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 ?

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EDIT : it does I made a calculation mistake

Answer 1

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

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

Answer 2

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.

Answer 3

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.