The characteristic eq. of A is

$\begin{vmatrix} \frac{3}{2} - \lambda & \frac{1}{2} \\ \frac{1}{2} & \frac{3}{2} - \lambda \end{vmatrix} = 0 \\ \bigg(\frac{3}{2} - \lambda \bigg)^2 - \frac{1}{4} = 0 \\ \therefore \frac{9}{4} - 3\lambda + \lambda^2 - \frac{1}{4} = 0 \\ \lambda^2 - 3\lambda + 2 = 0 \\ (\lambda - 1)(\lambda - 2) = 0 \\ \lambda = 1,2$

For $\lambda = 1, [A - \lambda I]X = 0$ given

$\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

By $R_{2} - R_{1} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ \frac{x_{1}}{2} + \frac{x_{2}}{2} = 0$

If x = -1 $x_{1} = 1$

Hence eigen vector is [1, -1]

For $\lambda = 2, [A - \lambda I]X = 0$ given

$\begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

By $R_{2} + R_{1} \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \\ \frac{-x_{1}}{2} + \frac{x_{2}}{2} = 0 \\ \therefore x_{1} = x_{2}$

If $x_{2} = 1, x_{1} = 1$

Hence eigen vector is [1,1]

$M = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \therefore |M| = 2 \\ M^{-1} = \frac{adj M}{|M|} = \frac{1}{2} \begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix}$

Now $D = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}$

If $f(A) = e^A \hspace{0.5cm} f(D) = e^D = \begin{bmatrix} e^1 & 0 \\ 0 & e^2 \end{bmatrix}$

If $f(A) = 4^A \hspace{0.5cm} f(D) = 4^D = \begin{bmatrix} 4^1 & 0 \\ 0 & 4^2 \end{bmatrix} \\ \therefore e^A = Mf(D)M^{-1} = \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} e^1 & 0 \\ 0 & e^2 \end{bmatrix} \frac{1}{2} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \frac{1}{2} \begin{bmatrix} e & e^2 \\ -e & e^2 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \\ e^A = \frac{1}{2} \begin{bmatrix} e + e^2 & -e + e^2 \\ -e + e^2 & e + e^2 \end{bmatrix}$

Replacing e by 4 we get

$4^A = \frac{1}{2} \begin{bmatrix} 20 & 12 \\ 12 & 20 \end{bmatrix} = \begin{bmatrix} 10 & 6 \\ 6 & 10 \end{bmatrix}$