Question

In Exercises19through 24 , find the matrix Bof the linear transformation $\mathit{T}\left(\stackrel{\rightharpoonup }{x}\right)\mathbf{=}\mathit{A}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}$ with respect to the basis $\mathfrak{I}\mathbf{=}\left(\stackrel{\rightharpoonup }{v_1},\stackrel{\rightharpoonup }{v_2}\right)$. For practice, solve each problem in three ways: (a) Use the formula $\mathit{B}\mathbf{=}{\mathit{S}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathit{S}$ , (b) use a commutative diagram (as in Examples 3 and 4), and (c) construct “column by column.”

localid="1664194720187" $\mathit{A}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{;}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{1}\end{array}\mathbf{\right]}\mathbf{;}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{2}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{-}\mathbf{1}\end{array}\mathbf{\right]}$

Short answer

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

(b) The matrix is, localid="1664195466605" $B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$.

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

Step-by-step solution

Consider the vectors.

The vectors are,

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

Compute the matrix using formula.

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

Compute the matrix

The inverse of the matrix is,

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

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

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

Substitute these values in the formula.

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

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

Compute the matrix using a commutative diagram.

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" $$\begin{gathered} B=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) \\ \therefore B=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right) \end{gathered}$$

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

Compute the matrix by constructing columns.

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

Compute the matrices,

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

Substitute these values in the formula.

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

Final answer.

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