Question

In Exercises 25through 30 , find the matrix Bof the linear transformation with respect to the basis $\mathfrak{J}\mathbf{=}\left({\overrightarrow{v}}_1,.......,{\overrightarrow{v}}_m\right)$ .

Short answer

The matrix is, $B=\left[\begin{array}{ccc}0& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]$.

Step-by-step solution

Consider the vectors.

The matrix and vectors are,

$A=\left[\begin{array}{ccc}5& -4& -2\\ -4& 5& -2\\ -2& -2& 8\end{array}\right];{\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}0\\ 1\\ -2\end{array}\right]$

Compute the matrix using formula.

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

Compute the matrix and the inverse of S.

$$\begin{gathered} S=\left({\overrightarrow{v}}_1,{\overrightarrow{v}}_2,{\overrightarrow{v}}_3\right) \\ =\left[\begin{array}{ccc}2& 1& 0\\ 2& -1& 1\\ 1& 0& -2\end{array}\right] \\ \Rightarrow S^{-1}=\frac{1}{9}\left[\begin{array}{ccc}2& 2& 1\\ 5& -4& -2\\ 1& 1& -4\end{array}\right] \end{gathered}$$

Substitute these values in the formula.

$$\begin{gathered} B=S^{-1}AS \\ \Rightarrow B=\left[\begin{array}{ccc}5& -4& -2\\ -4& 5& -2\\ -2& -2& 8\end{array}\right];{\overrightarrow{v}}_1=\left[\begin{array}{c}2\\ 2\\ 1\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}1\\ -1\\ 0\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}0\\ 1\\ -2\end{array}\right] \\ \therefore B=\left[\begin{array}{ccc}0& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right] \end{gathered}$$

Final answer.

The matrix is, $B=\left[\begin{array}{ccc}0& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]$.