Question

Find the inverse of the matrix
A = $\left[\begin{array}{ccc}0& 1& 2\\ 1& 2& 3\\ 3& 1& 1\end{array}\right]$

Answer 1

$|A|=0-1(1-2)+3(3-4)=1-2=-2\ne$ 0.

Hence matrix A is non-singular.

Co-factors of the 1st, 2nd and 3rd rows of | A | are

-1, 8, -5; 1, -6, 3; -1, 2,-1

Therefore the matrix formed by co-of factors |A| is

C = $\left[\begin{array}{ccc}-1& 8& -5\\ 1& -6& 3\\ -1& 2& -1\end{array}\right]$

$\therefore$ Adj. A = Transpose of C

$\therefore$ Adj. A = $\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]$

$\therefore A^{-1}=\frac{1}{|A|}Adj.A=-\frac{1}{2}\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]$

Multiply each element of the matrix by $-\frac{1}{2}$

or $A^{-1}=\left[\begin{array}{ccc}\frac{1}{2}& -\frac{1}{2}& \frac{1}{2}\\ -4& 3& -1\\ \frac{5}{2}& -\frac{3}{2}& \frac{1}{2}\end{array}\right]$

Verification : $A.A^{-1}=A^{-1}A=1$

$\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]\times \frac{1}{2}\left[\begin{array}{ccc}-1& 1& -1\\ 8& -6& 2\\ -5& 3& -1\end{array}\right]$

= $-\frac{1}{2}\left[\begin{array}{ccc}-2& 0& 0\\ 0& -2& 0\\ 0& 0& -2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=I_3$