In Exercises 11 and 12, find the \(B\)-matrix for the transformation \({\bf{x}} \mapsto A{\bf{x}}\), when \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2}} \right\}\).

12.\(A = \left( {\begin{aligned}{ - 1}&{}&4\\{ - 2}&{}&3\end{aligned}} \right)\), \({{\bf{b}}_1} = \left( {\begin{aligned}3\\2\end{aligned}} \right)\), \({{\bf{b}}_2} = \left( {\begin{aligned}{ - 1}\\1\end{aligned}} \right)\)

Diagonal Matrix Representation:Assume that \(A = PD{P^{ - 1}}\), then, \(D\) is the \(B\)-matrix for the transformation \({\bf{x}} \mapsto A{\bf{x}}\), and\(B\) is the basis for \({\mathbb{R}^n}\) formed from the columns of \(P\), where \(D\) is the \(n \times n\) diagonal matrix.

The given matrix is \(A = \left( {\begin{aligned}{ - 1}&{}&4\\{ - 2}&{}73\end{aligned}} \right)\), and the given vectors are \({{\bf{b}}_1} = \left( {\begin{aligned}2\\{ - 1}\end{aligned}} \right)\), \({{\bf{b}}_2} = \left( {\begin{aligned}{ - 1}\\1\end{aligned}} \right)\).

Form a matrix \(P\) by using the vectors \({{\bf{b}}_1} = \left( {\begin{aligned}3\\2\end{aligned}} \right)\) and \({{\bf{b}}_2} = \left( {\begin{aligned}{ - 1}\\1\end{aligned}} \right)\) as its columns.

\(\begin{aligned}{c}P = \left( {\begin{aligned}{{{\bf{b}}_1}}&{}&{{{\bf{b}}_2}}\end{aligned}} \right)\\ = \left( {\begin{aligned}3&{}&{ - 1}\\2&{}&1\end{aligned}} \right)\end{aligned}\)

The \(B\)-matrix can be found by finding \({P^{ - 1}}AP\), for which first find \({P^{ - 1}}\).

For any \(2 \times 2\) matrix \(A = \left( {\begin{aligned}a&{}&b\\c&d\end{aligned}} \right)\), \({A^{ - 1}}\) is given by,

\({A^{ - 1}} = \frac{1}{{\det \left( A \right)}}\left( {\begin{aligned}d&{}&{ - b}\\{ - c}&{}&a\end{aligned}} \right)\), where \(\det \left( A \right) = ad - bc\).

According to the formula of inverse, the determinant of the matrix is required. So determine the determinant of the matrix \(P\).

\(\begin{aligned}\det \left( P \right) = 3 \cdot 1 - 2\left( { - 1} \right)\\ = 3 + 2\\ = 5\end{aligned}\)

Now, find \({P^{ - 1}}\) by using the inverse formula:

\({P^{ - 1}} = \frac{1}{5}\left( {\begin{aligned}1&{}&1\\{ - 2}&{}&3\end{aligned}} \right)\)

Find \({P^{ - 1}}AP\).

\(\begin{aligned}{P^{ - 1}}AP &= \frac{1}{5}\left( {\begin{aligned}1&{}&1\\{ - 2}&{}&3\end{aligned}} \right)\left( {\begin{aligned}{ - 1}&{}&4\\{ - 2}&{}&3\end{aligned}} \right)\left( {\begin{aligned}3&{}&{ - 1}\\2&{}&1\end{aligned}} \right)\\ &= \frac{1}{5}\left( {\begin{aligned}1&{}&1\\{ - 2}&{}&3\end{aligned}} \right)\left( {\left( {\begin{aligned}{ - 1}&{}&4\\{ - 2}&{}&3\end{aligned}} \right)\left( {\begin{aligned}3&{}&{ - 1}\\2&{}&1\end{aligned}} \right)} \right)\\ &= \frac{1}{5}\left( {\begin{aligned}1&{}&1\\{ - 2}&{}&3\end{aligned}} \right)\left( {\begin{aligned}{ - 3 + 8}&{}&{1 + 4}\\{ - 6 + 6}&{}&{2 + 3}\end{aligned}} \right)\\ &= \frac{1}{5}\left( {\begin{aligned}1&{}&1\\{ - 2}&{}&3\end{aligned}} \right)\left( {\begin{aligned}5&{}&5\\0&{}&5\end{aligned}} \right)\\ &= \frac{1}{5}\left( {\begin{aligned}5&{}&{10}\\{ - 10}&{}&5\end{aligned}} \right)\\ &= \left( {\begin{aligned}1&{}&2\\{ - 2}&{}&1\end{aligned}} \right)\end{aligned}\)

So, the required \(B\)-matrix is \(\left( {\begin{aligned}1&{}&2\\{ - 2}&{}&1\end{aligned}} \right)\).