Question

Let $A$ be a $3\times 3$ matrix such that
$A\times \left[\begin{array}{ccc}1& 2& 3\\ 0& 2& 3\\ 0& 1& 1\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$
Then $A^{-1}$ is:

Answer 1

We are given,

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

Now multiplying both sides with $A^{-1}$ we get, (pre-multiplication)

or, $\left[\begin{array}{ccc}1& 2& 3\\ 0& 2& 3\\ 0& 1& 1\end{array}\right]=A^{-1}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]=A^{-1}.I_3$ [ Since $A^{-1}.A=I_3$ where $I_3$ is the identity matrix of order $3$.]

or, $\left[\begin{array}{ccc}1& 2& 3\\ 0& 2& 3\\ 0& 1& 1\end{array}\right]=A^{-1}$