Question

For two matrices \(A\) and \(B\), given that \( A^{-1} = \frac{1}{8}B \) then inverse of \( (8A) \) is

• A. \( \frac{1}{8}B \)
• B. \( 8B \)
• C. \( \frac{1}{64}B \)
• D. \( B \)

Hint:

Always remember: \( (kA)^{-1} = \frac{1}{k}A^{-1} \). Substitute carefully when \(A^{-1}\) is already given.

Answer 1

The Correct Option is C

Step 1: Use property of inverse of scalar multiple.
For any matrix \(A\) and scalar \(k\):
\[ (kA)^{-1} = \frac{1}{k}A^{-1}. \]

Step 2: Apply the formula.
\[ (8A)^{-1} = \frac{1}{8}A^{-1}. \]

Step 3: Substitute the given value of \(A^{-1}\).
Given:
\[ A^{-1} = \frac{1}{8}B. \]
So,
\[ (8A)^{-1} = \frac{1}{8} \times \frac{1}{8}B. \]

Step 4: Multiply the constants.
\[ = \frac{1}{64}B. \]

Step 5: Simplify the expression.
Thus, the inverse becomes:
\[ (8A)^{-1} = \frac{1}{64}B. \]

Step 6: Verify the concept.
Scaling a matrix scales its inverse inversely, hence factor appears squared here due to substitution.

Step 7: Final conclusion.
Thus, the correct answer is option (C).
Final Answer:
\[ \boxed{\frac{1}{64}B}. \]