NCERT Solutions for Miscellaneous Exercise Chapter 3 Class 12 - Matrices

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

Answer:

Given :

$A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

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

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

The result is true for n=1.

Let result be true for n=k,

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

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

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

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

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

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

$A^2=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

$A^2=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$

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

$(aI+bA{)}^{k+1}$$=a^{k+1}I+(k+1)a^{k}bAI+k{a}^{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 $(aI+bA{)}^n=a^{n}I+n{a}^{n-1}bA$ where $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$,$n\in N$.

Question 2. If $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$ then show that $A^n=\left[\begin{array}{ccc}{3}^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{array}\right]$, $n\in N$.

Answer:

Given :

$A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$

To prove:

$A^n=\left[\begin{array}{ccc}{3}^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{array}\right]$

For n=1, we have

$A^1=\left[\begin{array}{ccc}{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{array}\right]$$=\left[\begin{array}{ccc}{3}^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\\ 3^0& 3^0& 3^0\end{array}\right]$$=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]=A$

Thus, the result is true for n=1.

Now, take n=k,

$A^k=\left[\begin{array}{ccc}{3}^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\\ 3^{k-1}& 3^{k-1}& 3^{k-1}\end{array}\right]$

For, n=k+1,

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

$=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$$\left[\begin{array}{ccc}{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{array}\right]$

$=\left[\begin{array}{ccc}3.3^{k-1}& {3.3}^{k-1}& {3.3}^{k-1}\\ 3.3^{k-1}& {3.3}^{k-1}& {3.3}^{k-1}\\ 3.3^{k-1}& {3.3}^{k-1}& 3.3^{k-1}\end{array}\right]$

$=\left[\begin{array}{ccc}{3}^{(K+1)-1}& 3^{(K+1)-1}& 3^{(K+1)-1}\\ 3^{(K+1)-1}& 3^{(K+1)-1}& 3^{(K+1)-1}\\ 3^{(K+1)-1}& 3^{(K+1)-1}& 3^{(K+1)-1}\end{array}\right]$

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

Hence, we have $A^n=\left[\begin{array}{ccc}{3}^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\\ 3^{n-1}& 3^{n-1}& 3^{n-1}\end{array}\right]$, $n\in N$ where $A=\left[\begin{array}{ccc}1& 1& 1\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]$.

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

Answer:

Given :

$A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$

To prove:

$A^n=\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right]$

For n=1, we have

$A^1=\left[\begin{array}{cc}1+2\times 1& -4\times 1\\ 1& 1-2\times 1\end{array}\right]$$=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]=A$

Thus, result is true for n=1.

Now, take result is true for n=k,

$A^k=\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]$

For, n=k+1,

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

$=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$$\left[\begin{array}{cc}1+2k& -4k\\ k& 1-2k\end{array}\right]$

$=\left[\begin{array}{cc}3(1+2k)-4k& -12k-4(1-2k)\\ (1+2k)-k& -4k-(1-2k)\end{array}\right]$

$=\left[\begin{array}{cc}3+6k-4k& -12k-4k+8k\\ 1+k& -4k-1+2k\end{array}\right]$

$=\left[\begin{array}{cc}3+2k& -4k-4k\\ 1+k& -2k-1\end{array}\right]$

$=\left[\begin{array}{cc}1+2(k+1)& -4(k+1)\\ 1+k& 1-2(k+1)\end{array}\right]$

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

Hence, we have $A^n=\left[\begin{array}{cc}1+2n& -4n\\ n& 1-2n\end{array}\right]$, where $A=\left[\begin{array}{cc}3& -4\\ 1& -1\end{array}\right]$.

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

$A^{\prime }=A$ and $B^{\prime }=B$

we have, ${(AB-BA)}^{\prime }={\left(AB\right)}^{\prime }-{\left(BA\right)}^{\prime }=B^{\prime }{A}^{\prime }-A^{\prime }{B}^{\prime }$

$=BA-AB$

$=-(AB-BA)$

Hence, we have $(AB-BA)=-(AB-BA{)}^{\prime }$

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 $A^{\prime }=A$

Consider, $(B^{\prime }AB{)}^{\prime }={B^{\prime }(AB)}^{\prime }$

$={(AB)}^{\prime }(B^{\prime }{)}^{\prime }$

$=B^{\prime }{A}^{\prime }(B)$

$=B^{\prime }(A^{\prime }B)$

Replace $A^{\prime }$ by $A$

$=B^{\prime }(AB)$

i.e. $(B^{\prime }AB{)}^{\prime }$ $=B^{\prime }(AB)$

Thus, if A is a symmetric matrix than $B^{\prime }(AB)$ is a symmetric matrix.

Now, let A be a skew-symmetric matrix, then $A^{\prime }=-A$.

$(B^{\prime }AB{)}^{\prime }={B^{\prime }(AB)}^{\prime }$

$={(AB)}^{\prime }(B^{\prime }{)}^{\prime }$

$=B^{\prime }{A}^{\prime }(B)$

$=B^{\prime }(A^{\prime }B)$

Replace $A^{\prime }$ by -$A$,

$=B^{\prime }(-AB)$

$=-B^{\prime }AB$

i.e. $(B^{\prime }AB{)}^{\prime }$ $=-B^{\prime }AB$.

