frac{113}{200} > 0 \checkmark

System is viable.

Step 4: Find (I−A)−1(I-A)^{-1}:
adj(I−A)=[4/5amp;1/43/10amp;4/5]\text{adj}(I-A) = \begin{bmatrix} 4/5 & 1/4 \\ 3/10 & 4/5 \end{bmatrix}

(I−A)−1=200113[4/5amp;1/43/10amp;4/5]=1113[160amp;5060amp;160]( I-A)^{-1} = \frac{200}{113}\begin{bmatrix} 4/5 & 1/4 \\ 3/10 & 4/5 \end{bmatrix} = \frac{1}{113}\begin{bmatrix} 160 & 50 \\ 60 & 160 \end{bmatrix}

Step 5: Find output X=(I−A)−1DX = (I-A)^{-1}D where D=[812]D = \begin{bmatrix}8\\12\end{bmatrix}:
X=1113[160amp;5060amp;160][812]=1113[1280+600480+1920]=1113[18802400]X = \frac{1}{113}\begin{bmatrix}160&50\\60&160\end{bmatrix}\begin{bmatrix}8\\12\end{bmatrix} = \frac{1}{113}\begin{bmatrix}1280+600\\480+1920\end{bmatrix} = \frac{1}{113}\begin{bmatrix}1880\\2400\end{bmatrix}

X1=1880113≈16.64,X2=2400113≈21.24X_1 = \frac{1880}{113} \approx 16.64, \quad X_2 = \frac{2400}{113} \approx 21.24

∴\therefore Sector 1 must produce 1880113\dfrac{1880}{113} units and Sector 2 must produce 2400113\dfrac{2400}{113} units.
2Solve the following problem using Leontief input-output model:
| | FI | AI | Total |
|---|---|---|---|
| Food industry | 20 | 10 | 40 |
| Agricultural industry | 30 | 20 | 60 |
If the system is viable, find the output for new demand 80 and 120 from FI and AI respectively.Show solution
Step 1: Input-output coefficient matrix:
A=[20/40amp;10/6030/40amp;20/60]=[1/2amp;1/63/4amp;1/3]A = \begin{bmatrix} 20/40 & 10/60 \\ 30/40 & 20/60 \end{bmatrix} = \begin{bmatrix} 1/2 & 1/6 \\ 3/4 & 1/3 \end{bmatrix}

Step 2: I−AI - A:
I−A=[1/2amp;−1/6−3/4amp;2/3]I - A = \begin{bmatrix} 1/2 & -1/6 \\ -3/4 & 2/3 \end{bmatrix}

Step 3: Hawkins-Simon conditions:
- Diagonal elements: 1/2 > 0, 2/3 > 0 ✓
- |I-A| = (1/2)(2/3) - (-1/6)(-3/4) = 1/3 - 1/8 = 8/24 - 3/24 = 5/24 > 0 ✓

System is viable.

Step 4: (I−A)−1(I-A)^{-1}:
adj(I−A)=[2/3amp;1/63/4amp;1/2]\text{adj}(I-A) = \begin{bmatrix} 2/3 & 1/6 \\ 3/4 & 1/2 \end{bmatrix}

(I−A)−1=245[2/3amp;1/63/4amp;1/2]=[16/5amp;4/518/5amp;12/5](I-A)^{-1} = \frac{24}{5}\begin{bmatrix} 2/3 & 1/6 \\ 3/4 & 1/2 \end{bmatrix} = \begin{bmatrix} 16/5 & 4/5 \\ 18/5 & 12/5 \end{bmatrix}

Step 5: X=(I−A)−1DX = (I-A)^{-1}D, D=[80120]D = \begin{bmatrix}80\\120\end{bmatrix}:
X=[16/5amp;4/518/5amp;12/5][80120]=[1280+48051440+14405]=[352576]X = \begin{bmatrix}16/5 & 4/5\\18/5 & 12/5\end{bmatrix}\begin{bmatrix}80\\120\end{bmatrix} = \begin{bmatrix}\frac{1280+480}{5}\\\frac{1440+1440}{5}\end{bmatrix} = \begin{bmatrix}352\\576\end{bmatrix}

