Let $P$ be a $3 \times 3$ matrix with $det P = - 5$ . Find the determinant of the matrix $Q = (- 2 P)^{- 1}$ .

medium 2 marks Class 12 · Mathematics · Determinants

Answer

\frac{1}{40}

The correct answer is $\frac{1}{40}$ — Thus the determinant of $Q$ is $\frac{1}{40}$ .

Step-by-step solution

First compute the determinant of the scalar multiple $- 2 P$ . For a $3 \times 3$ matrix, $det (k A) = k^{3} det A$ . Here $k = - 2$ , so $det (- 2 P) = (- 2)^{3} det P = - 8 \times (- 5) = 40$ .
The matrix $Q$ is the inverse of $- 2 P$ . For any invertible matrix $M$ , $det (M^{- 1}) = \frac{1}{det M}$ .
Hence $det Q = \frac{1}{det ( - 2 P )} = \frac{1}{40}$ .
Thus the determinant of $Q$ is $\frac{1}{40}$ .

Common mistake: Using

(- 2)^{2}

instead of

(- 2)^{3}

for the scaling factor, or taking the reciprocal of

- 5

directly without accounting for the scalar.

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