Does $\det(A^2) = \det(A)*\det(A)$?

Question

I've tried it with numbers and both operation don't equal each other, while $\det(AB)$ equal $\det(A)*\det(B)$

$A$ = \begin{bmatrix}1&2&-2&1\\1&2&-1&0\\2&5&-6&4\\-2&-4&4&-1\end{bmatrix}

$\det(A)$ = -1

but $\det(A^2) = 535$

what I am doing wrong ?

EDIT : it does I made a calculation mistake

You should add what you have tried so that others can help you find and fix your mistake. — an4s, Mar 14 '18 at 14:38
Welcome to stackexchange. Please include your attempt with numbers, because there's probably an error in your calculation. Use mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — Ethan Bolker, Mar 14 '18 at 14:38
@RootOfMinusOne I've added a derivation by using the property that det doesn't change adding multimple of a row to Others row, in this way you can simplify a lot and avoid errors — user, Mar 14 '18 at 15:08
Possible duplicate of Is $;\det(A^n) =\left(\det (A)\right)^n;$? — GNUSupporter 8964民主女神地下教會, Mar 14 '18 at 15:17

user · Accepted Answer · 2018-03-14T15:06:19.263

4

Yes it is a theorem/property of determinant also known as Binet's theorem and $\det(A^2)=\det(A)\cdot \det(A)$ is just a special case of $\det(AB)=\det(A)\cdot \det(B)$ with $B=A$.

Indeed we find

$$A^2=\begin{bmatrix}-3&-8&12&-8\\1&1&2&-3\\-13&-32&43&-26\\4&12&-20&15\end{bmatrix}$$

and

$$\det(A^2)=\begin{vmatrix}-3&-8&12&-8\\1&1&2&-3\\-13&-32&43&-26\\4&12&-20&15\end{vmatrix}= \begin{vmatrix}0&-5&18&-17\\1&1&2&-3\\0&-19&69&-65\\0&8&-28&27\end{vmatrix}= \begin{vmatrix}0&-5&18&-17\\1&1&2&-3\\0&1&-3&3\\0&-2&8&-7\end{vmatrix}=\begin{vmatrix}0&0&3&-2\\1&1&2&-3\\0&1&-3&3\\0&0&2&-1\end{vmatrix}=1$$

edited Mar 14 '18 at 15:06

answered Mar 14 '18 at 14:40

user

154,566

3

it should be a comment. – Surb Mar 14 '18 at 14:41
2

@Surb I've precisely answered to the OP. The problem is indeed that very often answers are posted as comment. – user Mar 14 '18 at 14:45

score 1 · Answer 2 · answered Mar 14 '18 at 14:58

You correctly computed the $\det A = -1$.

The square of $A$ is \begin{bmatrix}-3&-8&12&-8\\1&1&2&-3\\-13&-32&43&-26\\4&12&-20&15\end{bmatrix} and its determinant is $1$.

So you either squared the matrix incorrectly, or computed the second determinant incorrectly.

score 0 · Answer 3 · answered Mar 14 '18 at 14:51

You can check your answer using Wolfram Alpha.

It agrees that $det(A)=-1$.

It finds that $$A^2=\begin{bmatrix}-3&-8&12&-8\\1&1&2&-3\\-13&-32&43&-26\\4&12&-20&15\end{bmatrix}$$

And it confirms that $det(A^2)=1$.

Looking over the above, perhaps you can find the error in your hand calculations.

Does $\det(A^2) = \det(A)*\det(A)$?

3 Answers3