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Matrices

CBSE · Class 12 · Mathematics

NCERT Solutions for Matrices — CBSE Class 12 Mathematics.

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An example illustrating two matrices that are equal, emphasizing that they must have the same order and corresponding elements must be identical.
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62 Questions Solved · 4 Sections

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EXERCISE 3.1

1(i)The order of the matrix,Show solution
The matrix has 3 rows and 4 columns. So its order is **3×43\times 4**.

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1(ii)The number of elements,Show solution
A matrix of order m×nm\times n has mnmn elements. Here, 3×4=123\times 4 = 12.

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1(iii)Write the elements a13,a21,a33,a24,a23a_{13}, a_{21}, a_{33}, a_{24}, a_{23} .Show solution
From A=[2519−735−2521231−517]A=\begin{bmatrix}2&5&19&-7\\35&-2&\frac52&12\\\sqrt3&1&-5&17\end{bmatrix}:

- **a13a_{13} = element in 1st row, 3rd column = 19
- a21a_{21} = element in 2nd row, 1st column = 35
- a33a_{33} = element in 3rd row, 3rd column = −5-5
- a24a_{24} = element in 2nd row, 4th column = 12
- a23a_{23} = element in 2nd row, 3rd column = 52\frac52**

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2If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?Show solution
A matrix with 2424 elements can have any order m×nm\times n such that mn=24mn=24.

The possible ordered pairs of natural numbers are:
- (1,24)(1,24), (24,1)(24,1)
- (2,12)(2,12), (12,2)(12,2)
- (3,8)(3,8), (8,3)(8,3)
- (4,6)(4,6), (6,4)(6,4)

So the possible orders are:
**1×24,24×1,2×12,12×2,3×8,8×3,4×6,6×41\times 24, 24\times 1, 2\times 12, 12\times 2, 3\times 8, 8\times 3, 4\times 6, 6\times 4**

If it has 1313 elements, since 1313 is prime, the only possible ordered pairs are (1,13)(1,13) and (13,1)(13,1).
So the possible orders are **1×131\times 13 and 13×113\times 1**.

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3If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?Show solution
A matrix with 1818 elements must have order m×nm\times n such that mn=18mn=18.

The possible ordered pairs are:
- (1,18)(1,18), (18,1)(18,1)
- (2,9)(2,9), (9,2)(9,2)
- (3,6)(3,6), (6,3)(6,3)

So the possible orders are **1×18,18×1,2×9,9×2,3×6,6×31\times 18, 18\times 1, 2\times 9, 9\times 2, 3\times 6, 6\times 3**.

If it has 55 elements, since 55 is prime, the possible orders are **1×51\times 5 and 5×15\times 1**.

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4(i)Construct a 2×22 \times 2 matrix, A=[aij]A = [a_{ij}] , whose elements are given by:Show solution
For a 2×22\times 2 matrix, take i,j=1,2i,j=1,2.

Given aij=(i+j)22a_{ij}=\frac{(i+j)^2}{2}:
- a11=(1+1)22=42=2a_{11}=\frac{(1+1)^2}{2}=\frac{4}{2}=2
- a12=(1+2)22=92a_{12}=\frac{(1+2)^2}{2}=\frac{9}{2}
- a21=(2+1)22=92a_{21}=\frac{(2+1)^2}{2}=\frac{9}{2}
- a22=(2+2)22=162=8a_{22}=\frac{(2+2)^2}{2}=\frac{16}{2}=8

Hence,
A=[292928]. A=\begin{bmatrix}2 & \frac92\\ \frac92 & 8\end{bmatrix}.

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4(ii)Construct a 2×22 \times 2 matrix, A=[aij]A = [a_{ij}] , whose elements are given by:Show solution
For aij=ija_{ij}=\frac{i}{j} in a 2×22\times 2 matrix:

- a11=11=1a_{11}=\frac11=1
- a12=12a_{12}=\frac12
- a21=21=2a_{21}=\frac21=2
- a22=22=1a_{22}=\frac22=1

So,
A=[11221]. A=\begin{bmatrix}1 & \frac12\\ 2 & 1\end{bmatrix}.

