The matrix \(A = \begin{bmatrix}4 & 1\\ -2 & 1\end{bmatrix}\) satisfies \(A^{2} - 5A + 6I = O\). Then \(A^{-1}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{1}{6}(5I - A)\)
Since \(|A| = 4 \cdot 1 - 1 \cdot (-2) = 6 \ne 0\), \(A^{-1}\) exists. Rearranging the given relation:
\(A^{2} - 5A = -6I\)
Multiplying both sides by \(A^{-1}\) (using \(A^{-1}A^{2} = A\) and \(A^{-1}A = I\)):
\(A - 5I = -6A^{-1}\)
\( \displaystyle A^{-1} = \frac{1}{6}(5I - A) \)
Check:
\( \displaystyle \frac{1}{6}(5I - A) = \frac{1}{6}\begin{bmatrix}1 & -1\\ 2 & 4\end{bmatrix} \)
This agrees with \( \displaystyle \frac{1}{|A|}\,\mathrm{adj}\,A = \frac{1}{6}\begin{bmatrix}1 & -1\\ 2 & 4\end{bmatrix} \).
Why the other options are incorrect
Option (2) \(\displaystyle\frac{1}{6}(A - 5I)\) — Incorrect. This has the wrong sign; it equals \(-A^{-1}\).
Option (3) \(\displaystyle\frac{1}{5}(6I - A)\) — Incorrect. The roles of 5 and 6 are exchanged; this gives \(\frac{1}{5}\begin{bmatrix}2 & -1\\ 2 & 5\end{bmatrix}\), which is not \(A^{-1}\).
Option (4) \(\displaystyle\frac{1}{6}(5A - I)\) — Incorrect. This gives \(\frac{1}{6}\begin{bmatrix}19 & 5\\ -10 & 4\end{bmatrix}\), which is not \(A^{-1}\).