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

It is given that $A=\left(\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right)$ We shall prove the result by using the principle of mathematical induction. For $n=1$, we have: $$ P(1):(a I+b A)=a I+b a^{0} A=a I+b A $$ Therefore, the result is true for $n=1$. Let the result be true for $n=k$ That is, $P(k):(a I+b A)^{k}=a^{k} I+k a^{k-1} b A$ Now, we have to prove that the result is true for $n=k+1$. Consider, $$ \begin{align*} (a I+b A)^{k+1} & =(a I+b A)^{k}(a I+b A) \\ & =\left(a^{k} I+k a^{k-1} b A\right)(a I+b A) \\ & =a^{k+1} I+k a^{k} b A I+a^{k} b I A+k a^{k-1} b^{2} A^{2} \\ & =a^{k+1} I+(k+1) a^{k} b A+k a^{k-1} b^{2} A^{2} \tag{1} \end{align*} $$ Now, $$ A^{2}=\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right)\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right)=\left(\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right)=0 $$ From (1), we have $$ \begin{aligned} (a I+b A)^{k+1} & =a^{k+1} I+(k+1) a^{k} b A+0 \\ & =a^{k+1} I+(k+1) a^{k} b A \end{aligned} $$ Therefore, the result is true for $n=k+1$. Thus, by the principle of mathematical induction, we have: $$ (a I+b A)^{n}=a^{n} I+n a^{n-1} b A \text { where } A=\left(\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right)_{, n \in N} $$

Q: $$ \text { If } A=\left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right) \text {, prove that } A^{n}=\left(\begin{array}{lll} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{array}\right), n \in N $$

It is given that $A=\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array}\right)$ We shall prove the result by using the principle of mathematical induction. For $n=1$, we have: $$ P(1):\left(\begin{array}{lll} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{array}\right)=\left(\begin{array}{lll} 3^{0} & 3^{0} & 3^{0} \\ 3^{0} & 3^{0} & 3^{0} \\ 3^{0} & 3^{0} & 3^{0} \end{array}\right)=\left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right)=A $$ Therefore, the result is true for $n=1$. Let the result be true for $n=k$. $$ P(k): A^{k}=\left(\begin{array}{lll} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{array}\right) $$ Now, we have to prove that the result is true for $n=k+1$. Since, $$ \begin{aligned} A^{k+1} & =A \cdot A^{k} \\ & =\left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right)\left(\begin{array}{lll} 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \\ 3^{k-1} & 3^{k-1} & 3^{k-1} \end{array}\right) \\ & =\left(\begin{array}{lll} 3 \cdot 3^{k-1} & 3 \cdot 3^{k-1} & 3 \cdot 3^{k-1} \\ 3 \cdot 3^{k-1} & 3 \cdot 3^{k-1} & 3 \cdot 3^{k-1} \\ 3 \cdot 3^{k-1} & 3 \cdot 3^{k-1} & 3 \cdot 3^{k-1} \end{array}\right) \\ & =\left(\begin{array}{lll} 3^{(k+1)-1} & 3^{(k+1)-1} & 3^{(k+1)-1} \\ 3^{(k+1)-1} & 3^{(k+1)-1} & 3^{(k+1)-1} \\ 3^{(k+1)-1} & 3^{(k+1)-1} & 3^{(k+1)-1} \end{array}\right) \end{aligned} $$ Therefore, the result is true for $n=k+1$. Thus, by the principle of mathematical induction, we have: $$ A^{n}=\left(\begin{array}{ccc} 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \\ 3^{n-1} & 3^{n-1} & 3^{n-1} \end{array}\right), n \in N $$

Q: If $A=\left(\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right)$, prove that $A^{n}=\left(\begin{array}{cc}1+2 n & -4 n \\ n & 1-2 n\end{array}\right)$, where $n$ is any positive integer.

