Question

Let A = [ 0 1 2

1 0 3

4 -3 8 ]

If A is an invertible matrix, then show that.

det (A-1) = 1/det ( A )

Answer 1

$A^{-1}=\frac{A^T_*}{|A|}$

$|A|=4-2\cdot3=-2$

Minor matrix: each element $m_{ij}$ is the determinant of 2x2 matrix, if we remove i-th row and j-th column in matrix $A$.

$M=\begin{pmatrix} 9 & -4&-3 \\ 14 & -8&-4\\ 3&-2&-1 \\ \end{pmatrix}$

Matrix of cofactors: change signs of $m_{12},m_{21},m_{23},m_{32}.$

$A_*=\begin{pmatrix} 9 & 4&-3 \\ -14 & -8&4\\ 3&2&-1 \\ \end{pmatrix}$

Transpose matrix of cofactors:

$A^T_*=\begin{pmatrix} 9 & -14&3 \\ -4 & -8&2\\ -3&4&1 \\ \end{pmatrix}$

$A^{-1}=-\frac{1}{2}\begin{pmatrix} 9 & -14&3 \\ -4 & -8&2\\ -3&4&1 \\ \end{pmatrix}$

$|A^{-1}|=4.5\cdot4-7\cdot0.5-1.5\cdot10=-0.5=1/|A|$