# How do you solve 4x-y=-3 and 5x-3y=-23 using matrices?

Mar 24, 2016

$x = 2 \mathmr{and} y = 11$

#### Explanation:

There are 3 ways using matrices and linear algebra :

Method number 1 - Gauss-Jordan Elimination

Write the linear system in the form $A x = b$ where

$A = \left[\begin{matrix}4 & - 1 \\ 5 & - 3\end{matrix}\right]$ is the coefficient matrix of the system,

$x = \left[\begin{matrix}x \\ y\end{matrix}\right]$ is the column vector of variables, and

$b = \left[\begin{matrix}- 3 \\ - 23\end{matrix}\right]$ is the column vector of solutions.

Now augment the coefficient matrix with the column vector of solutions and perform elementary row operations until you obtain the $2 \times 2$ identity matrix ${I}_{2}$ on the left hand side and then what remains on the right hand side will be the solution set for $x$. I will attach a sketch at the end to show these steps.
Eventually we get :

$\left[\begin{matrix}4 & - 1 \\ 5 & - 3\end{matrix}\right] | \left[\begin{matrix}- 3 \\ - 23\end{matrix}\right] \to \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right] | \left[\begin{matrix}2 \\ 11\end{matrix}\right]$

and hence $x = 2 \mathmr{and} y = 11$.

Method number 2 - Using the inverse matrix method

Since $A x = b$ it implies that $x = {A}^{- 1} b$.

We may find ${A}^{- 1}$ by using elementary row operations in augmenting the coefficient matrix A with the identity matrix ${I}_{2}$, alternatively, we may use the formula :
${A}^{- 1} = \frac{1}{\det \left(A\right)} \cdot a \mathrm{dj} \left(A\right)$, where $\det \left(A\right) = \Delta$.

$= \frac{1}{- 12 - \left(- 5\right)} \cdot \left[\begin{matrix}- 3 & 1 \\ - 5 & 4\end{matrix}\right]$

$= \left[\begin{matrix}\frac{3}{7} & - \frac{1}{7} \\ \frac{5}{7} & - \frac{4}{7}\end{matrix}\right]$

Hence $x = {A}^{- 1} b = \left[\begin{matrix}\frac{3}{7} & - \frac{1}{7} \\ \frac{5}{7} & - \frac{4}{7}\end{matrix}\right] \left[\begin{matrix}- 3 \\ - 23\end{matrix}\right] = \left[\begin{matrix}2 \\ 11\end{matrix}\right]$

and so $x = 2 \mathmr{and} y = 11$.

Method number 3 - Kramer's Rule

Find the determinants of the matrices formed by replacing each column vector in the coefficient matrix with the column vector of solutions :

${\Delta}_{x} = | \left(- 3 , - 1\right) , \left(- 23 , - 3\right) | = \left(- 3\right) \left(- 3\right) - \left(- 1\right) \left(- 23\right) = - 14$.

${\Delta}_{y} = | \left(4 , - 3\right) , \left(5 , - 23\right) | = - 77$.

Then : $x = {\Delta}_{x} / \Delta = - \frac{14}{-} 7 = 2$

and $y = {\Delta}_{y} / \Delta = - \frac{77}{-} 7 = 11$.