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Expert-verified 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 16, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.16. $$\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{4}}\\{\bf{0}}&{ - {\bf{3}}}&{\bf{1}}\\{\bf{0}}&{\bf{0}}&{ - {\bf{2}}}\end{array}} \right)$$

The adjugate matrix is $$\left( {\begin{array}{*{20}{c}}6&4&{14}\\0&{ - 2}&{ - 1}\\0&0&{ - 3}\end{array}} \right)$$, and the inverse matrix is See the step by step solution

## Step 1: First, find the determinant

Let $$A = \left( {\begin{array}{*{20}{c}}1&2&4\\0&{ - 3}&1\\0&0&{ - 2}\end{array}} \right)$$. Then,

$$\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&2&4\\0&{ - 3}&1\\0&0&{ - 2}\end{array}} \right|\\ = 1\left| {\begin{array}{*{20}{c}}{ - 3}&1\\0&{ - 2}\end{array}} \right| + 0 + 0\\\det A = 6 \ne 0\end{array}$$

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

## Step 2: Compute the adjugate matrix

The nine cofactors are:

$$\begin{array}{c}{C_{11}} = {\left( { - 1} \right)^2}\left| {\begin{array}{*{20}{c}}{ - 3}&1\\0&{ - 2}\end{array}} \right|\\ = 6\end{array}$$

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

$$\begin{array}{c}{C_{13}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}0&{ - 3}\\0&0\end{array}} \right|\\ = 0\end{array}$$

$$\begin{array}{c}{C_{21}} = {\left( { - 1} \right)^3}\left| {\begin{array}{*{20}{c}}2&4\\0&{ - 2}\end{array}} \right|\\ = - \left( { - 4} \right)\\ = 4\end{array}$$

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

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

$$\begin{array}{c}{C_{31}} = {\left( { - 1} \right)^4}\left| {\begin{array}{*{20}{c}}2&4\\{ - 3}&1\end{array}} \right|\\ = 14\end{array}$$

$$\begin{array}{c}{C_{32}} = {\left( { - 1} \right)^5}\left| {\begin{array}{*{20}{c}}1&4\\0&1\end{array}} \right|\\ = - 1\end{array}$$

$$\begin{array}{c}{C_{33}} = {\left( { - 1} \right)^6}\left| {\begin{array}{*{20}{c}}1&2\\0&{ - 3}\end{array}} \right|\\ = - 3\end{array}$$

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

$$\begin{array}{c}{\rm{adj}}\,A = \left( {\begin{array}{*{20}{c}}{{C_{11}}}&{{C_{21}}}&{{C_{31}}}\\{{C_{12}}}&{{C_{22}}}&{{C_{32}}}\\{{C_{13}}}&{{C_{23}}}&{{C_{33}}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}6&4&{14}\\0&{ - 2}&{ - 1}\\0&0&{ - 3}\end{array}} \right)\end{array}$$

## Step 3: Use Theorem 8 to find $${A^{ - {\bf{1}}}}$$

By Theorem 8,  ### Want to see more solutions like these? 