First slide

Applications of determinant

Question

$(- A)^{- 1} is always equal to (where A is n^{th} -order square matrix)$

Moderate

A

${(- 1)}^{n} A^{- 1}$
B

$- A^{- 1}$
C

${(- 1)}^{n - 1} A^{- 1}$
D

none of these

Solution

${(- A)}^{- 1} = \frac{a d j (- A)}{| - A |}$
$\begin{array}{l} = \frac{{(- 1)}^{n - 1} a d j (A)}{{(- 1)}^{n} | A |} \\ = \frac{a d j (A)}{- | A |} \\ = - A^{- 1} \end{array}$

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