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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