Question

Prove: If A is an nxn matrix then det[adj(A)] = [det(A)]^(n-1)

Answer 1

@Lyrae help me

Answer 2

One way to find the inverse if by:\[A^{-1}=\frac{1}{\text{det}(A)}\text{adj}(A)\]
so, \[ \text{det}(A)A^{-1}=\text{adj}(A)\]
Now, if you multiply one row of a determinant by a constant \(k\), then the result is equal to \(k\cdot \text{det}(A)\)
If you multiply the 2nd as well, then mean you are multiplying the previous determinant \(k\cdot \text{det}(A)\), by one row by a factor \(k\) , which gives \(k^2 \cdot \text{det}(A)\)
.. by induction, doing this on all \(n\) rows rows yields \(k^n \cdot \text{det}(A)\)

which implies that \(kA=k^n\text{det}(A)\)

because \(kA\) essentially means you are multiplying all \(n\) rows of \(A\) by \(k\).
So, back to \[ \text{det}(A)A^{-1}=\text{adj}(A)\]
applying the determinant to both sides :
\[ \text{det}\left( \text{det}(A)A^{-1}\right) =\text{det}\left(\text{adj}(A) \right)\\
\left(\text{det}(A) \right)^n \text{det}(A^{-1})=\text{det}\left(\text{adj}(A) \right)\]
using the property above since determinant is just a number. And simply recall that \[ \text{det}(A^{-1})=\frac{1}{\text{det}(A)} \]

Answer 3

which implies \(\text{det}(kA)=k^n\text{det}(A)\) *** typo there sorry

Answer 4

if there is any sub-result I used above that you haven't seen yet, then you might have to prove those individually to make the proof complete

Answer 5

thank you!!

Answer 6

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