If \( A=\begin{bmatrix}5 & 1\\4 & 3\end{bmatrix} \) satisfies the equation \( A^2-8A+11I_2=0 \), where \( I_2 \) is identity matrix of order 2, then \( A^{-1} \) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \begin{bmatrix}\displaystyle\frac{3}{11} & -\displaystyle\frac{1}{11}\\-\displaystyle\frac{4}{11} & \displaystyle\frac{5}{11}\end{bmatrix} \)
Here the 0 on the right side is the zero matrix O. Since |A| = 15 − 4 = 11 ≠ 0, A−1 exists. Multiplying the given equation by A−1:
\( \displaystyle A-8I_2+11A^{-1}=O \)
\( \displaystyle A^{-1}=\frac{1}{11}\left(8I_2-A\right) \)
\( \displaystyle =\frac{1}{11}\begin{bmatrix}8-5 & 0-1\\0-4 & 8-3\end{bmatrix} \)
\( \displaystyle =\frac{1}{11}\begin{bmatrix}3 & -1\\-4 & 5\end{bmatrix} \)
\( \displaystyle A^{-1}=\begin{bmatrix}\frac{3}{11} & -\frac{1}{11}\\-\frac{4}{11} & \frac{5}{11}\end{bmatrix} \)
Shortcut: For a 2 × 2 matrix, swap the diagonal entries, change the signs of the other two and divide by |A| = 11; this gives the same matrix directly.
Why the other options are incorrect
Option (2) \( \begin{bmatrix}-\displaystyle\frac{3}{11} & \displaystyle\frac{1}{11}\\\displaystyle\frac{4}{11} & -\displaystyle\frac{5}{11}\end{bmatrix} \) — Incorrect. This is \( -A^{-1} \); every sign is reversed.
Option (3) \( \begin{bmatrix}\displaystyle\frac{3}{11} & \displaystyle\frac{1}{11}\\\displaystyle\frac{4}{11} & \displaystyle\frac{5}{11}\end{bmatrix} \) — Incorrect. The off-diagonal entries must change sign in the inverse; here they keep the signs of A.
Option (4) \( \begin{bmatrix}-\displaystyle\frac{3}{11} & -\displaystyle\frac{1}{11}\\\displaystyle\frac{4}{11} & -\displaystyle\frac{5}{11}\end{bmatrix} \) — Incorrect. Its (1, 1) entry is negative, but \( A^{-1} \) has \( \frac{3}{11} \) there.