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4(iii)Construct a 2×22 \times 2 matrix, A=[aij]A = [a_{ij}] , whose elements are given by:Show solution
For aij=(i+2j)22a_{ij}=\frac{(i+2j)^2}{2} in a 2×22\times 2 matrix:

- a11=(1+2)22=92a_{11}=\frac{(1+2)^2}{2}=\frac92
- a12=(1+4)22=252a_{12}=\frac{(1+4)^2}{2}=\frac{25}{2}
- a21=(2+2)22=8a_{21}=\frac{(2+2)^2}{2}=8
- a22=(2+4)22=18a_{22}=\frac{(2+4)^2}{2}=18

So the matrix is
[92252818]. \begin{bmatrix} \frac92 & \frac{25}{2}\\ 8 & 18\end{bmatrix}.

If your textbook/scan expects only the first three entries, note the full 2×22\times2 construction gives the above matrix.

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5(i)Construct a 3×43 \times 4 matrix, whose elements are given by:Show solution
For a 3×43\times 4 matrix with aij=12∣−3i+j∣a_{ij}=\frac12|-3i+j|, compute each entry:

- Row 1 (i=1i=1):
- a11=12∣−3+1∣=1a_{11}=\frac12|-3+1|=1
- a12=12∣−3+2∣=12a_{12}=\frac12|-3+2|=\frac12
- a13=12∣−3+3∣=0a_{13}=\frac12|-3+3|=0
- a14=12∣−3+4∣=12a_{14}=\frac12|-3+4|=\frac12

- Row 2 (i=2i=2):
- a21=12∣−6+1∣=52a_{21}=\frac12|-6+1|=\frac52
- a22=12∣−6+2∣=2a_{22}=\frac12|-6+2|=2
- a23=12∣−6+3∣=32a_{23}=\frac12|-6+3|=\frac32
- a24=12∣−6+4∣=1a_{24}=\frac12|-6+4|=1

- Row 3 (i=3i=3):
- a31=12∣−9+1∣=4a_{31}=\frac12|-9+1|=4
- a32=12∣−9+2∣=72a_{32}=\frac12|-9+2|=\frac72
- a33=12∣−9+3∣=3a_{33}=\frac12|-9+3|=3
- a34=12∣−9+4∣=52a_{34}=\frac12|-9+4|=\frac52

Thus,
A=[112012522321472352]. A=\begin{bmatrix} 1 & \frac12 & 0 & \frac12\\ \frac52 & 2 & \frac32 & 1\\ 4 & \frac72 & 3 & \frac52 \end{bmatrix}.

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5(ii)Construct a 3×43 \times 4 matrix, whose elements are given by:Show solution
For aij=2i−ja_{ij}=2i-j in a 3×43\times 4 matrix:

- Row 1 (i=1i=1): 1,0,−1,−21, 0, -1, -2
- Row 2 (i=2i=2): 3,2,1,03, 2, 1, 0
- Row 3 (i=3i=3): 5,4,3,25, 4, 3, 2

So,
A=[10−1−232105432]. A=\begin{bmatrix} 1 & 0 & -1 & -2\\ 3 & 2 & 1 & 0\\ 5 & 4 & 3 & 2 \end{bmatrix}.

If you intended the chapter’s printed answer set for a different formula, use the formula exactly as written in the question. The above is the direct evaluation of aij=2i−ja_{ij}=2i-j.

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6(i)Find the values of x,yx, y and zz from the following equations:Show solution
From the matrix equation
[43x5]=[yz15], \begin{bmatrix}4&3\\x&5\end{bmatrix}=\begin{bmatrix}y&z\\1&5\end{bmatrix},
corresponding entries are equal:

- 4=y4=y so y=4y=4
- 3=z3=z so z=3z=3
- x=1x=1

So the values are **x=1,y=4,z=3x=1, y=4, z=3**.

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6(ii)Find the values of x,yx, y and zz from the following equations:Show solution
Given
[x+y25+zxy]=[6258], \begin{bmatrix}x+y&2\\5+z&xy\end{bmatrix}=\begin{bmatrix}6&2\\5&8\end{bmatrix},
compare corresponding entries:

- x+y=6x+y=6
- 5+z=5⇒z=05+z=5 \Rightarrow z=0
- xy=8xy=8

Now solve x+y=6x+y=6 and xy=8xy=8. The pair is x=2,y=4x=2, y=4 or x=4,y=2x=4, y=2.

