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Matrices

Uttarakhand Board · Class 12 · Mathematics

NCERT Solutions for Matrices — Uttarakhand Board Class 12 Mathematics.

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56 Questions Solved · 5 Sections

Exercise 3.1

1In the matrix A=[2amp;5amp;19amp;−735amp;−2amp;52amp;123amp;1amp;−5amp;17]A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}, write: (i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a13,a21,a33,a24,a23a_{13}, a_{21}, a_{33}, a_{24}, a_{23}.Show solution
Given: Matrix A=[2amp;5amp;19amp;−735amp;−2amp;52amp;123amp;1amp;−5amp;17]A = \begin{bmatrix} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{bmatrix}

(i) Order of the matrix:
The matrix has 3 rows and 4 columns.
∴Order=3×4\therefore \text{Order} = 3 \times 4

(ii) Number of elements:
Number of elements =m×n=3×4=12= m \times n = 3 \times 4 = \mathbf{12}

(iii) Elements:
- a13a_{13} = element in row 1, column 3 =19= 19
- a21a_{21} = element in row 2, column 1 =35= 35
- a33a_{33} = element in row 3, column 3 =−5= -5
- a24a_{24} = element in row 2, column 4 =12= 12
- a23a_{23} = element in row 2, column 3 =52= \dfrac{5}{2}
2If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?Show solution
Concept: For a matrix of order m×nm \times n, the number of elements =m×n= m \times n.

For 24 elements: We need all pairs (m,n)(m, n) such that m×n=24m \times n = 24.

The possible orders are:
1×24,  24×1,  2×12,  12×2,  3×8,  8×3,  4×6,  6×41\times24,\; 24\times1,\; 2\times12,\; 12\times2,\; 3\times8,\; 8\times3,\; 4\times6,\; 6\times4

For 13 elements: We need m×n=13m \times n = 13. Since 13 is prime, the only factor pairs are 1×131 \times 13 and 13×113 \times 1.

The possible orders are:
1×13,  13×11\times13,\; 13\times1
3If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?Show solution
Concept: For a matrix of order m×nm \times n, the number of elements =m×n= m \times n.

For 18 elements: We need all pairs (m,n)(m, n) such that m×n=18m \times n = 18.

The possible orders are:
1×18,  18×1,  2×9,  9×2,  3×6,  6×31\times18,\; 18\times1,\; 2\times9,\; 9\times2,\; 3\times6,\; 6\times3

For 5 elements: Since 5 is prime, m×n=5m \times n = 5 gives only:
1×5,  5×11\times5,\; 5\times1
4Construct a 2×22 \times 2 matrix, A=[aij]A = [a_{ij}], whose elements are given by: (i) aij=(i+j)22a_{ij} = \frac{(i+j)^2}{2}, (ii) aij=ija_{ij} = \frac{i}{j}, (iii) aij=(i+2j)22a_{ij} = \frac{(i+2j)^2}{2}Show solution
A 2×22\times2 matrix has elements a11,a12,a21,a22a_{11}, a_{12}, a_{21}, a_{22} where i,j∈{1,2}i,j \in \{1,2\}.

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

a11=(1+1)22=42=2,a12=(1+2)22=92a_{11}=\frac{(1+1)^2}{2}=\frac{4}{2}=2,\quad a_{12}=\frac{(1+2)^2}{2}=\frac{9}{2}
a21=(2+1)22=92,a22=(2+2)22=162=8a_{21}=\frac{(2+1)^2}{2}=\frac{9}{2},\quad a_{22}=\frac{(2+2)^2}{2}=\frac{16}{2}=8
∴A=[2amp;9292amp;8]\therefore A = \begin{bmatrix} 2 & \frac{9}{2} \\ \frac{9}{2} & 8 \end{bmatrix}

(ii) aij=ija_{ij} = \dfrac{i}{j}

a11=11=1,a12=12,a21=21=2,a22=22=1a_{11}=\frac{1}{1}=1,\quad a_{12}=\frac{1}{2},\quad a_{21}=\frac{2}{1}=2,\quad a_{22}=\frac{2}{2}=1
∴A=[1amp;122amp;1]\therefore A = \begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix}

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

a11=(1+2)22=92,a12=(1+4)22=252a_{11}=\frac{(1+2)^2}{2}=\frac{9}{2},\quad a_{12}=\frac{(1+4)^2}{2}=\frac{25}{2}
a21=(2+2)22=162=8,a22=(2+4)22=362=18a_{21}=\frac{(2+2)^2}{2}=\frac{16}{2}=8,\quad a_{22}=\frac{(2+4)^2}{2}=\frac{36}{2}=18
∴A=[92amp;2528amp;18]\therefore A = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} \\ 8 & 18 \end{bmatrix}
5Construct a 3×43 \times 4 matrix, whose elements are given by: (i) aij=12∣−3i+j∣a_{ij} = \frac{1}{2}|-3i+j|, (ii) aij=2i−ja_{ij} = 2i - jShow solution
A 3×43\times4 matrix has i∈{1,2,3}i \in \{1,2,3\} and j∈{1,2,3,4}j \in \{1,2,3,4\}.

(i) aij=12∣−3i+j∣a_{ij} = \dfrac{1}{2}|-3i+j|

Computing each element:

a11=12∣−3+1∣=1,  a12=12∣−3+2∣=12,  a13=12∣−3+3∣=0,  a14=12∣−3+4∣=12a_{11}=\frac{1}{2}|-3+1|=1,\; a_{12}=\frac{1}{2}|-3+2|=\frac{1}{2},\; a_{13}=\frac{1}{2}|-3+3|=0,\; a_{14}=\frac{1}{2}|-3+4|=\frac{1}{2}

a21=12∣−6+1∣=52,  a22=12∣−6+2∣=2,  a23=12∣−6+3∣=32,  a24=12∣−6+4∣=1a_{21}=\frac{1}{2}|-6+1|=\frac{5}{2},\; a_{22}=\frac{1}{2}|-6+2|=2,\; a_{23}=\frac{1}{2}|-6+3|=\frac{3}{2},\; a_{24}=\frac{1}{2}|-6+4|=1

a31=12∣−9+1∣=4,  a32=12∣−9+2∣=72,  a33=12∣−9+3∣=3,  a34=12∣−9+4∣=52a_{31}=\frac{1}{2}|-9+1|=4,\; a_{32}=\frac{1}{2}|-9+2|=\frac{7}{2},\; a_{33}=\frac{1}{2}|-9+3|=3,\; a_{34}=\frac{1}{2}|-9+4|=\frac{5}{2}

∴A=[1amp;12amp;0amp;1252amp;2amp;32amp;14amp;72amp;3amp;52]\therefore A = \begin{bmatrix} 1 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & 2 & \frac{3}{2} & 1 \\ 4 & \frac{7}{2} & 3 & \frac{5}{2} \end{bmatrix}

(ii) aij=2i−ja_{ij} = 2i - j

a11=2−1=1,  a12=2−2=0,  a13=2−3=−1,  a14=2−4=−2a_{11}=2-1=1,\; a_{12}=2-2=0,\; a_{13}=2-3=-1,\; a_{14}=2-4=-2

a21=4−1=3,  a22=4−2=2,  a23=4−3=1,  a24=4−4=0a_{21}=4-1=3,\; a_{22}=4-2=2,\; a_{23}=4-3=1,\; a_{24}=4-4=0

a31=6−1=5,  a32=6−2=4,  a33=6−3=3,  a34=6−4=2a_{31}=6-1=5,\; a_{32}=6-2=4,\; a_{33}=6-3=3,\; a_{34}=6-4=2

∴A=[1amp;0amp;−1amp;−23amp;2amp;1amp;05amp;4amp;3amp;2]\therefore A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}
6Find the values of x,yx, y and zz from the following equations: (i) [4amp;3xamp;5]=[yamp;z1amp;5]\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}, (ii) [x+yamp;25+zamp;xy]=[6amp;25amp;8]\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}, (iii) [x+y+zx+zy+z]=[957]\begin{bmatrix} x+y+z \\ x+z \\ y+z \end{bmatrix} = \begin{bmatrix} 9 \\ 5 \\ 7 \end{bmatrix}Show solution
Concept: Two matrices are equal if and only if their corresponding elements are equal.

