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MHT CET · Maths · Matrices

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. A \(\frac{1}{7}\left[\begin{array}{ll}3 & 2 \\ 4 & 5\end{array}\right]\)
  2. B \(7\left[\begin{array}{cc}3 & 2 \\ 4 & 5\end{array}\right]\)
  3. C \(\frac{1}{7}\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
  4. D \(7\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{7}\left[\begin{array}{cc}3 & -2 \\ -4 & 5\end{array}\right]\)

Step-by-step Solution

Detailed explanation

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