NCERT Solutions for Miscellaneous Exercise Chapter 3 Class 12 - Matrices

Question:1 Let $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$, show that $(aI + bA)^n = a^n I + na^{n-1} bA$, where I is the identity matrix of order 2 and $n \in N$.

Answer:

Given :

$A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

To prove : $(aI + bA)^n = a^n I + na^{n-1} bA$

For n=1, $aI + bA = a I + a^{0} bA =a I + bA$

The result is true for n=1.

Let result be true for n=k,

$(aI + bA)^k = a^k I + ka^{k-1} bA$

Now, we prove that the result is true for n=k+1,

$(aI + bA)^{k+1} = (aI + bA)^k (aI + bA)$

$= (a^k I + ka^{k-1} bA)$$(aI + bA)$

$=a^{k+1}I+Ka^{k}bAI+a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$A^{2} = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 0 &0 \\ 0 & 0 \end{bmatrix}=0$

Put the value of $A^{2}$ in above equation,

$(aI + bA)^{k+1}$$=a^{k+1}I+(k+1)a^{k}bAI+ka^{k-1}b^{2}A^{2}$

$(aI + bA)^{k+1}$$=a^{k+1}I+(k+1)a^{k}bAI+0$

$=a^{k+1}I+(k+1)a^{k}bAI$

Hence, the result is true for n=k+1.

Thus, we have $(aI + bA)^n = a^n I + na^{n-1} bA$ where $A = \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}$,$n \in N$.

Question 2. If $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$ then show that $A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$, $n\in N$.

Answer:

Given :

$A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$

To prove:

$A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$

For n=1, we have

$A^1 =\begin{bmatrix} 3^{1-1} & 3^{1-1} &3^{1-1} \\ 3^{1-1}& 3^{1-1} & 3^{1-1}\\ 3^{1-1} & 3^{1-1}& 3^{1-1} \end{bmatrix}$$=\begin{bmatrix} 3^{0} & 3^{0} &3^{0} \\ 3^{0}& 3^{0} & 3^{0}\\ 3^{0} & 3^{0}& 3^{0} \end{bmatrix}$$= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}=A$

Thus, the result is true for n=1.

Now, take n=k,

$A^k =\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

For, n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$$\begin{bmatrix} 3^{k-1} & 3^{k-1} &3^{k-1} \\ 3^{k-1}& 3^{k-1} & 3^{k-1}\\ 3^{k-1} & 3^{k-1}& 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix}3. 3^{k-1} & 3.3^{k-1} &3.3^{k-1} \\3. 3^{k-1}& 3.3^{k-1} & 3.3^{k-1}\\3. 3^{k-1} & 3.3^{k-1}&3. 3^{k-1} \end{bmatrix}$

$=\begin{bmatrix} 3^{(K+1)-1} &3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1}&3^{(K+1)-1} &3^{(K+1)-1}\\ 3^{(K+1)-1} & 3^{(K+1)-1}& 3^{(K+1)-1}\end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n =\begin{bmatrix} 3^{n-1} & 3^{n-1} &3^{n-1} \\ 3^{n-1}& 3^{n-1} & 3^{n-1}\\ 3^{n-1} & 3^{n-1}& 3^{n-1} \end{bmatrix}$, $n\in N$ where $A = \begin{bmatrix} 1 & 1 & 1\\ 1& 1& 1\\ 1& 1& 1 \end{bmatrix}$.

Question 3. If $A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$, then prove that $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$, where n is any positive integer.

Answer:

Given :

$A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$

To prove:

$A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$

For n=1, we have

$A^1 = \begin{bmatrix} 1+2\times 1 & -4\times 1\\ 1 & 1-2\times 1 \end{bmatrix}$$= \begin{bmatrix} 3 & -4\\ 1 & -1 \end{bmatrix}=A$

Thus, result is true for n=1.

