Q85 / 180 · COMEDK 2026 (9 May Shift 2) · Mathematics
Let A be a square matrix of order $3 \times 3$. If $|A| = -4$, then the value of $\left|\dfrac{A^{-1}}{-2}\right|$ is:
- A$-1$
- B$-\dfrac{1}{16}$
- Correct answerC$\dfrac{1}{32}$
- D$2$
Solution
Given $A$ is a $3 \times 3$ matrix and $|A| = -4$.
Using the property of determinants $|kA| = k^n |A|$ where $n$ is the order of the matrix, we get:
$\left|\dfrac{A^{-1}}{-2}\right| = \left|-\dfrac{1}{2} A^{-1}\right| = \left(-\dfrac{1}{2}\right)^3 |A^{-1}|$
Since $|A^{-1}| = \dfrac{1}{|A|}$:
Substituting $|A| = -4$, we get $|A^{-1}| = -\dfrac{1}{4}$.
Therefore, $\left|\dfrac{A^{-1}}{-2}\right| = \left(-\dfrac{1}{8}\right) \times \left(-\dfrac{1}{4}\right) = \dfrac{1}{32}$.
Answer: $\dfrac{1}{32}$
Using the property of determinants $|kA| = k^n |A|$ where $n$ is the order of the matrix, we get:
$\left|\dfrac{A^{-1}}{-2}\right| = \left|-\dfrac{1}{2} A^{-1}\right| = \left(-\dfrac{1}{2}\right)^3 |A^{-1}|$
Since $|A^{-1}| = \dfrac{1}{|A|}$:
Substituting $|A| = -4$, we get $|A^{-1}| = -\dfrac{1}{4}$.
Therefore, $\left|\dfrac{A^{-1}}{-2}\right| = \left(-\dfrac{1}{8}\right) \times \left(-\dfrac{1}{4}\right) = \dfrac{1}{32}$.
Answer: $\dfrac{1}{32}$
Hence, the correct answer is option: 'C'.