# solving a matrix equation for a,b,c, and d

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• Feb 3rd 2010, 10:52 AM
cottekr
solving a matrix equation for a,b,c, and d
[a-b b+c] = [8 1]
[3d+c 2a-4d] [7 6]

I know that since the matrices are equal:

a-b=8
b+c=1
3d+c=7
2a-4d=6

but not sure where to go from here
• Feb 3rd 2010, 12:16 PM
pickslides
Quote:

Originally Posted by cottekr
[a-b b+c] = [8 1]
[3d+c 2a-4d] [7 6]

I know that since the matrices are equal:

a-b=8
b+c=1
3d+c=7
2a-4d=6

but not sure where to go from here

You have made a good start.

From the first equation $\displaystyle a = 8+b$

From the second $\displaystyle c = 1-b$

From the third (and second) $\displaystyle d = \frac{7-c}{3} \implies \frac{7-(1-b)}{3} = \frac{6+b}{3}$

Now looking at the 4th equation $\displaystyle 2a-4d=6$ we have new information for $\displaystyle a$ and $\displaystyle d$ so lets use it in the 4th.

$\displaystyle 2a-4d=6\implies 2(8+b)-4\left(\frac{6+b}{3}\right)=6$

Now expand and solve for $\displaystyle b$
• Feb 3rd 2010, 08:08 PM
math2009
$\displaystyle A\begin{bmatrix}a\\b\\c\\d\end{bmatrix}= \begin{bmatrix}a-b\\b+c\\3d+c\\2a-4d\end{bmatrix}= a\begin{bmatrix}1\\0\\0\\2\end{bmatrix}+ b\begin{bmatrix}-1\\1\\0\\0\end{bmatrix}+ c\begin{bmatrix}0\\1\\1\\0\end{bmatrix}+ d\begin{bmatrix}0\\0\\3\\-4\end{bmatrix} \\$

$\displaystyle =\begin{bmatrix} 1 &-1 &0 &0 \\ 0 &1 &1 &0 \\ 0 &0 &1 &3 \\ 2 &0 &0 &-4 \\ \end{bmatrix} \begin{bmatrix}a\\b\\c\\d\end{bmatrix}= \begin{bmatrix}8\\1\\7\\6\end{bmatrix}~,~ \begin{bmatrix}a\\b\\c\\d\end{bmatrix}= A^{-1}\begin{bmatrix}8\\1\\7\\6\end{bmatrix}= \begin{bmatrix}5\\-3\\4\\1\end{bmatrix}$