(i) Equating corresponding elements:
y=4,z=3,x=1y = 4,\quad z = 3,\quad x = 1
x=1,  y=4,  z=3\boxed{x=1,\; y=4,\; z=3}

(ii) Equating corresponding elements:
x+y=6⋯(1)x + y = 6 \quad \cdots(1)
5+z=5⇒z=0⋯(2)5 + z = 5 \Rightarrow z = 0 \quad \cdots(2)
xy=8⋯(3)xy = 8 \quad \cdots(3)

From (1): y=6−xy = 6 - x. Substituting in (3):
x(6−x)=8⇒6x−x2=8⇒x2−6x+8=0x(6-x) = 8 \Rightarrow 6x - x^2 = 8 \Rightarrow x^2 - 6x + 8 = 0
(x−4)(x−2)=0⇒x=4 or x=2(x-4)(x-2) = 0 \Rightarrow x = 4 \text{ or } x = 2

If x=4x = 4, then y=2y = 2. If x=2x = 2, then y=4y = 4.
x=4,  y=2,  z=0orx=2,  y=4,  z=0\boxed{x=4,\; y=2,\; z=0 \quad \text{or} \quad x=2,\; y=4,\; z=0}

(iii) Equating corresponding elements:
x+y+z=9⋯(1)x + y + z = 9 \quad \cdots(1)
x+z=5⋯(2)x + z = 5 \quad \cdots(2)
y+z=7⋯(3)y + z = 7 \quad \cdots(3)

From (1) and (2): y=9−5=4y = 9 - 5 = 4

From (3): z=7−y=7−4=3z = 7 - y = 7 - 4 = 3

From (2): x=5−z=5−3=2x = 5 - z = 5 - 3 = 2
x=2,  y=4,  z=3\boxed{x=2,\; y=4,\; z=3}
7Find the value of a,b,ca, b, c and dd from the equation: [a−bamp;2a+c2a−bamp;3c+d]=[−1amp;50amp;13]\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}Show solution
Given: [a−bamp;2a+c2a−bamp;3c+d]=[−1amp;50amp;13]\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}

Equating corresponding elements:
a−b=−1⋯(1)a - b = -1 \quad \cdots(1)
2a+c=5⋯(2)2a + c = 5 \quad \cdots(2)
2a−b=0⋯(3)2a - b = 0 \quad \cdots(3)
3c+d=13⋯(4)3c + d = 13 \quad \cdots(4)

Subtracting (1) from (3):
(2a−b)−(a−b)=0−(−1)(2a - b) - (a - b) = 0 - (-1)
a=1a = 1

From (1): b=a+1=1+1=2b = a + 1 = 1 + 1 = 2

From (2): c=5−2a=5−2=3c = 5 - 2a = 5 - 2 = 3

From (4): d=13−3c=13−9=4d = 13 - 3c = 13 - 9 = 4

a=1,  b=2,  c=3,  d=4\boxed{a = 1,\; b = 2,\; c = 3,\; d = 4}
8A=[aij]m×nA = [a_{ij}]_{m \times n} is a square matrix, if (A) m < n (B) m > n (C) m=nm = n (D) None of theseShow solution
Correct Answer: (C) m=nm = n

A matrix is called a square matrix when the number of rows equals the number of columns, i.e., m=nm = n.
9Which of the given values of xx and yy make the following pair of matrices equal: [3x+7amp;5y+1amp;2−3x]\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix}, [0amp;y−28amp;4]\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}? (A) x=−13,y=7x = \frac{-1}{3}, y = 7 (B) Not possible to find (C) y=7,x=−23y = 7, x = \frac{-2}{3} (D) x=−13,y=−23x = \frac{-1}{3}, y = \frac{-2}{3}Show solution
Correct Answer: (B) Not possible to find

Equating corresponding elements:
3x+7=0⇒x=−73⋯(1)3x + 7 = 0 \Rightarrow x = \frac{-7}{3} \quad \cdots(1)
2−3x=4⇒3x=−2⇒x=−23⋯(2)2 - 3x = 4 \Rightarrow 3x = -2 \Rightarrow x = \frac{-2}{3} \quad \cdots(2)

From (1) and (2), we get two different values of xx, which is a contradiction. Hence it is not possible to find values of xx and yy that satisfy all conditions simultaneously.
10The number of all possible matrices of order 3×33 \times 3 with each entry 0 or 1 is: (A) 27 (B) 18 (C) 81 (D) 512Show solution
Correct Answer: (D) 512

A 3×33 \times 3 matrix has 3×3=93 \times 3 = 9 entries. Each entry can be filled in 2 ways (either 0 or 1).

Total number of matrices=29=512\text{Total number of matrices} = 2^9 = 512

Exercise 3.2

1Let A=[2amp;43amp;2]A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}, B=[1amp;3−2amp;5]B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}, C=[−2amp;53amp;4]C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}. Find each of the following: (i) A+B, (ii) A-B, (iii) 3A-C, (iv) AB, (v) BAShow solution
Given: A=[2amp;43amp;2]A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}, B=[1amp;3−2amp;5]B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}, C=[−2amp;53amp;4]C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}

(i) A+BA + B:
A+B=[2+1amp;4+33+(−2)amp;2+5]=[3amp;71amp;7]A+B = \begin{bmatrix} 2+1 & 4+3 \\ 3+(-2) & 2+5 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}

(ii) A−BA - B:
A−B=[2−1amp;4−33−(−2)amp;2−5]=[1amp;15amp;−3]A-B = \begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}

(iii) 3A−C3A - C:
3A=[6amp;129amp;6]3A = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}
3A−C=[6−(−2)amp;12−59−3amp;6−4]=[8amp;76amp;2]3A - C = \begin{bmatrix} 6-(-2) & 12-5 \\ 9-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}

(iv) ABAB:
AB=[2amp;43amp;2][1amp;3−2amp;5]AB = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}
=[2(1)+4(−2)amp;2(3)+4(5)3(1)+2(−2)amp;3(3)+2(5)]=[2−8amp;6+203−4amp;9+10]=[−6amp;26−1amp;19]= \begin{bmatrix} 2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5) \end{bmatrix} = \begin{bmatrix} 2-8 & 6+20 \\ 3-4 & 9+10 \end{bmatrix} = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}

(v) BABA:
BA=[1amp;3−2amp;5][2amp;43amp;2]BA = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}
=[1(2)+3(3)amp;1(4)+3(2)−2(2)+5(3)amp;−2(4)+5(2)]=[2+9amp;4+6−4+15amp;−8+10]=[11amp;1011amp;2]= \begin{bmatrix} 1(2)+3(3) & 1(4)+3(2) \\ -2(2)+5(3) & -2(4)+5(2) \end{bmatrix} = \begin{bmatrix} 2+9 & 4+6 \\ -4+15 & -8+10 \end{bmatrix} = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}
2Compute the following: (i) [aamp;b−bamp;a]+[aamp;bbamp;a]\begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix}, (ii) [a2+b2amp;b2+c2a2+c2amp;a2+b2]+[2abamp;2bc−2acamp;−2ab]\begin{bmatrix} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix}, (iii) [−1amp;4amp;−68amp;5amp;162amp;8amp;5]+[12amp;7amp;68amp;0amp;53amp;2amp;4]\begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix}, (iv) [cos⁡2xamp;sin⁡2xsin⁡2xamp;cos⁡2x]+[sin⁡2xamp;cos⁡2xcos⁡2xamp;sin⁡2x]\begin{bmatrix} \cos^2 x & \sin^2 x \\ \sin^2 x & \cos^2 x \end{bmatrix} + \begin{bmatrix} \sin^2 x & \cos^2 x \\ \cos^2 x & \sin^2 x \end{bmatrix}Show solution
(i)
[aamp;b−bamp;a]+[aamp;bbamp;a]=[2aamp;2b0amp;2a]\begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}

(ii)
[a2+b2+2abamp;b2+c2+2bca2+c2−2acamp;a2+b2−2ab]=[(a+b)2amp;(b+c)2(a−c)2amp;(a−b)2]\begin{bmatrix} a^2+b^2+2ab & b^2+c^2+2bc \\ a^2+c^2-2ac & a^2+b^2-2ab \end{bmatrix} = \begin{bmatrix} (a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2 \end{bmatrix}

(iii)
[−1+12amp;4+7amp;−6+68+8amp;5+0amp;16+52+3amp;8+2amp;5+4]=[11amp;11amp;016amp;5amp;215amp;10amp;9]\begin{bmatrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix} = \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}

(iv)
[cos⁡2x+sin⁡2xamp;sin⁡2x+cos⁡2xsin⁡2x+cos⁡2xamp;cos⁡2x+sin⁡2x]=[1amp;11amp;1]\begin{bmatrix} \cos^2x+\sin^2x & \sin^2x+\cos^2x \\ \sin^2x+\cos^2x & \cos^2x+\sin^2x \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}
(Using sin⁡2x+cos⁡2x=1\sin^2x + \cos^2x = 1)
3Compute the indicated products: (i) [aamp;b−bamp;a][aamp;−bbamp;a]\begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix}, (ii) [123][2amp;3amp;4]\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\begin{bmatrix} 2 & 3 & 4 \end{bmatrix}, (iii) [1amp;−22amp;3][1amp;2amp;32amp;3amp;1]\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}, (iv) [2amp;3amp;43amp;4amp;54amp;5amp;6][1amp;−3amp;50amp;2amp;43amp;0amp;5]\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}, (v) [2amp;13amp;2−1amp;1][1amp;0amp;1−1amp;2amp;1]\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}, (vi) [3amp;−1amp;3−1amp;0amp;2][2amp;−31amp;03amp;1]\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}Show solution
(i)
[aamp;b−bamp;a][aamp;−bbamp;a]=[a2+b2amp;−ab+ab−ab+abamp;b2+a2]=[a2+b2amp;00amp;a2+b2]\begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix} = \begin{bmatrix} a^2+b^2 & -ab+ab \\ -ab+ab & b^2+a^2 \end{bmatrix} = \begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix}

(ii)
[123]3×1[2amp;3amp;4]1×3=[1⋅2amp;1⋅3amp;1⋅42⋅2amp;2⋅3amp;2⋅43⋅2amp;3⋅3amp;3⋅4]=[2amp;3amp;44amp;6amp;86amp;9amp;12]\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}_{3\times1}\begin{bmatrix} 2 & 3 & 4 \end{bmatrix}_{1\times3} = \begin{bmatrix} 1\cdot2 & 1\cdot3 & 1\cdot4 \\ 2\cdot2 & 2\cdot3 & 2\cdot4 \\ 3\cdot2 & 3\cdot3 & 3\cdot4 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}

(iii)
[1amp;−22amp;3][1amp;2amp;32amp;3amp;1]\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}
=[1−4amp;2−6amp;3−22+6amp;4+9amp;6+3]=[−3amp;−4amp;18amp;13amp;9]= \begin{bmatrix} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}

