If $d$ is the determinant of a square matrix A of order n, then the determinant of its adjoint is:

d^{n}

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d^{n - 1}

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d^{n - 2}

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d

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Solution

The correct option is B $d^{n - 1}$
$| A d j A | = {| A |}^{n - 1}$
$∴ | a d j A | = d^{n - 1}$

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Similar questions

If A be square matrix of order n and if |A| = D and |adj A| = D' , then [RPET 2000]

D = d i a g [d_{1}, d_{2}, . . . ., d_{n}]

D^{- 1} = d i a g [d_{1}^{- 1}, d_{2}^{- 1} . . . . ., d_{n}^{- 1}]

D = d i a g [d_{1}, d_{2}, . . . . . . d_{n}],

D^{n} = d i a g [d_{1}^{n}, d_{2}^{n} . . . . . ., d_{n}^{n}] .

The expression for $n^{t h}$ term of an AP, with first term $a$ and common difference $d$ , is

1, d_{1}, d_{2}, d_{3} \dots, d_{n - 1}

n^{t h}

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