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# Find the inverse of following matrix :$\begin{bmatrix} 1 & 0 & 3 \\2 & 1 & -1 \\1 & -1 & 1 \end{bmatrix}$

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 $adjA$ ( or adjoint of the matrix A) is given by $adjA=[A_{ij}]^T$
• A determinant can be expanded by using the elements of any row or column.
• The inverse of a non-singular square matrix A is given by $A^{-1} = \frac{1}{|A|} adj\: A.$ A non-singular matrix is one whose determinant value is nonzero.
Step 1
$A = \begin{bmatrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix}$
$|A| = \begin{vmatrix} 1 & 0 & 3 \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix} = 1(1-1)+0+3(-2-1)$
$= 0-9 = -9 \neq 0$
A is non-singular and so $A^{-1}$ exists.
Step 2
To find $adj\: A$
$[ A_{ij} ] = \begin{bmatrix} (1-1) & -(2+1) & (-2-1) \\ -(0+3) & (1-3) & -(-1-0) \\ (0-3) & -(-1-6) & (1-0) \end{bmatrix}$
$= \begin{bmatrix} 0 & -3 & -3 \\ -3 & -2 & 1 \\ -3 & 7 & 1 \end{bmatrix}$
$adj\: A = [ A_{ij} ]^T = \begin{bmatrix} 0 & -3 & -3 \\ -3 & -2 & 7 \\-3 & 1 & 1 \end{bmatrix}$
Step 3
$A^{-1} = \large\frac{1}{|A|} \: adj\: A = \large\frac{1}{-9} \begin{bmatrix} 0 & -3 & -3 \\ -3 & -2 & 7 \\ -3 & 1 & 1 \end{bmatrix}$
$= \large\frac{1}{9} \begin{bmatrix} 0 & 3 & 3 \\ 3 & 2 & -7 \\ 3 & -1 & -1 \end{bmatrix}$

edited Jun 3, 2013