Question

In Exercises 9through 24, find the matrix Bof the linear transformation $\mathit{T}\mathbf{\left(}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}\mathbf{\right)}\mathbf{=}\mathit{A}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{x}}$ 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.”

role="math" localid="1664342971664" $\mathit{A}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{1}& \mathbf{2}\\ \mathbf{3}& \mathbf{6}\end{array}\mathbf{\right)}\mathbf{;}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{1}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{3}\end{array}\mathbf{\right]}\mathbf{;}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{v}}_{\mathbf{2}}}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{-}\mathbf{2}\\ \mathbf{1}\end{array}\mathbf{\right]}$

Short answer

(a) The matrix is, $B=\left(\begin{array}{cc}7& 0\\ 0& 0\end{array}\right)$.

(b) The matrix is,$B=\left(\begin{array}{cc}7& 0\\ 0& 0\end{array}\right)$.

(c) The matrix is, $B=\left(\begin{array}{cc}7& 0\\ 0& 0\end{array}\right)$.

Step-by-step solution

Consider the vectors.

The vectors are,

$A=\left(\begin{array}{cc}1& 2\\ 3& 6\end{array}\right);\stackrel{\rightharpoonup }{{\mathrm{v}}_1}=\left[\begin{array}{c}1\\ 3\end{array}\right];\stackrel{\rightharpoonup }{{\mathrm{v}}_2}=\left[\begin{array}{c}-2\\ 1\end{array}\right]$

Compute the matrix using the formula

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

Compute the matrix $S$.

The inverse of the matrix is,$S=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\Rightarrow S^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$

$$\begin{gathered} S=\left(\stackrel{\rightharpoonup }{v_1},\stackrel{\rightharpoonup }{v_1}\right)=\left(\begin{array}{cc}1& -2\\ 3& 1\end{array}\right) \\ \Rightarrow S^{-1}=\frac{1}{7}\left(\begin{array}{cc}1& 2\\ -3& 1\end{array}\right) \end{gathered}$$

Substitute these values in the formula.

$$\begin{gathered} B=S^{-1}AS \\ \Rightarrow B=\frac{1}{7}\left(\begin{array}{cc}1& 2\\ -3& 1\end{array}\right)\left(\begin{array}{cc}1& 2\\ 3& 6\end{array}\right)\left(\begin{array}{cc}1& -2\\ 3& 1\end{array}\right) \\ B=\left(\begin{array}{cc}7& 0\\ 0& 0\end{array}\right) \end{gathered}$$

Compute the matrix using a commutative diagram.

The matrix is,

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

$a=7,b=0,c=0,d=0$

Substitute these values in the formula.

$$\begin{gathered} B=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) \\ \therefore B=\left(\begin{array}{cc}7& 0\\ 0& 0\end{array}\right) \end{gathered}$$

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 matrix

$$\begin{gathered} T\left(\stackrel{\rightharpoonup }{v_1}\right)=A\stackrel{\rightharpoonup }{v_1}\left(\begin{array}{cc}1& 2\\ 3& 6\end{array}\right)\left(\begin{array}{c}1\\ 3\end{array}\right)=\left(\begin{array}{c}7\\ 0\end{array}\right)=\stackrel{\rightharpoonup }{v_1} \\ T\left(\stackrel{\rightharpoonup }{v_2}\right)=A\stackrel{\rightharpoonup }{v_2}\left(\begin{array}{cc}1& 2\\ 3& 6\end{array}\right)\left(\begin{array}{c}-2\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)=-\stackrel{\rightharpoonup }{v_2} \end{gathered}$$

Substitute these values in the formula.

$$\begin{gathered} T\left(\stackrel{\rightharpoonup }{v_1}\right)=\left(\begin{array}{c}7\\ 0\end{array}\right),T\left(\stackrel{\rightharpoonup }{v_2}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right) \\ \therefore B=\left(\begin{array}{cc}7& 0\\ 0& 0\end{array}\right) \end{gathered}$$

Final answer.

(a) The matrix is, $B=\left(\begin{array}{cc}7& 0\\ 0& 0\end{array}\right)$.

(b) The matrix is,$B=\left(\begin{array}{cc}7& 0\\ 0& 0\end{array}\right)$.

(c) The matrix is, $B=\left(\begin{array}{cc}7& 0\\ 0& 0\end{array}\right)$.