For the matrices A in Exercises 1 through 12, find closed formulas for , where t is an arbitrary positive integer. Follow the strategy outlined in Theorem 7.4.2 and illustrated in Example 2. In Exercises 9 though 12, feel free to use technology.

$1.A=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$

The given matrix is$A=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$.Since the given matrix is upper triangular, its eigenvalues are their main diagonals. It is${\lambda }_1=1,{\lambda }_2=3$

Since, there are two distinct real eigenvalues, the matrix is diagonalizable and its diagonalization will be

$B=\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right]$

$B^t=\left[\begin{array}{cc}1& 0\\ 0& 3^t\end{array}\right]$

First we are going to find the eigenvalues of matrix A. The characteristic polynomial of A is,

$det(A-1))=\mathrm{de}t\begin{array}{cc}1-\lambda & 2\\ 0& 3-\lambda \end{array}$

To find the matrix S, we are going to find the eigenspaces of matrix A. the eigenspaces associated with $\lambda$, is denoted by $E_{\lambda }$and it is given by$E_{\lambda }=ker(A-\lambda l)$,

$$\begin{gathered} \left[\begin{array}{cc}1-\lambda & 2\\ 0& 3-\lambda \end{array}\right]\times \left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \\ (1-\lambda )x_1+2x_2=0 \\ (3-\lambda )x_2=0 \end{gathered}$$

Now we can solve for $\lambda -1$ and by substitute , we get $\lambda -2$

$\left[\left.\begin{array}{c}{X}_1\\ X_2\end{array}\right]=\left[\begin{array}{c}X\\ 0\end{array}\right.\right]$

So,. $E_1=sp\mathrm{an}\left[\begin{array}{c}1\\ 0\end{array}\right]$

Now we can solve for localid="1668432270468" $\lambda =3$and by substitute , we get

localid="1668432279313" $\left[\left.\begin{array}{c}{X}_1\\ X_2\end{array}\right]=\left[\begin{array}{c}X\\ X\end{array}\right.\right]$

So, localid="1668432282790" $E_2=span\left[\begin{array}{c}1\\ 1\end{array}\right]$

So finally,$S=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$

Now we can find the matrix.

$S=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right],S^{-1}=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]$

Finally now, we can calculate,

$$\begin{gathered} A^t=S{B}^t{S}^{-1} \\ =\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\times \left[\begin{array}{cc}1& 0\\ 0& 3^t\end{array}\right]\times \left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right] \\ {A}^t=\left[\begin{array}{cc}1& 3^t-1\\ 0& 3^t\end{array}\right] \end{gathered}$$

Hence the closed formulas for