If A - - 1 2 1- 0 1 1- 1 2 0 - then Determinant - AdjAdj A - is

The correct option is D 1 #160;(1) | A |= 1 #160;Adj (Adj A) = #160; | A |^3 2A= | A |A #160;Det #160;Adj (Adj A) = det #160;|A| A ...

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If A = [ 1 2...

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If A $= ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & 1 1 & 2 & 0 \end{matrix} ⎤ ⎥ ⎦,$ then Determinant $[A d j (A d j A)]$ is ________

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[\begin{matrix} 4 & 2 5 & 3 \end{matrix}]

| a d j (a d j A) | =

A = ∣ ∣ ∣ ∣ \begin{matrix} 3 & 2 & 1 4 & - 1 & 2 0 & 1 & 2 \end{matrix} ∣ ∣ ∣ ∣

a d j (a d j A)

A

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| a d j (a d j A) | = \pm 1

| a d j (a d j A) | = {| A |}^{n^{2} - 1}

If A is a n x n matrix, then adj(adj A) =

A = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 3 & 4 2 & - 3 & 4 0 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦

a d j (a d j A) = A

a d j (a d j A) = {| A |}^{n - 2} A

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