Question

For a square matrix $A$, $(3A)^{-1}$ is

• A. $3A^{-1}$
• B. $9A^{-1}$
• C. $\dfrac{1}{3}A^{-1}$
• D. $\dfrac{1}{9}A^{-1}$

Hint:

Always remember the matrix rule: \((kA)^{-1}=\dfrac{1}{k}A^{-1}\).

Answer 1

The Correct Option is C

Step 1: Recall the property of inverse of scalar multiple of a matrix.
If \(A\) is a non-singular square matrix and \(k\) is a scalar, then the inverse of \(kA\) is given by the identity \[ (kA)^{-1} = \frac{1}{k}A^{-1} \] This is a standard result in matrix algebra.
Step 2: Apply the formula to the given expression.
Here the scalar value is \[ k = 3 \] Therefore \[ (3A)^{-1} = \frac{1}{3}A^{-1} \] Step 3: Verification using matrix inverse definition.
For inverse matrices \[ AA^{-1}=I \] Now multiply \[ (3A)\left(\frac13 A^{-1}\right) \] \[ =3A \cdot \frac13 A^{-1} \] \[ =AA^{-1} \] \[ =I \] Thus the result satisfies the inverse definition.
Step 4: Conclusion.
Hence the inverse of \(3A\) is \[ \frac13 A^{-1} \] Final Answer: $\boxed{\frac13 A^{-1}}$