Thus, if A is a skew-symmetric matrix then $-B^{\prime }AB$ 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 $A=\left[\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right]$ satisfy the equation $A^{\prime }A=I$

Answer:

$A=\left[\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right]$

$A^{\prime }=\left[\begin{array}{ccc}0& x& x\\ 2y& y& -y\\ z& -z& z\end{array}\right]$

$A^{\prime }A=I$

$\left[\begin{array}{ccc}0& x& x\\ 2y& y& -y\\ z& -z& z\end{array}\right]$$\left[\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right]$$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\left[\begin{array}{ccc}{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{array}\right]$$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\left[\begin{array}{ccc}2x^2& 0& 0\\ 0& 6y^2& 0\\ 0& 0& 3z^2\end{array}\right]$$=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

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: $\left[\begin{array}{ccc}1& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 0\\ 2& 0& 1\\ 1& 0& 2\end{array}\right]\left[\begin{array}{c}0\\ 2\\ x\end{array}\right]=O$?

Answer:

$\left[\begin{array}{ccc}1& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 2& 0\\ 2& 0& 1\\ 1& 0& 2\end{array}\right]\left[\begin{array}{c}0\\ 2\\ x\end{array}\right]=O$

$\left[\begin{array}{ccc}1+4+1& 2+0+0& 0+2+2\end{array}\right]\left[\begin{array}{c}0\\ 2\\ x\end{array}\right]=O$

$\left[\begin{array}{ccc}6& 2& 4\end{array}\right]\left[\begin{array}{c}0\\ 2\\ x\end{array}\right]=O$

$\left[\begin{array}{c}0+4+4x\end{array}\right]=O$

$4+4x=0$

$4x=-4$

$x=-1$

Thus, value of x is -1.

Question 8. If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, show that $A^2-5A+7I=0$.

Answer:

$A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$

$A^2=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$$\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$

$A^2=\left[\begin{array}{cc}9-1& 3+2\\ -3-2& -1+4\end{array}\right]$

$A^2=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$

$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

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

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

$=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$$-5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$$+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}8-15+7& 5-5+0\\ -5+5+0& 3-10+7\end{array}\right]$

$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0=R.H.S$

Hence, we proved that

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

Question 9. Find x, if $\left[\begin{array}{ccc}x& -5& -1\end{array}\right]\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0$.

Answer:

$\left[\begin{array}{ccc}x& -5& -1\end{array}\right]\left[\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0$

$\left[\begin{array}{ccc}x+0-2& 0-10+0& 2x-5-3\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0$

$\left[\begin{array}{ccc}x-2& -10& 2x-8\end{array}\right]\left[\begin{array}{c}x\\ 4\\ 1\end{array}\right]=0$

$\left[\begin{array}{c}x(x-2)-40+(2x-8)\end{array}\right]=0$

$\left[\begin{array}{c}{x}^2-2x-40+2x-8\end{array}\right]=0$

$\therefore x^2-48=0$

$x^2=48$

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 :

$\left[\begin{array}{ccc}10000& 2000& 18000\end{array}\right]\left[\begin{array}{c}2.50\\ 1.50\\ 1.00\end{array}\right]$

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

$=25000+3000+18000$

$=46000$

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

$\left[\begin{array}{ccc}6000& 20000& 8000\end{array}\right]\left[\begin{array}{c}2.50\\ 1.50\\ 1.00\end{array}\right]$

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

$=15000+30000+8000$

$=53000$

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 :

$\left[\begin{array}{ccc}10000& 2000& 18000\end{array}\right]\left[\begin{array}{c}2.00\\ 1.00\\ 0.50\end{array}\right]$

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

$=20000+2000+9000$

$=31000$

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 :

$\left[\begin{array}{ccc}6000& 20000& 8000\end{array}\right]\left[\begin{array}{c}2.00\\ 1.00\\ 0.50\end{array}\right]$

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

$=12000+20000+4000$

$=36000$

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 $X\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$

Answer:

$X\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$

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 $\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]$

$\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]$$\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$

$\left[\begin{array}{ccc}a+4c& 2a+5c& 3a+6c\\ b+4d& 2b+5d& 3b+6d\end{array}\right]=\left[\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right]$

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

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

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 $\left[\begin{array}{cc}1& -2\\ 2& 0\end{array}\right]$.

Question 12. If A and B are square matrices of the same order such that $AB=BA$, then prove by induction that $A{B}^n=B^{n}A$. 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 : $A{B}^n=B^{n}A$, $n\in N$

For n=1, we have $A{B}^1=B^{1}A$

Thus, the result is true for n=1.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 13 Choose the correct answer in the following questions:

If $A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]$ 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:

$A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]$

$A^2=I$

$\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]$$\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]$$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\left[\begin{array}{cc}{\alpha }^2+\beta \gamma & \alpha \beta -\alpha \beta \\ \alpha \gamma -\alpha \gamma & \beta \gamma +{\alpha }^2\end{array}\right]$$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\left[\begin{array}{cc}{\alpha }^2+\beta \gamma & 0\\ 0& \beta \gamma +{\alpha }^2\end{array}\right]$$=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

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

$A^{\prime }=A$ and $A^{\prime }=-A$

$A^{\prime }=A^{\prime }$

$\Rightarrow A=-A$

$\Rightarrow A+A=0$

$\Rightarrow 2A=0$

$\Rightarrow A=0$

Hence, A is a zero matrix.

Option B is correct.

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

(A) A
(B) I – A
(C) I
(D) 3A

Answer:

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

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

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

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

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

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

$=I+A-A$

$=I$

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

Option C is correct.