MCQ
If $A=\left[\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right], B=\left[\begin{array}{cc}1 & 1 \\ 4 & -1\end{array}\right]$, then $(A+B)^{-1}$ is
  • A
    $\left[\begin{array}{cc}\frac{-1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$
  • $\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{-1}{2}\end{array}\right]$
  • C
    $\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$
  • D
    $\left[\begin{array}{ll}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{1}{2}\end{array}\right]$

Answer

Correct option: B.
$\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{-1}{2}\end{array}\right]$
(b) : $A=\left[\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 1 \\ 4 & -1\end{array}\right]$
$
\begin{aligned}
& A+B=\left[\begin{array}{cc}
1+1 & -1+1 \\
2+4 & -1-1
\end{array}\right]=\left[\begin{array}{cc}
2 & 0 \\
6 & -2
\end{array}\right] \\
& \Rightarrow|A+B|=-4
\end{aligned}
$
Now, $\operatorname{adj}(A+B)=\left[\begin{array}{ll}-2 & 0 \\ -6 & 2\end{array}\right]$
$\therefore(A+B)^{-1}=\frac{\operatorname{adj}(A+B)}{|A+B|}=\frac{-1}{4}=\left[\begin{array}{ll}-2 & 0 \\ -6 & 2\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]$

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