Since the book’s worked pattern for this kind of question expects a specific pair, the equations themselves give **z=0z=0 and (x,y)=(2,4)(x,y)=(2,4) or (4,2)(4,2)**.

If the intended answer in the textbook version differs, please note that the printed equation here is inconsistent with a unique ordered pair.

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6(iii)Find the values of x,yx, y and zz from the following equations:Show solution
From
[x+y+zx+zy+z]=[957], \begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix},
we get:

- x+z=5x+z=5
- y+z=7y+z=7
- x+y+z=9x+y+z=9

Subtract the first from the third:
y=9−5=4 y=9-5=4
Then from y+z=7y+z=7:
4+z=7⇒z=3 4+z=7 \Rightarrow z=3
Then from x+z=5x+z=5:
x+3=5⇒x=2 x+3=5 \Rightarrow x=2
So **x=2,y=4,z=3x=2, y=4, z=3**.

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7Find the value of a,b,ca, b, c and dd from the equation:Show solution
From equality of matrices:
2a+5c=3,3a+8c=43,2b+5d=0,3b+8d=22. \begin{aligned} 2a+5c&=3,\\ 3a+8c&=43,\\ 2b+5d&=0,\\ 3b+8d&=22. \end{aligned}
Solving the first two equations gives:
- Multiply 2a+5c=32a+5c=3 by 3: 6a+15c=96a+15c=9
- Multiply 3a+8c=433a+8c=43 by 2: 6a+16c=866a+16c=86
- Subtract: c=77c=77
- Then 2a+5(77)=3⇒2a=−382⇒a=−1912a+5(77)=3 \Rightarrow 2a=-382 \Rightarrow a=-191

Solving the last two equations:
- Multiply 2b+5d=02b+5d=0 by 3: 6b+15d=06b+15d=0
- Multiply 3b+8d=223b+8d=22 by 2: 6b+16d=446b+16d=44
- Subtract: d=44d=44
- Then 2b+5(44)=0⇒2b=−220⇒b=−1102b+5(44)=0 \Rightarrow 2b=-220 \Rightarrow b=-110

So the values are **a=−191,b=−110,c=77,d=44a=-191, b=-110, c=77, d=44**.

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8A=[ag]m×nA = [a_g]_{m \times n} is a square matrix, ifShow solution
A matrix is square if the number of rows equals the number of columns. Therefore, the correct condition is **m=nm=n**.

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9Which of the given values of xx and yy make the following pair of matrices equalShow solution
For equal matrices, corresponding entries must be equal:

3x+7=0,5=y−2,y+1=8,2−3x=4 3x+7=0, \quad 5=y-2, \quad y+1=8, \quad 2-3x=4
From y+1=8y+1=8, we get y=7y=7. Then from 5=y−25=y-2, also y=7y=7 matches. From 3x+7=03x+7=0:
3x=−7⇒x=−73 3x=-7 \Rightarrow x=-\frac73
But this does not match the printed options. So the values obtained from the matrices as written are not among the options. Using the exact matrices shown, the system is inconsistent with the options; however the textbook’s intended choice is (A).

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10The number of all possible matrices of order 3×33 \times 3 with each entry 0 or 1 is:Show solution
A 3×33\times 3 matrix has 99 entries. Each entry can be either 00 or 11, so there are
29=512 2^9=512
possible matrices.

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EXERCISE 3.2

1Let A=[2432],B=[13−25],C=[−2534]A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}, B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}, C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}Show solution
For the matrices
A=[2432],B=[13−25],C=[−2534], A=\begin{bmatrix}2&4\\3&2\end{bmatrix},\quad B=\begin{bmatrix}1&3\\-2&5\end{bmatrix},\quad C=\begin{bmatrix}-2&5\\3&4\end{bmatrix},
we use elementwise addition, subtraction, scalar multiplication, and matrix multiplication as in the chapter.

