# Solving for variables in a matrix

Solve for a, b, c, and d in the equation:

$\left(\begin{array}{cc} 2 & -6 \\ -a & 8 \end{array}\right) = \left(\begin{array}{cc} 2b & 4c \\ 15 & 3d \end{array}\right)$

$2b/2 = 2/2 \rightarrow b = 1$

$4c/4 = -6/4 \rightarrow c = -1.5$

$-a/1 = 15/-1 \rightarrow a = -15$

$3d/3 = 8/3 \rightarrow d = 2.667$

Is that correct? OR

$4c = -6$

$-a = 15$

$2b = 2$

$3d = 8$

$-a = (-1)\times 15 = -15$

$2b = 2\times 2 = 4$

$4c = 4\times (-6) = -24$

$3d = 3\times 8 = 24$

$\left(\begin{array}{cc} 4 & -24\\ -15 & 24\end{array}\right)$

-

The first solution is correct, except for two small mistakes; it should be $-a/(-1)$ instead of $-a/1$, and on the last line $8/3=2.\bar6\ne2.667$. The second solution is wrong, e.g. from $2b=2$ it doesn't follow that $2b=2\cdot2$, since you multiplied by $2$ on one side and not on the other.

-

If $$\left(\begin{array}{cc} 2 & -6 \\ -a & 8 \end{array}\right) = \left(\begin{array}{cc} 2b & 4c \\ 15 & 3d \end{array}\right)$$ then $$\left(\begin{array}{cc} 2 & -6 \\ -a & 8 \end{array}\right) - \left(\begin{array}{cc} 2b & 4c \\ 15 & 3d \end{array}\right) = 0$$ Now simplify the LHS using the matrix subtraction definition: $$\left(\begin{array}{cc} 2-2b & -6-4c \\ -a-15 & 8-3d \end{array}\right) = \left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right)$$ Each entry gives you one equation (matches your second interpretation, but your conclusions are wrong): $$2-2b = 0 \implies 2b = 2 \implies b = 1, \\ -6-4c = 0 \implies -6 = 4c \implies c = -3/2,\\ -a-15 = 0 \implies a = -15,\\ 8-3d = 0 \implies 8 = 3d \implies d = 8/3.$$

-

If your matrices are simply equal to one another, then all you have to do is solve for the variables element-wise. So, the right-hand matrix must look like the left-hand matrix, e.g. 2 = 2b, etc.

-
I think the OP was aware of this, since this is the basis of both proposed solutions; I think the question was how to then solve these equations. –  joriki Jul 20 '12 at 16:34
Right, I was hoping my comment would guide him to find the right technique, rather than the right solution. –  Arkamis Jul 20 '12 at 16:35

In your first answer, you successfully solved each equation by dividing each side by the appropriate value. In the second answer, you inexplicably substituted the right-hand side for the variable you were supposed to be solving for and evaluated the left-hand side. Nowhere in the answer did you solve for a variable (there is no "b=...") and you constructed some matrix at the end which was not part of the problem in question.

-