# How do you solve 5x+3y = -5 and 7x+5y = -11 using matrices?

Feb 29, 2016

5x+3y=−5
7x+5y=−11

Convert to matrix equation

$\left(\begin{matrix}5 & 3 \\ 7 & 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 5 \\ - 11\end{matrix}\right)$

Use definition of inverse to get
$A = \left(\begin{matrix}5 & 3 \\ 7 & 5\end{matrix}\right)$

${A}^{- 1} = \frac{1}{5 \cdot 5 - 3 \cdot 7} \left(\begin{matrix}5 & - 7 \\ - 3 & 5\end{matrix}\right) = \frac{1}{4} \left(\begin{matrix}5 & - 7 \\ - 3 & 5\end{matrix}\right)$

This becomes

$x = \frac{1}{4} \cdot 52 = 13$
$y = \frac{1}{4} \cdot 70 = 17.5$.

#### Explanation:

5x+3y=−5
7x+5y=−11

$\left(\begin{matrix}5 & 3 \\ 7 & 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 5 \\ - 11\end{matrix}\right)$

Our linear system is now a matrix equation of the form.
$A v = c$ with

$A = \left(\begin{matrix}5 & 3 \\ 7 & 5\end{matrix}\right)$

$v = \left(\begin{matrix}x \\ y\end{matrix}\right)$ and

$c = \left(\begin{matrix}- 5 \\ - 11\end{matrix}\right)$.

We want to find v, we need to rearrange this into something of the form $v =$ something.

Just like solving any other equation we need to move $A$ from the left hand side to the right. If a was a real number we'd do this my dividing both side by $A$, such that $\frac{1}{A} A v = \frac{c}{A}$ this works because $\frac{1}{A} A = 1$. $A$ and $\frac{1}{A}$ are inverses, they make 1, but what is 1 for a matrix and how can we find an inverse?

For matrices, 1 is the identify matrix, a matrix with 1's along the diagonal and zeros everywhere else, for a 2 by 2 matrix it is given by:

$I = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$.

It is useful because I times any vector is just that vector, $I v = v$, like $1 \times a = a$. A inverse of a matrix, ${A}^{- 1}$ is a matrix that when multiplied by A produces the identity matrix ${A}^{- 1} A = I$ or $A {A}^{- 1} = I$ . So solve $A v = c$ we multiply both
side by the inverse of A.

${A}^{- 1} A v = {A}^{- 1} c$ since ${A}^{- 1} A = I$
$I v = {A}^{t} c$
$v = {A}^{t} c$

At this point you just have to memorize something, the inverse of a 2 by 2 matrix.

If
$A = \left(\begin{matrix}a & b \\ c & d\end{matrix}\right)$

${A}^{-} 1 = \frac{1}{a d - c b} \left(\begin{matrix}d & - c \\ - b & a\end{matrix}\right)$

Here
$A = \left(\begin{matrix}5 & 3 \\ 7 & 5\end{matrix}\right)$
${A}^{- 1} = \frac{1}{5 \cdot 5 - 3 \cdot 7} \left(\begin{matrix}5 & - 7 \\ - 3 & 5\end{matrix}\right) = \frac{1}{4} \left(\begin{matrix}5 & - 7 \\ - 3 & 5\end{matrix}\right)$

$\left(\begin{matrix}5 & 3 \\ 7 & 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- 5 \\ - 11\end{matrix}\right)$

Multiple both side by the inverse we have

$\frac{1}{4} \left(\begin{matrix}5 & - 7 \\ - 3 & 5\end{matrix}\right) \left(\begin{matrix}5 & 3 \\ 7 & 5\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{4} \left(\begin{matrix}5 & - 7 \\ - 3 & 5\end{matrix}\right) \left(\begin{matrix}- 5 \\ - 11\end{matrix}\right)$

Using the definition of the inverse and some matrix multiplication, we are left with

$\left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{4} \left(\begin{matrix}5 \cdot \left(- 5\right) - 7 \cdot \left(- 11\right) \\ - 3 \cdot \left(5\right) + 5 \cdot \left(- 11\right)\end{matrix}\right)$

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{4} \left(\begin{matrix}- 25 + 77 \\ - 15 - 55\end{matrix}\right)$

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \frac{1}{4} \left(\begin{matrix}52 \\ - 70\end{matrix}\right)$
$x = \frac{1}{4} \cdot 52 = 13$
$y = \frac{1}{4} \cdot 70 = 17.5$