It is given that $A=\left(\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right)$ We shall prove the result by using the principle of mathematical induction. For $n=1$, we have: $$ \begin{aligned} P(1) & : A^{1}=\left(\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right) \\ & =\left(\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right) \\ & =A \end{aligned} $$ Therefore, the result is true for $n=1$. Let the result be true for $n=k$. $$ P(k): A^{k}=\left(\begin{array}{cc} 1+2 k & -4 k \\ k & 1-2 k \end{array}\right), n \in N $$ Now, we have to prove that the result is true for $n=k+1$. Since, $$ \begin{aligned} A^{k+1} & =A \cdot A^{k} \\ & =\left(\begin{array}{cc} 1+2 k & -4 k \\ k & 1-2 k \end{array}\right)\left(\begin{array}{ll} 3 & -4 \\ 1 & -1 \end{array}\right) \\ & =\left(\begin{array}{cc} 3(1+2 k)-4 k & -4(1+2 k)+4 k \\ 3 k+1-2 k & -4 k-1(1-2 k) \end{array}\right) \\ & =\left(\begin{array}{cc} 3+6 k-4 k & -4-8 k+4 k \\ 3 k+1-2 k & -4 k-1+2 k \end{array}\right) \\ & =\left(\begin{array}{cc} 3+2 k & -4-4 k \\ 1+k & -1-2 k \end{array}\right) \\ & =\left(\begin{array}{cc} 1+2(k+1) & -4(k+1) \\ 1+k & 1-2(k+1) \end{array}\right) \end{aligned} $$ Therefore, the result is true for $n=k+1$. Thus, by the principle of mathematical induction, we have: $$ A^{n}=\left(\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right) ; n \in N $$

Q: Find the values of $x, y, z$ if the matrix $A=\left(\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right)$ satisfy the equation $A^{\prime} A=I$.

It is given that $A=\left(\begin{array}{ccc}0 & 2 y & z \\ x & y & -z \\ x & -y & z\end{array}\right)$ Therefore, $$ A^{\prime}=\left(\begin{array}{ccc} 0 & x & x \\ 2 y & y & -y \\ z & -z & z \end{array}\right) $$ Now, $A^{\prime} A=I$ Hence, $$ \begin{aligned} & \Rightarrow\left(\begin{array}{ccc} 0 & x & x \\ 2 y & y & -y \\ z & -z & z \end{array}\right)\left(\begin{array}{ccc} 0 & 2 y & z \\ x & y & -z \\ x & -y & z \end{array}\right)=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \\ & \Rightarrow\left(\begin{array}{ccc} 0+x^{2}+x^{2} & 0+x y-x y & 0-x z+x z \\ 0+x y-x y & 4 y^{2}+y^{2}+y^{2} & 2 y z-y z-y z \\ 0-x z+z x & 2 y z-y z-y z & z^{2}+z^{2}+z^{2} \end{array}\right)=\left(\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \\ & \Rightarrow\left(\begin{array}{ccc} 2 x^{2} & 0 & 0 \\ 0 & 6 y^{2} & 0 \\ 0 & 0 & 3 z^{2} \end{array}\right)=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right) \end{aligned} $$ On comparing the corresponding elements, we have: $$ \begin{aligned} & 2 x^{2}=1 \\ & \Rightarrow x= \pm \frac{1}{\sqrt{2}} \\ & 6 y^{2}=1 \\ & \Rightarrow y= \pm \frac{1}{\sqrt{6}} \\ & 3 z^{2}=1 \\ & \Rightarrow z= \pm \frac{1}{\sqrt{3}} \end{aligned} $$ Thus, $x= \pm \frac{1}{\sqrt{2}}, y= \pm \frac{1}{\sqrt{6}}$ and $z= \pm \frac{1}{\sqrt{3}}$

Q: For what values of $$ x:\left[\begin{array}{lll} 1 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}\right]\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]=0 $$

We have: $$ \left[\begin{array}{lll} 1 & 2 & 1 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{array}\right]\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]=0 $$ Hence, $$ \begin{aligned} & \Rightarrow\left[\begin{array}{lll} 1+4+1 & 2+0+0 & 0+2+2 \end{array}\right]\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]=0 \\ & \Rightarrow\left[\begin{array}{lll} 6 & 2 & 4 \end{array}\right]\left[\begin{array}{l} 0 \\ 2 \\ x \end{array}\right]=0 \\ & \Rightarrow[6(0)+2(2)+4(x)]=0 \\ & \Rightarrow[4+4 x]=0 \\ & \Rightarrow 4 x=-4 \\ & \Rightarrow x=-1 \end{aligned} $$ Thus, the required value of $x=-1$.

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

It is given that $A=\left(\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right)$ Therefore, $$ \begin{aligned} A^{2} & =A \cdot A \\ & =\left(\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right)\left(\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right) \\ & =\left(\begin{array}{cc} 3(3)+1(-1) & 3(1)+1(2) \\ -1(3)+2(-1) & -1(1)+2(2) \end{array}\right) \\ & =\left(\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right) \end{aligned} $$ Now, $$ \begin{aligned} L H S & =A^{2}-5 A+7 I \\ & =\left(\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right)-5\left(\begin{array}{cc} 3 & 1 \\ -1 & 2 \end{array}\right)+7\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right) \\ & =\left(\begin{array}{cc} 8 & 5 \\ -5 & 3 \end{array}\right)-\left(\begin{array}{cc} 15 & 5 \\ -5 & 10 \end{array}\right)+\left(\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}\right) \\ & =\left(\begin{array}{cc} -7 & 0 \\ 0 & -7 \end{array}\right)+\left(\begin{array}{cc} 7 & 0 \\ 0 & 7 \end{array}\right) \\ & =0 \\ & =R H S \end{aligned} $$ Thus, $A^{2}-5 A+7 I=0$

Question 1

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

Accepted Answer

Given: $A$ and $B$ are square matrices of the same order such that $A B=B A$.

