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Q12E

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Found in: Page 165

### Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

# Question: In Exercise 12, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.12. $$\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{1}}&{\bf{3}}\\{ - {\bf{2}}}&{\bf{2}}&{\bf{1}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}\end{array}} \right)$$

The adjugate matrix is $$\left( {\begin{array}{*{20}{c}}1&2&{ - 5}\\2&1&{ - 7}\\{ - 2}&{ - 1}&4\end{array}} \right)$$, and the inverse matrix is

See the step by step solution

## Step 1: First, find the determinant

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

$$\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&1&3\\{ - 2}&2&1\\0&1&1\end{array}} \right|\\ = 0 - 1\left| {\begin{array}{*{20}{c}}1&3\\{ - 2}&1\end{array}} \right| + 1\left| {\begin{array}{*{20}{c}}1&1\\{ - 2}&2\end{array}} \right|\\ = - \left( 7 \right) + 4\\\det A = - 3 \ne 0\end{array}$$

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

## Step 2: Compute the adjugate matrix

The nine cofactors are:

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

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

$$\begin{array}{c}{C_{13}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}{ - 2}&2\\0&1\end{array}} \right|\\ = - 2\end{array}$$

$$\begin{array}{c}{C_{21}} = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}}1&3\\1&1\end{array}} \right|\\ = - \left( { - 2} \right)\\ = 2\end{array}$$

$$\begin{array}{c}{C_{22}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}1&3\\0&1\end{array}} \right|\\ = 1\end{array}$$

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

$$\begin{array}{c}{C_{31}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}1&3\\2&1\end{array}} \right|\\ = - 5\end{array}$$

$$\begin{array}{c}{C_{32}} = {\left( { - 1} \right)^5}\left| {\begin{array}{*{20}{c}}1&3\\{ - 2}&1\end{array}} \right|\\ = - 7\end{array}$$

$$\begin{array}{c}{C_{33}} = {\left( { - 1} \right)^6}\left| {\begin{array}{*{20}{c}}1&1\\{ - 2}&2\end{array}} \right|\\ = 4\end{array}$$

The adjugate matrix is the transpose of the matrix of cofactors. Hence,

$$\begin{array}{c}{\rm{adj}}\,A = \left( {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{21}}}&{{C_{31}}}\\{{C_{12}}}&{{C_{22}}}&{{C_{32}}}\\{{C_{13}}}&{{C_{23}}}&{{C_{33}}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&2&{ - 5}\\2&1&{ - 7}\\{ - 2}&{ - 1}&4\end{array}} \right)\end{array}$$

By Theorem 8,