# How do you solve [(2x+y),(x-3y)]=[(5), (13)]?

Dec 29, 2016

Create and augmented matrix and the perform row operations until you obtain an identity matrix on the left of the divide.

#### Explanation:

The matrix of coefficients is:

A = [ (2,1), (1,-3) ]

The matrix of constant terms is:

B = [ (5), (13) ]

The method that I use is to make an Augmented Matrix

(A|B) = [ (2,1,|,5), (1,-3,|,13) ]

And then perform Elementary Row Operations until I obtain an Identity Matrix on the left side of the divide.

${R}_{2} \leftrightarrow {R}_{1}$

[ (1,-3,|,13), (2,1,|,5) ]

${R}_{2} - 2 {R}_{1} \to {R}_{2}$

[ (1,-3,|,13), (0,7,|,-21) ]

${R}_{2} / 7 \to {R}_{2}$

[ (1,-3,|,13), (0,1,|,-3) ]

${R}_{1} + 3 {R}_{2} \to {R}_{1}$

[ (1,0,|,4), (0,1,|,-3) ]

We have obtained an identity matrix.

The values of the unknowns can be read on the right:

$x = 4 \mathmr{and} y = - 3$

Check:

$2 \left(4\right) + - 3 = 5$
$4 - 3 \left(- 3\right) = 13$

$5 = 5$
$13 = 13$

This checks

Dec 29, 2016

The answer is $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}4 \\ - 3\end{matrix}\right)$

#### Explanation:

We rewrite the equation in matrix form

$\left(\begin{matrix}2 & 1 \\ 1 & - 3\end{matrix}\right) \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}5 \\ 13\end{matrix}\right)$

$\left(\begin{matrix}x \\ y\end{matrix}\right) = {\left(\begin{matrix}2 & 1 \\ 1 & - 3\end{matrix}\right)}^{- 1} \left(\begin{matrix}5 \\ 13\end{matrix}\right)$

Let $A = \left(\begin{matrix}2 & 1 \\ 1 & - 3\end{matrix}\right)$

We calculate

$\det A = | \left(2 , 1\right) , \left(1 , - 3\right) | = - 6 - 1 = - 7$

$\det A \ne 0$, therefore the matrix is invertible

We start by calculating

the matrix of cofactor

$C = \left(\begin{matrix}- 3 & - 1 \\ - 1 & 2\end{matrix}\right)$

Then, we calculate the transpose

${C}^{T} = \left(\begin{matrix}- 3 & - 1 \\ - 1 & 2\end{matrix}\right)$

The inverse is

${A}^{- 1} = \frac{1}{-} 7 \left(\begin{matrix}- 3 & - 1 \\ - 1 & 2\end{matrix}\right)$

$= \left(\begin{matrix}\frac{3}{7} & \frac{1}{7} \\ \frac{1}{7} & - \frac{2}{7}\end{matrix}\right)$

Verification

$A \cdot {A}^{- 1} = \left(\begin{matrix}2 & 1 \\ 1 & - 3\end{matrix}\right) \cdot \left(\begin{matrix}\frac{3}{7} & \frac{1}{7} \\ \frac{1}{7} & - \frac{2}{7}\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right)$

Therefore,

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\frac{3}{7} & \frac{1}{7} \\ \frac{1}{7} & - \frac{2}{7}\end{matrix}\right) \left(\begin{matrix}5 \\ 13\end{matrix}\right)$

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}4 \\ - 3\end{matrix}\right)$