# How do you solve  6x+5y = 8 and 3x-y = 7 using matrices?

$x = \frac{43}{21}$ & $y = - \frac{6}{7}$

#### Explanation:

Given equations can be written as follows

$6 x + 5 y - 8 = 0 \setminus \ldots \ldots \left(1\right)$

$3 x - y - 7 = 0 \setminus \ldots \ldots \left(2\right)$

Using matrix method, the solution can be written as

$\setminus \frac{x}{\left(5\right) \left(- 7\right) - \left(- 8\right) \left(- 1\right)} = \setminus \frac{y}{\left(- 8\right) \left(3\right) - \left(6\right) \left(- 7\right)} = \setminus \frac{1}{\left(6\right) \left(- 1\right) - \left(5\right) \left(3\right)}$

$\setminus \frac{x}{- 43} = \setminus \frac{y}{18} = - \frac{1}{21}$

$\setminus \implies \setminus \frac{x}{- 43} = - \frac{1}{21}$

$x = \frac{43}{21}$

$\setminus \implies \setminus \frac{y}{18} = - \frac{1}{21}$

$y = - \frac{6}{7}$

Jul 22, 2018

$\textcolor{b l u e}{x = \frac{43}{21} , y = - \frac{6}{7}}$

#### Explanation:

Put the coefficients of the equations into a matrix, and put the solutions into 1 column vector and the variables into another vector:

$\boldsymbol{A} = \left[\begin{matrix}6 & 5 \\ 3 & - 1\end{matrix}\right]$

$\boldsymbol{b} = \left[\begin{matrix}8 \\ 7\end{matrix}\right]$

$\boldsymbol{X} = \left[\begin{matrix}x \\ y\end{matrix}\right]$

We now have the matrix equation:

$\boldsymbol{A X} = \boldsymbol{b}$

If $\boldsymbol{A}$ in invertible, then $\boldsymbol{{A}^{-} 1}$ exists, so:

$\boldsymbol{{A}^{-} 1 A X} = \boldsymbol{{A}^{-} 1} \boldsymbol{b}$

$\boldsymbol{I X} = \boldsymbol{{A}^{-} 1} \boldsymbol{b}$

We need to find $\boldsymbol{{A}^{-} 1}$

First find the determinant of $\boldsymbol{A}$

Determinant of $\boldsymbol{A} = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right] = a d - b c$

So determinant of:

$\boldsymbol{A} = \left[\begin{matrix}6 & 5 \\ 3 & - 1\end{matrix}\right] = 6 \left(- 1\right) - \left(5\right) \left(3\right) = - 21$

Swap the elements on the leading diagonal and change the signs on the non-leading diagonal:

$\boldsymbol{A} = \left[\begin{matrix}- 1 & - 5 \\ - 3 & 6\end{matrix}\right]$

Next multiply by the reciprocal of the determinant:

$\boldsymbol{{A}^{-} 1} = \left[\begin{matrix}\frac{1}{21} & \frac{5}{21} \\ \frac{1}{7} & - \frac{6}{21}\end{matrix}\right]$

We now have:

$\left[\begin{matrix}x \\ y\end{matrix}\right] = \left[\begin{matrix}\frac{1}{21} & \frac{5}{21} \\ \frac{1}{7} & - \frac{6}{21}\end{matrix}\right] \left[\begin{matrix}8 \\ 7\end{matrix}\right] = \left[\begin{matrix}\frac{8}{21} + \frac{35}{21} \\ \frac{8}{7} - 2\end{matrix}\right] = \left[\begin{matrix}\frac{43}{21} \\ - \frac{6}{7}\end{matrix}\right]$

Hence:

$x = \frac{43}{21}$

$y = - \frac{6}{7}$