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Show that the adjoint of $A=\begin{bmatrix} -4 & -3 & -3 \\1 & 0 & 1 \\4 & 4 & 3 \end{bmatrix}$ is$\;A\;$ it self

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  • 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
To find $ adj\: A$. The matrix of cofactors
$ A_{ij} = \begin{bmatrix} (0-4) & -(3-4) & (4-0) \\ -(-9+12) & (-12+12) & -(-16+12) \\ (-3+0) & -(-4+3) & (0+3) \end{bmatrix} = \begin{bmatrix} -4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3 \end{bmatrix}$
$ adj\: A =\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$
Step 2
$ A =\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ -4 & 4 & 3 \end{bmatrix}$
From step 1, $ A = adj\: A$

 

answered May 23, 2013 by thanvigandhi_1
edited Jun 3, 2013 by thanvigandhi_1
 

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