(iv)
[2amp;3amp;43amp;4amp;54amp;5amp;6][1amp;−3amp;50amp;2amp;43amp;0amp;5]\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}

Row 1: (2+0+12,  −6+6+0,  10+12+20)=(14,0,42)(2+0+12,\; -6+6+0,\; 10+12+20) = (14, 0, 42)

Row 2: (3+0+15,  −9+8+0,  15+16+25)=(18,−1,56)(3+0+15,\; -9+8+0,\; 15+16+25) = (18, -1, 56)

Row 3: (4+0+18,  −12+10+0,  20+20+30)=(22,−2,70)(4+0+18,\; -12+10+0,\; 20+20+30) = (22, -2, 70)

=[14amp;0amp;4218amp;−1amp;5622amp;−2amp;70]= \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix}

(v)
[2amp;13amp;2−1amp;1][1amp;0amp;1−1amp;2amp;1]\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}

Row 1: (2−1,  0+2,  2+1)=(1,2,3)(2-1,\; 0+2,\; 2+1) = (1, 2, 3)

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

Row 3: (−1−1,  0+2,  −1+1)=(−2,2,0)(-1-1,\; 0+2,\; -1+1) = (-2, 2, 0)

=[1amp;2amp;31amp;4amp;5−2amp;2amp;0]= \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix}

(vi)
[3amp;−1amp;3−1amp;0amp;2][2amp;−31amp;03amp;1]\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}

Row 1: (6−1+9,  −9+0+3)=(14,−6)(6-1+9,\; -9+0+3) = (14, -6)

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

=[14amp;−64amp;5]= \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}
4If A=[1amp;2amp;−35amp;0amp;21amp;−1amp;1]A = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix}, B=[3amp;−1amp;24amp;2amp;52amp;0amp;3]B = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} and C=[4amp;1amp;20amp;3amp;21amp;−2amp;3]C = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix}, then compute (A+B)(A+B) and (B−C)(B-C). Also, verify that A+(B−C)=(A+B)−CA+(B-C) = (A+B)-C.Show solution
Computing A+BA+B:
A+B=[1+3amp;2−1amp;−3+25+4amp;0+2amp;2+51+2amp;−1+0amp;1+3]=[4amp;1amp;−19amp;2amp;73amp;−1amp;4]A+B = \begin{bmatrix} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix} = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix}

Computing B−CB-C:
B−C=[3−4amp;−1−1amp;2−24−0amp;2−3amp;5−22−1amp;0−(−2)amp;3−3]=[−1amp;−2amp;04amp;−1amp;31amp;2amp;0]B-C = \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{bmatrix} = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix}

LHS: A+(B−C)A+(B-C):
=[1amp;2amp;−35amp;0amp;21amp;−1amp;1]+[−1amp;−2amp;04amp;−1amp;31amp;2amp;0]=[0amp;0amp;−39amp;−1amp;52amp;1amp;1]= \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} + \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}

RHS: (A+B)−C(A+B)-C:
=[4amp;1amp;−19amp;2amp;73amp;−1amp;4]−[4amp;1amp;20amp;3amp;21amp;−2amp;3]=[0amp;0amp;−39amp;−1amp;52amp;1amp;1]= \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} - \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix}

Since LHS = RHS, A+(B−C)=(A+B)−CA+(B-C) = (A+B)-C is verified.
5If A=[23amp;1amp;5313amp;23amp;4373amp;2amp;23]A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} and B=[25amp;35amp;115amp;25amp;4575amp;65amp;25]B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}, then compute 3A−5B3A - 5B.Show solution
Computing 3A3A:
3A=[2amp;3amp;51amp;2amp;47amp;6amp;2]3A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}

Computing 5B5B:
5B=[2amp;3amp;51amp;2amp;47amp;6amp;2]5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}

Therefore:
3A−5B=[2−2amp;3−3amp;5−51−1amp;2−2amp;4−47−7amp;6−6amp;2−2]=[0amp;0amp;00amp;0amp;00amp;0amp;0]3A - 5B = \begin{bmatrix} 2-2 & 3-3 & 5-5 \\ 1-1 & 2-2 & 4-4 \\ 7-7 & 6-6 & 2-2 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
6Simplify cos⁡θ[cos⁡θamp;sin⁡θ−sin⁡θamp;cos⁡θ]+sin⁡θ[sin⁡θamp;−cos⁡θcos⁡θamp;sin⁡θ]\cos\theta\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix}Show solution
=[cos⁡2θamp;cos⁡θsin⁡θ−sin⁡θcos⁡θamp;cos⁡2θ]+[sin⁡2θamp;−sin⁡θcos⁡θsin⁡θcos⁡θamp;sin⁡2θ]= \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ -\sin\theta\cos\theta & \cos^2\theta \end{bmatrix} + \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix}

=[cos⁡2θ+sin⁡2θamp;cos⁡θsin⁡θ−sin⁡θcos⁡θ−sin⁡θcos⁡θ+sin⁡θcos⁡θamp;cos⁡2θ+sin⁡2θ]= \begin{bmatrix} \cos^2\theta+\sin^2\theta & \cos\theta\sin\theta-\sin\theta\cos\theta \\ -\sin\theta\cos\theta+\sin\theta\cos\theta & \cos^2\theta+\sin^2\theta \end{bmatrix}

=[1amp;00amp;1]=I= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I
7Find X and Y, if (i) X+Y=[7amp;02amp;5]X+Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} and X−Y=[3amp;00amp;3]X-Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}, (ii) 2X+3Y=[2amp;34amp;0]2X+3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} and 3X+2Y=[2amp;−2−1amp;5]3X+2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}Show solution
(i) Adding the two equations:
2X=[7+3amp;0+02+0amp;5+3]=[10amp;02amp;8]2X = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}
X=[5amp;01amp;4]X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}

Subtracting:
2Y=[7−3amp;02−0amp;5−3]=[4amp;02amp;2]2Y = \begin{bmatrix} 7-3 & 0 \\ 2-0 & 5-3 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix}
Y=[2amp;01amp;1]Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}

(ii) Let 2X+3Y=P=[2amp;34amp;0]2X+3Y = P = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} and 3X+2Y=Q=[2amp;−2−1amp;5]3X+2Y = Q = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}.

Multiply first equation by 2 and second by 3:
4X+6Y=[4amp;68amp;0]4X + 6Y = \begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix}
9X+6Y=[6amp;−6−3amp;15]9X + 6Y = \begin{bmatrix} 6 & -6 \\ -3 & 15 \end{bmatrix}

Subtracting:
5X=[2amp;−12−11amp;15]⇒X=15[2amp;−12−11amp;15]5X = \begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix} \Rightarrow X = \frac{1}{5}\begin{bmatrix} 2 & -12 \\ -11 & 15 \end{bmatrix}

Multiply first equation by 3 and second by 2:
6X+9Y=[6amp;912amp;0]6X + 9Y = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix}
6X+4Y=[4amp;−4−2amp;10]6X + 4Y = \begin{bmatrix} 4 & -4 \\ -2 & 10 \end{bmatrix}

Subtracting:
5Y=[2amp;1314amp;−10]⇒Y=15[2amp;1314amp;−10]5Y = \begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix} \Rightarrow Y = \frac{1}{5}\begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix}
8Find X, if Y=[3amp;21amp;4]Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} and 2X+Y=[1amp;0−3amp;2]2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}Show solution
Given: 2X+Y=[1amp;0−3amp;2]2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}

2X=[1amp;0−3amp;2]−Y=[1amp;0−3amp;2]−[3amp;21amp;4]=[−2amp;−2−4amp;−2]2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}

X=12[−2amp;−2−4amp;−2]=[−1amp;−1−2amp;−1]X = \frac{1}{2}\begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix}
9Find xx and yy, if 2[1amp;30amp;x]+[yamp;01amp;2]=[5amp;61amp;8]2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}Show solution
[2amp;60amp;2x]+[yamp;01amp;2]=[5amp;61amp;8]\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}

[2+yamp;61amp;2x+2]=[5amp;61amp;8]\begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}

Equating corresponding elements:
2+y=5⇒y=32 + y = 5 \Rightarrow y = 3
2x+2=8⇒2x=6⇒x=32x + 2 = 8 \Rightarrow 2x = 6 \Rightarrow x = 3

x=3,  y=3\boxed{x = 3,\; y = 3}
10Solve the equation for x,y,zx, y, z and tt, if 2[xamp;zyamp;t]+3[1amp;−10amp;2]=3[3amp;54amp;6]2\begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}Show solution
[2xamp;2z2yamp;2t]+[3amp;−30amp;6]=[9amp;1512amp;18]\begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}

[2x+3amp;2z−32yamp;2t+6]=[9amp;1512amp;18]\begin{bmatrix} 2x+3 & 2z-3 \\ 2y & 2t+6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}

Equating corresponding elements:
2x+3=9⇒x=32x+3=9 \Rightarrow x=3
2z−3=15⇒z=92z-3=15 \Rightarrow z=9
2y=12⇒y=62y=12 \Rightarrow y=6
2t+6=18⇒t=62t+6=18 \Rightarrow t=6

x=3,  y=6,  z=9,  t=6\boxed{x=3,\; y=6,\; z=9,\; t=6}
11If x[23]+y[−11]=[105]x\begin{bmatrix} 2 \\ 3 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}, find the values of xx and yy.Show solution
[2x3x]+[−yy]=[105]\begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}

[2x−y3x+y]=[105]\begin{bmatrix} 2x-y \\ 3x+y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}