Now, take result is true for n=k,

$A^k = \begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

For, n=k+1,

$A^{K+1}=A.A^K$

$= \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$$\begin{bmatrix} 1+2k & -4k\\ k & 1-2k \end{bmatrix}$

$=\begin{bmatrix} 3(1+2k)-4k & -12k-4(1-2k)\\ (1+2k)-k &-4k-(1-2k) \end{bmatrix}$

$=\begin{bmatrix} 3+6k-4k & -12k-4k+8k\\ 1+k &-4k-1+2k \end{bmatrix}$

$=\begin{bmatrix} 3+2k & -4k-4k\\ 1+k &-2k-1 \end{bmatrix}$

$=\begin{bmatrix} 1+2(k+1)& -4(k+1)\\ 1+k &1-2(k+1) \end{bmatrix}$

Thus, the result is true for n=k+1.

Hence, we have $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$, where $A = \begin{bmatrix} 3 & -4\\ 1& -1 \end{bmatrix}$.

Question 4. If A and B are symmetric matrices, prove that $AB - BA$ is a skew symmetric matrix.

Answer:

If A, B are symmetric matrices then

$A'=A$ and $B' = B$

we have, $\left ( AB-BA \right )'=\left ( AB \right )'-\left ( BA \right )'=B'A'-A'B'$

$=BA-AB$

$= -(AB-BA)$

Hence, we have $(AB-BA) = -(AB-BA)'$

Thus,( AB-BA)' is skew symmetric.

Question 5. Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Answer:

Let be a A is symmetric matrix , then $A'=A$

Consider, $(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by $A$

$=B'(AB)$

i.e. $(B'AB)'$ $=B'(AB)$

Thus, if A is a symmetric matrix than $B'(AB)$ is a symmetric matrix.

Now, let A be a skew-symmetric matrix, then $A'=-A$.

$(B'AB)' ={B'(AB)}'$

$={(AB)}'(B')'$

$= B'A'(B)$

$= B'(A'B)$

Replace $A'$ by -$A$,

$=B'(-AB)$

$= - B'AB$

i.e. $(B'AB)'$ $= - B'AB$.

Thus, if A is a skew-symmetric matrix then $- B'AB$ is a skew-symmetric matrix.

Hence, the matrix B′AB is symmetric or skew-symmetric according to as A is symmetric or skew-symmetric.

Question 6. Find the values of x, y, z if the matrix $A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$ satisfy the equation $A'A = I$

Answer:

$A = \begin{bmatrix} 0 & 2y & z\\ x & y & -z\\ x & -y &z \end{bmatrix}$

$A' = \begin{bmatrix} 0 & x & x\\ 2y & y & -y\\ z & -z &z \end{bmatrix}$

$A'A = I$

$\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}$

$\begin{bmatrix} x^{2}+x^{2} & xy-xy& -xz+xz\\ xy-xy& 4y^{2}+y^{2}+y^{2} & 2yz-yz-yz\\ -zx+zx & 2yz-yz-yz &z^{2}+z^{2}+z^{2}\end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

$\begin{bmatrix} 2x^{2} & 0& 0\\ 0& 6y^{2} & 0\\ 0 & 0 &3z^{2}\end{bmatrix}$$= \begin{bmatrix} 1 & 0& 0\\ 0 & 1 & 0\\ 0 & 0 &1\end{bmatrix}$

Thus equating the terms element wise

$2x^{2} = 1$ $6y^{2} = 1$ $3z^{2} = 1$

$x^{2} = \frac{1}{2}$ $y^{2} = \frac{1}{6}$ $z^{2}=\frac{1}{3}$

$x = \pm \frac{1}{\sqrt{2}}$ $y= \pm \frac{1}{\sqrt{6}}$ $z=\pm \frac{1}{\sqrt{3}}$

Question 7. For what values of x: $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$?

Answer:

$\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 &1 \\ 1& 0& 2\end{bmatrix}\begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 6& 2& 4 \end{bmatrix} \begin{bmatrix} 0\\ 2 \\ x \end{bmatrix} = O$

$\begin{bmatrix} 0+4+4x \end{bmatrix} = O$

$4+4x=0$

$4x=-4$

$x=-1$

Thus, value of x is -1.

Question 8. If $A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 -5A + 7I= 0$.

