If matrix $\left[\begin{array}{ccc}0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0\end{array}\right]$ is a skew-symmetric matrix, then find the values of $a, b$ and $c$.
Let $A=\left[\begin{array}{ccc}0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0\end{array}\right]$ Since, $A$ is skew-symmetric matrix.
$$\therefore \quad A^{\prime}=-A$$
$$\begin{aligned} &\begin{array}{ll} \Rightarrow \quad {\left[\begin{array}{ccc} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{array}\right]=-\left[\begin{array}{ccc} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{array}\right]} \\ \Rightarrow \quad\left[\begin{array}{ccc} 0 & 2 & c \\ a & b & 1 \\ 3 & -1 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -a & -3 \\ -2 & -b & +1 \\ -c & -1 & 0 \end{array}\right] \end{array}\\ &\text { By equality of matrices, we get }\\ &\begin{array}{ll} & a=-2, c=-3 \text { and } b=-b \Rightarrow b=0 \\ \therefore & a=-2, b=0 \text { and } c=-3 \end{array} \end{aligned}$$
If $\quad P(x)=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$, then show that $P(x) \cdot P(y)=P(x+y)$ $=P(y) \cdot P(x)$.
$$\begin{aligned} &\text { We have, }\\ &\begin{array}{ll} & P(x)=\left[\begin{array}{lll} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right] \\ \therefore & P(y)=\left[\begin{array}{ll} \cos y & \sin y \\ -\sin y & \cos y \end{array}\right] \end{array} \end{aligned}$$
$$\begin{aligned} \text { Now, } \quad P(x) \cdot P(y) & =\left[\begin{array}{ll} \cos x & \sin x \\ -\sin x & \cos x \end{array}\right]\left[\begin{array}{cc} \cos y & \sin y \\ -\sin y & \cos y \end{array}\right] \\ & =\left[\begin{array}{ll} \cos x \cdot \cos y-\sin x \cdot \sin y & \cos x \cdot \sin y+\sin x \cdot \cos y \\ -\sin x \cdot \cos y-\cos x \cdot \sin y & -\sin x \cdot \sin y+\cos x \cdot \cos y \end{array}\right]\quad\text{.... (i)} \end{aligned}$$
$$\begin{aligned} & \quad\left[\begin{array}{l} \because \cos (x+y)=\cos x \cdot \cos y-\sin x \cdot \sin y \\ \text { and } \sin (x+y)=\sin x \cdot \cos y+\cos x \cdot \sin y \end{array}\right] \\ & \text { and } \quad P(x+y)=\left[\begin{array}{lr} \cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y) \end{array}\right]\quad\text{.... (ii)} \end{aligned}$$
Also, $\quad P(y) \cdot P(x)=\left[\begin{array}{ll}\cos y & \sin y \\ -\sin y & \cos y\end{array}\right]\left[\begin{array}{lc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]$
$$\begin{aligned} & =\left[\begin{array}{lr} \cos y \cdot \cos x-\sin y \cdot \sin x & \cos y \cdot \sin x+\sin y \cdot \cos x \\ -\sin y \cdot \cos x-\sin x \cdot \cos y & -\sin y \cdot \sin x+\cos y \cdot \cos x \end{array}\right] \\ & =\left[\begin{array}{ll} \cos (x+y) & \sin (x+y) \\ -\sin (x+y) & \cos (x+y) \end{array}\right]\quad\text{.... (iii)} \end{aligned}$$
Thus, we see from the Eqs. (i), (ii) and (iii) that,
$$P(x) \cdot P(y)=P(x+y)=P(y) \cdot P(x)$$ Hence proved.
If $A$ is square matrix such that $A^2=A$, then show that $(I+A)^3=7 A+I$.
$$\begin{aligned} &\begin{aligned} \text { Since, } A^2=A \text { and }(I+A) \cdot(I+A)=I^2 & +I A+A I+A^2 \\ & =I^2+2 A I+A^2 \\ & =I+2 A+A=I+3 A \end{aligned} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { and } \quad (I+A) \cdot(I+A)(I+A) & =(I+A)(I+3 A) \\ & =I^2+3 A I+A I+3 A^2 \\ & =I+4 A I+3 A \\ & =I+7 A=7 A+I\quad\text{Hence proved.} \end{aligned} \end{aligned}$$
If $A, B$ are square matrices of same order and $B$ is a skew-symmetric matrix, then show that $A^{\prime} B A$ is skew-symmetric.
Since, $A$ and $B$ are square matrices of same order and $B$ is a skew-symmetric matrix i.e., $B^{\prime}=-B$
$$ \begin{aligned} &\text { Now, we have to prove that } A^{\prime} B A \text { is a skew-symmetric matrix. }\\ &\begin{array}{rlrl} \therefore \quad A^{\prime} B A^{\prime} & =A^{\prime} B A^{\prime}=B A^{\prime} A^{\prime} \quad [\because AB'=B'A']\\ & =A^{\prime} B^{\prime} A=A^{\prime}-B A=-A^{\prime} B A \end{array} \end{aligned}$$
Hence, $A'BA$ is a skew-symmetric matrix.
If $A B=B A$ for any two square matrices, then prove by mathematical induction that $(A B)^n=A^n B^n$.
$$\begin{aligned} &\begin{aligned} \text { Let } \quad & P(n):(A B)^n=A^n B^n \\ \therefore\quad & P(1):(A B)^1=A^1 B^1 \Rightarrow A B=A B \end{aligned} \end{aligned}$$
So, $P(1)$ is true.
Now, $$P(k):(A B)^k=A^k B^k, k \in N$$
So, $P(K)$ is true, whenever $P(k+1)$ is true.
$$\therefore \quad P(K+1: A B)^{K+1}=A^{k+1} B^{k+1}\quad\text{.... (i)}$$
$$\begin{array}{ll} \Rightarrow & A B^k \cdot A B^1 \quad [\because AB=BA]\\ \Rightarrow & A^k B^k \cdot B A \Rightarrow A^k B^{k+1} A \\ \Rightarrow & A^k \cdot A \cdot B^{k+1} \Rightarrow A^{k+1} B^{k+1} \\ \Rightarrow & (A \cdot B)^{k+1}=A^{k+1} B^{k+1} \end{array}$$
So, $P(k+1)$ is true for all $n \in N$, whenever $P(k)$ is true.
By mathematical induction $(A B)=A^n B^n$ is true for all $n \in N$.