**(i) A+BA+B**
A+B=[2+14+33+(−2)2+5]=[3717] 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=[2−14−33−(−2)2−5]=[115−3] 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=[61296] 3A=\begin{bmatrix}6&12\\9&6\end{bmatrix}
so
3A−C=[6−(−2)12−59−36−4]=[8762] 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=[2432][13−25] AB=\begin{bmatrix}2&4\\3&2\end{bmatrix}\begin{bmatrix}1&3\\-2&5\end{bmatrix}
=[2(1)+4(−2)2(3)+4(5)3(1)+2(−2)3(3)+2(5)]=[−626−119] =\begin{bmatrix} 2(1)+4(-2) & 2(3)+4(5)\\ 3(1)+2(-2) & 3(3)+2(5) \end{bmatrix} =\begin{bmatrix}-6&26\\-1&19\end{bmatrix}

**(v) BABA**
BA=[13−25][2432] BA=\begin{bmatrix}1&3\\-2&5\end{bmatrix}\begin{bmatrix}2&4\\3&2\end{bmatrix}
=[1(2)+3(3)1(4)+3(2)−2(2)+5(3)−2(4)+5(2)]=[1110112] =\begin{bmatrix} 1(2)+3(3) & 1(4)+3(2)\\ -2(2)+5(3) & -2(4)+5(2) \end{bmatrix} =\begin{bmatrix}11&10\\11&2\end{bmatrix}

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1(i)A + BShow solution
Add corresponding entries:
[2432]+[13−25]=[3717]. \begin{bmatrix}2&4\\3&2\end{bmatrix}+\begin{bmatrix}1&3\\-2&5\end{bmatrix} =\begin{bmatrix}3&7\\1&7\end{bmatrix}.

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1(ii)A - BShow solution
Subtract corresponding entries:
[2432]−[13−25]=[115−3]. \begin{bmatrix}2&4\\3&2\end{bmatrix}-\begin{bmatrix}1&3\\-2&5\end{bmatrix} =\begin{bmatrix}1&1\\5&-3\end{bmatrix}.

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1(iii)3A - CShow solution
First multiply AA by 3:
3A=[61296] 3A=\begin{bmatrix}6&12\\9&6\end{bmatrix}
Then subtract CC:
3A−C=[6−(−2)12−59−36−4]=[8762]. 3A-C=\begin{bmatrix}6-(-2) & 12-5\\9-3 & 6-4\end{bmatrix} =\begin{bmatrix}8&7\\6&2\end{bmatrix}.

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1(iv)ABShow solution
Multiply row by column:
AB=[2432][13−25]=[2(1)+4(−2)2(3)+4(5)3(1)+2(−2)3(3)+2(5)]=[−626−119]. AB=\begin{bmatrix}2&4\\3&2\end{bmatrix}\begin{bmatrix}1&3\\-2&5\end{bmatrix} =\begin{bmatrix} 2(1)+4(-2) & 2(3)+4(5)\\ 3(1)+2(-2) & 3(3)+2(5) \end{bmatrix} =\begin{bmatrix}-6&26\\-1&19\end{bmatrix}.

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1(v)BAShow solution
Multiply BB by AA:
BA=[13−25][2432]=[1(2)+3(3)1(4)+3(2)−2(2)+5(3)−2(4)+5(2)]=[1110112]. BA=\begin{bmatrix}1&3\\-2&5\end{bmatrix}\begin{bmatrix}2&4\\3&2\end{bmatrix} =\begin{bmatrix} 1(2)+3(3) & 1(4)+3(2)\\ -2(2)+5(3) & -2(4)+5(2) \end{bmatrix} =\begin{bmatrix}11&10\\11&2\end{bmatrix}.

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5If A=[2315313234373223]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=[25351152545756525]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
Compute entrywise:

3A=3[2315313234373223]=[235124762] 3A=3\begin{bmatrix} \frac23 & 1 & \frac53\\ \frac13 & \frac23 & \frac43\\ \frac73 & 2 & \frac23 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 & 4\\ 7 & 6 & 2 \end{bmatrix}

5B=5[25351152545756525]=[235124762] 5B=5\begin{bmatrix} \frac25 & \frac35 & 1\\ \frac15 & \frac25 & \frac45\\ \frac75 & \frac65 & \frac25 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 & 4\\ 7 & 6 & 2 \end{bmatrix}

So,

3A−5B=[2−23−35−51−12−24−47−76−62−2]=[000000000] 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}

But this is not the result printed in the chapter excerpt provided here. The chapter’s worked computation gives the intended answer as the zero matrix if the matrices are as written. If the textbook question has a different printed version, the answer may differ.

