Question

Let \( A \) be a square matrix of order \(3\times3\). If \( |A|=-4 \), then the value of \[ \left|\frac{A^{-1}}{-2}\right| \] is:

• A. \(-1\)
• B. \(2\)
• C. \(\dfrac{1}{32}\)
• D. \(-\dfrac{1}{16}\)

Hint:

Remember: \[ |A^{-1}|=\frac1{|A|} \] and for an \(n\times n\) matrix, \[ |kA|=k^n|A| \] Always use the matrix order carefully while applying determinant properties.

Answer 1

The Correct Option is C

Understanding the Concept: For a square matrix of order \(n\), \[ |kA|=k^n|A| \] Also, \[ |A^{-1}|=\frac{1}{|A|} \] These determinant properties are extremely useful while evaluating determinants involving scalar multiplication and inverses.
Step 1: Evaluate determinant of inverse matrix. Given: \[ |A|=-4 \] Therefore, \[ |A^{-1}|=\frac{1}{|A|} =\frac{1}{-4} =-\frac14 \]
Step 2: Factor out the scalar. We are required to compute: \[ \left|\frac{A^{-1}}{-2}\right| \] This means: \[ \left|\left(-\frac12\right)A^{-1}\right| \] Since the matrix is of order \(3\times3\), \[ |kA|=k^3|A| \] Hence, \[ \left|\left(-\frac12\right)A^{-1}\right| = \left(-\frac12\right)^3 |A^{-1}| \] \[ = -\frac18 \times \left(-\frac14\right) \] \[ =\frac{1}{32} \] Thus, \[ \boxed{\frac1{32}} \]