Question

In Exercises 25through 30, find the matrix B of the linear transformation$\mathit{T}\mathbf{(}\overrightarrow{\mathbf{x}}\mathbf{}\mathbf{)}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$ with respect to the basis $\mathbf{؏}\mathbf{=}\left(\overrightarrow{V_1},--\overrightarrow{V_m}\right)$ .

$\mathit{A}\mathbf{=}\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\mathbf{;}{\overrightarrow{\mathbf{v}}}_{\mathbf{1}}\mathbf{=}\left[\left.\begin{array}{c}1\\ 1\end{array}\right],{\overrightarrow{V}}_2=\left|\begin{array}{c}1\\ 2\end{array}\right.\right]$

Short answer

The matrix is, $B=\left(\begin{array}{cc}-1& -1\\ 4& 6\end{array}\right)$.

Step-by-step solution

Consider the vectors.

The matrix and vectors are,

$A=\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right);\overrightarrow{v_1}=\left[\begin{array}{c}1\\ 1\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}1\\ 2\end{array}\right]$

Compute the matrix using formula.

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

Compute the matrix S.

The inverse of the matrix is,

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

For calculating S.

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

Substitute these values in the formula.

$$\begin{gathered} B=S^{-1}AS \\ \Rightarrow B=\frac{1}{1}\left(\begin{array}{cc}2& -1\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& 2\\ 3& 4\end{array}\right)\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right) \\ \therefore B=\left(\begin{array}{cc}-1& -1\\ 4& 6\end{array}\right) \end{gathered}$$

Final answer.

The matrix is, $B=\left(\begin{array}{cc}-1& -1\\ 4& 6\end{array}\right)$.