A 3×3 matrix K satisfies det K = -8. Find det((2K^T)^-1 (3K)).

A $3 \times 3$ matrix $K$ satisfies $det K = - 8$ . Find $det ((2 K^{T})^{- 1} (3 K))$ .

medium 3 marks Class 12 · Mathematics · Determinants

Answer

\frac{27}{8}

The correct answer is $\frac{27}{8}$ — Multiply the two results: $(- \frac{1}{64}) \times (- 216) = \frac{216}{64} = \frac{27}{8}$ .

Use $det (A B) = det A det B$ ; compute each determinant separately.
For a $3 \times 3$ matrix, $det (K^{T}) = det K = - 8$ .
Scalar multiplication by $2$ multiplies the determinant by $2^{3} = 8$ , so $det (2 K^{T}) = 8 \times (- 8) = - 64$ .
Hence $det ((2 K^{T})^{- 1}) = \frac{1}{d e t ( 2 K ^{T} )} = \frac{1}{- 64} = - \frac{1}{64}$ .
Similarly, $det (3 K) = 3^{3} det K = 27 \times (- 8) = - 216$ .
Multiply the two results: $(- \frac{1}{64}) \times (- 216) = \frac{216}{64} = \frac{27}{8}$ .

Common mistake: Using

2

instead of

2^{3}

(or

3

instead of

3^{3}

) for the scalar factor in a

3 \times 3

determinant.

Concept tested: Determinant of product with scalar multiples and transpose