Linear Algebra

Question

If $I$ is the unit matrix of order $n$ , where  $k!=0$ is a constant then  $adj  \ kI$ is

Comments

Is it $I k^{n-1}$?

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where  k!=0 is a constant

What this statement mean?

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$k \text{ should not equal to } 0$

:)

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I'm thinking how k factorial would be 0 no possiblity at all!! :P

Yes it is $k^{n-1}I$

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Haha, at first glance I had also considered it as "k factorial". :)

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can u explain please

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$Adj (kI) = (kI)^{-1}  |kI|$

$Adj (kI) = \frac{1}{k} * I^{-1} * k^n |I|$

Note :- $I^{-1} = I \text{ and } |I| = 1$

Plugging these values we get

$Adj (kI) = \frac{1}{k}*I * k^{n}$

$Adj (kI) = Ik^{n-1}$

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Adj(kI)=(kI)⁻¹|kI|

Shouldn't it be

Adj(kI)=(kI)⁻¹|kI| I

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Though the answer will remain same. @  Shubhanshu

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@srivivek95 I took simple matrix inverse operation, which is:-

$A^{-1} = \frac{Adj(A)}{|A|}$

and then Cross multiplication.