Question:
For a square matrix $A$ of order $n$, the inverse matrix $(3A)^{-1}$ is equal to:
For a square matrix $A$ of order $n$, the inverse matrix $(3A)^{-1}$ is equal to:
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Be careful not to confuse this with the determinant property! For determinants, $\det(kA) = k^n \det(A)$, but for inverses, it is simply a linear reciprocal scalar: $(3A)^{-1} = \frac{1}{3}A^{-1}$.
Updated On: Jul 4, 2026
- $3A^{-1}$
- $9A^{-1}$
- $\frac{1}{3}A^{-1}$
- $\frac{1}{9}A^{-1}$
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The Correct Option is C
Solution and Explanation
Concept:
A fundamental property of matrix inversion states that for any non-zero scalar $k$ and an invertible square matrix $A$:
\[
(kA)^{-1} = \frac{1}{k} A^{-1}
\]
Step 1: Apply the scalar property of matrix inverses.
We are looking for the expression equal to $(3A)^{-1}$. Here, the scalar multiplier is $k = 3$. According to the property $(kA)^{-1} = \frac{1}{k} A^{-1}$, we can substitute $k=3$ directly: \[ (3A)^{-1} = \frac{1}{3} A^{-1} \] This matches option (C).
Step 1: Apply the scalar property of matrix inverses.
We are looking for the expression equal to $(3A)^{-1}$. Here, the scalar multiplier is $k = 3$. According to the property $(kA)^{-1} = \frac{1}{k} A^{-1}$, we can substitute $k=3$ directly: \[ (3A)^{-1} = \frac{1}{3} A^{-1} \] This matches option (C).
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