To prove: $P(n): A B^{n}=B^{n} A, n \in N$

For $\mathrm{n}=1$, we have:

$$

\begin{aligned}

& P(1): A B=B A \quad 	ext { [Given] } \

& \quad \Rightarrow A B^{1}=B^{1} A

\end{aligned}

$$

Therefore, the result is true for $n=1$.

Let the result be true for $n=k$.

$$

\begin{equation*}

P(k)=A B^{k}=B^{k} A 	ag{1}

\end{equation*}

$$

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

$$

\begin{aligned}

A B^{k+1} & =A B^{k} \cdot B & & \

& =\left(B^{k} Aight) B & & {[B y(1)] } \

& =B^{k}(A B) & & {[	ext { Associative law }] } \

& =B^{k}(B A) & & {[A B=B A(	ext { Given })] } \

& =\left(B^{k} Bight) A & & {[	ext { Associative law }] } \

& =B^{k+1} A & &

\end{aligned}

$$

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

Thus, by the principle of mathematical induction, we have $A B^{n}=B^{n} A, n \in N$

Now, we have to prove that $(A B)^{n}=A^{n} B^{n}$ for all $n \in N$

For $n=1$, we have:

$$

(A B)^{1}=A^{1} B^{1}=A B

$$

Therefore, the result is true for $n=1$.

Let the result be true for $n=k$.

$$

\begin{equation*}

(A B)^{k}=A^{k} B^{k} 	ag{2}

\end{equation*}

$$

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

$$

\begin{align*}

A B^{k+1} & =(A B)^{k} \cdot(A B) & & \

& =\left(A^{k} B^{k}ight) \cdot(A B) & & {[	ext { By (2) }] }  	ag{2}\

& =A^{k}\left(B^{k} Aight) B & & {[	ext { Associative law }] } \

& =A^{k}\left(A B^{k}ight) B & & {\left[A B^{n}=B^{n} A, n \in Night] } \

& =\left(A^{k} Aight) \cdot\left(B^{k} Bight) & & {[	ext { Associative law }] } \

& =A^{k+1} B^{k+1} & &

\end{align*}

$$

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

Thus, by the principle of mathematical induction, we have $(A B)^{n}=A^{n} B^{n}, n \in N$

Question 2

Let $A=\left(\begin{array}{ll}0 & 1 \ 0 & 0\end{array}ight)$, show that $(a I+b A)^{n}=a^{n} I+n a^{n-1} b A$, where $I$ is the identity matrix of order 2 and $n \in N$.

Accepted Answer

It is given that $A=\left(\begin{array}{ll}0 & 1 \ 0 & 0\end{array}ight)$

We shall prove the result by using the principle of mathematical induction.

For $n=1$, we have:

$$

P(1):(a I+b A)=a I+b a^{0} A=a I+b A

$$

Therefore, the result is true for $n=1$.

Let the result be true for $n=k$

That is, $P(k):(a I+b A)^{k}=a^{k} I+k a^{k-1} b A$

Now, we have to prove that the result is true for $n=k+1$.

Consider,

$$

\begin{align*}

(a I+b A)^{k+1} & =(a I+b A)^{k}(a I+b A) \

& =\left(a^{k} I+k a^{k-1} b Aight)(a I+b A) \

& =a^{k+1} I+k a^{k} b A I+a^{k} b I A+k a^{k-1} b^{2} A^{2} \

& =a^{k+1} I+(k+1) a^{k} b A+k a^{k-1} b^{2} A^{2} 	ag{1}

\end{align*}

$$

Now,

$$

A^{2}=\left(\begin{array}{ll}

0 & 1 \

0 & 0

\end{array}ight)\left(\begin{array}{ll}

0 & 1 \

0 & 0

\end{array}ight)=\left(\begin{array}{ll}

0 & 0 \

0 & 0

\end{array}ight)=0

$$

From (1), we have

$$

\begin{aligned}

(a I+b A)^{k+1} & =a^{k+1} I+(k+1) a^{k} b A+0 \

& =a^{k+1} I+(k+1) a^{k} b A

\end{aligned}

$$

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

Thus, by the principle of mathematical induction, we have:

