First slide

Applications of determinant

Question

$Let A and B be two square matrices of order 3 such that \det . (A) = 3 and det. (B) = 2, then the value of$
$det. ({(adj \cdot (B^{- 1} A^{- 1}))}^{- 1}) is$

Moderate

Solution

$\begin{array}{l} {(adj \cdot (B^{- 1} A^{- 1}))}^{- 1} \\ = {(adj \cdot (A B)^{- 1})}^{- 1} \\ = {((adj \cdot (A B))^{- 1})}^{- 1} \\ = adj \cdot (A B) \end{array}$
$∴ \det ({(adj \cdot (B^{- 1} A^{- 1}))}^{- 1}) = | (adj \cdot (A B)) |$
$= | A B |^{2} = ⌊{A|}^{2} | B |^{2} = 9 \times 4 = 36$

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