# Thread: Find det(X) given BXA^2 + 2I = 0.

1. ## Find det(X) given BXA^2 + 2I = 0.

hello, this look rather simple, but I struggle, I need some help, or at least a final answer so I can check myself....thank you

A and B are 5X5 matrices. it is known that:
det(A)=2
det(B)=-3

X is a square matrix so BXA^2 + 2I = 0
(multiplication of B, X and the square of A, plus twice the identity matrix)

what is det(X) ?

2. ## Re: simple question...

the determinant function is multiplicative which means det(AB) =det(A)*det(B)

so first i suggest to add -2I to both sides then take the det

BXA^2 = -2I

can you continue ?

3. ## Re: Find det(X) given BXA^2 + 2I = 0.

thank you, I think I got it, I got det(X)=2.666, hope this is correct....

thanks again !

4. ## Re: Find det(X) given BXA^2 + 2I = 0.

$det(BXA^2) = det(-2I)$

$det(B) \; det(X) \; det(A)\; det(A) = -2$

$-3 det(X) (4) = -2$

$det(X) = \frac{-2}{-12}$

5. ## Re: Find det(X) given BXA^2 + 2I = 0.

Originally Posted by Amer
$det(BXA^2) = det(-2I)$

$det(B) \; det(X) \; det(A)\; det(A) = -2$

$-3 det(X) (4) = -2$

$det(X) = \frac{-2}{-12}$
No! These are 5x5 matrices, so $\det(-2I) = (-2)^5 = -32.$

6. ## Re: Find det(X) given BXA^2 + 2I = 0.

Originally Posted by Opalg
No! These are 5x5 matrices, so $\det(-2I) = (-2)^5 = -32.$
thanks I did not noticed that