$$

(a I+b A)^{n}=a^{n} I+n a^{n-1} b A 	ext { where } A=\left(\begin{array}{ll}

0 & 1 \

0 & 0

\end{array}ight)_{, n \in N}

$$

Question 3

$$

ext { If } A=\left(\begin{array}{lll}

1 & 1 & 1 \

1 & 1 & 1

\end{array}ight) 	ext {, prove that } A^{n}=\left(\begin{array}{lll}

3^{n-1} & 3^{n-1} & 3^{n-1} \

3^{n-1} & 3^{n-1} & 3^{n-1}

\end{array}ight), n \in N

$$

Accepted Answer

It is given that $A=\left(\begin{array}{lll}1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1\end{array}ight)$

We shall prove the result by using the principle of mathematical induction.

For $n=1$, we have:

$$

P(1):\left(\begin{array}{lll}

3^{n-1} & 3^{n-1} & 3^{n-1} \

3^{n-1} & 3^{n-1} & 3^{n-1}

\end{array}ight)=\left(\begin{array}{lll}

3^{0} & 3^{0} & 3^{0} \

3^{0} & 3^{0} & 3^{0}

\end{array}ight)=\left(\begin{array}{lll}

1 & 1 & 1 \

1 & 1 & 1

\end{array}ight)=A

$$

Therefore, the result is true for $n=1$.

Let the result be true for $n=k$.

$$

P(k): A^{k}=\left(\begin{array}{lll}

3^{k-1} & 3^{k-1} & 3^{k-1} \

3^{k-1} & 3^{k-1} & 3^{k-1}

\end{array}ight)

$$

Now, we have to prove that the result is true for $n=k+1$.

Since,

$$

\begin{aligned}

A^{k+1} & =A \cdot A^{k} \

& =\left(\begin{array}{lll}

1 & 1 & 1 \

1 & 1 & 1

\end{array}ight)\left(\begin{array}{lll}

3^{k-1} & 3^{k-1} & 3^{k-1} \

3^{k-1} & 3^{k-1} & 3^{k-1}

\end{array}ight) \

& =\left(\begin{array}{lll}

3 \cdot 3^{k-1} & 3 \cdot 3^{k-1} & 3 \cdot 3^{k-1} \

3 \cdot 3^{k-1} & 3 \cdot 3^{k-1} & 3 \cdot 3^{k-1}

\end{array}ight) \

& =\left(\begin{array}{lll}

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}ight)

\end{aligned}

$$

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

Thus, by the principle of mathematical induction, 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}

\end{array}ight), n \in N

$$

Question 4

If $A=\left(\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight)$, prove that $A^{n}=\left(\begin{array}{cc}1+2 n & -4 n \ n & 1-2 n\end{array}ight)$, where $n$ is any positive integer.

Accepted Answer

It is given that $A=\left(\begin{array}{ll}3 & -4 \ 1 & -1\end{array}ight)$

We shall prove the result by using the principle of mathematical induction.

For $n=1$, we have:

$$

\begin{aligned}

P(1) & : A^{1}=\left(\begin{array}{cc}

1+2 n & -4 n \

n & 1-2 n

\end{array}ight) \

& =\left(\begin{array}{ll}

3 & -4 \

1 & -1

\end{array}ight) \

& =A

\end{aligned}

$$

Therefore, the result is true for $n=1$.

Let the result be true for $n=k$.

$$

P(k): A^{k}=\left(\begin{array}{cc}

1+2 k & -4 k \

k & 1-2 k

\end{array}ight), n \in N

$$

Now, we have to prove that the result is true for $n=k+1$.

Since,

$$

\begin{aligned}

A^{k+1} & =A \cdot A^{k} \

& =\left(\begin{array}{cc}

1+2 k & -4 k \

k & 1-2 k

\end{array}ight)\left(\begin{array}{ll}

3 & -4 \

1 & -1

\end{array}ight) \

& =\left(\begin{array}{cc}

3(1+2 k)-4 k & -4(1+2 k)+4 k \

3 k+1-2 k & -4 k-1(1-2 k)

\end{array}ight) \

& =\left(\begin{array}{cc}

3+6 k-4 k & -4-8 k+4 k \

3 k+1-2 k & -4 k-1+2 k

\end{array}ight) \

& =\left(\begin{array}{cc}

3+2 k & -4-4 k \

1+k & -1-2 k

\end{array}ight) \

& =\left(\begin{array}{cc}

1+2(k+1) & -4(k+1) \

1+k & 1-2(k+1)

\end{array}ight)

\end{aligned}

$$

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

Thus, by the principle of mathematical induction, we have:

$$

A^{n}=\left(\begin{array}{cc}

1+2 n & -4 n \

n & 1-2 n

\end{array}ight) ; n \in N

$$

Question 5

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

Accepted Answer

It is given that $A$ and $B$ are symmetric matrices.