Equating corresponding elements:
2x−y=10⋯(1)2x - y = 10 \quad \cdots(1)
3x+y=5⋯(2)3x + y = 5 \quad \cdots(2)

Adding (1) and (2):
5x=15⇒x=35x = 15 \Rightarrow x = 3

From (1): y=2(3)−10=−4y = 2(3) - 10 = -4

x=3,  y=−4\boxed{x = 3,\; y = -4}
12Given 3[xamp;yzamp;w]=[xamp;6−1amp;2w]+[4amp;x+yz+wamp;3]3\begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix}, find the values of x,y,zx, y, z and ww.Show solution
[3xamp;3y3zamp;3w]=[x+4amp;6+x+y−1+z+wamp;2w+3]\begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix}

Equating corresponding elements:
3x=x+4⇒2x=4⇒x=23x = x+4 \Rightarrow 2x = 4 \Rightarrow x = 2
3y=6+x+y⇒2y=6+2=8⇒y=43y = 6+x+y \Rightarrow 2y = 6+2 = 8 \Rightarrow y = 4
3z=−1+z+w⇒2z−w=−1⋯(1)3z = -1+z+w \Rightarrow 2z - w = -1 \quad \cdots(1)
3w=2w+3⇒w=33w = 2w+3 \Rightarrow w = 3

From (1): 2z−3=−1⇒z=12z - 3 = -1 \Rightarrow z = 1

x=2,  y=4,  z=1,  w=3\boxed{x=2,\; y=4,\; z=1,\; w=3}
13If F(x)=[cos⁡xamp;−sin⁡xamp;0sin⁡xamp;cos⁡xamp;00amp;0amp;1]F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}, show that F(x)F(y)=F(x+y)F(x)F(y) = F(x+y).Show solution
Given: F(x)=[cos⁡xamp;−sin⁡xamp;0sin⁡xamp;cos⁡xamp;00amp;0amp;1]F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}

Computing F(x)F(y)F(x)F(y):
F(x)F(y)=[cos⁡xamp;−sin⁡xamp;0sin⁡xamp;cos⁡xamp;00amp;0amp;1][cos⁡yamp;−sin⁡yamp;0sin⁡yamp;cos⁡yamp;00amp;0amp;1]F(x)F(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix}

Element (1,1)(1,1): cos⁡xcos⁡y−sin⁡xsin⁡y=cos⁡(x+y)\cos x\cos y - \sin x\sin y = \cos(x+y)

Element (1,2)(1,2): −cos⁡xsin⁡y−sin⁡xcos⁡y=−sin⁡(x+y)-\cos x\sin y - \sin x\cos y = -\sin(x+y)

Element (1,3)(1,3): 00

Element (2,1)(2,1): sin⁡xcos⁡y+cos⁡xsin⁡y=sin⁡(x+y)\sin x\cos y + \cos x\sin y = \sin(x+y)

Element (2,2)(2,2): −sin⁡xsin⁡y+cos⁡xcos⁡y=cos⁡(x+y)-\sin x\sin y + \cos x\cos y = \cos(x+y)

Element (2,3)(2,3): 00

Element (3,1)(3,1): 00, Element (3,2)(3,2): 00, Element (3,3)(3,3): 11

∴F(x)F(y)=[cos⁡(x+y)amp;−sin⁡(x+y)amp;0sin⁡(x+y)amp;cos⁡(x+y)amp;00amp;0amp;1]=F(x+y)\therefore F(x)F(y) = \begin{bmatrix} \cos(x+y) & -\sin(x+y) & 0 \\ \sin(x+y) & \cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix} = F(x+y)

Hence proved.
14Show that (i) [5amp;−16amp;7][2amp;13amp;4]≠[2amp;13amp;4][5amp;−16amp;7]\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}, (ii) [1amp;2amp;30amp;1amp;01amp;1amp;0][−1amp;1amp;00amp;−1amp;12amp;3amp;4]≠[−1amp;1amp;00amp;−1amp;12amp;3amp;4][1amp;2amp;30amp;1amp;01amp;1amp;0]\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}\begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix}Show solution
(i) Computing LHS:
[5amp;−16amp;7][2amp;13amp;4]=[10−3amp;5−412+21amp;6+28]=[7amp;133amp;34]\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 10-3 & 5-4 \\ 12+21 & 6+28 \end{bmatrix} = \begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}

Computing RHS:
[2amp;13amp;4][5amp;−16amp;7]=[10+6amp;−2+715+24amp;−3+28]=[16amp;539amp;25]\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} = \begin{bmatrix} 10+6 & -2+7 \\ 15+24 & -3+28 \end{bmatrix} = \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}

Since [7amp;133amp;34]≠[16amp;539amp;25]\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix} \neq \begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}, the result is proved.

(ii) Let P=[1amp;2amp;30amp;1amp;01amp;1amp;0]P = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} and Q=[−1amp;1amp;00amp;−1amp;12amp;3amp;4]Q = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix}

Computing PQ:
Row 1: (−1+0+6,  1−2+9,  0+2+12)=(5,8,14)(-1+0+6,\; 1-2+9,\; 0+2+12) = (5, 8, 14)
Row 2: (0+0+0,  0−1+0,  0+1+0)=(0,−1,1)(0+0+0,\; 0-1+0,\; 0+1+0) = (0, -1, 1)
Row 3: (−1+0+0,  1−1+0,  0+1+0)=(−1,0,1)(-1+0+0,\; 1-1+0,\; 0+1+0) = (-1, 0, 1)
PQ=[5amp;8amp;140amp;−1amp;1−1amp;0amp;1]PQ = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}

Computing QP:
Row 1: (−1+0+0,  −2+1+0,  −3+0+0)=(−1,−1,−3)(-1+0+0,\; -2+1+0,\; -3+0+0) = (-1, -1, -3)
Row 2: (0+1+1,  0−1+1,  0+0+0)=(2,0,0)(0+1+1,\; 0-1+1,\; 0+0+0) = (2, 0, 0) — let me recalculate:
Row 1: (−1)(1)+(1)(0)+(0)(1),  (−1)(2)+(1)(1)+(0)(1),  (−1)(3)+(1)(0)+(0)(0)=(−1,−1,−3)(-1)(1)+(1)(0)+(0)(1),\; (-1)(2)+(1)(1)+(0)(1),\; (-1)(3)+(1)(0)+(0)(0) = (-1, -1, -3)
Row 2: (0)(1)+(−1)(0)+(1)(1),  (0)(2)+(−1)(1)+(1)(1),  (0)(3)+(−1)(0)+(1)(0)=(1,0,0)(0)(1)+(-1)(0)+(1)(1),\; (0)(2)+(-1)(1)+(1)(1),\; (0)(3)+(-1)(0)+(1)(0) = (1, 0, 0)
Row 3: (2)(1)+(3)(0)+(4)(1),  (2)(2)+(3)(1)+(4)(1),  (2)(3)+(3)(0)+(4)(0)=(6,11,6)(2)(1)+(3)(0)+(4)(1),\; (2)(2)+(3)(1)+(4)(1),\; (2)(3)+(3)(0)+(4)(0) = (6, 11, 6)
QP=[−1amp;−1amp;−31amp;0amp;06amp;11amp;6]QP = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}

Since PQ≠QPPQ \neq QP, the result is proved.
15Find A2−5A+6IA^2 - 5A + 6I, if A=[2amp;0amp;12amp;1amp;31amp;−1amp;0]A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}Show solution
Step 1: Compute A2=A⋅AA^2 = A \cdot A
A2=[2amp;0amp;12amp;1amp;31amp;−1amp;0][2amp;0amp;12amp;1amp;31amp;−1amp;0]A^2 = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}

Row 1: (4+0+1,  0+0−1,  2+0+0)=(5,−1,2)(4+0+1,\; 0+0-1,\; 2+0+0) = (5, -1, 2)
Row 2: (4+2+3,  0+1−3,  2+3+0)=(9,−2,5)(4+2+3,\; 0+1-3,\; 2+3+0) = (9, -2, 5)
Row 3: (2−2+0,  0−1+0,  1−3+0)=(0,−1,−2)(2-2+0,\; 0-1+0,\; 1-3+0) = (0, -1, -2)

A2=[5amp;−1amp;29amp;−2amp;50amp;−1amp;−2]A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}

Step 2: Compute 5A5A
5A=[10amp;0amp;510amp;5amp;155amp;−5amp;0]5A = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}

Step 3: Compute 6I6I
6I=[6amp;0amp;00amp;6amp;00amp;0amp;6]6I = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}

Step 4: A2−5A+6IA^2 - 5A + 6I
=[5−10+6amp;−1−0+0amp;2−5+09−10+0amp;−2−5+6amp;5−15+00−5+0amp;−1+5+0amp;−2−0+6]=[1amp;−1amp;−3−1amp;−1amp;−10−5amp;4amp;4]= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2-0+6 \end{bmatrix} = \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}
16If A=[1amp;0amp;20amp;2amp;12amp;0amp;3]A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}, prove that A3−6A2+7A+2I=0A^3 - 6A^2 + 7A + 2I = 0.Show solution
Step 1: Compute A2A^2
A2=[1amp;0amp;20amp;2amp;12amp;0amp;3][1amp;0amp;20amp;2amp;12amp;0amp;3]A^2 = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}

Row 1: (1+0+4,  0+0+0,  2+0+6)=(5,0,8)(1+0+4,\; 0+0+0,\; 2+0+6) = (5, 0, 8)
Row 2: (0+0+2,  0+4+0,  0+2+3)=(2,4,5)(0+0+2,\; 0+4+0,\; 0+2+3) = (2, 4, 5)
Row 3: (2+0+6,  0+0+0,  4+0+9)=(8,0,13)(2+0+6,\; 0+0+0,\; 4+0+9) = (8, 0, 13)

