If A = $\begin{bmatrix} \pi & \pi / 4 \\ 0 & \pi / 2 \\ \end{bmatrix}$ , Find $cosA$

Given, A = $\begin{bmatrix} \pi & \pi / 4 \\ 0 & \pi / 2 \\ \end{bmatrix}$

Characteristic equation is given by

$\lvert A - \lambda I \lvert = 0$

$\begin{align} \begin{bmatrix} \pi- \lambda & \pi / 4 \\ 0 & \pi / 2 - \lambda\\ \end{bmatrix} = 0 \\ (\pi - \lambda) (\pi /2 - \lambda) - 0 = 0 \\ (\pi - \lambda) (\pi /2 - \lambda) = 0 \\ \therefore \lambda = \pi , \ \lambda = \pi /2 \\ since \ the \ matrix \ is \ of \ order \ 2 , \ we \ consider \\ \phi(A) = cos A = \alpha A + \beta I - - - - 1 \\ since \ \lambda \ satisfies \ the above \ equation \\ cos \lambda = \alpha \lambda A + \beta - - - - 2 \\ put \lambda = \pi in \ equation \ 2 \\ cos\pi = \alpha \lambda + \beta \\ -1 = \alpha \lambda + \beta - - - - 3 \\ put \lambda = - \pi /2 in \ equation \ 2\\ cos\pi /2 = 2 \pi /2 + \beta \\ 0 = 2 \pi /2 + \beta - - - - 4 \\ solve\ equation \ 3 \ and \ 4 \ simultaneously \\ \alpha = -2/\pi , \beta = 1 \\ put\ 2\ and \beta values\ in \ equation \ 1 \\ cosA = -2/\pi A + \beta I \\ since \ A = \begin{bmatrix} \pi & \pi / 4 \\ 0 & \pi / 2 \\ \end{bmatrix} and I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\\ \therefore cos A = \frac{-2}{\pi} \begin{bmatrix} \pi & \pi / 4 \\ 0 & \pi / 2 \\ \end{bmatrix} + 1 \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\\ cos A = \begin{bmatrix} -2 & 1/2 \\ 0 & -1 / 2 \\ \end{bmatrix} + 1 \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\\ cos A = \begin{bmatrix} -1 & 1-/2 / \\ 0 & -1 / 2 \\ \end{bmatrix} \\ \end{align}$

Given, A = $\begin{bmatrix} \pi & \pi / 4 \\ 0 & \pi / 2 \\ \end{bmatrix}$

Characteristic equation is given by

$\lvert A - \lambda I \lvert = 0$

$\begin{align} \begin{bmatrix} \pi- \lambda & \pi / 4 \\ 0 & \pi / 2 - \lambda\\ \end{bmatrix} = 0 \\ (\pi - \lambda) (\pi /2 - \lambda) - 0 = 0 \\ (\pi - \lambda) (\pi /2 - \lambda) = 0 \\ \therefore \lambda = \pi , \ \lambda = \pi /2 \\ since \ the \ matrix \ is \ of \ order \ 2 , \ we \ consider \\ \phi(A) = cos A = \alpha A + \beta I - - - - 1 \\ since \ \lambda \ satisfies \ the above \ equation \\ cos \lambda = \alpha \lambda A + \beta - - - - 2 \\ put \lambda = \pi in \ equation \ 2 \\ cos\pi = \alpha \lambda + \beta \\ -1 = \alpha \lambda + \beta - - - - 3 \\ put \lambda = - \pi /2 in \ equation \ 2\\ cos\pi /2 = 2 \pi /2 + \beta \\ 0 = 2 \pi /2 + \beta - - - - 4 \\ solve\ equation \ 3 \ and \ 4 \ simultaneously \\ \alpha = -2/\pi , \beta = 1 \\ put\ 2\ and \beta values\ in \ equation \ 1 \\ cosA = -2/\pi A + \beta I \\ since \ A = \begin{bmatrix} \pi & \pi / 4 \\ 0 & \pi / 2 \\ \end{bmatrix} and I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\\ \therefore cos A = \frac{-2}{\pi} \begin{bmatrix} \pi & \pi / 4 \\ 0 & \pi / 2 \\ \end{bmatrix} + 1 \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\\ cos A = \begin{bmatrix} -2 & 1/2 \\ 0 & -1 / 2 \\ \end{bmatrix} + 1 \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\\ cos A = \begin{bmatrix} -1 & 1-/2 / \\ 0 & -1 / 2 \\ \end{bmatrix} \\ \end{align}$