Hi, I'm trying to solve the following problem:

"Solve
XA=Bfor A using matrix algebra, where Bis a n x 1
vector,
X is a n x k matrix, and A is a k x 1 vector. You can assume that

any square matrices that you encounter have full rank."

Usually this could be solved easily just by multiplying by the inverse of X, but since X is not a square matrix, I don't know what to do.

Thanks!

2. Originally Posted by willis
Hi, I'm trying to solve the following problem:

"Solve
XA=Bfor A using matrix algebra, where Bis a n x 1
vector,
X is a n x k matrix, and A is a k x 1 vector. You can assume that

any square matrices that you encounter have full rank."

Usually this could be solved easily just by multiplying by the inverse of X, but since X is not a square matrix, I don't know what to do.

Thanks!
You need to manufacture a square matrix from X.

To do this, pre-multiply both sides by the TRANSPOSE of X.

Then you can pre-multiply both sides of the equation by the inverse of the new square matrix.

$X^TXA = X^TB$

$(X^TX)^{-1}X^TXA = (X^TX)^{-1}X^TB$

$IA = (X^TX)^{-1}X^TB$

$A = (X^TX)^{-1}X^TB$.

3. Thank you so much!!! I get it now!