Question

Find the inverse of the matrix (if it exists)

$A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 2& 4\\ 0& 0& 5\end{array}\right]$

Answer 1

$A=\left[\begin{array}{ccc}1& 2& 3\\ 0& 2& 4\\ 0& 0& 5\end{array}\right]$

Here, $|A|=1(10-0)=10$ (expanding along the first column)

Since, $|A|\ne 0$

Hence, $A^{-1}$ exists.

We know that, $adjA=C^T$

So, we will find out co-factors of each element of A.

$C_{11}=(-1{)}^{1+1}\left|\begin{array}{cc}2& 4\\ 0& 5\end{array}\right|$

$\Rightarrow C_{11}=10$

$C_{12}=(-1{)}^{1+2}\left|\begin{array}{cc}0& 4\\ 0& 5\end{array}\right|$

$\Rightarrow C_{12}=0$

$C_{13}=(-1{)}^{1+3}\left|\begin{array}{cc}0& 2\\ 0& 0\end{array}\right|$

$\Rightarrow C_{13}=0$

$C_{21}=(-1{)}^{2+1}\left|\begin{array}{cc}2& 3\\ 0& 5\end{array}\right|$

$\Rightarrow C_{21}=-(10-0)=-10$

$C_{22}=(-1{)}^{2+2}\left|\begin{array}{cc}1& 3\\ 0& 5\end{array}\right|$

$\Rightarrow C_{22}=5-0=5$

$C_{23}=(-1{)}^{2+3}\left|\begin{array}{cc}1& 2\\ 0& 0\end{array}\right|$

$\Rightarrow C_{23}=0$

$C_{31}=(-1{)}^{3+1}\left|\begin{array}{cc}2& 3\\ 2& 4\end{array}\right|$

$\Rightarrow C_{31}=8-6=2$

$C_{32}=(-1{)}^{3+2}\left|\begin{array}{cc}1& 3\\ 0& 4\end{array}\right|$

$\Rightarrow C_{32}=-(4-0)=-4$

$C_{33}=(-1{)}^{3+3}\left|\begin{array}{cc}1& 2\\ 0& 2\end{array}\right|$

$\Rightarrow C_{33}=2-0=2$

So, the cofactor matrix $C=\left[\begin{array}{ccc}10& 0& 0\\ -10& 5& 0\\ 2& -4& 2\end{array}\right]$

$\Rightarrow adjA=C^T=\left[\begin{array}{ccc}10& -10& 2\\ 0& 5& -4\\ 0& 0& 2\end{array}\right]$

$\Rightarrow A^{-1}=\frac{adjA}{|A|}=\frac{1}{10}\left[\begin{array}{ccc}10& -10& 2\\ 0& 5& -4\\ 0& 0& 2\end{array}\right]$

$\Rightarrow A^{-1}=\left[\begin{array}{ccc}1& -1& \frac{1}{5}\\ 0& \frac{1}{2}& \frac{-2}{5}\\ 0& 0& \frac{1}{5}\end{array}\right]$