If det P = -3 for a 3×3 matrix P, find det(5 P^-1).

If $det P = - 3$ for a $3 \times 3$ matrix $P$ , find $det (5 P^{- 1})$ .

medium 2 marks Class 12 · Mathematics · Determinants

Answer

d i s pl a y s t y l e - f r a c 5^{3} 3 = - f r a c 1253

The correct answer is $d i s pl a y s t y l e - f r a c 5^{3} 3 = - f r a c 1253$ — Thus the determinant of $5 P^{- 1}$ is $- \frac{125}{3}$ .

Recall the property $det (P^{- 1}) = \frac{1}{d e t P}$ .
Given $det P = - 3$ , so $det (P^{- 1}) = \frac{1}{- 3} = - \frac{1}{3}$ .
For a scalar multiple $k P$ , $det (k P) = k^{n} det P$ where $n$ is the order of the matrix (here $n = 3$ ).
Apply this to $5 P^{- 1}$ : $det (5 P^{- 1}) = 5^{3} det (P^{- 1})$ .
Compute $5^{3} = 125$ .
Multiply by $det (P^{- 1})$ : $125 \times (- \frac{1}{3}) = - \frac{125}{3}$ .
Thus the determinant of $5 P^{- 1}$ is $- \frac{125}{3}$ .

Common mistake: Using

det (k A) = k det A

instead of

k^{n} det A

for

n \times n

matrices

Concept tested: Determinant of scalar multiple and inverse