Therefore, we have:

$$

\begin{equation*}

A^{\prime}=A 	ext { and } B^{\prime}=B 	ag{1}

\end{equation*}

$$

Now,

$$

\begin{array}{rlrl}

(A B-B A)^{\prime} & =(A B)^{\prime}-(B A)^{\prime} & & {\left[(A-B)^{\prime}=A^{\prime}-B^{\prime}ight]} \

& =B^{\prime} A^{\prime}-A^{\prime} B^{\prime} & & {\left[(A B)^{\prime}=B^{\prime} A^{\prime}ight]} \

& =B A-A B & & {[	ext { Using (1) }]} \

& =-(A B-B A) &

\end{array}

$$

Hence,

$$

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

$$

Thus, $A B-B A$ is a skew symmetric matrix.

Question 6

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

Accepted Answer

We suppose that $A$ is a symmetric matrix, then

$$

\begin{equation*}

A^{\prime}=A 	ag{1}

\end{equation*}

$$

Consider,

$$

\begin{array}{rlrl}

\left(B^{\prime} A Bight)^{\prime} & =\left\{B^{\prime}(A B)ight\}^{\prime} & & \

& =(A B)^{\prime}\left(B^{\prime}ight)^{\prime} & & {\left[\because(A B)^{\prime}=B^{\prime} A^{\prime}ight]} \

& =B^{\prime} A^{\prime}(B) & & {\left[\because\left(B^{\prime}ight)^{\prime}=Bight]} \

& =B^{\prime}\left(A^{\prime} Bight) & \

& =B^{\prime}(A B) & {[	ext { Using }(1)]}

\end{array}

$$

Therefore,

$$

\left(B^{\prime} A Bight)^{\prime}=B^{\prime} A B

$$

Thus, if $A$ is symmetric matrix, then $B^{\prime} A B$ is a symmetric matrix.

Now, we suppose that $A$ is a skew symmetric matrix, then

$$

\begin{equation*}

A^{\prime}=-A 	ag{2}

\end{equation*}

$$

Consider,

$$

\begin{array}{rlr}

\left(B^{\prime} A Bight)^{\prime} & =\left\{B^{\prime}(A B)ight\}^{\prime} & \

& =(A B)^{\prime}\left(B^{\prime}ight)^{\prime} & \

& =\left(B^{\prime} A^{\prime}ight) B & \

& =B^{\prime}(-A) B & {[	ext { Using }(2)]} \

& =-B^{\prime} A B &

\end{array}

$$

Therefore,

$$

\left(B^{\prime} A Bight)^{\prime}=-B^{\prime} A B

$$

Thus, if $A$ is a skew symmetric matrix, then $B^{\prime} A B$ is a skew symmetric matrix.

Hence, if $A$ is symmetric or skew symmetric matrix, then $B^{\prime} A B$ is symmetric or skew symmetric accordingly.

Question 7

Find the values of $x, y, z$ if the matrix $A=\left(\begin{array}{ccc}0 & 2 y & z \ x & y & -z \ x & -y & z\end{array}ight)$ satisfy the equation $A^{\prime} A=I$.

Accepted Answer

It is given that $A=\left(\begin{array}{ccc}0 & 2 y & z \ x & y & -z \ x & -y & z\end{array}ight)$

Therefore,

$$

A^{\prime}=\left(\begin{array}{ccc}

0 & x & x \

2 y & y & -y \

z & -z & z

\end{array}ight)

$$

Now, $A^{\prime} A=I$

Hence,

$$

\begin{aligned}

& \Rightarrow\left(\begin{array}{ccc}

0 & x & x \

2 y & y & -y \

z & -z & z

\end{array}ight)\left(\begin{array}{ccc}

0 & 2 y & z \

x & y & -z \

x & -y & z

\end{array}ight)=\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ccc}

0+x^{2}+x^{2} & 0+x y-x y & 0-x z+x z \

0+x y-x y & 4 y^{2}+y^{2}+y^{2} & 2 y z-y z-y z \

0-x z+z x & 2 y z-y z-y z & z^{2}+z^{2}+z^{2}

\end{array}ight)=\left(\begin{array}{lll}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight) \