A2=[5amp;0amp;82amp;4amp;58amp;0amp;13]A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}

Step 2: Compute A3=A2⋅AA^3 = A^2 \cdot A
A3=[5amp;0amp;82amp;4amp;58amp;0amp;13][1amp;0amp;20amp;2amp;12amp;0amp;3]A^3 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}

Row 1: (5+0+16,  0+0+0,  10+0+24)=(21,0,34)(5+0+16,\; 0+0+0,\; 10+0+24) = (21, 0, 34)
Row 2: (2+0+10,  0+8+0,  4+4+15)=(12,8,23)(2+0+10,\; 0+8+0,\; 4+4+15) = (12, 8, 23)
Row 3: (8+0+26,  0+0+0,  16+0+39)=(34,0,55)(8+0+26,\; 0+0+0,\; 16+0+39) = (34, 0, 55)

A3=[21amp;0amp;3412amp;8amp;2334amp;0amp;55]A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}

Step 3: Compute A3−6A2+7A+2IA^3 - 6A^2 + 7A + 2I
6A2=[30amp;0amp;4812amp;24amp;3048amp;0amp;78],7A=[7amp;0amp;140amp;14amp;714amp;0amp;21],2I=[2amp;0amp;00amp;2amp;00amp;0amp;2]6A^2 = \begin{bmatrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{bmatrix},\quad 7A = \begin{bmatrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{bmatrix},\quad 2I = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}

A3−6A2+7A+2I=[21−30+7+2amp;0amp;34−48+14+012−12+0+0amp;8−24+14+2amp;23−30+7+034−48+14+0amp;0amp;55−78+21+2]A^3 - 6A^2 + 7A + 2I = \begin{bmatrix} 21-30+7+2 & 0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+2 & 23-30+7+0 \\ 34-48+14+0 & 0 & 55-78+21+2 \end{bmatrix}

=[0amp;0amp;00amp;0amp;00amp;0amp;0]=O= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O

Hence proved.
17If A=[3amp;−24amp;−2]A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} and I=[1amp;00amp;1]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, find kk so that A2=kA−2IA^2 = kA - 2I.Show solution
Step 1: Compute A2A^2
A2=[3amp;−24amp;−2][3amp;−24amp;−2]A^2 = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix}
=[9−8amp;−6+412−8amp;−8+4]=[1amp;−24amp;−4]= \begin{bmatrix} 9-8 & -6+4 \\ 12-8 & -8+4 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix}

Step 2: Compute kA−2IkA - 2I
kA−2I=k[3amp;−24amp;−2]−[2amp;00amp;2]=[3k−2amp;−2k4kamp;−2k−2]kA - 2I = k\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}

Step 3: Equate A2=kA−2IA^2 = kA - 2I
[1amp;−24amp;−4]=[3k−2amp;−2k4kamp;−2k−2]\begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix}

From element (1,2)(1,2): −2k=−2⇒k=1-2k = -2 \Rightarrow k = 1

Verification: 3k−2=13k-2 = 1 ✓, 4k=44k = 4 ✓, −2k−2=−4-2k-2 = -4 ✓

k=1\boxed{k = 1}
18If A=[0amp;−tan⁡α2tan⁡α2amp;0]A = \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix} and II is the identity matrix of order 2, show that I+A=(I−A)[cos⁡αamp;−sin⁡αsin⁡αamp;cos⁡α]I + A = (I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}.Show solution
Let t=tan⁡α2t = \tan\dfrac{\alpha}{2}. Then:
I+A=[1amp;−ttamp;1],I−A=[1amp;t−tamp;1]I + A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix}, \quad I - A = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}

Using the identities:
cos⁡α=1−t21+t2,sin⁡α=2t1+t2\cos\alpha = \frac{1-t^2}{1+t^2}, \quad \sin\alpha = \frac{2t}{1+t^2}

Computing RHS = (I−A)[cos⁡αamp;−sin⁡αsin⁡αamp;cos⁡α](I-A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}:

=[1amp;t−tamp;1][1−t21+t2amp;−2t1+t22t1+t2amp;1−t21+t2]= \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}\begin{bmatrix} \frac{1-t^2}{1+t^2} & \frac{-2t}{1+t^2} \\ \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2} \end{bmatrix}

Element (1,1)(1,1): 1−t21+t2+2t21+t2=1+t21+t2=1\dfrac{1-t^2}{1+t^2} + \dfrac{2t^2}{1+t^2} = \dfrac{1+t^2}{1+t^2} = 1

Element (1,2)(1,2): −2t1+t2+t(1−t2)1+t2=−2t+t−t31+t2=−t(1+t2)1+t2=−t\dfrac{-2t}{1+t^2} + \dfrac{t(1-t^2)}{1+t^2} = \dfrac{-2t+t-t^3}{1+t^2} = \dfrac{-t(1+t^2)}{1+t^2} = -t

Element (2,1)(2,1): −t(1−t2)1+t2+2t1+t2=−t+t3+2t1+t2=t(1+t2)1+t2=t\dfrac{-t(1-t^2)}{1+t^2} + \dfrac{2t}{1+t^2} = \dfrac{-t+t^3+2t}{1+t^2} = \dfrac{t(1+t^2)}{1+t^2} = t

Element (2,2)(2,2): 2t21+t2+1−t21+t2=1+t21+t2=1\dfrac{2t^2}{1+t^2} + \dfrac{1-t^2}{1+t^2} = \dfrac{1+t^2}{1+t^2} = 1

RHS=[1amp;−ttamp;1]=I+A=LHS\text{RHS} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} = I + A = \text{LHS}

Hence proved.
19A trust fund has ₹30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) ₹1800, (b) ₹2000Show solution
Let the amount invested in the first bond be ₹xx and in the second bond be ₹(30000−x)(30000 - x).

Using matrix multiplication, the interest earned is:
[x(30000−x)][0.050.07]=Total Interest[x \quad (30000-x)]\begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix} = \text{Total Interest}

(a) Total interest = ₹1800:
0.05x+0.07(30000−x)=18000.05x + 0.07(30000-x) = 1800
0.05x+2100−0.07x=18000.05x + 2100 - 0.07x = 1800
−0.02x=−300-0.02x = -300
x=15000x = 15000

∴ ₹15,000 in the first bond and ₹15,000 in the second bond.

(b) Total interest = ₹2000:
0.05x+0.07(30000−x)=20000.05x + 0.07(30000-x) = 2000
0.05x+2100−0.07x=20000.05x + 2100 - 0.07x = 2000
−0.02x=−100-0.02x = -100
x=5000x = 5000

∴ ₹5,000 in the first bond and ₹25,000 in the second bond.
20The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹80, ₹60 and ₹40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.Show solution
Number of books (in dozens converted to units):
- Chemistry: 10×12=12010 \times 12 = 120
- Physics: 8×12=968 \times 12 = 96
- Economics: 10×12=12010 \times 12 = 120

Represent as row matrix: A=[12096120]A = [120 \quad 96 \quad 120]

Selling prices as column matrix: B=[806040]B = \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}

Total amount = ABAB:
AB=[120×80+96×60+120×40]AB = [120 \times 80 + 96 \times 60 + 120 \times 40]
=[9600+5760+4800]= [9600 + 5760 + 4800]
=[20160]= [20160]

∴Total amount=₹20,160\therefore \text{Total amount} = \mathbf{₹20,160}
21The restriction on n,kn, k and pp so that PY+WYPY + WY will be defined are: (A) k=3,p=nk=3, p=n (B) kk is arbitrary, p=2p=2 (C) pp is arbitrary, k=3k=3 (D) k=2,p=3k=2, p=3Show solution
Correct Answer: (A) k=3,p=nk = 3, p = n

Given orders: X:2×nX: 2\times n, Y:3×kY: 3\times k, Z:2×pZ: 2\times p, W:n×3W: n\times 3, P:p×kP: p\times k.

For PYPY: PP is p×kp\times k and YY is 3×k3\times k. For multiplication PYPY to be defined, number of columns of PP must equal number of rows of YY: k=3k = 3. Result is p×kp\times k.

For WYWY: WW is n×3n\times 3 and YY is 3×k3\times k. This is defined (columns of WW = rows of YY = 3). Result is n×kn\times k.

For PY+WYPY + WY: Both must have the same order, so p=np = n.

Therefore k=3k = 3 and p=np = n.
22If n=pn = p, then the order of the matrix 7X−5Z7X - 5Z is: (A) p×2p\times 2 (B) 2×n2\times n (C) n×3n\times 3 (D) p×np\times nShow solution
Correct Answer: (B) 2×n2 \times n

Given: XX is of order 2×n2\times n and ZZ is of order 2×p2\times p.

Since n=pn = p, both XX and ZZ are of order 2×n2\times n.

Therefore 7X−5Z7X - 5Z is also of order 2×n\mathbf{2 \times n}.

Exercise 3.3

1Find the transpose of each of the following matrices: (i) [512−1]\begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}, (ii) [1amp;−12amp;3]\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}, (iii) [−1amp;5amp;63amp;5amp;62amp;3amp;−1]\begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}Show solution
Concept: Transpose of a matrix is obtained by interchanging its rows and columns.

