MCQ
If matrix $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ such that $A X=I,$ then $X=$
  • A
    $\frac{1}{5}\left[\begin{array}{rr}1 & 3 \\ 2 & -1\end{array}\right]$
  • B
    $\frac{1}{5}\left[\begin{array}{rr}4 & 2 \\ 4 & -1\end{array}\right]$
  • $\frac{1}{5}\left[\begin{array}{lr}-3 & 2 \\ 4 & -1\end{array}\right]$
  • D
    $\frac{1}{5}\left[\begin{array}{ll}-1 & 2 \\ -1 & 4\end{array}\right]$

Answer

Correct option: C.
$\frac{1}{5}\left[\begin{array}{lr}-3 & 2 \\ 4 & -1\end{array}\right]$
We have, $A=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ such that $A X=I$
$ \Rightarrow X=A^{-1} I$
$\because |A|=3-8=-5 \neq 0$
$ \Rightarrow A^{-1} \text { exists. }$
$A^{-1}=\frac{1}{-5}\left[\begin{array}{cc} 3 & -2 \\ -4 & 1 \end{array}\right]=\frac{1}{5}\left[\begin{array}{cc}
-3 & 2 \\ 4 & -1 \end{array}\right] $
$\therefore  X=\frac{1}{5}\left[\begin{array}{cc} -3 & 2 \\ 4 & -1 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right]$
$=\frac{1}{5}\left[\begin{array}{cc} -3 & 2 \\ 4 & -1 \end{array}\right]$

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