# How do you solve 2x - 4y = 7 and -3x + y = 12 using matrices?

$x = - 5.5$
$y = - 4.5$

#### Explanation:

From the given equations

$2 x - 4 y = 7 \text{ }$first equation
$- 3 x + y = 12 \text{ }$second equation

Equations take the form:

${a}_{11} x + {b}_{12} y = {c}_{13}$
${a}_{21} x + {b}_{22} y = {c}_{23}$

So that we have
${a}_{11} = 2$ and ${a}_{21} = - 3$
${b}_{12} = - 4$ and ${b}_{22} = 1$
${c}_{13} = 7$ and ${c}_{23} = 12$

The formulas

$\Delta = \left[\begin{matrix}{a}_{11} & {b}_{12} \\ {a}_{21} & {b}_{22}\end{matrix}\right] = {a}_{11} \cdot {b}_{22} - {a}_{21} \cdot {b}_{12}$

$x = \frac{\left[\begin{matrix}{c}_{13} & {b}_{12} \\ {c}_{23} & {b}_{22}\end{matrix}\right]}{\left[\begin{matrix}{a}_{11} & {b}_{12} \\ {a}_{21} & {b}_{22}\end{matrix}\right]} = \frac{{c}_{13} \cdot {b}_{22} - {c}_{23} \cdot {b}_{12}}{{a}_{11} \cdot {b}_{22} - {a}_{21} \cdot {b}_{12}}$

$y = \frac{\left[\begin{matrix}{a}_{11} & {c}_{13} \\ {a}_{21} & {c}_{23}\end{matrix}\right]}{\left[\begin{matrix}{a}_{11} & {b}_{12} \\ {a}_{21} & {b}_{22}\end{matrix}\right]} = \frac{{a}_{11} \cdot {c}_{23} - {a}_{21} \cdot {c}_{13}}{{a}_{11} \cdot {b}_{22} - {a}_{21} \cdot {b}_{12}}$

Let us solve for x and y

$\Delta = \left[\begin{matrix}{a}_{11} & {b}_{12} \\ {a}_{21} & {b}_{22}\end{matrix}\right] = {a}_{11} \cdot {b}_{22} - {a}_{21} \cdot {b}_{12}$

$\Delta = \left[\begin{matrix}2 & - 4 \\ - 3 & 1\end{matrix}\right] = 2 \cdot 1 - \left(- 3\right) \cdot \left(- 4\right) = 2 - 12 = - 10$

$x = \frac{\left[\begin{matrix}{c}_{13} & {b}_{12} \\ {c}_{23} & {b}_{22}\end{matrix}\right]}{\left[\begin{matrix}{a}_{11} & {b}_{12} \\ {a}_{21} & {b}_{22}\end{matrix}\right]} = \frac{{c}_{13} \cdot {b}_{22} - {c}_{23} \cdot {b}_{12}}{{a}_{11} \cdot {b}_{22} - {a}_{21} \cdot {b}_{12}}$

$x = \frac{\left[\begin{matrix}7 & - 4 \\ 12 & 1\end{matrix}\right]}{\left[\begin{matrix}2 & - 4 \\ - 3 & 1\end{matrix}\right]} = \frac{7 \cdot 1 - 12 \cdot - 4}{2 \cdot 1 - \left(- 3\right) \cdot \left(- 4\right)} = \frac{7 + 48}{- 10} = \frac{55}{-} 10$

$x = - 5.5$

$y = \frac{\left[\begin{matrix}{a}_{11} & {c}_{13} \\ {a}_{21} & {c}_{23}\end{matrix}\right]}{\left[\begin{matrix}{a}_{11} & {b}_{12} \\ {a}_{21} & {b}_{22}\end{matrix}\right]} = \frac{{a}_{11} \cdot {c}_{23} - {a}_{21} \cdot {c}_{13}}{{a}_{11} \cdot {b}_{22} - {a}_{21} \cdot {b}_{12}}$

$y = \frac{\left[\begin{matrix}2 & 7 \\ - 3 & 12\end{matrix}\right]}{\left[\begin{matrix}2 & - 4 \\ - 3 & 1\end{matrix}\right]} = \frac{2 \cdot 12 - \left(- 3\right) \cdot \left(7\right)}{2 \cdot 1 - \left(- 3\right) \cdot \left(- 4\right)} = \frac{24 + 21}{- 10} = \frac{45}{- 10}$

$y = - 4.5$

God bless...I hope the explanation is useful.