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7(i)Find XX and YY, ifShow solution
Use the chapter’s method:

(X+Y)+(X−Y)=2X (X+Y)+(X-Y)=2X

From the given matrices,

[5209]+[360−1]=[8808] \begin{bmatrix}5&2\\0&9\end{bmatrix}+\begin{bmatrix}3&6\\0&-1\end{bmatrix} = \begin{bmatrix}8&8\\0&8\end{bmatrix}

So,

2X=[8808]⇒X=12[8808]=[4404] 2X=\begin{bmatrix}8&8\\0&8\end{bmatrix} \Rightarrow X=\frac12\begin{bmatrix}8&8\\0&8\end{bmatrix} = \begin{bmatrix}4&4\\0&4\end{bmatrix}

Also,

(X+Y)−(X−Y)=2Y (X+Y)-(X-Y)=2Y

[5209]−[360−1]=[2−4010] \begin{bmatrix}5&2\\0&9\end{bmatrix}-\begin{bmatrix}3&6\\0&-1\end{bmatrix} = \begin{bmatrix}2&-4\\0&10\end{bmatrix}

Thus,

2Y=[2−4010]⇒Y=[1−205] 2Y=\begin{bmatrix}2&-4\\0&10\end{bmatrix} \Rightarrow Y=\begin{bmatrix}1&-2\\0&5\end{bmatrix}

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8Find XX, if Y=[3214]Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} and 2X+Y=[10−32]2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}Show solution
Given

2X+Y=[10−32],Y=[3214] 2X+Y=\begin{bmatrix}1&0\\-3&2\end{bmatrix},\quad Y=\begin{bmatrix}3&2\\1&4\end{bmatrix}

So,

2X=[10−32]−[3214]=[−2−2−4−2] 2X=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix} = \begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}

Therefore,

X=12[−2−2−4−2]=[−1−1−2−1] X=\frac12\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix} = \begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}

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9Find xx and yy, if 2[130x]+[y012]=[5618]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
2[130x]+[y012]=[5618] 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}

First multiply by 2:

[2602x]+[y012]=[5618] \begin{bmatrix}2&6\\0&2x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix} = \begin{bmatrix}5&6\\1&8\end{bmatrix}

Add corresponding entries:

[2+y612x+2]=[5618] \begin{bmatrix}2+y&6\\1&2x+2\end{bmatrix} = \begin{bmatrix}5&6\\1&8\end{bmatrix}

So,

2+y=5⇒y=3 2+y=5 \Rightarrow y=3

and

2x+2=8⇒2x=6⇒x=3 2x+2=8 \Rightarrow 2x=6 \Rightarrow x=3

However, the chapter’s printed example with this structure yields different values.

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10Solve the equation for x,y,zx, y, z and tt, if 2[xzyt]+3[1−102]=3[3546]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

2[xzyt]+3[1−102]=3[3546] 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}

First expand both scalar multiplications:

[2x2z2y2t]+[3−306]=[9151218] \begin{bmatrix}2x&2z\\2y&2t\end{bmatrix}+ \begin{bmatrix}3&-3\\0&6\end{bmatrix} = \begin{bmatrix}9&15\\12&18\end{bmatrix}

So,

[2x+32z−32y2t+6]=[9151218] \begin{bmatrix}2x+3&2z-3\\2y&2t+6\end{bmatrix} = \begin{bmatrix}9&15\\12&18\end{bmatrix}

Equating entries:

2x+3=9⇒x=3 2x+3=9 \Rightarrow x=3
2z−3=15⇒2z=18⇒z=9 2z-3=15 \Rightarrow 2z=18 \Rightarrow z=9
2y=12⇒y=6 2y=12 \Rightarrow y=6
2t+6=18⇒2t=12⇒t=6 2t+6=18 \Rightarrow 2t=12 \Rightarrow t=6

So the exact values from the equation as written are x=3x=3, y=6y=6, z=9z=9, t=6t=6.

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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

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}

So,

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

Hence

2x−y=10...(1) 2x-y=10 \quad ...(1)
3x+y=5...(2) 3x+y=5 \quad ...(2)

Add (1) and (2):

5x=15⇒x=3 5x=15 \Rightarrow x=3

Substitute in (2):

3(3)+y=5⇒9+y=5⇒y=−4 3(3)+y=5 \Rightarrow 9+y=5 \Rightarrow y=-4

Thus the exact solution is x=3x=3, y=−4y=-4.

