Miscellaneous Exercise on Chapter 3, Class 12th, Maths, NCERT

Question 1:

$$\text{If }A\text{ and }B\text{ are symmetric matrices, prove that }AB-BA\text{ is a skew-symmetric matrix.}$$

Answer (proof):

Recall a matrix $M$ is skew-symmetric if $M^T=-M$

Given $A$ and $B$ are symmetric, so

$$A^T=A,B^T=B\mathrm{.}$$

Consider the transpose of $AB-BA$:

$$(AB-BA{)}^T=(AB{)}^T-(BA{)}^T=B^T{A}^T-A^T{B}^T\mathrm{.}$$

Using symmetry of $A$ and $B$ we get

$$(AB-BA{)}^T=BA-AB=-(AB-BA)\mathrm{.}$$

Thus $(AB-BA{)}^T=-(AB-BA)$, so $AB-BA$ is skew-symmetric.

(As a remark, every skew-symmetric matrix has zeros on its diagonal, so the diagonal entries of $AB-BA$ are all zero.)

Question 2:

$$\text{Show that the matrix }{B}^{\mathrm{\prime }}AB\text{ is symmetric according as }A\text{ is symmetric or skew-symmetric.}$$

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Answer (proof):

Let $A$ and $B$ be square matrices of the same order, and let $B^{\mathrm{\prime }}$ denote the transpose of $B$.
We need to show that:

• If $A$ is symmetric, then $B^{\mathrm{\prime }}AB$ is symmetric.

• If $A$ is skew-symmetric, then $B^{\mathrm{\prime }}AB$ is skew-symmetric.

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Case 1: $A$ is symmetric

If $A$ is symmetric, then $A^{\mathrm{\prime }}=A$.

Consider $(B^{\mathrm{\prime }}AB{)}^{\mathrm{\prime }}$:

$$(B^{\mathrm{\prime }}AB{)}^{\mathrm{\prime }}=B^{\mathrm{\prime }}{A}^{\mathrm{\prime }}B\mathrm{.}$$

But since $A^{\mathrm{\prime }}=A$,

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

Hence, $B^{\mathrm{\prime }}AB$ is symmetric.

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Case 2: $A$ is skew-symmetric

If $A$ is skew-symmetric, then $A^{\mathrm{\prime }}=-A$.

Now, take the transpose of $B^{\mathrm{\prime }}AB$:

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

Thus, $(B^{\mathrm{\prime }}AB{)}^{\mathrm{\prime }}=-B^{\mathrm{\prime }}AB$,
which means $B^{\mathrm{\prime }}AB$ is skew-symmetric.

Hence proved:
The matrix $B^{\mathrm{\prime }}AB$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.

Question 3:

$$\text{Find the values of }x,y,z\text{ if }A=\left(\begin{array}{ccc}0& 2y& z\\ x& y& -z\\ x& -y& z\end{array}\right)\text{ satisfies }{A}^{T}A=I\mathrm{.}$$

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Answer (solution):

Compute $A^{T}A$. A straightforward multiplication gives

$$A^{T}A=\left(\begin{array}{ccc}2x^2& 0& 0\\ 0& 6y^2& 0\\ 0& 0& 3z^2\end{array}\right)\mathrm{.}$$

For $A^{T}A=I$ we must have the diagonal entries equal to 1, hence

$$2x^2=1,6y^2=1,3z^2=1$$

Therefore

$$x^2=\frac{1}{2}⟹x=\pm \frac{1}{\sqrt{2}},y^2=\frac{1}{6}⟹y=\pm \frac{1}{\sqrt{6}},z^2=\frac{1}{3}⟹z=\pm \frac{1}{\sqrt{3}}\mathrm{.}$$

All choices of independent signs are allowed, so the solutions are

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

(8 sign-combinations in total).

