Question

Question: In Exercise 15, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.

15. \(\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{0}}&{\bf{0}}\\{ - {\bf{1}}}&{\bf{1}}&{\bf{0}}\\{ - {\bf{2}}}&{\bf{3}}&{ - {\bf{1}}}\end{array}} \right)\)

Short answer

The adjugate matrix is \(\left( {\begin{array}{*{20}{c}}{ - 1}&0&0\\{ - 1}&{ - 5}&0\\{ - 1}&{ - 15}&5\end{array}} \right)\), and the inverse matrix is

Step-by-step solution

First, find the determinant

Let \(A = \left( {\begin{array}{*{20}{c}}5&0&0\\{ - 1}&1&0\\{ - 2}&3&{ - 1}\end{array}} \right)\). Then,

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}5&0&0\\{ - 1}&1&0\\{ - 2}&3&{ - 1}\end{array}} \right|\\ = 0 + 0 - 1\left| {\begin{array}{*{20}{c}}5&0\\{ - 1}&1\end{array}} \right|\\\det A = - 5 \ne 0\end{array}\)

Here, \(\det A \ne 0\). Hence, the inverse of A exists.

Compute the adjugate matrix  

The nine cofactorsare:

\(\begin{array}{c}{C_{11}} = {\left( { - 1} \right)^2}\left| {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right|\\ = - 1\end{array}\)

\(\begin{array}{c}{C_{12}} = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}}{ - 1}&0\\{ - 2}&{ - 1}\end{array}} \right|\\ = - 1\end{array}\)

\(\begin{array}{c}{C_{13}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}{ - 1}&1\\{ - 2}&3\end{array}} \right|\\ = - 1\end{array}\)

\(\begin{array}{c}{C_{21}} = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}}0&0\\3&{ - 1}\end{array}} \right|\\ = 0\end{array}\)

\(\begin{array}{c}{C_{22}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}5&0\\{ - 2}&{ - 1}\end{array}} \right|\\ = - 5\end{array}\)

\(\begin{array}{c}{C_{23}} = {\left( { - 1} \right)^5}\left| {\begin{array}{*{20}{c}}5&0\\{ - 2}&3\end{array}} \right|\\ = - 15\end{array}\)

\(\begin{array}{c}{C_{31}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}0&0\\1&0\end{array}} \right|\\ = 0\end{array}\)

\(\begin{array}{c}{C_{32}} = {\left( { - 1} \right)^5}\left| {\begin{array}{*{20}{c}}5&0\\{ - 1}&0\end{array}} \right|\\ = 0\end{array}\)

\(\begin{array}{c}{C_{33}} = {\left( { - 1} \right)^6}\left| {\begin{array}{*{20}{c}}5&0\\{ - 1}&1\end{array}} \right|\\ = 5\end{array}\)

Theadjugate matrix is the transpose of the matrix of cofactors. Hence,

\(\begin{array}{c}{\rm{adj}}\,A = \left( {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{21}}}&{{C_{31}}}\\{{C_{12}}}&{{C_{22}}}&{{C_{32}}}\\{{C_{13}}}&{{C_{23}}}&{{C_{33}}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 1}&0&0\\{ - 1}&{ - 5}&0\\{ - 1}&{ - 15}&5\end{array}} \right)\end{array}\)

Use Theorem 8 to find \({A^{ - {\bf{1}}}}\)

By Theorem 8,