Question

inverse of 1 2 3

4 5 3

7 8 9

Answer 1

The inverse of matrix A can be computed using the inverse of matrix formula, by dividing the adjoint of a matrix by the determinant of the matrix.

For a matrix A, its inverse

$A^{-1}=1/|A| Adj A$

$|A|=1(-1)^{1+1}\begin{vmatrix} 5 & 3 \\ 8 & 9 \end{vmatrix} +2(-1)^{1+2}\begin{vmatrix} 4 & 3 \\ 7 & 9 \end{vmatrix} +3(-1)^{1+3}\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} =$

$=45-24-2(36-21)+3(32-35)=21-72+42+96-105=-18$

AdjA=Transpose of its cofactor matrix

$C=\begin{pmatrix} +\begin{vmatrix} 5 & 3 \\ 8 & 9 \end{vmatrix}& -\begin{vmatrix} 4 & 3 \\ 7 & 9 \end{vmatrix} & +\begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix} \\ -\begin{vmatrix} 2& 3 \\ 8 & 9 \end{vmatrix}& +\begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} & -\begin{vmatrix} 1 & 2 \\ 7 & 8 \end{vmatrix}\\ +\begin{vmatrix} 2& 3 \\ 5 & 3 \end{vmatrix} & -\begin{vmatrix} 1 & 3 \\ 4 & 3 \end{vmatrix} &+\begin{vmatrix} 1 & 2 \\ 4 & 5 \end{vmatrix} \end{pmatrix}$ $=\begin{pmatrix} 21 & -15 &-3 \\ 6 & -12&6\\ -9 & 9&-3 \end{pmatrix}$

$AdjA=\begin{pmatrix} 21 & 6 &-9\\ -15 & -12&9\\ -3&6&-3 \end{pmatrix}$

Thus, $A^{-1}= \begin{pmatrix} \frac {21}{-18} & \frac 6 {-18}&\frac {-9} {-18}\\ \frac {-15} {-18} & \frac {-12} {-18}&\frac 9 {-18}\\ \frac {-3} {-18}&\frac 6 {-18}& \frac {-3} {-18} \end{pmatrix}=\begin{pmatrix} -\frac 7 6 & -\frac 1 3 & \frac 1 2 \\ \frac 5 6 & \frac 2 3 & -\frac 1 2 \\ \frac 1 6 & -\frac 1 3 & \frac 1 6 \end{pmatrix}$