(i) A=[512−1]3×1A = \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}_{3\times1}
A′=[5amp;12amp;−1]1×3A' = \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}_{1\times3}

(ii) A=[1amp;−12amp;3]A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}
A′=[1amp;2−1amp;3]A' = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}

(iii) A=[−1amp;5amp;63amp;5amp;62amp;3amp;−1]A = \begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}
A′=[−1amp;3amp;25amp;5amp;36amp;6amp;−1]A' = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}
2If A=[−1amp;2amp;35amp;7amp;9−2amp;1amp;1]A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} and B=[−4amp;1amp;−51amp;2amp;01amp;3amp;1]B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}, then verify that (i) (A+B)′=A′+B′(A+B)' = A'+B', (ii) (A−B)′=A′−B′(A-B)' = A'-B'Show solution
Computing A′A' and B′B':
A′=[−1amp;5amp;−22amp;7amp;13amp;9amp;1],B′=[−4amp;1amp;11amp;2amp;3−5amp;0amp;1]A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}, \quad B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}

(i) Verify (A+B)′=A′+B′(A+B)' = A'+B':
A+B=[−5amp;3amp;−26amp;9amp;9−1amp;4amp;2]A+B = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}
LHS: (A+B)′=[−5amp;6amp;−13amp;9amp;4−2amp;9amp;2]\text{LHS: }(A+B)' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}
RHS: A′+B′=[−1−4amp;5+1amp;−2+12+1amp;7+2amp;1+33−5amp;9+0amp;1+1]=[−5amp;6amp;−13amp;9amp;4−2amp;9amp;2]\text{RHS: }A'+B' = \begin{bmatrix} -1-4 & 5+1 & -2+1 \\ 2+1 & 7+2 & 1+3 \\ 3-5 & 9+0 & 1+1 \end{bmatrix} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}
LHS = RHS ✓

(ii) Verify (A−B)′=A′−B′(A-B)' = A'-B':
A−B=[3amp;1amp;84amp;5amp;9−3amp;−2amp;0]A-B = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}
LHS: (A−B)′=[3amp;4amp;−31amp;5amp;−28amp;9amp;0]\text{LHS: }(A-B)' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}
RHS: A′−B′=[−1+4amp;5−1amp;−2−12−1amp;7−2amp;1−33+5amp;9−0amp;1−1]=[3amp;4amp;−31amp;5amp;−28amp;9amp;0]\text{RHS: }A'-B' = \begin{bmatrix} -1+4 & 5-1 & -2-1 \\ 2-1 & 7-2 & 1-3 \\ 3+5 & 9-0 & 1-1 \end{bmatrix} = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}
LHS = RHS ✓
3If A′=[3amp;4−1amp;20amp;1]A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} and B=[−1amp;2amp;11amp;2amp;3]B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}, then verify that (i) (A+B)′=A′+B′(A+B)' = A'+B', (ii) (A−B)′=A′−B′(A-B)' = A'-B'Show solution
Finding AA from A′A':
A=(A′)′=[3amp;−1amp;04amp;2amp;1]A = (A')' = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}

Finding B′B':
B′=[−1amp;12amp;21amp;3]B' = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}

(i) Verify (A+B)′=A′+B′(A+B)' = A'+B':
A+B=[3−1amp;−1+2amp;0+14+1amp;2+2amp;1+3]=[2amp;1amp;15amp;4amp;4]A+B = \begin{bmatrix} 3-1 & -1+2 & 0+1 \\ 4+1 & 2+2 & 1+3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}
LHS: (A+B)′=[2amp;51amp;41amp;4]\text{LHS: }(A+B)' = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}
RHS: A′+B′=[3amp;4−1amp;20amp;1]+[−1amp;12amp;21amp;3]=[2amp;51amp;41amp;4]\text{RHS: }A'+B' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}
LHS = RHS ✓

(ii) Verify (A−B)′=A′−B′(A-B)' = A'-B':
A−B=[4amp;−3amp;−13amp;0amp;−2]A-B = \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix}
LHS: (A−B)′=[4amp;3−3amp;0−1amp;−2]\text{LHS: }(A-B)' = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}
RHS: A′−B′=[3amp;4−1amp;20amp;1]−[−1amp;12amp;21amp;3]=[4amp;3−3amp;0−1amp;−2]\text{RHS: }A'-B' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}
LHS = RHS ✓
4If A′=[−2amp;31amp;2]A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} and B=[−1amp;01amp;2]B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}, then find (A+2B)′(A+2B)'.Show solution
Finding AA:
A=(A′)′=[−2amp;13amp;2]A = (A')' = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}

Computing A+2BA + 2B:
2B=[−2amp;02amp;4]2B = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}
A+2B=[−2−2amp;1+03+2amp;2+4]=[−4amp;15amp;6]A + 2B = \begin{bmatrix} -2-2 & 1+0 \\ 3+2 & 2+4 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}

Computing (A+2B)′(A+2B)':
(A+2B)′=[−4amp;51amp;6](A+2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}
5For the matrices AA and BB, verify that (AB)′=B′A′(AB)' = B'A', where (i) A=[1−43]A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}, B=[−1amp;2amp;1]B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}, (ii) A=[012]A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}, B=[1amp;5amp;7]B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}Show solution
(i)
AB=[1−43][−1amp;2amp;1]=[−1amp;2amp;14amp;−8amp;−4−3amp;6amp;3]AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}\begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}

LHS: (AB)′=[−1amp;4amp;−32amp;−8amp;61amp;−4amp;3]\text{LHS: }(AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}

B′=[−121],A′=[1amp;−4amp;3]B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix},\quad A' = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}

RHS: B′A′=[−121][1amp;−4amp;3]=[−1amp;4amp;−32amp;−8amp;61amp;−4amp;3]\text{RHS: }B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}\begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}

LHS = RHS ✓

(ii)
AB=[012][1amp;5amp;7]=[0amp;0amp;01amp;5amp;72amp;10amp;14]AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}

LHS: (AB)′=[0amp;1amp;20amp;5amp;100amp;7amp;14]\text{LHS: }(AB)' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}

B′=[157],A′=[0amp;1amp;2]B' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix},\quad A' = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}

RHS: B′A′=[157][0amp;1amp;2]=[0amp;1amp;20amp;5amp;100amp;7amp;14]\text{RHS: }B'A' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}\begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}

LHS = RHS ✓
6If (i) A=[cos⁡αamp;sin⁡α−sin⁡αamp;cos⁡α]A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}, then verify that A′A=IA'A = I. (ii) If A=[sin⁡αamp;cos⁡α−cos⁡αamp;sin⁡α]A = \begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}, then verify that A′A=IA'A = I.Show solution
(i)
A′=[cos⁡αamp;−sin⁡αsin⁡αamp;cos⁡α]A' = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}

A′A=[cos⁡αamp;−sin⁡αsin⁡αamp;cos⁡α][cos⁡αamp;sin⁡α−sin⁡αamp;cos⁡α]A'A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}

=[cos⁡2α+sin⁡2αamp;cos⁡αsin⁡α−sin⁡αcos⁡αsin⁡αcos⁡α−cos⁡αsin⁡αamp;sin⁡2α+cos⁡2α]=[1amp;00amp;1]=I✓= \begin{bmatrix} \cos^2\alpha+\sin^2\alpha & \cos\alpha\sin\alpha-\sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha-\cos\alpha\sin\alpha & \sin^2\alpha+\cos^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \checkmark

(ii)
A′=[sin⁡αamp;−cos⁡αcos⁡αamp;sin⁡α]A' = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}

A′A=[sin⁡αamp;−cos⁡αcos⁡αamp;sin⁡α][sin⁡αamp;cos⁡α−cos⁡αamp;sin⁡α]A'A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}\begin{bmatrix} \sin\alpha & \cos\alpha \\ -\cos\alpha & \sin\alpha \end{bmatrix}

=[sin⁡2α+cos⁡2αamp;sin⁡αcos⁡α−cos⁡αsin⁡αcos⁡αsin⁡α−sin⁡αcos⁡αamp;cos⁡2α+sin⁡2α]=[1amp;00amp;1]=I✓= \begin{bmatrix} \sin^2\alpha+\cos^2\alpha & \sin\alpha\cos\alpha-\cos\alpha\sin\alpha \\ \cos\alpha\sin\alpha-\sin\alpha\cos\alpha & \cos^2\alpha+\sin^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \checkmark
7(i) Show that the matrix A=[1amp;−1amp;5−1amp;2amp;15amp;1amp;3]A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} is a symmetric matrix. (ii) Show that the matrix A=[0amp;1amp;−1−1amp;0amp;11amp;−1amp;0]A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} is a skew symmetric matrix.Show solution
(i) A matrix is symmetric if A′=AA' = A.

A′=[1amp;−1amp;5−1amp;2amp;15amp;1amp;3]=AA' = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} = A

Since A′=AA' = A, the matrix is symmetric. ✓

(ii) A matrix is skew symmetric if A′=−AA' = -A.

