Let A be a 4×4 matrix with det A = -6. Define B = (3 A^-1)^T. What is det B?

Let $A$ be a $4 \times 4$ matrix with $d e t A = - 6$ . Define $B = (3 A^{- 1})^{T}$ . What is $d e tB$ ?

hard MCQ 3 marks Class 12 · Mathematics · Determinants

Answer

- \frac{27}{2}

The correct answer is $- \frac{27}{2}$ — Therefore $d e tB = - \frac{27}{2}$ .

Use the fact that $d e t (A^{- 1}) = 1/ det A$ . Hence $d e t (A^{- 1}) = 1/ (- 6) = - \frac{1}{6}$ .
A scalar multiple $k A$ of an $n \times n$ matrix multiplies its determinant by $k^{n}$ . Here $n = 4$ , so $d e t (3 A^{- 1}) = 3^{4} det (A^{- 1}) = 81 \times (- \frac{1}{6}) = - \frac{81}{6} = - \frac{27}{2}$ .
The transpose of a matrix does not change its determinant, i.e., $d e t ((3 A^{- 1})^{T}) = det (3 A^{- 1})$ .
Therefore $d e tB = - \frac{27}{2}$ .

Common mistake: Using

3^{3}

instead of

3^{4}

for a

4 \times 4

matrix

Concept tested: Determinant of a scalar multiple of an inverse matrix