If $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right]$, then find $A^2+2 A+7 I$
We have, $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right]$
$\therefore \quad A^2=\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 4 & 1\end{array}\right] \quad\left[\because A^2=A \cdot A\right]$
$$ \begin{aligned} & =\left[\begin{array}{lll} 1+8 & 2+2 \\ 4+4 & 8+1 \end{array}\right]=\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right] \\ \therefore\quad A^2+2 A+7 I & =\left[\begin{array}{ll} 9 & 4 \\ 8 & 9 \end{array}\right]+\left[\begin{array}{ll} 2 & 4 \\ 8 & 2 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]=\left[\begin{array}{cc} 18 & 8 \\ 16 & 18 \end{array}\right] \end{aligned}$$
If $A=\left[\begin{array}{rr}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]$ and $A^{-1}=A^{\prime}$, then find the value of $\alpha$.
We have, $A=\left[\begin{array}{cc}\cos a & \sin a \\ -\sin a & \cos a\end{array}\right]$ and $A^{\prime}=\left[\begin{array}{cc}\cos a & -\sin a \\ \sin a & \cos a\end{array}\right]$
$$\begin{array}{lrl} \text { Also, } & A^{-1} & =A^{\prime} \\ \Rightarrow & A A^{-1} & =A A^{\prime} \end{array}$$
$\left.\begin{array}{lll}\Rightarrow & I & =\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right]\end{array} \begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
$\Rightarrow \quad\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{lc}\cos ^2 \alpha+\sin ^2 \alpha & 0 \\ 0 & \sin ^2 \alpha+\cos ^2 \alpha\end{array}\right]$
By using equality of matrices, we get
$$\cos ^2 \alpha+\sin ^2 \alpha=1$$
which is true for all real values of $\alpha$.
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 $(I+A)^3=7 A+I$.
$$\begin{aligned} &\begin{aligned} \text { Since, } A^2=A \text { and }(I+A) \cdot(I+A)=I^2 & +I A+A I+A^2 \\ & =I^2+2 A I+A^2 \\ & =I+2 A+A=I+3 A \end{aligned} \end{aligned}$$
$$\begin{aligned} &\begin{aligned} \text { and } \quad (I+A) \cdot(I+A)(I+A) & =(I+A)(I+3 A) \\ & =I^2+3 A I+A I+3 A^2 \\ & =I+4 A I+3 A \\ & =I+7 A=7 A+I\quad\text{Hence proved.} \end{aligned} \end{aligned}$$