Answer:

$A = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$$\begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 9-1 &3+2 \\ -3-2 & -1+4 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}$

$I= \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

To prove: $A^2 -5A + 7I= 0$

L.H.S : $A^2 -5A + 7I$

$= \begin{bmatrix} 8 &5 \\ -5 & 3 \end{bmatrix}$$-5 \begin{bmatrix} 3 &1 \\ -1 & 2 \end{bmatrix}$$+ 7 \begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}$

$=\begin{bmatrix} 8-15+7 &5-5+0 \\ -5+5+0& 3-10+7 \end{bmatrix}$

$=\begin{bmatrix} 0 &0 \\ 0& 0 \end{bmatrix} =0=R.H.S$

Hence, we proved that

$A^2 -5A + 7I= 0$.

Question 9. Find x, if $\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$.

Answer:

$\begin{bmatrix} x & -5 & -1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1\\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix} x +0-2& 0-10+0 & 2x-5-3 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix} x -2& -10 & 2x-8 \end{bmatrix} \begin{bmatrix} x\\ 4\\ 1 \end{bmatrix} = 0$

$\begin{bmatrix}x (x -2)-40+(2x-8) \end{bmatrix} = 0$

$\begin{bmatrix}x ^{2}-2x-40+2x-8\end{bmatrix} = 0$

$\therefore \, \, x ^{2}-48= 0$

$x ^{2}=48$

thus the value of x is

$x =\pm 4\sqrt{3}$

Question 10(a) A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

Answer:

The unit sale prices of x, y and z are ` 2.50, ` 1.50 and ` 1.00, respectively.

The total revenue in the market I with the help of matrix algebra can be represented as :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

$= 10000\times 2.50+2000\times 1.50+18000\times 1.00$

$= 25000+3000+18000$

$= 46000$

The total revenue in market II with the help of matrix algebra can be represented as :

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50\\ 1.50\\ 1.00 \end{bmatrix}$

$= 6000\times 2.50+20000\times 1.50+8000\times 1.00$

$= 15000+30000+8000$

$= 53000$

Hence, total revenue in the market I is 46000 and total revenue in market II is 53000.

Question 10(b). A manufacturer produces three products x, y, z which he sells in two markets.
Annual sales are indicated below:

Market Products
I 10,000 2,000 18,000
II 6,000 20,000 8,000

If the unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively. Find the gross profit.

Answer:

The unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively.

The total cost price in market I with the help of matrix algebra can be represented as :

$\begin{bmatrix} 10000& 2000 & 18000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

$= 10000\times 2.00+2000\times 1.00+18000\times 0.50$

$= 20000+2000+9000$

$= 31000$

Total revenue in the market I is 46000 , gross profit in the market is $= 46000-31000$$=Rs. 15000$

The total cost price in market II with the help of matrix algebra can be represented as :

$\begin{bmatrix} 6000& 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.00\\ 1.00\\ 0.50 \end{bmatrix}$

$= 6000\times 2.0+20000\times 1.0+8000\times 0.50$

$= 12000+20000+4000$

$= 36000$

Total revenue in market II is 53000, gross profit in the market is$= 53000-36000= Rs. 17000$

Question 11. Find the matrix X so that $X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

Answer:

$X\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

The matrix given on R.H.S is $2\times 3$ matrix and on LH.S is $2\times 3$ matrix.Therefore, X has to be $2\times 2$ matrix.

Let X be $\begin{bmatrix} a & c\\ b & d \end{bmatrix}$

$\begin{bmatrix} a & c\\ b & d \end{bmatrix}$$\begin{bmatrix} 1 & 2 &3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

$\begin{bmatrix} a+4c & 2a+5c &3a+6c \\ b+4d & 2b+5d & 3b+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9\\ 2 &4 & 6 \end{bmatrix}$

$a+4c=-7$ $2a+5c=-8$ $3a+6c=-9$

$b+4d=2$ $2b+5d=4$ $3b+6d=6$

Taking, $a+4c=-7$

$a=-4c-7$

$2a+5c=-8$

$-8c-14+5c=-8$

$-3c=6$

$c=-2$

$a=-4\times -2-7$

$a=8-7=1$

$b+4d=2$

$b=-4d+2$

$2b+5d=4$

$\Rightarrow$ $-8d+4+5d=4$

$\Rightarrow -3d=0$

$\Rightarrow d=0$

$b=-4d+2$

$\Rightarrow b=-4\times 0+2=2$

Hence, we have $a=1, b=2,c=-2,d=0$

Matrix X is $\begin{bmatrix} 1 & -2\\ 2 & 0 \end{bmatrix}$.

