91Ó°ÊÓ

A function is defined as a relationship between a set of inputs that each have one output.

Given,

$$\begin{gathered} A=\left[1         0         -1 \\ -2     -1     -2 \\ 1            1          3\right] \end{gathered}$$

$$\begin{gathered} \left|1-\mathrm{λ}         0                   -1 \\ 2            -1-\mathrm{λ}           -2 \\ 1                  1               3-\mathrm{λ}\right|=0 \end{gathered}$$

$$\begin{gathered} \left(1-\mathrm{λ} \right)\left(-1-\mathrm{λ}\right)+2+2\left(1-\mathrm{λ} \right)+\left(-1-\mathrm{λ} \right)=0 \\ -{\mathrm{λ}}^2+3{\mathrm{λ}}^2-2\mathrm{λ}=0 \end{gathered}$$

$-\mathrm{λ}\left(\mathrm{λ}-1\right)\left(\mathrm{λ}-2\right)=0$

${\mathrm{λ}}_1=0,{\mathrm{λ}}_2=1,{\mathrm{λ}}_3=2$

We have three distinct real eigenvalues of a matrix, so there exists an eigen basis in which the diagonalization of A is

$$\begin{gathered} B=\left[1           0         0 \\ 0          2          0 \\ 0          0         0\right] \end{gathered}$$

$ \mathrm{λ}=0$we solved,

$Ax=0$

$$\begin{gathered} \left[1            0             -1 \\ -2       -1       -2 \\ 1            0             3\right]  \left[ x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right]    \end{gathered}$$

$x_1-x_3=0,  -2x_1-x_2-2x_2=0,x_1+x_2+3x_3=0$

$$\begin{gathered} E_0=span\left(\left[1 \\ -4 \\ 1\right]\right) \end{gathered}$$

$ \mathrm{λ}=1$similarly,

solved,

$\left(A-I\right)x=0$

$$\begin{gathered} \left[0           0             -1 \\ -2       -2      -2 \\ 1            0             2\right]  \left[ x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right]    \end{gathered}$$

$x_3=0,  x_2+x_3=0, x_1+x_2+2x_3=0$

$$\begin{gathered} E_0=span\left(\left[1 \\ -1 \\ 0\right]\right) \end{gathered}$$

Finally,

$ \mathrm{λ}=2$solved,

$\left(A-2I\right)x=0$

$$\begin{gathered} \left[-1        0       -1 \\ -2    -3      -2 \\ 1           1           1\right]  \left[x_1 \\  x_2 \\ {x}_3\right]=\left[0 \\ 0 \\ 0\right]    \end{gathered}$$

$x_1+x_3=0,  -2x_1-3x_2-2 x_2=0,   x_1+x_2+x_3=0$

$$\begin{gathered} E_0=span\left(\left[-1 \\ 0 \\ 1\right]\right) \end{gathered}$$

$$\begin{gathered} S=\left[\begin{array}{ccc}-1& 1& -1\\ 0& -4& 1\\ 1& 1& 0\end{array}\right] \\ {S}^{-1}=\left[\begin{array}{ccc}-2& -1& -2\\ 1/2& 1/2& 3/2\\ -1/2& -1/2& -1/2\end{array}\right] \end{gathered}$$

$A^t=S{B}^t{S}^{-1}=\left[\begin{array}{ccc}-1& -1& 1\\ 1& 0& -4\\ 0& 1& 1\end{array}\right]×\left[\begin{array}{ccc}1& 0& 0\\ 0& 2^t& 0\\ 0& 0& 0\end{array}\right]×\frac{1}{2}\left[\begin{array}{ccc}-4& -2& -4\\ 1& 1& 3\\ -1& -1& -1\end{array}\right]$

$=\frac{1}{2}\left[\begin{array}{ccc}-1& -1& 1\\ 1& 0& -4\\ 0& 1& 1\end{array}\right]×\left[\begin{array}{ccc}1& 0& 0\\ 0& 2^t& 0\\ 0& 0& 0\end{array}\right]×\left[\begin{array}{ccc}-4& -2& -4\\ 1& 1& 3\\ -1& -1& -1\end{array}\right]$

$=\frac{1}{2}\left[\begin{array}{ccc}4-2^t& 2-2^t& 4-3×2^t\\ -4& -2& -4\\ 2^t& 2^t& 3×2^t\end{array}\right]$

Hence,

$A^t=\frac{1}{2}\left[\begin{array}{ccc}4-2^t& 2-2^t& 4-3×2^t\\ -4& -2& -4\\ 2^t& 2^t& 3×2^t\end{array}\right]$