# Find the inverse of following matrix : $\begin{bmatrix} 8 & -1 &- 3 \\-5 & 1 & 2 \\10 & -1 & -4 \end{bmatrix}$

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 $adj\: A$ ( or adjoint of the matrix A) is given by $adj\: A=[A_{ij}]^T$
• 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
$A = \begin{bmatrix} 8 & -1 & 3 \\ -5 & 1 & 2 \\ 10 & -1 & -4 \end{bmatrix}$
$|A| = \begin{vmatrix} 8 & -1 & -3 \\ -5 & 1 & 2 \\ 10 & -1 & -4 \end{vmatrix} = 8(-4+2)+1(20-20)-3(5-10)$
$= -16+0+15= -1 \neq 0$
$A^{-1}$ is exists
Step 2
To find $adj\: A$
$[A_{ij}] = \begin{bmatrix} (-4+2) & -(-20+20) & (5-10) \\ -(4-3) & (-32+30) & -(-8+10) \\ (-2+3) & -(16-15) & (8-5) \end{bmatrix}$
$= \begin{bmatrix} -2 & 0 & -5 \\ -1 & -2 & -2 \\ 1 & -1 & 3 \end{bmatrix}$
$adj\:A = [A_{ij}]^T = \begin{bmatrix} -2 & -1 & 1 \\ 0 & -2 & -1 \\ 5 & -2 & 3 \end{bmatrix}$
Step 3
$A^{-1} = \frac{1}{|A|} adj\:A = \frac{1}{-1} \begin{bmatrix} 2 & -1 & 1 \\ 0 & -2 & -1 \\ -5 & -2 & 3 \end{bmatrix} = \begin{bmatrix} -2 & 1 & -1 \\ 0 & 2 & 1 \\ 5 & 2 & -3 \end{bmatrix}$

edited Jun 3, 2013