Elementary Linear Algebra 7th Edition

Published by Cengage Learning

Chapter 3 - Determinants - 3.4 Applications of Determinants - 3.4 Exercises - Page 136: 4

Answer

$$\operatorname{adj}(A)=\left[ \begin {array}{ccc} 4&2&-5\\ -2&-4&1 \\ -2&2&1\end {array} \right] .$$ $$A^{-1}= -\frac{1}{6}\left[ \begin {array}{ccc} 4&2&-5\\ -2&-4&1 \\ -2&2&1\end {array} \right].$$

Work Step by Step

The matrix is given by $A=\left[ \begin {array}{ccc} 1&2&3\\ 0&1&-1 \\ 2&2&2\end {array} \right] .$ To find $\operatorname{adj}(A)$, we calculate first the cofactor matrix of $A$ as follows $$\left[ \begin {array}{ccc} 4&-2&-2\\ 2&-4&2 \\ -5&1&1\end {array} \right] .$$ Now, the adjoint of $A$ is $$\operatorname{adj}(A)=\left[ \begin {array}{ccc} 4&2&-5\\ -2&-4&1 \\ -2&2&1\end {array} \right] .$$ To find $A^{-1}$, we have to calculate $\det(A)$ which is given by $$\det(A)=-6.$$ Finally, we have $$A^{-1}=\frac{1}{\det(A)}\operatorname{adj}(A)=-\frac{1}{6}\left[ \begin {array}{ccc} 4&2&-5\\ -2&-4&1 \\ -2&2&1\end {array} \right].$$

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