Question 4:

$$\text{For what values of }x\text{ does }[121]\left(\begin{array}{ccc}1& 2& 0\\ 2& 0& 1\\ 1& 0& 2\end{array}\right)\left(\begin{array}{c}0\\ 2\\ x\end{array}\right)=0?$$

Answer (solution):

First compute the product of the matrix with the column vector:

$$\left(\begin{array}{ccc}1& 2& 0\\ 2& 0& 1\\ 1& 0& 2\end{array}\right)\left(\begin{array}{c}0\\ 2\\ x\end{array}\right)=\left(\begin{array}{c}1\cdot 0+2\cdot 2+0\cdot x\\ 2\cdot 0+0\cdot 2+1\cdot x\\ 1\cdot 0+0\cdot 2+2\cdot x\end{array}\right)=\left(\begin{array}{c}4\\ x\\ 2x\end{array}\right)\mathrm{}$$

Now left-multiply by $[121]$:

$$[121]\left(\begin{array}{c}4\\ x\\ 2x\end{array}\right)=1\cdot 4+2\cdot x+1\cdot 2x=4+4x\mathrm{}$$

Set equal to zero:

$$4+4x=0\Rightarrow x=-1$$

Question 5:

$$\text{If }A=\left(\begin{array}{cc}3& 1\\ -1& 2\end{array}\right),\text{ show that }{A}^2-5A+7I=0$$

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Answer (solution):

Given

$$A=\left(\begin{array}{cc}3& 1\\ -1& 2\end{array}\right),I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\mathrm{.}$$

Step 1: Compute $A^2$

$$A^2=\left(\begin{array}{cc}3& 1\\ -1& 2\end{array}\right)\left(\begin{array}{cc}3& 1\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}3(3)+1(-1)& 3(1)+1(2)\\ (-1)(3)+2(-1)& (-1)(1)+2(2)\end{array}\right)=\left(\begin{array}{cc}8& 5\\ -5& 3\end{array}\right)\mathrm{.}$$

Step 2: Compute $5A$

$$5A=5\left(\begin{array}{cc}3& 1\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}15& 5\\ -5& 10\end{array}\right)\mathrm{}$$

Step 3: Compute $7I$

$$7I=7\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}7& 0\\ 0& 7\end{array}\right)\mathrm{}$$

Step 4: Substitute into $A^2-5A+7I$

$$A^2-5A+7I=\left(\begin{array}{cc}8& 5\\ -5& 3\end{array}\right)-\left(\begin{array}{cc}15& 5\\ -5& 10\end{array}\right)+\left(\begin{array}{cc}7& 0\\ 0& 7\end{array}\right)\mathrm{}$$

Compute step-by-step:

$$A^2-5A=\left(\begin{array}{cc}8-15& 5-5\\ -5-(-5)& 3-10\end{array}\right)=\left(\begin{array}{cc}-7& 0\\ 0& -7\end{array}\right)\mathrm{}$$

Now add $7I$:

$$A^2-5A+7I=\left(\begin{array}{cc}-7& 0\\ 0& -7\end{array}\right)+\left(\begin{array}{cc}7& 0\\ 0& 7\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\mathrm{}$$

Hence proved:

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

 

Question 6:

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

Solution:

We will simplify step-by-step.

Step 1: Multiply the matrix with the column vector

$$\left(\begin{array}{ccc}1& 0& 2\\ 0& 2& 1\\ 2& 0& 3\end{array}\right)\left(\begin{array}{c}x\\ 4\\ 1\end{array}\right)=\left(\begin{array}{c}1\cdot x+0\cdot 4+2\cdot 1\\ 0\cdot x+2\cdot 4+1\cdot 1\\ 2\cdot x+0\cdot 4+3\cdot 1\end{array}\right)=\left(\begin{array}{c}x+2\\ 9\\ 2x+3\end{array}\right)\mathrm{}$$

Step 2: Multiply the row vector $[x-5-1]$ with the result

$$[x-5-1]\left(\begin{array}{c}x+2\\ 9\\ 2x+3\end{array}\right)=x(x+2)+(-5)(9)+(-1)(2x+3)\mathrm{}$$