A′=[0amp;−1amp;11amp;0amp;−1−1amp;1amp;0]A' = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}

−A=[0amp;−1amp;11amp;0amp;−1−1amp;1amp;0]-A = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}

Since A′=−AA' = -A, the matrix is skew symmetric. ✓
8For the matrix A=[1amp;56amp;7]A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}, verify that (i) (A+A′)(A+A') is a symmetric matrix, (ii) (A−A′)(A-A') is a skew symmetric matrix.Show solution
A′=[1amp;65amp;7]A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}

(i) A+A′A + A':
A+A′=[2amp;1111amp;14]A + A' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}

Let P=A+A′P = A + A'. Then:
P′=[2amp;1111amp;14]=PP' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} = P

Since P′=PP' = P, (A+A′)(A+A') is a symmetric matrix. ✓

(ii) A−A′A - A':
A−A′=[0amp;−11amp;0]A - A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}

Let Q=A−A′Q = A - A'. Then:
Q′=[0amp;1−1amp;0]=−QQ' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -Q

Since Q′=−QQ' = -Q, (A−A′)(A-A') is a skew symmetric matrix. ✓
9Find 12(A+A′)\frac{1}{2}(A+A') and 12(A−A′)\frac{1}{2}(A-A'), when A=[0amp;aamp;b−aamp;0amp;c−bamp;−camp;0]A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}Show solution
A′=[0amp;−aamp;−baamp;0amp;−cbamp;camp;0]A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}

Computing 12(A+A′)\frac{1}{2}(A+A'):
A+A′=[0amp;0amp;00amp;0amp;00amp;0amp;0]A + A' = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
12(A+A′)=[0amp;0amp;00amp;0amp;00amp;0amp;0]\frac{1}{2}(A+A') = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

Computing 12(A−A′)\frac{1}{2}(A-A'):
A−A′=[0amp;2aamp;2b−2aamp;0amp;2c−2bamp;−2camp;0]A - A' = \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}
12(A−A′)=[0amp;aamp;b−aamp;0amp;c−bamp;−camp;0]=A\frac{1}{2}(A-A') = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} = A

(Note: Since AA is already skew symmetric, 12(A+A′)=O\frac{1}{2}(A+A') = O and 12(A−A′)=A\frac{1}{2}(A-A') = A.)
10Express the following matrices as the sum of a symmetric and a skew symmetric matrix: (i) [3amp;51amp;−1]\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}, (ii) [6amp;−2amp;2−2amp;3amp;−12amp;−1amp;3]\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}, (iii) [3amp;3amp;−1−2amp;−2amp;1−4amp;−5amp;2]\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}, (iv) [1amp;5−1amp;2]\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}Show solution
Concept: Any square matrix AA can be written as A=P+QA = P + Q where P=12(A+A′)P = \frac{1}{2}(A+A') is symmetric and Q=12(A−A′)Q = \frac{1}{2}(A-A') is skew symmetric.

(i) A=[3amp;51amp;−1]A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}, A′=[3amp;15amp;−1]A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}

P=12(A+A′)=12[6amp;66amp;−2]=[3amp;33amp;−1]P = \frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}

Q=12(A−A′)=12[0amp;4−4amp;0]=[0amp;2−2amp;0]Q = \frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}

A=[3amp;33amp;−1]+[0amp;2−2amp;0]A = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}

(ii) A=[6amp;−2amp;2−2amp;3amp;−12amp;−1amp;3]A = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}

Since A′=AA' = A, the matrix is already symmetric.
P=A=[6amp;−2amp;2−2amp;3amp;−12amp;−1amp;3],Q=OP = A = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}, \quad Q = O

(iii) A=[3amp;3amp;−1−2amp;−2amp;1−4amp;−5amp;2]A = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}, A′=[3amp;−2amp;−43amp;−2amp;−5−1amp;1amp;2]A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}

P=12(A+A′)=12[6amp;1amp;−51amp;−4amp;−4−5amp;−4amp;4]=[3amp;12amp;−5212amp;−2amp;−2−52amp;−2amp;2]P = \frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix}

Q=12(A−A′)=12[0amp;5amp;3−5amp;0amp;6−3amp;−6amp;0]=[0amp;52amp;32−52amp;0amp;3−32amp;−3amp;0]Q = \frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix}

A=[3amp;12amp;−5212amp;−2amp;−2−52amp;−2amp;2]+[0amp;52amp;32−52amp;0amp;3−32amp;−3amp;0]A = \begin{bmatrix} 3 & \frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & -2 & -2 \\ -\frac{5}{2} & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & \frac{5}{2} & \frac{3}{2} \\ -\frac{5}{2} & 0 & 3 \\ -\frac{3}{2} & -3 & 0 \end{bmatrix}

(iv) A=[1amp;5−1amp;2]A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}, A′=[1amp;−15amp;2]A' = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}

P=12(A+A′)=12[2amp;44amp;4]=[1amp;22amp;2]P = \frac{1}{2}(A+A') = \frac{1}{2}\begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}

Q=12(A−A′)=12[0amp;6−6amp;0]=[0amp;3−3amp;0]Q = \frac{1}{2}(A-A') = \frac{1}{2}\begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}

A=[1amp;22amp;2]+[0amp;3−3amp;0]A = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}
11If A, B are symmetric matrices of same order, then AB – BA is a (A) Skew symmetric matrix (B) Symmetric matrix (C) Zero matrix (D) Identity matrixShow solution
Correct Answer: (A) Skew symmetric matrix

Since AA and BB are symmetric: A′=AA' = A and B′=BB' = B.

Let P=AB−BAP = AB - BA. Then:
P′=(AB−BA)′=(AB)′−(BA)′=B′A′−A′B′=BA−AB=−(AB−BA)=−PP' = (AB-BA)' = (AB)' - (BA)' = B'A' - A'B' = BA - AB = -(AB-BA) = -P

Since P′=−PP' = -P, AB−BAAB - BA is a skew symmetric matrix.
12If A=[cos⁡αamp;−sin⁡αsin⁡αamp;cos⁡α]A = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}, and A+A′=IA + A' = I, then the value of α\alpha is (A) π6\frac{\pi}{6} (B) π3\frac{\pi}{3} (C) π\pi (D) 3π2\frac{3\pi}{2}Show solution
Correct Answer: (B) π3\dfrac{\pi}{3}

A′=[cos⁡αamp;sin⁡α−sin⁡αamp;cos⁡α]A' = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}

A+A′=[2cos⁡αamp;00amp;2cos⁡α]=[1amp;00amp;1]=IA + A' = \begin{bmatrix} 2\cos\alpha & 0 \\ 0 & 2\cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

Equating: 2cos⁡α=1⇒cos⁡α=12⇒α=π32\cos\alpha = 1 \Rightarrow \cos\alpha = \dfrac{1}{2} \Rightarrow \alpha = \dfrac{\pi}{3}

Exercise 3.4

1Matrices A and B will be inverse of each other only if (A) AB=BAAB = BA (B) AB=BA=0AB = BA = 0 (C) AB=0AB = 0, BA=IBA = I (D) AB=BA=IAB = BA = IShow solution
Correct Answer: (D) AB=BA=IAB = BA = I

By definition, matrix BB is the inverse of matrix AA if and only if AB=BA=IAB = BA = I, where II is the identity matrix of the same order. This is the standard definition of the inverse of a matrix.

Miscellaneous Exercise on Chapter 3

1If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.Show solution
Given: AA and BB are symmetric matrices, so A′=AA' = A and B′=BB' = B.

To prove: AB−BAAB - BA is skew symmetric, i.e., (AB−BA)′=−(AB−BA)(AB-BA)' = -(AB-BA).

Proof:
(AB−BA)′=(AB)′−(BA)′(AB - BA)' = (AB)' - (BA)'
=B′A′−A′B′[using (PQ)′=Q′P′]= B'A' - A'B' \quad [\text{using } (PQ)' = Q'P']
=BA−AB[since A′=A,B′=B]= BA - AB \quad [\text{since } A'=A, B'=B]
=−(AB−BA)= -(AB - BA)

Since (AB−BA)′=−(AB−BA)(AB-BA)' = -(AB-BA), the matrix AB−BAAB - BA is skew symmetric. Hence proved.
2Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.Show solution
Case 1: A is symmetric, i.e., A′=AA' = A.

Let P=B′ABP = B'AB. Then:
P′=(B′AB)′=B′A′(B′)′=B′A′B=B′AB=PP' = (B'AB)' = B'A'(B')' = B'A'B = B'AB = P
(since A′=AA' = A and (B′)′=B(B')' = B)

Since P′=PP' = P, B′ABB'AB is symmetric.

Case 2: A is skew symmetric, i.e., A′=−AA' = -A.

P′=(B′AB)′=B′A′(B′)′=B′A′B=B′(−A)B=−B′AB=−PP' = (B'AB)' = B'A'(B')' = B'A'B = B'(-A)B = -B'AB = -P

Since P′=−PP' = -P, B′ABB'AB is skew symmetric.

