Question

For a square matrix \(A\), \[ (3A)^{-1}= \]

• A. \( 3A^{-1} \)
• B. \( 9A^{-1} \)
• C. \( \frac{1}{3} A^{-1} \)
• D. \( \frac{1}{9} A^{-1} \)

Hint:

Don't confuse \( (kA)^{-1} = \frac{1}{k} A^{-1} \) with the determinant property \( |kA| = k^n |A| \).
Inverses work by reciprocating the scalar, while determinants scale by the power of the matrix order.

Answer 1

The Correct Option is C

Concept:

• Inverse of a scalar multiple of a matrix: If \( k \) is a non-zero scalar and \( A \) is an invertible matrix, then \( (kA)^{-1} = \frac{1}{k} A^{-1} \).
• This property arises because \( (kA) \cdot (\frac{1}{k} A^{-1}) = (k \cdot \frac{1}{k}) (A \cdot A^{-1}) = 1 \cdot I = I \).

Step 1: Identify the scalar and apply the property
In the expression \( (3A)^{-1} \), the scalar \( k \) is \( 3 \).
Applying the property \( (kA)^{-1} = \frac{1}{k} A^{-1} \):
\[ (3A)^{-1} = \frac{1}{3} A^{-1} \]

Step 2: Verify with the definition of inverse
Let \( B = \frac{1}{3} A^{-1} \).
Check if \( (3A)B = I \):
\[ (3A) \left( \frac{1}{3} A^{-1} \right) = \left( 3 \cdot \frac{1}{3} \right) (A \cdot A^{-1}) \]
\[ = 1 \cdot I = I \]
Since the product is the identity matrix, \( \frac{1}{3} A^{-1} \) is indeed the inverse of \( 3A \).