# 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:

$$A^4+2A=0$$

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

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

$$A^3=-2I$$

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

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

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

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

$$|A|^3 = -2$$

$$|A|=-1.2599$$

$$A^4+2A = 0 implies A^4 = -2A implies det(A^4) = det(-2A^4) implies det(A)^4 = (-2)^6det(A)$$. 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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