& \Rightarrow\left(\begin{array}{ccc}

2 x^{2} & 0 & 0 \

0 & 6 y^{2} & 0 \

0 & 0 & 3 z^{2}

\end{array}ight)=\left(\begin{array}{ccc}

1 & 0 & 0 \

0 & 1 & 0 \

0 & 0 & 1

\end{array}ight)

\end{aligned}

$$

On comparing the corresponding elements, we have:

$$

\begin{aligned}

& 2 x^{2}=1 \

& \Rightarrow x= \pm \frac{1}{\sqrt{2}} \

& 6 y^{2}=1 \

& \Rightarrow y= \pm \frac{1}{\sqrt{6}} \

& 3 z^{2}=1 \

& \Rightarrow z= \pm \frac{1}{\sqrt{3}}

\end{aligned}

$$

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

Question 8

For what values of

$$

x:\left[\begin{array}{lll}

1 & 2 & 1

\end{array}ight]\left[\begin{array}{lll}

1 & 2 & 0 \

2 & 0 & 1 \

1 & 0 & 2

\end{array}ight]\left[\begin{array}{l}

0 \

2 \

x

\end{array}ight]=0

$$

Accepted Answer

We have:

$$

\left[\begin{array}{lll}

1 & 2 & 1

\end{array}ight]\left[\begin{array}{lll}

1 & 2 & 0 \

2 & 0 & 1 \

1 & 0 & 2

\end{array}ight]\left[\begin{array}{l}

0 \

2 \

x

\end{array}ight]=0

$$

Hence,

$$

\begin{aligned}

& \Rightarrow\left[\begin{array}{lll}

1+4+1 & 2+0+0 & 0+2+2

\end{array}ight]\left[\begin{array}{l}

0 \

2 \

x

\end{array}ight]=0 \

& \Rightarrow\left[\begin{array}{lll}

6 & 2 & 4

\end{array}ight]\left[\begin{array}{l}

0 \

2 \

x

\end{array}ight]=0 \

& \Rightarrow[6(0)+2(2)+4(x)]=0 \

& \Rightarrow[4+4 x]=0 \

& \Rightarrow 4 x=-4 \

& \Rightarrow x=-1

\end{aligned}

$$

Thus, the required value of $x=-1$.

Question 9

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

Accepted Answer

It is given that $A=\left(\begin{array}{cc}3 & 1 \ -1 & 2\end{array}ight)$

Therefore,

$$

\begin{aligned}

A^{2} & =A \cdot A \

& =\left(\begin{array}{cc}

3 & 1 \

-1 & 2

\end{array}ight)\left(\begin{array}{cc}

3 & 1 \

-1 & 2

\end{array}ight) \

& =\left(\begin{array}{cc}

3(3)+1(-1) & 3(1)+1(2) \

-1(3)+2(-1) & -1(1)+2(2)

\end{array}ight) \

& =\left(\begin{array}{cc}

8 & 5 \

-5 & 3

\end{array}ight)

\end{aligned}

$$

Now,

$$

\begin{aligned}

L H S & =A^{2}-5 A+7 I \

& =\left(\begin{array}{cc}

8 & 5 \

-5 & 3

\end{array}ight)-5\left(\begin{array}{cc}

3 & 1 \

-1 & 2

\end{array}ight)+7\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight) \

& =\left(\begin{array}{cc}

8 & 5 \

-5 & 3

\end{array}ight)-\left(\begin{array}{cc}

15 & 5 \

-5 & 10

\end{array}ight)+\left(\begin{array}{cc}

7 & 0 \

0 & 7

\end{array}ight) \

& =\left(\begin{array}{cc}

-7 & 0 \

0 & -7

\end{array}ight)+\left(\begin{array}{cc}

7 & 0 \

0 & 7

\end{array}ight) \

& =0 \

& =R H S

\end{aligned}

$$

Thus, $A^{2}-5 A+7 I=0$

Question 10

Find $x$, if $\left[\begin{array}{lll}x & -5 & -1\end{array}ight]\left[\begin{array}{lll}1 & 0 & 2 \ 0 & 2 & 1 \ 2 & 0 & 3\end{array}ight]\left[\begin{array}{l}x \ 4 \ 1\end{array}ight]=0$

