If $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], \quad B=\left[\begin{array}{ll}4 & 1 \\ 3 &…

If $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right], \quad B=\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right], \quad$ then $(A+B)^{-1}=$
  1. $\frac{1}{7}\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]$
  2. $7\left[\begin{array}{cc}3 & 2 \\ 4 & 5\end{array}\right]$
  3. $\frac{1}{7}\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]$
  4. $7\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]$

Solution

$A+B=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]+\left[\begin{array}{ll}4 & 1 \\ 3 & 1\end{array}\right]=\left[\begin{array}{ll}5 & 2 \\ 4 & 3\end{array}\right]$ $|A+B|=15-8=7$ and adj $(A+B)=\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]$ $\therefore(A+B)^{-1}=\frac{1}{7}\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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