If \(A = \begin{bmatrix}5 & 3\\ 4 & 2\end{bmatrix}\), then \(6A^{-1}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-3\,(\mathrm{adj}\,A)\)
For an invertible matrix, \( \displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A \).
\(|A| = 5 \cdot 2 - 3 \cdot 4 = -2\)
So
\( \displaystyle A^{-1} = -\frac{1}{2}\,\mathrm{adj}\,A \)
\( \displaystyle 6A^{-1} = 6 \times \left(-\frac{1}{2}\right)\mathrm{adj}\,A = -3\,(\mathrm{adj}\,A) \)
Why the other options are incorrect
Option (1) \(3\,(\mathrm{adj}\,A)\) — Incorrect. The sign of \(|A| = -2\) is lost.
Option (3) \(12\,(\mathrm{adj}\,A)\) — Incorrect. This multiplies by \(|A|\) without its sign instead of dividing by \(|A|\).
Option (4) \(-12\,(\mathrm{adj}\,A)\) — Incorrect. This multiplies by \(|A| = -2\) instead of dividing by it.