Accepted Answer

\section*{We have}

$$

\left[\begin{array}{lll}

x & -5 & -1

\end{array}ight]\left[\begin{array}{lll}

1 & 0 & 2 \

0 & 2 & 1 \

2 & 0 & 3

\end{array}ight]\left[\begin{array}{l}

x \

4 \

1

\end{array}ight]=0

$$

Hence,

$$

\begin{aligned}

& \Rightarrow\left[\begin{array}{lll}

x+0-2 & 0-10+0 & 2 x-5-3

\end{array}ight]\left[\begin{array}{l}

x \

4 \

1

\end{array}ight]=0 \

& \Rightarrow\left[\begin{array}{lll}

x-2 & -10 & 2 x-8

\end{array}ight]\left[\begin{array}{l}

x \

4 \

1

\end{array}ight]=0 \

& \Rightarrow[x(x-2)-40+2 x-8]=0 \

& \Rightarrow\left[x^{2}-2 x-40+2 x-8ight]=0 \

& \Rightarrow\left[x^{2}-48ight]=0 \

& \Rightarrow x^{2}-48=0 \

& \Rightarrow x^{2}=48 \

& \Rightarrow x= \pm 4 \sqrt{3}

\end{aligned}

$$

Thus, $x= \pm 4 \sqrt{3}$.

Question 11

A manufacturer produces three products $x, y, z$ which he sells in two markets. Annual sales are indicated below:

\begin{tabular}{cccc}

Market & \multicolumn{3}{c}{ Products } \

I & 10000 & 2000 & 18000 \

II & 6000 & 20000 & 8000

\end{tabular}

(a) 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.

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

Accepted Answer

(a) The unit sale prices of $x, y$ and $z$ are $₹ 2.50$, ₹ 1.50 and ₹ 1.00 respectively.

Consequently, the total revenue in market I can be represented in the form of a matrix as:

$$

\begin{aligned}

{\left[\begin{array}{lll}

10000 & 2000 & 18000

\end{array}ight]\left[\begin{array}{l}

2.50 \

1.50 \

1.00

\end{array}ight] } & =10000 	imes 2.50+2000 	imes 1.50+18000 	imes 1.00 \

& =25000+3000+18000 \

& =46000

\end{aligned}

$$

The total revenue in market II can be represented in the form of a matrix as:

$$

\begin{aligned}

{\left[\begin{array}{lll}

6000 & 20000 & 8000

\end{array}ight]\left[\begin{array}{l}

2.50 \

1.50 \

1.00

\end{array}ight] } & =6000 	imes 2.50+20000 	imes 1.50+8000 	imes 1.00 \

& =15000+30000+8000 \

& =53000

\end{aligned}

$$

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

(b) The unit costs of $x, y$ and $z$ are $₹ 2.00$, $₹ 1.00$ and 50 paise respectively.

Consequently, the total cost prices of all the products in market I can be represented in the form of a matrix as:

$$

\begin{aligned}

{\left[\begin{array}{lll}

10000 & 2000 & 18000

\end{array}ight]\left[\begin{array}{l}

2.00 \

1.00 \

0.50

\end{array}ight] } & =10000 	imes 2.00+2000 	imes 1.00+18000 	imes 0.50 \

& =20000+2000+9000 \

& =31000

\end{aligned}

$$

Since the total revenue in market I is ₹ 46000 , the gross profit in this market in ₹ is

$$

46000-31000=15000

$$

The total cost prices of all the products in market II can be represented in the form of a matrix as:

$$

\begin{aligned}

{\left[\begin{array}{lll}

6000 & 20000 & 8000

\end{array}ight]\left[\begin{array}{l}

2.00 \

1.00 \

0.50

\end{array}ight] } & =6000 	imes 2.00+20000 	imes 1.00+8000 	imes 0.50 \

& =12000+20000+4000 \

& =36000

\end{aligned}

$$

Since the total revenue in market I is ₹ 53000 , the gross profit in this market in ₹ is

$$

53000-36000=17000

$$

Thus, the gross profit in market I is ₹ 15000 and in market II is ₹ 17000 .

Question 12

Find the matrix $X$ so that $X\left[\begin{array}{lll}1 & 2 & 3 \ 4 & 5 & 6\end{array}ight]=\left[\begin{array}{ccc}-7 & -8 & -9 \ 2 & 4 & 6\end{array}ight]$

Accepted Answer

It is given that $X\left[\begin{array}{lll}1 & 2 & 3 \ 4 & 5 & 6\end{array}ight]=\left[\begin{array}{ccc}-7 & -8 & -9 \ 2 & 4 & 6\end{array}ight]$

The matrix given on the R.H.S. of the equation is a $2 	imes 3$ matrix and the one given on the L.H.S. of the equation is a $2 	imes 3$ matrix.

Therefore, $X$ has to be a $2 	imes 2$ matrix.

