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# Show that the adjoint of $A=\begin{bmatrix} -1 & -2 & -2 \\2 & 1 & -2 \\2 & -2 & 1 \end{bmatrix}$ is$\; 3A^T.$

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Toolbox:
• Let $A = [ a_{ij} ]$ be a square matrix x. Let $A_{ij}$ be the cofactor of $a_{ij}$. Then $[ A_{ij}]$ is the matrix of cofactors and $adj\: A$ ( or adjoint of the matrix A) is given by $adj\: A=[A_{ij}]^T$
Step 1
$A = \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$
The cofactor matrix $[ A_{ij}] = \begin{bmatrix} (1-4) & -(2+4) & (-4-2) \\ -(2-4) & (-1+4) & -(2+4) \\ (4+2) & -(2+4) & (-1+4) \end{bmatrix}$
$= \begin{bmatrix} -3 & -6 & -6 \\ 6 & 3 & -6 \\ 6 & -6 & 3 \end{bmatrix}= 3\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$
$adj\: A = [A_{ij}]^T = 3\begin{bmatrix} -1 & 2 & 2 \\ -2 & 1 & -2 \\ -2 & -2 & 1 \end{bmatrix}$
Step 2
$3\: A^T = 3\begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix} = 3\begin{bmatrix} -1 & 2 & 2 \\ -2 & 1 & -2 \\ -2 & -2 & 1 \end{bmatrix}$
From step 1 $adj\: A = 3A^T$
answered May 23, 2013

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