91影视

The vectors are,

$\mathrm{A}=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right);\stackrel{鈬赌}{{\mathrm{v}}_1}=\left[\begin{array}{c}1\\ 1\end{array}\right];\stackrel{鈬赌}{{\mathrm{v}}_2}=\left[\begin{array}{c}1\\ -1\end{array}\right]$

The formula is, $B=S^{-1}AS$.

Compute the matrix

The inverse of the matrix is,

$\begin{array}{l}s=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\\ 鈬�s^{-1}=\frac{1}{ad-bc}-\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)\end{array}$

Where$S=\left(\stackrel{鈫�}{v_1},\stackrel{鈫�}{v_2}\right)$, now put the values of $\stackrel{鈫�}{v_1}$ and $\stackrel{鈫�}{v_2}$ in $S=\left(\stackrel{鈫�}{v_1},\stackrel{鈫�}{v_2}\right)$.

$$\begin{gathered} S=\left(\stackrel{鈬赌}{v_1},\stackrel{鈬赌}{v_2}\right) \\ =\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right) \\ 鈬�S^{-1}=\frac{1}{-2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right) \end{gathered}$$

Substitute these values in the formula.

role="math" localid="1664195157574" $\begin{array}{l}B=S^{-1}AS\\ 鈬�B=\frac{1}{-2}\left(\begin{array}{cc}-1& -1\\ -1& 1\end{array}\right)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& -1\end{array}\right)\\ B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\end{array}$

Hence the matrix is $B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

The matrix is,

$$\begin{gathered} {\left(\begin{array}{c}{c}_1\\ -c_2\end{array}\right)}_{\mathfrak{I}}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right){\left(\begin{array}{c}{c}_1\\ c_2\end{array}\right)}_{\mathfrak{I}} \\ =\left(\begin{array}{c}a{c}_1+b{c}_2\\ c{c}_1+d{c}_2\end{array}\right) \end{gathered}$$

After comparing it gives,

$a=1,b=0,c=0,d=-1$

Substitute these values in the formula.

role="math" localid="1664195387240" $$\begin{gathered} B=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) \\ 鈭�B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \end{gathered}$$

Hence the matrix is$B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

The formula is, $T\left(\stackrel{鈬赌}{v_1}\right)=A\stackrel{鈬赌}{v_1},T\left(\stackrel{鈬赌}{v_2}\right)=A\stackrel{鈬赌}{v_2}$

Compute the matrices,

$$\begin{gathered} T\left(\stackrel{鈬赌}{v_1}\right)=A\stackrel{鈬赌}{v_1} \\ =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right) \\ =\left(\begin{array}{c}1\\ 1\end{array}\right) \\ =\stackrel{鈬赌}{v_1} \\ T\left(\stackrel{鈬赌}{v_2}\right)=A\stackrel{鈬赌}{v_2} \\ =\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\left(\begin{array}{c}1\\ -1\end{array}\right) \\ =\left(\begin{array}{c}-1\\ 1\end{array}\right) \\ =\stackrel{鈬赌}{v_2} \end{gathered}$$

Substitute these values in the formula.

$$\begin{gathered} T\left(\stackrel{鈬赌}{v_1}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right),T\left(\stackrel{鈬赌}{v_2}\right)=\left(\begin{array}{c}0\\ -1\end{array}\right) \\ B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \end{gathered}$$

(a) The matrix is, $B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

(b) The matrix is, $B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

(c) The matrix is, $B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.