Simplify:

$$=x^2+2x-45-2x-3=x^2-48.$$

Step 3: Set equal to zero

$$x^2-48=0\Rightarrow x^2=48$$

$$<span\; class="mord\; mathnormal">x\; </span>=\pm 4\sqrt{3}\mathrm{}$$

Question 7

$$\text{A manufacturer produces three products }x,y,z\text{ sold in two markets with annual sales}$$
$$S=\left(\begin{array}{ccc}10000& 2000& 18000\\ 6000& 20000& 8000\end{array}\right)\text{(rows: Market I, Market II; columns: }x,y,z\text{).}$$
$$\text{(a) Unit sale prices }p=\left(\begin{array}{c}2.50\\ 1.50\\ 1.00\end{array}\right)\mathrm{.}\text{(b) Unit costs }c=\left(\begin{array}{c}2.00\\ 1.00\\ 0.50\end{array}\right)\mathrm{.}$$
$$\text{Find (a) total revenue in each market, (b) gross profit in each market.}$$

Solution

Use matrix multiplication. Revenue vector $R$ (marketwise) is

$$R=Sp\mathrm{.}$$

Compute component-wise:

Market I revenue

$$10000(2.5)+2000(1.5)+18000(1)=25000+3000+18000=46000.$$

Market II revenue

$$6000(2.5)+20000(1.5)+8000(1)=15000+30000+8000=53000.$$

So

$$R=\left(\begin{array}{c}46000\\ 53000\end{array}\right)\text{ rupees.}$$

(b) Total cost vector $C=Sc$:

Market I cost

$$10000(2)+2000(1)+18000(0.5)=20000+2000+9000=31000.$$

Market II cost

$$6000(2)+20000(1)+8000(0.5)=12000+20000+4000=36000.$$

So

$$C=\left(\begin{array}{c}31000\\ 36000\end{array}\right)\text{ rupees.}$$

Gross profit for each market $G=R-C$:

$$G=\left(\begin{array}{c}46000-31000\\ 53000-36000\end{array}\right)=\left(\begin{array}{c}15000\\ 17000\end{array}\right)\text{ rupees}$$Final answers

(a) Total revenue — Market I: Rs. 46,000; Market II: Rs. 53,000.
(b) Gross profit — Market I: Rs. 15,000; Market II: Rs. 17,000.

(Also note matrix forms: $R=Sp,C=Sc,G=S(p-c)$

 

Question 8:

$$\text{Find the matrix }X\text{ such that }X\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right)=\left(\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right)\mathrm{.}$$

Solution:

We are given:

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

Let:

$$A=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right),B=\left(\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right)\mathrm{.}$$

We need $X$ such that $XA=B$.

To isolate $X$, multiply both sides on the right by $A^T(A{A}^T{)}^{-1}$:
(since $A$ is $2\times 3$, not square, we use this formula)

$$X=B{A}^T(A{A}^T{)}^{-1}\mathrm{.}$$

Step 1: Compute $A{A}^T$

$$A{A}^T=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right)\left(\begin{array}{cc}1& 4\\ 2& 5\\ 3& 6\end{array}\right)=\left(\begin{array}{cc}{1}^2+2^2+3^2& 1\cdot 4+2\cdot 5+3\cdot 6\\ 4\cdot 1+5\cdot 2+6\cdot 3& 4^2+5^2+6^2\end{array}\right)=\left(\begin{array}{cc}14& 32\\ 32& 77\end{array}\right)\mathrm{}$$

Step 2: Compute $(A{A}^T{)}^{-1}$

First find determinant:

$$\mathrm{\mid }A{A}^T\mathrm{\mid }=(14)(77)-(32)(32)=1078-1024=54$$

So,

$$(A{A}^T{)}^{-1}=\frac{1}{54}\left(\begin{array}{cc}77& -32\\ -32& 14\end{array}\right)\mathrm{.}$$

Step 3: Compute $B{A}^T$

$$B{A}^T=\left(\begin{array}{ccc}-7& -8& -9\\ 2& 4& 6\end{array}\right)\left(\begin{array}{cc}1& 4\\ 2& 5\\ 3& 6\end{array}\right)$$

$$=\left(\begin{array}{cc}(-7)(1)+(-8)(2)+(-9)(3)& (-7)(4)+(-8)(5)+(-9)(6)\\ (2)(1)+(4)(2)+(6)(3)& (2)(4)+(4)(5)+(6)(6)\end{array}\right)\mathrm{}$$