Question 12. If A and B are square matrices of the same order such that $AB = BA$, then prove by induction that $AB^n = B^n A$. Further, prove that $(AB)^n = A^n B^n$for all $n \in N$.

Answer:

A and B are square matrices of the same order such that $AB = BA$,

To prove : $AB^n = B^n A$, $n \in N$

For n=1, we have $AB^1 = B^1 A$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $AB^k = B^k A$

Now, taking n=k+1 , we have $AB^{k+1} = AB^k .B$

$AB^{k+1} = (B^kA) .B$

$AB^{k+1} = (B^k) .AB$

$AB^{k+1} = (B^k) .BA$

$AB^{k+1} = (B^k.B) .A$

$AB^{k+1} = (B^{k+1}) .A$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$, $n \in N$.

To prove: $(AB)^n = A^n B^n$

For n=1, we have $(AB)^1 = A^1 B^1$

Thus, the result is true for n=1.

Let the result be true for n=k,then we have $(AB)^k = A^k B^k$

Now, taking n=k+1 , we have $(AB)^{k+1} = (A B)^k.(AB)$

$(AB)^{k+1} = A^k B^k.(AB)$

$(AB)^{k+1} = A^{K}( B^kA)B$

$(AB)^{k+1} = A^{K}( AB^k)B$

$(AB)^{k+1} = (A^{K}A)(B^kB)$

$(AB)^{k+1} = (A^{k+1})(B^{k+1})$

Thus, the result is true for n=k+1.

Hence, we have $AB^n = B^n A$ and $(AB)^n = A^n B^n$for all $n \in N$.

Question 13 Choose the correct answer in the following questions:

If $A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$ is such that $A^2 = I$

(A) $1 + \alpha^2 + \beta \gamma = 0$

(B) $1 - \alpha^2 + \beta \gamma = 0$

(C) $1 - \alpha^2 - \beta \gamma = 0$

(D) $1 + \alpha^2 - \beta \gamma = 0$

Answer:

$A = \begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$

$A^2 = I$

$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$$\begin{bmatrix} \alpha &\beta\\ \gamma &-\alpha \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&\alpha \beta-\alpha \beta\\\alpha \gamma-\alpha \gamma&\beta \gamma+\alpha^{2} \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix} \alpha^{2} +\beta \gamma&0\\0&\beta \gamma+\alpha^{2} \end{bmatrix}$$= \begin{bmatrix} 1 &0\\0&1 \end{bmatrix}$

Thus we obtained that

$\alpha^{2} +\beta \gamma=1$

$\Rightarrow 1-\alpha^{2} -\beta \gamma=0$

Option C is correct.

Question 14. If 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 these

Answer:

If the matrix A is both symmetric and skew-symmetric, then

$A'=A$ and $A'=-A$

$A'=A'$

$\Rightarrow \, \, \, \, \, \, \, A=-A$

$\Rightarrow \, \, \, \, \, \, \, A+A=0$

$\Rightarrow \, \, \, \, \, \, \, 2A=0$

$\Rightarrow \, \, \, \, \, \, \, A=0$

Hence, A is a zero matrix.

Option B is correct.

Question 15. If A is square matrix such that $A^{2}=A$, then $(I + A)^3 - 7 A$ is equal to

(A) A
(B) I – A
(C) I
(D) 3A

Answer:

A is a square matrix such that $A^{2}=A$

$(I + A)^3 - 7 A$

$=I^{3}+A^{3}+3I^{2}A+3IA^{2}-7A$

$=I+A^{2}.A+3A+3A^{2}-7A$

$=I+A.A+3A+3A-7A$ (Replace $A^{2}$ by $A$)

$=I+A^{2}+6A-7A$

$=I+A-A$

$=I$

Hence, we have $(I + A)^3 - 7 A=I$

Option C is correct.