A and B are real 4x4 matrices, if det(B)=1 and B=2A^3(A)T (T means transpose), what is the det(A)? Thank you very much in advance for your time and effort.
From how I interpreted your problem:
$\displaystyle det(B) = 1$
$\displaystyle B = 2A^3A^T $
If that's the case, remember determinants can be broken up over multiplication.
So if I re-write B:
$\displaystyle B = 2A^3A^T = 2 A A A A^T $
Now that you have that, be careful when you multiply by a scalar (in this case 2). Think about how that changes the matrix A in terms of row operations.
Hi eXist,
Thanks very much for your reply. Could you please explain further as I am a newbie to matrix. from det(B)=1, I can see B is an identity matrix and det(2AAA(A)T)=1. so how can I solve det(A)? Do I need to find the matrix A? Thank you and please excuse me for my poor knowledge on this.
cheers,
pingping
From what you have been told you should see that:
$\displaystyle 2^n [\det(A)]^3\det(A^T)=1$
also you should know that $\displaystyle \det(A^T)=\det(A)$ so:
$\displaystyle 2^n [\det(A)]^4=1$
where $\displaystyle n$ is the size of the matrix in this case $\displaystyle n=4$
(when you multiply a matrix by a scalar every element is multiplied by the scalar, and for an $\displaystyle n \times n$ matrix the determinant can be written as a sum of $\displaystyle \pm$ the product of $\displaystyle n$ terms at a time from the matrix, this is where the $\displaystyle 2^n$ term above comes from)
CB