in Linear Algebra
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What is the value of determinant of adjoint( A). What I am not getting in this is that since $AA\ ^{-1}=\mathbb{I}_{n}$ .Now if I take determinant on both sides I will get $\mid A\mid \mid A^{-1}\mid=\mathbb{I}_{n}=1$  ,now $\mid A^{-1}\mid=1/\mid A\mid$ ,then $\mid A^{-1}\mid=\mid Adjoint A\mid/ \mid A\mid$. By substituting the above value I get $\mid Adj A\mid =1$. What is wrong in this statement ? since value of $\text{ det(adj(A))}=det(A)^{n-1}$

in Linear Algebra
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in your approach also I got the same answer

you did wrong here

A−1∣=∣AdjointA∣/∣A

it should be

A−1∣=∣ AdjointA / ∣A∣ |

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1 Answer

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Best answer

A A-1  = In

A (adj A) = |A| In

take determinants Both side

| A (adj A) | = |  |A| In |

==>We know that if A be an n-rowed square matrix  and ' k'  be any scalar then

| K A | = K|A|

Now

| A | | adj A| = | A || I|

| A | | adj A| = | A |n

| adj A| | A |n-1

2 Comments

I have gone through this proof ,I just wanted to ask what is wrong in my approach ?
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∣A-1∣=|adj(A) / ∣A∣|

        =|adj(A)| / |A|n

and not |adj(A)| / ∣A|

 

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