Question

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

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

Short answer

The \(B\)-matrix is \(\left( {\begin{aligned}1&{}&5\\0&{}&1\end{aligned}} \right)\).

Step-by-step solution

Theorem

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.

Determine \(P\)- matrix

The given matrix is \(A = \left( {\begin{aligned}3&{}&4\\{ - 1}&{}&{ - 1}\end{aligned}} \right)\), and the given vectors are \({{\bf{b}}_1} = \left( {\begin{aligned}2\\{ - 1}\end{aligned}} \right)\), \({{\bf{b}}_2} = \left( {\begin{aligned}1\\2\end{aligned}} \right)\).

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

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

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

Formula to find inverse of a matrix

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 \(\det \left( A \right) = ad - bc\).

Determine \({P^{ - 1}}\)

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

\(\begin{aligned}\det \left( P \right) = 2 \cdot 2 - \left( { - 1} \right)1\\ = 4 + 1\\ = 5\end{aligned}\)

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

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

Determine \(B\)- matrix

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

\(\begin{aligned}{P^{ - 1}}AP &= \frac{1}{5}\left( {\begin{aligned}2&{}&1\\{ - 1}&{}&2\end{aligned}} \right)\left( {\begin{aligned}3&{}&4\\{ - 1}&{}&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}2&{}&1\\{ - 1}&{}&2\end{aligned}} \right)\\ &= \frac{1}{5}\left( {\begin{aligned}2&{}&1\\{ - 1}&{}&2\end{aligned}} \right)\left( {\left( {\begin{aligned}3&{}&4\\{ - 1}&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}2&{}&1\\{ - 1}&2\end{aligned}} \right)} \right)\\ &= \frac{1}{5}\left( {\begin{aligned}2&{}&1\\{ - 1}&2\end{aligned}} \right)\left( {\begin{aligned}{6 - 4}&{}&{3 + 8}\\{ - 2 + 1}&{}&{ - 1 - 2}\end{aligned}} \right)\\ &= \frac{1}{5}\left( {\begin{aligned}2&{}&1\\{ - 1}&{}&2\end{aligned}} \right)\left( {\begin{aligned}2&{}&{11}\\{ - 1}&{}&{ - 3}\end{aligned}} \right)\\ &= \frac{1}{5}\left( {\begin{aligned}5&{}&{25}\\0&{}&5\end{aligned}} \right)\\ &= \left( {\begin{aligned}1&{}&5\\0&{}&1\end{aligned}} \right)\end{aligned}\)

So, the required \(B\)-matrix is \(\left( {\begin{aligned}1&{}&5\\0&{}&1\end{aligned}} \right)\).