If $A=\left[\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right], B=\left[\begin{array}{cc}1 & 1 \\ 4 &…

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
  1. $\left[\begin{array}{cc}\frac{-1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$
  2. $\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{-1}{2}\end{array}\right]$
  3. $\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$
  4. $\left[\begin{array}{ll}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{1}{2}\end{array}\right]$

Solution

$\begin{aligned} A+B & =\left[\begin{array}{cc}1 & -1 \\ 2 & -1\end{array}\right]+\left[\begin{array}{cc}1 & 1 \\ 4 & -1\end{array}\right]=\left[\begin{array}{cc}2 & 0 \\ 6 & -2\end{array}\right] \\ \therefore \quad|A+B| & =\left|\begin{array}{cc}2 & 0 \\ 6 & -2\end{array}\right|=-4 \neq 0 \\ \text { If } A & =\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \text { and ad }-\mathrm{bc} \neq 0, \\ \text { then } A^{-1} & =\frac{1}{(a d-b c)}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right] \\ (A+B)^{-1} & =\frac{1}{-4}\left[\begin{array}{cc}-2 & 0 \\ -6 & 2\end{array}\right] \\ \therefore \quad & =\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & -\frac{1}{2}\end{array}\right]\end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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