If $A=\begin{bmatrix} \cfrac { \pi }{ 2 } & \pi \\ 0 & \cfrac { 3\pi }{ 2 } \end{bmatrix}$, find sin A.

Solution:

$A=\begin{bmatrix} \cfrac { \pi }{ 2 } & \pi \\ 0 & \cfrac { 3\pi }{ 2 } \end{bmatrix}$

C.E. is given by, $|A-\lambda I|=0$

$\begin{vmatrix} \cfrac { \pi }{ 2 } & \pi \\ 0 & \cfrac { 3\pi }{ 2 } \end{vmatrix}=0$

$\left( \cfrac{\pi}{2} - \lambda \right ) \left ( \cfrac { 3\pi }{ 2 } -\lambda \right )-0=0$

$\therefore \lambda =\cfrac{\pi}{2} \ \& \ \lambda =\cfrac{3\pi}{2} $

Let $\phi(A)=sin \ A = \alpha A+ \beta I$ ------------(I)

Since $\lambda$ satisfies the above equation,

$sin \ \lambda = \alpha \lambda +\beta$ --------------(2)

Put $\lambda = \cfrac{\pi}{2}$ in (2),

$sin \ \cfrac{\pi}{2} = \alpha \cfrac{\pi}{2} + \beta$

$1= \alpha \cfrac{\pi}{2} + \beta$ -------------(3)

Put $\lambda = \cfrac{3 \pi}{2}$ in (2),

$sin \ \cfrac{3 \pi}{2} = \alpha \cfrac{3 \pi}{2} + \beta$

$-1= \alpha \cfrac{3 \pi}{2} + \beta$ -------------(4)

Solving equations (3) and (4) simultaneously, we get,

$\alpha = - \cfrac{2}{\pi}$ -----------(5)

$\beta = 2$ --------------(6)

Put (5) and (6) in (1),

$sin \ A= - \cfrac{2}{\pi} \begin{bmatrix} \cfrac { \pi }{ 2 } & \pi \\ 0 & \cfrac { 3\pi }{ 2 } \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} -1 & -2 \\ 0 & -3 \end{bmatrix}+\begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 0 & -1 \end{bmatrix}$