Question

The inverse of the matrix $\left[\begin{array}{ccc}2& 3& 4\\ 4& 3& 1\\ 1& 2& 4\end{array}\right]$ is

• A. $$\left[\begin{array}{ccc}2& -\frac{4}{5}& -\frac{9}{5}\\ -3& \frac{4}{5}& \frac{14}{5}\\ 1& -\frac{1}{5}& -\frac{6}{5}\end{array}\right]$$
• B. $$\left[\begin{array}{ccc}-2& \frac{4}{5}& \frac{9}{5}\\ 3& -\frac{4}{5}& -\frac{14}{5}\\ -1& \frac{1}{5}& \frac{6}{5}\end{array}\right]$$
• C. $$\left[\begin{array}{ccc}10& -4& -9\\ -15& 4& 14\\ 5& -1& -6\end{array}\right]$$
• D. $$\left[\begin{array}{ccc}-10& 4& 9\\ 15& -4& -14\\ -5& 1& 6\end{array}\right]$$

Answer 1

The correct option is B $$\left[\begin{array}{ccc}-2& \frac{4}{5}& \frac{9}{5}\\ 3& -\frac{4}{5}& -\frac{14}{5}\\ -1& \frac{1}{5}& \frac{6}{5}\end{array}\right]$$

A = $\left[\begin{array}{ccc}2& 3& 4\\ 4& 3& 1\\ 1& 2& 4\end{array}\right]$
Det.of matrix, |A| = $2(12-2)-3(16-1)+4(8-3)$
=(20 -45+20)
$=-5$
$A^{-1}=\frac{1}{|A|}(adj.A)$
$cofA=\left[\begin{array}{ccc}10& -15& 5\\ -4& 4& -1\\ -9& 14& -6\end{array}\right]$
$adjA=(cofA{)}^T$
$=\left[\begin{array}{ccc}10& -4& -9\\ -15& 4& 14\\ 5& -1& -6\end{array}\right]$
$A^{-1}=-\frac{1}{5}\left[\begin{array}{ccc}10& -15& 5\\ -4& 4& -1\\ -9& 14& -6\end{array}\right]$
=$\left[\begin{array}{ccc}-2& 4/5& 9/5\\ 3& -4/5& -14/5\\ -1& 1/5& 6/5\end{array}\right]$