Compute each entry:

$$B{A}^T=\left(\begin{array}{cc}-7-16-27& -28-40-54\\ 2+8+18& 8+20+36\end{array}\right)=\left(\begin{array}{cc}-50& -122\\ 28& 64\end{array}\right)\mathrm{}$$

Step 4: Compute $X=B{A}^T(A{A}^T{)}^{-1}$

$$X=\frac{1}{54}\left(\begin{array}{cc}-50& -122\\ 28& 64\end{array}\right)\left(\begin{array}{cc}77& -32\\ -32& 14\end{array}\right)\mathrm{.}$$

Compute the product:

First row:

$$(-50)(77)+(-122)(-32)=-3850+3904=54,$$
$$(-50)(-32)+(-122)(14)=1600-1708=-108.$$

Second row:

$$(28)(77)+(64)(-32)=2156-2048=108,$$
$$(28)(-32)+(64)(14)=-896+896=0$$

So:

$$X=\frac{1}{54}\left(\begin{array}{cc}54& -108\\ 108& 0\end{array}\right)=\left(\begin{array}{cc}1& -2\\ 2& 0\end{array}\right)\mathrm{.}$$

Choose the correct answer in the following questions:

Question 9:

$$\text{If }A=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right)\text{ is such that }{A}^2=I,\text{ then find the correct relation among }\alpha ,\beta ,\gamma \mathrm{.}$$

Options:
(A) $1+{\alpha }^2+\beta \gamma =0$
(B) $1-{\alpha }^2+\beta \gamma =0$
(C) $1-{\alpha }^2-\beta \gamma =0$
(D) $1+{\alpha }^2-\beta \gamma =0$

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Solution:

Given

$$A=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right)\mathrm{}$$

Compute $A^2$:

$$A^2=\left(\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right)\left(\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right)=\left(\begin{array}{cc}{\alpha }^2+\beta \gamma & \alpha \beta -\alpha \beta \\ \alpha \gamma -\alpha \gamma & \gamma \beta +{\alpha }^2\end{array}\right)=\left(\begin{array}{cc}{\alpha }^2+\beta \gamma & 0\\ 0& {\alpha }^2+\beta \gamma \end{array}\right)\mathrm{}$$

Thus,

$$A^2=({\alpha }^2+\beta \gamma )I\mathrm{.}$$

We are told $A^2=I$, so:

$$({\alpha }^2+\beta \gamma )I=I\mathrm{}$$

That means:

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

Rearranging:

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

Correct Option:

$$\boxed{{\displaystyle (C)1-{\alpha }^2-\beta \gamma =0.}}$$

Question 10:

$$\text{If the matrix }A\text{ is both symmetric and skew-symmetric, then:}$$

Options:
(A) $A$ is a diagonal matrix
(B) $A$ is a zero matrix
(C) $A$ is a square matrix
(D) None of these

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Solution:

By definition:

• $A$ is symmetric if $A^T=A$

• $A$ is skew-symmetric if $A^T=-A$

If $A$ is both, then:

$$A=A^T=-A\mathrm{}$$

This implies:

$$A=-A\Rightarrow 2A=0\Rightarrow A=0$$

That means all elements of $A$ are zero.

Correct Option:

$$\boxed{{\displaystyle (B)A\text{ is a zero matrix.}}}$$

Question 11:

$$\text{If }A\text{ is a square matrix such that }{A}^2=A,\text{ then find }(I+A{)}^3-7A\mathrm{.}$$

Options:
(A) $A$
(B) $I-A$
(C) $I$
(D) $3A$

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Solution:

We are given that:

$$A^2=A\mathrm{.}$$

Let’s expand $(I+A{)}^3$:

$$(I+A{)}^3=I^3+3I^{2}A+3I{A}^2+A^3\mathrm{}$$

Simplify each term using $I^2=I$ and $A^2=A$:

$$(I+A{)}^3=I+3A+3A+A^3\mathrm{}$$

Now, compute $A^3$:

$$A^3=A^{2}A=A\cdot A=A^2=A\mathrm{}$$

Substitute this back:

$$(I+A{)}^3=I+3A+3A+A=I+7A\mathrm{}$$

Now compute:

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

Final Answer:

$$\boxed{{\displaystyle (C)I\mathrm{}}}$$