Now, let $X=\left[\begin{array}{ll}a & c \ b & d\end{array}ight]$

Therefore,

$$

\begin{aligned}

& \Rightarrow\left[\begin{array}{ll}

a & c \

b & d

\end{array}ight]\left[\begin{array}{lll}

1 & 2 & 3 \

4 & 5 & 6

\end{array}ight]=\left[\begin{array}{ccc}

-7 & -8 & -9 \

2 & 4 & 6

\end{array}ight] \

& \Rightarrow\left|\begin{array}{ccc}

a+4 c & 2 a+5 c & 3 a+6 c \

b+4 d & 2 b+5 d & 3 b+6 d

\end{array}ight|=\left|\begin{array}{ccc}

-7 & -8 & -9 \

2 & 4 & 6

\end{array}ight|

\end{aligned}

$$

Equating the corresponding elements of the two matrices, we have:

$$

\begin{array}{lll}

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

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

\end{array}

$$

Now,

$$

\begin{aligned}

& a+4 c=-7 \

& \Rightarrow a=-7-4 c

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

& 2 a+5 c=-8 \

& \Rightarrow 2(-7-4 c)+5 c=-8 \

& \Rightarrow-14-8 c+5 c=-8 \

& \Rightarrow-3 c=6 \

& \Rightarrow c=-2

\end{aligned}

$$

Hence,

$$

\begin{aligned}

& \Rightarrow a=-7-4(-2) \

& \Rightarrow a=-7+8 \

& \Rightarrow a=1

\end{aligned}

$$

Now,

$$

\begin{aligned}

& b+4 d=2 \

& \Rightarrow b=2-4 d

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

& 2 b+5 d=4 \

& \Rightarrow 2(2-4 d)+5 d=4 \

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

& \Rightarrow-3 d=0 \

& \Rightarrow d=0

\end{aligned}

$$

Hence,

$$

\begin{aligned}

& b=2-4 d \

& \Rightarrow b=2

\end{aligned}

$$

Thus, $a=1, b=2, c=-2$ and $d=0$

Hence, the required matrix $X=\left[\begin{array}{cc}1 & -2 \ 2 & 0\end{array}ight]$

Question 13

If $A=\left[\begin{array}{cc}\alpha & \beta \ \gamma & -\alpha\end{array}ight]$ is such that $A^{2}=I$ then,

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

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

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

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

Accepted Answer

It is given that $A=\left[\begin{array}{cc}\alpha & \beta \ \gamma & -\alpha\end{array}ight]$

Therefore,

$$

\begin{aligned}

A^{2} & =A \cdot A \

& =\left(\begin{array}{cc}

\alpha & \beta \

\gamma & -\alpha

\end{array}ight)\left(\begin{array}{cc}

\alpha & \beta \

\gamma & -\alpha

\end{array}ight) \

& =\left(\begin{array}{cc}

\alpha^{2}+\beta \gamma & \alpha \beta-\alpha \beta \

\alpha \gamma-\alpha \gamma & \beta \gamma+\alpha^{2}

\end{array}ight) \

& =\left(\begin{array}{cc}

\alpha^{2}+\beta \gamma & 0 \

0 & \beta \gamma+\alpha^{2}

\end{array}ight)

\end{aligned}

$$

Now, $A^{2}=I$

Hence,

$$

\left(\begin{array}{cc}

\alpha^{2}+\beta \gamma & 0 \

0 & \beta \gamma+\alpha^{2}

\end{array}ight)=\left(\begin{array}{ll}

1 & 0 \

0 & 1

\end{array}ight)

$$

On comparing the corresponding elements, we have:

$$

\begin{aligned}

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

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

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

\end{aligned}

$$

Thus, the correct option is C.

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

Accepted Answer

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

$$

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

$$

Hence,

$$

\begin{aligned}

& \Rightarrow A=-A \

& \Rightarrow A+A=0 \

& \Rightarrow 2 A=0 \

& \Rightarrow A=0

\end{aligned}

$$

Therefore, $A$ is a zero matrix.

Thus, the correct option is B.

Question 15

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

(A) $A$

(B) $I-A$

(C) $I$

(D) $3 A$

Accepted Answer

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

Now,

$$

\begin{array}{rlr}

(I+A)^{3}-7 A & =I^{3}+A^{3}+3 I^{2} A+3 A^{2} I-7 A & \

& =I+A^{2} \cdot A+3 A+3 A^{2}-7 A & \

& =I+A \cdot A+3 A+3 A-7 A & {\left[\because A^{2}=Aight]} \

& =I+A^{2}-A & \

& =I+A-A & {\left[\because A^{2}=Aight]} \

& =I &

\end{array}

$$

Hence,

$$

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

$$

Thus, the correct option is C.