Hence proved.
3Find the values of x,y,zx, y, z if the matrix A=[0amp;2yamp;zxamp;yamp;−zxamp;−yamp;z]A = \begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} satisfy the equation A′A=IA'A = I.Show solution
Computing A′A':
A′=[0amp;xamp;x2yamp;yamp;−yzamp;−zamp;z]A' = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}

Computing A′A=IA'A = I:
A′A=[0amp;xamp;x2yamp;yamp;−yzamp;−zamp;z][0amp;2yamp;zxamp;yamp;−zxamp;−yamp;z]=[1amp;0amp;00amp;1amp;00amp;0amp;1]A'A = \begin{bmatrix} 0 & x & x \\ 2y & y & -y \\ z & -z & z \end{bmatrix}\begin{bmatrix} 0 & 2y & z \\ x & y & -z \\ x & -y & z \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

Element (1,1)(1,1): 0+x2+x2=1⇒2x2=1⇒x2=12⇒x=±120 + x^2 + x^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x^2 = \dfrac{1}{2} \Rightarrow x = \pm\dfrac{1}{\sqrt{2}}

Element (2,2)(2,2): 4y2+y2+y2=1⇒6y2=1⇒y2=16⇒y=±164y^2 + y^2 + y^2 = 1 \Rightarrow 6y^2 = 1 \Rightarrow y^2 = \dfrac{1}{6} \Rightarrow y = \pm\dfrac{1}{\sqrt{6}}

Element (3,3)(3,3): z2+z2+z2=1⇒3z2=1⇒z2=13⇒z=±13z^2 + z^2 + z^2 = 1 \Rightarrow 3z^2 = 1 \Rightarrow z^2 = \dfrac{1}{3} \Rightarrow z = \pm\dfrac{1}{\sqrt{3}}

Verification of off-diagonal elements confirms these values are consistent.

x=±12,  y=±16,  z=±13\boxed{x = \pm\frac{1}{\sqrt{2}},\; y = \pm\frac{1}{\sqrt{6}},\; z = \pm\frac{1}{\sqrt{3}}}
4For what values of xx: [121][1amp;2amp;02amp;0amp;11amp;0amp;2][02x]=0[1\quad 2\quad 1]\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = 0?Show solution
Step 1: Compute [121][1amp;2amp;02amp;0amp;11amp;0amp;2][1\quad 2\quad 1]\begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}

=[1+4+1,  2+0+0,  0+2+2]=[6,  2,  4]= [1+4+1,\; 2+0+0,\; 0+2+2] = [6,\; 2,\; 4]

Step 2: Multiply by [02x]\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix}:

[624][02x]=0+4+4x=4+4x[6\quad 2\quad 4]\begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = 0 + 4 + 4x = 4 + 4x

Step 3: Set equal to 0:
4+4x=0⇒x=−14 + 4x = 0 \Rightarrow x = -1

x=−1\boxed{x = -1}
5If A=[3amp;1−1amp;2]A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}, show that A2−5A+7I=0A^2 - 5A + 7I = 0.Show solution
Step 1: Compute A2A^2
A2=[3amp;1−1amp;2][3amp;1−1amp;2]=[9−1amp;3+2−3−2amp;−1+4]=[8amp;5−5amp;3]A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}

Step 2: Compute 5A5A
5A=[15amp;5−5amp;10]5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}

Step 3: Compute 7I7I
7I=[7amp;00amp;7]7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}

Step 4:
A2−5A+7I=[8−15+7amp;5−5+0−5+5+0amp;3−10+7]=[0amp;00amp;0]=OA^2 - 5A + 7I = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O

Hence proved.
6Find xx, if [xamp;−5amp;−1][1amp;0amp;20amp;2amp;12amp;0amp;3][x41]=0\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = 0Show solution
Step 1: Compute [xamp;−5amp;−1][1amp;0amp;20amp;2amp;12amp;0amp;3]\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}

=[x+0−2,  0−10+0,  2x−5−3]=[x−2,  −10,  2x−8]= [x+0-2,\; 0-10+0,\; 2x-5-3] = [x-2,\; -10,\; 2x-8]

Step 2: Multiply by [x41]\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix}:

[x−2,  −10,  2x−8][x41]=x(x−2)−40+(2x−8)=0[x-2,\; -10,\; 2x-8]\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = x(x-2) - 40 + (2x-8) = 0

x2−2x−40+2x−8=0x^2 - 2x - 40 + 2x - 8 = 0
x2−48=0x^2 - 48 = 0
x2=48x^2 = 48
x=±43x = \pm 4\sqrt{3}
7A manufacturer produces three products x,y,zx, y, z which he sells in two markets. Annual sales are indicated below: Market I: 10,000; 2,000; 18,000 and Market II: 6,000; 20,000; 8,000. (a) If unit sale prices of x,yx, y and zz are ₹2.50, ₹1.50 and ₹1.00 respectively, find the total revenue in each market with the help of matrix algebra. (b) If the unit costs of the above three commodities are ₹2.00, ₹1.00 and 50 paise respectively. Find the gross profit.Show solution
Sales matrix:
S=[10000amp;2000amp;180006000amp;20000amp;8000]S = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}

(a) Revenue:

Price matrix: P=[2.501.501.00]P = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}

Revenue=SP=[10000amp;2000amp;180006000amp;20000amp;8000][2.501.501.00]\text{Revenue} = SP = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}\begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}

Market I: 10000(2.50)+2000(1.50)+18000(1.00)=25000+3000+18000=4600010000(2.50) + 2000(1.50) + 18000(1.00) = 25000 + 3000 + 18000 = 46000

Market II: 6000(2.50)+20000(1.50)+8000(1.00)=15000+30000+8000=530006000(2.50) + 20000(1.50) + 8000(1.00) = 15000 + 30000 + 8000 = 53000

Revenue=[4600053000]\text{Revenue} = \begin{bmatrix} 46000 \\ 53000 \end{bmatrix}

Total revenue: Market I = ₹46,000, Market II = ₹53,000.

(b) Gross Profit:

Cost matrix: C=[2.001.000.50]C = \begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}

Cost=SC=[10000amp;2000amp;180006000amp;20000amp;8000][2.001.000.50]\text{Cost} = SC = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}\begin{bmatrix} 2.00 \\ 1.00 \\ 0.50 \end{bmatrix}

Market I: 10000(2)+2000(1)+18000(0.5)=20000+2000+9000=3100010000(2) + 2000(1) + 18000(0.5) = 20000 + 2000 + 9000 = 31000

Market II: 6000(2)+20000(1)+8000(0.5)=12000+20000+4000=360006000(2) + 20000(1) + 8000(0.5) = 12000 + 20000 + 4000 = 36000

Gross Profit:
- Market I: 46000−31000=46000 - 31000 = ₹15,000
- Market II: 53000−36000=53000 - 36000 = ₹17,000
8Find the matrix X so that X[1amp;2amp;34amp;5amp;6]=[−7amp;−8amp;−92amp;4amp;6]X\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}Show solution
Let X=[aamp;bcamp;d]X = \begin{bmatrix} a & b \\ c & d \end{bmatrix} (since XX must be 2×22\times2 for the product to be 2×32\times3).

[aamp;bcamp;d][1amp;2amp;34amp;5amp;6]=[a+4bamp;2a+5bamp;3a+6bc+4damp;2c+5damp;3c+6d]=[−7amp;−8amp;−92amp;4amp;6]\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}

From row 1:
a+4b=−7⋯(1)a + 4b = -7 \quad \cdots(1)
2a+5b=−8⋯(2)2a + 5b = -8 \quad \cdots(2)

From (2) −2×- 2\times(1): −3b=6⇒b=−2-3b = 6 \Rightarrow b = -2

From (1): a=−7+8=1a = -7 + 8 = 1

From row 2:
c+4d=2⋯(3)c + 4d = 2 \quad \cdots(3)
2c+5d=4⋯(4)2c + 5d = 4 \quad \cdots(4)

From (4) −2×- 2\times(3): −3d=0⇒d=0-3d = 0 \Rightarrow d = 0

From (3): c=2c = 2

X=[1amp;−22amp;0]\boxed{X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}}
9If A=[αamp;βγamp;−α]A = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} is such that A2=IA^2 = I, then (A) 1+α2+βγ=01+\alpha^2+\beta\gamma=0 (B) 1−α2+βγ=01-\alpha^2+\beta\gamma=0 (C) 1−α2−βγ=01-\alpha^2-\beta\gamma=0 (D) 1+α2−βγ=01+\alpha^2-\beta\gamma=0Show solution
Correct Answer: (C) 1−α2−βγ=01 - \alpha^2 - \beta\gamma = 0

A2=[αamp;βγamp;−α][αamp;βγamp;−α]=[α2+βγamp;αβ−αβαγ−αγamp;βγ+α2]=[α2+βγamp;00amp;α2+βγ]A^2 = \begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \\ \gamma & -\alpha \end{bmatrix} = \begin{bmatrix} \alpha^2+\beta\gamma & \alpha\beta-\alpha\beta \\ \alpha\gamma-\alpha\gamma & \beta\gamma+\alpha^2 \end{bmatrix} = \begin{bmatrix} \alpha^2+\beta\gamma & 0 \\ 0 & \alpha^2+\beta\gamma \end{bmatrix}

For A2=IA^2 = I:
α2+βγ=1⇒1−α2−βγ=0\alpha^2 + \beta\gamma = 1 \Rightarrow 1 - \alpha^2 - \beta\gamma = 0
10If the matrix A is both symmetric and skew symmetric, then (A) A is a diagonal matrix (B) A is a zero matrix (C) A is a square matrix (D) None of theseShow solution
Correct Answer: (B) A is a zero matrix

If AA is symmetric: A′=AA' = A ...(1)

If AA is skew symmetric: A′=−AA' = -A ...(2)

From (1) and (2): A=−A⇒2A=O⇒A=OA = -A \Rightarrow 2A = O \Rightarrow A = O

Therefore AA must be the zero matrix.
11If A is square matrix such that A2=AA^2 = A, then (I+A)3−7A(I+A)^3 - 7A is equal to (A) A (B) I-A (C) I (D) 3AShow solution
Correct Answer: (C) I

Given: A2=AA^2 = A (A is idempotent)

Note: A2=A⇒A3=A2⋅A=A⋅A=A2=AA^2 = A \Rightarrow A^3 = A^2 \cdot A = A \cdot A = A^2 = A

Expanding (I+A)3(I+A)^3:
(I+A)3=I3+3I2A+3IA2+A3(I+A)^3 = I^3 + 3I^2A + 3IA^2 + A^3
=I+3A+3A2+A3= I + 3A + 3A^2 + A^3
=I+3A+3A+A[since A2=A3=A]= I + 3A + 3A + A \quad [\text{since } A^2 = A^3 = A]
=I+7A= I + 7A

Therefore:
(I+A)3−7A=I+7A−7A=I(I+A)^3 - 7A = I + 7A - 7A = I

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