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Find the determinant of a matrix $A$, such that $A^4 + 2A = 0$

Mathematics Asked by avivgood2 on December 10, 2020

Find the determinant of an revertible, $6 times 6$ matrix $A$, such that $A^4 + 2A = 0$


This question seems odd to me, because when I tried to solve it:

$A^4 + 2A = O$

$(A^3+2I)A = Odet$

$|(A^3+2I)||A| = |O|$

But I know that $|A|$ is revertible, thus $|A| != 0$, which must mean that $|(A^3+2I)| = 0$.

But if I try:

$|(A^3+2I)||A| = |O|$

$0|A| = 0$
which means it true for every $6 * 6$ matrices, but the question specified $1$.

What am I getting wrong here, and how should I solve this?

edit:

Here is another thing I have tried:

$A^4+2A=0$

$(A^3+2I)A = 0$

$A^3+2I=0$

$A^3=-2I$

$A^3=-2A*A^-1$

$A^2=-2*A^-1det$

$|A^2|/2=-2*(1/|A|)$

$|A| * |A| / 1 = -2/|A|$

$|A|^3 = -2 $

$|A|=-1.2599$

2 Answers

$A^4+2A = 0 implies A^4 = -2A implies det(A^4) = det(-2A^4) implies det(A)^4 = (-2)^6det(A)$. Let $det(A) = x$, then the last equation reduces to $x^4 = (-2)^6x$. So $$x(x^3+2^6) = 0$$

Since $A$ is invertible, the only possibility is $x = -2^2$.

Correct answer by weierstrash on December 10, 2020

We have that

$$A^4+2A=0implies A^{-1}A^4+2A^{-1}A=0implies A^3=-2I $$

$$implies det (A^3) = det (-2I)=2^6$$

then by $det(XY)=det X :det Y$ we have

$$det (A^3)=(det A)^3=2^6 implies det A=2^2=4$$

Answered by user on December 10, 2020

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