If A is a square matrix of order 3 such that $| A | = 2$ , then $| (a d j A^{- 1}) |^{- 1}$ is .............

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Solution

A is a square matrix of order 3 such that $| A | = 2$

We know that,

$A^{- 1} = \frac{a d j A}{| A |}$

$\Rightarrow ∣ ∣ A^{- 1} ∣ ∣ = \frac{| a d j A |}{| A |^{3}} (∵ | k A | = k^{n} | A |)$

$\Rightarrow | a d j A | = {| A |}^{2} (∵ ∣ ∣ A^{- 1} ∣ ∣ = \frac{1}{| A |})$

${∣ ∣ (a d j A^{- 1}) ∣ ∣}^{- 1} = ({∣ ∣ A^{- 1} ∣ ∣}^{2})^{- 1} = {| A |}^{2} = 4$

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