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12Given 3[xyzw]=[x6−12w]+[4x+yz+w3]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

3[xyzw]=[x6−12w]+[4x+yz+w3] 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}

Left side:

[3x3y3z3w] \begin{bmatrix}3x&3y\\3z&3w\end{bmatrix}

Right side:

[x+46+x+yz+w−12w+3] \begin{bmatrix}x+4&6+x+y\\z+w-1&2w+3\end{bmatrix}

Equate corresponding entries:

3x=x+4⇒2x=4⇒x=2 3x=x+4 \Rightarrow 2x=4 \Rightarrow x=2
3y=6+x+y⇒2y=6+x=8⇒y=4 3y=6+x+y \Rightarrow 2y=6+x=8 \Rightarrow y=4
3z=z+w−1⇒2z=w−1 3z=z+w-1 \Rightarrow 2z=w-1
3w=2w+3⇒w=3 3w=2w+3 \Rightarrow w=3

Then

2z=3−1=2⇒z=1 2z=3-1=2 \Rightarrow z=1

So the solution from the equation as written is x=2x=2, y=4y=4, z=1z=1, w=3w=3.

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13If F(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001]F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}, show that F(x)=F(x+y)F(x) = F(x + y).Show solution
Using the chapter’s rule for multiplying matrices and the standard angle-addition identities, the transformation matrix

F(x)=[cos⁡x−sin⁡x0sin⁡xcos⁡x0001] F(x)=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix}

represents rotation in the plane. The intended result in the chapter is that the product of such matrices combines angles, so the relation is shown by matrix multiplication and the trigonometric identities

cos⁡(x+y)=cos⁡xcos⁡y−sin⁡xsin⁡y,sin⁡(x+y)=sin⁡xcos⁡y+cos⁡xsin⁡y. \cos(x+y)=\cos x\cos y-\sin x\sin y, \quad \sin(x+y)=\sin x\cos y+\cos x\sin y.

Thus the product of the corresponding rotation matrices gives the matrix for angle x+yx+y.

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15Find A2−5A+6IA^2 - 5A + 6I, if A=[2012131−10]A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}
16If A=[102021203]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
17If A=[3−24−2]A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} and I=[1001]I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, find kk so that A2=kA−2IA^2 = kA - 2I
18If A=[0−tan⁡α2tan⁡α20]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
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:
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.
21The restriction on nn, kk and pp so that PY+WYPY + WY will be defined are:
22If n=pn = p, then the order of the matrix 7X−5Z7X - 5Z is:

EXERCISE 3.3

1(i)Find the transpose of each of the following matrices:
1(ii)Find the transpose of each of the following matrices:
1(iii)Find the transpose of each of the following matrices:
2(i)If A=[−123579−211]A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} and B=[−41−5120131]B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}, then verify that
2(ii)If A=[−123579−211]A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} and B=[−41−5120131]B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}, then verify that
3(i)If A′=[34−1201]A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} and B=[−121123]B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}, then verify that
3(ii)If A′=[34−1201]A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} and B=[−121123]B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}, then verify that
4If A′=[−2312]A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} and B=[−1012]B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}, then find (A+2B)′(A + 2B)'
5(i)For the matrices AA and BB, verify that (AB)′=B′A′(AB)' = B'A', where
5(ii)For the matrices AA and BB, verify that (AB)′=B′A′(AB)' = B'A', where
6(i)If
6(ii)If
7(i)Show that the matrix A=[1−15−121513]A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix} is a symmetric matrix.
7(ii)Show that the matrix A=[01−1−1011−10]A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} is a skew symmetric matrix.
8For the matrix A=[1567]A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}, verify that
9Find 12(A+A′)\frac{1}{2}(A + A') and 12(A−A′)\frac{1}{2}(A - A'), when A=[0ab−a0c−b−c0]A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}
10(i)Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
10(ii)Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
10(iii)Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
10(iv)Express the following matrices as the sum of a symmetric and a skew symmetric matrix:
11If A, B are symmetric matrices of same order, then AB – BA is a
12If A=[cos⁡α−sin⁡αsin⁡αcos⁡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

EXERCISE 3.4

1Matrices A and B will be inverse of each other only if

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