Let A be a 2 x 2 symmetric matrix such that A[1, 7] = [3, 7] and the determinant of A be 1.

Question

Let A be a \(2 \times 2\) symmetric matrix such that \(A\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right]\) and the determinant of A be 1.

If \(\mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}\), where \(\mathrm{I}\) is an identity matrix of order \(2 \times 2\), then \(\alpha+\beta\) equals ______.

Answer 1

Correct answer: 5

Let \(A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]\)

\(\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{b} & \mathrm{d}\end{array}\right]\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}3 \\ 7\end{array}\right], \mathrm{ad}-\mathrm{b}^{2}=1\)

\(\mathrm{a}+\mathrm{b}=3, \mathrm{~b}+\mathrm{d}=7,(3-\mathrm{b})(7-\mathrm{b})-\mathrm{b}^{2}=1\)

\(21-10 b=1 \rightarrow b=2, a=1, d=5\)

\(A=\left[\begin{array}{ll}1 & 2 \\ 2 & 5\end{array}\right], A^{-1}=\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]\)

\(\mathrm{A}^{-1}=\alpha \mathrm{A}+\beta \mathrm{I}\)

\(\left[\begin{array}{cc}5 & -2 \\ -2 & 1\end{array}\right]=\left[\begin{array}{cc}\alpha+\beta & 2 \alpha \\ 2 \alpha & 5 \alpha+\beta\end{array}\right]\)

\(\alpha=-1, \beta=6 \)

\(\Rightarrow \alpha+\beta=5\)