Find the \(B\) matrix for the transformation\({\rm{x}} \mapsto A{\rm{x}}\), when \(B = \left\{ {{b_1},{b_2},{b_3}} \right\}\).

\(A = \left( {\begin{aligned}{}{ - 14}&{}&4&{}&{ - 14}\\{ - 33}&{}&9&{}&{ - 31}\\{11}&{}&{ - 4}&{}&{11}\end{aligned}} \right)\), \({b_1} = \left( {\begin{aligned}{}{ - 1}\\{ - 2}\\1\end{aligned}} \right)\), \({b_2} = \left( {\begin{aligned}{}{ - 1}\\{ - 1}\\1\end{aligned}} \right)\), \({b_3} = \left( {\begin{aligned}{}{ - 1}\\{ - 2}\\0\end{aligned}} \right)\)

Let \(P\) is an invertible matrix and \(D\) is the diagonal matrix, such that the matrix A can be written as follows:

\(A = PD{P^{ - 1}}\)

Where the columns of the \(P\) matrix are the same as that of basis \(B\). So, the matrix \(P\) can be written as:

\(\begin{aligned}{}P &= \left( {{{\rm{b}}_1}\,\,{{\rm{b}}_2}\,\,{{\rm{b}}_3}} \right)\\ &= \left( {\begin{aligned}{}{ - 1}&{}&{ - 1}&{}&{ - 1}\\{ - 2}&{}&{ - 1}&{}&{ - 2}\\1&{}&1&{}&0\end{aligned}} \right)\end{aligned}\)

As \(A = PD{P^{ - 1}}\) then the diagonal matrix \(D\) can be obtained as \(D = {P^{ - 1}}AP\). To find \(D = {P^{ - 1}}AP\), first, find \({P^{ - 1}}\)using the cofactors and determinant method.

\(\begin{aligned}{}P = \left( {\begin{aligned}{}{ - 1}&{}&{ - 1}&{}&{ - 1}\\{ - 2}&{}&{ - 1}&{}&{ - 2}\\1&{}&1&{}&0\end{aligned}} \right)\\{P^{ - 1}} = \left( {\begin{aligned}{}2&{}&{ - 1}&{}&1\\{ - 2}&{}&1&{}&0\\{ - 1}&{}&0&{}&{ - 1}\end{aligned}} \right)\end{aligned}\)

Now find the matrix \(D\) as follows:

\(\begin{aligned}{}D &= {P^{ - 1}}AP\\ &= \left( {\begin{aligned}{}2&{}&{ - 1}&{}&1\\{ - 2}&{}&1&{}&0\\{ - 1}&{}&0&{}&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{}{ - 14}&{}&4&{}&{ - 14}\\{ - 33}&{}&9&{}&{ - 31}\\{11}&{}&{ - 4}&{}&{11}\end{aligned}} \right)\left( {\begin{aligned}{}{ - 1}&{}&{ - 1}&{}&{ - 1}\\{ - 2}&{}&{ - 1}&{}&{ - 2}\\1&{}&1&{}&0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{}8&{}&3&{}&{ - 6}\\0&{}&1&{}&3\\0&{}&0&{}&{ - 3}\end{aligned}} \right)\end{aligned}\)