1

I have to solve $B(Y+C)+3A=5(A+Y)$ for given matrices of $B$, $C$ and $A$. so what I do is $BY+BC+3A=5A+5Y$, and then $BY-5Y=2A-BC$ and then $Y(B-5)=2A-BC$. here I subtitute $B-5=M$ and $2A-BC=N$ and I know what I have to do next, but how do I find that $B-5$?

Herng Yi
  • 3,156
dgfddf
  • 71

1 Answers1

3

5 is not a matrix. The step $BY - 5Y = (B - 5)Y$ should really be $BY - 5Y = BY - 5IY = (B - 5I)Y$ where $I$ is the identity matrix. Then $Y = (B - 5I)^{-1}(2A - BC)$ (assuming $B - 5I$ comes out to be invertible).

Edit: Note that the $B - 5I$ term needs to be to the left.

Jim
  • 30,682
  • 1
    No, it should be $Y=(B-5I)^{-1}(2A-BC)$, because you need to cancel out a factor to the left of $Y$. Remember matrix multiplication is not commutative. (The same error appears in the question, where the OP writes $Y(B-5)$ instead of $(B-5)Y$ -- the $B$ was on the left of the $Y$ from the beginning). – hmakholm left over Monica Feb 12 '13 at 06:25
  • Ahh, yes. I was just working from the $Y(B - 5) = 2A - BC$ step. – Jim Feb 12 '13 at 06:31