∴\therefore FI must produce 352 units and AI must produce 576 units.
3Solve the following problem using Leontief input-output model:
| | Sector 1 | Sector 2 | Total |
|---|---|---|---|
| Sector 1 | 12 | 20 | 40 |
| Sector 2 | 15 | 20 | 30 |
If the system is viable, find the output for new demand 8 and 8 from sector 1 and sector 2 respectively.Show solution
Step 1: Input-output coefficient matrix:
A=[12/40amp;20/3015/40amp;20/30]=[3/10amp;2/33/8amp;2/3]A = \begin{bmatrix} 12/40 & 20/30 \\ 15/40 & 20/30 \end{bmatrix} = \begin{bmatrix} 3/10 & 2/3 \\ 3/8 & 2/3 \end{bmatrix}

Step 2: I−AI - A:
I−A=[7/10amp;−2/3−3/8amp;1/3]I - A = \begin{bmatrix} 7/10 & -2/3 \\ -3/8 & 1/3 \end{bmatrix}

Step 3: Hawkins-Simon conditions:
- Diagonal elements: 7/10 > 0, 1/3 > 0 ✓
- ∣I−A∣=(7/10)(1/3)−(−2/3)(−3/8)=7/30−6/24=7/30−1/4|I-A| = (7/10)(1/3) - (-2/3)(-3/8) = 7/30 - 6/24 = 7/30 - 1/4
= \frac{14}{60} - \frac{15}{60} = -\frac{1}{60} < 0

Since |I-A| < 0, the Hawkins-Simon condition is NOT satisfied. The system is not viable.
4Solve the following problem using Leontief input-output model:
| | Sector 1 | Sector 2 | Total |
|---|---|---|---|
| Sector 1 | 5 | 7 | 30 |
| Sector 2 | 6 | 14 | 21 |
If the system is viable, find the output for new demand 8 and 12 from sector 1 and sector 2 respectively.Show solution
Step 1: Input-output coefficient matrix:
A=[5/30amp;7/216/30amp;14/21]=[1/6amp;1/31/5amp;2/3]A = \begin{bmatrix} 5/30 & 7/21 \\ 6/30 & 14/21 \end{bmatrix} = \begin{bmatrix} 1/6 & 1/3 \\ 1/5 & 2/3 \end{bmatrix}

Step 2: I−AI - A:
I−A=[5/6amp;−1/3−1/5amp;1/3]I - A = \begin{bmatrix} 5/6 & -1/3 \\ -1/5 & 1/3 \end{bmatrix}

Step 3: Hawkins-Simon conditions:
- Diagonal elements: 5/6 > 0, 1/3 > 0 ✓
- ∣I−A∣=(5/6)(1/3)−(−1/3)(−1/5)=5/18−1/15|I-A| = (5/6)(1/3) - (-1/3)(-1/5) = 5/18 - 1/15
= \frac{25}{90} - \frac{6}{90} = \frac{19}{90} > 0 \checkmark

System is viable.

Step 4: (I−A)−1(I-A)^{-1}:
adj(I−A)=[1/3amp;1/31/5amp;5/6]\text{adj}(I-A) = \begin{bmatrix} 1/3 & 1/3 \\ 1/5 & 5/6 \end{bmatrix}

(I−A)−1=9019[1/3amp;1/31/5amp;5/6]=119[30amp;3018amp;75](I-A)^{-1} = \frac{90}{19}\begin{bmatrix} 1/3 & 1/3 \\ 1/5 & 5/6 \end{bmatrix} = \frac{1}{19}\begin{bmatrix} 30 & 30 \\ 18 & 75 \end{bmatrix}

Step 5: X=(I−A)−1DX = (I-A)^{-1}D, D=[812]D = \begin{bmatrix}8\\12\end{bmatrix}:
X=119[30amp;3018amp;75][812]=119[240+360144+900]=119[6001044]X = \frac{1}{19}\begin{bmatrix}30&30\\18&75\end{bmatrix}\begin{bmatrix}8\\12\end{bmatrix} = \frac{1}{19}\begin{bmatrix}240+360\\144+900\end{bmatrix} = \frac{1}{19}\begin{bmatrix}600\\1044\end{bmatrix}

X1=60019≈31.58,X2=104419=104419≈54.95X_1 = \frac{600}{19} \approx 31.58, \quad X_2 = \frac{1044}{19} = \frac{1044}{19} \approx 54.95

∴\therefore Sector 1 must produce 60019\dfrac{600}{19} units and Sector 2 must produce 104419\dfrac{1044}{19} units.

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