Question

Use a matrix to solve the system.

8.

2x + 6y = 38

(5x - y = 15

Answer 1

x=4 and y=5

Explanation:

The given system of equations are

`2x+6y=38`

`5x-y=15`

The matrix form is

`\begin{bmatrix}2&6\\ \:5&-1\end{bmatrix}\begin{bmatrix}x\\ \:y\end{bmatrix}=\begin{bmatrix}38\\ \:15\end{bmatrix}`

Let as assume

`A=\begin{bmatrix}2&6\\ \:5&-1\end{bmatrix}`

`X=\begin{bmatrix}x\\ \:y\end{bmatrix}`

`B=\begin{bmatrix}38\\ \:15\end{bmatrix}`

then,

`AX=B`

`X=A^(-1)B`

We know that,

`\begin{bmatrix}a\:&\:b\:\\ c\:&\:d\:\end{bmatrix}^(-1)=\frac{1}{\det \begin{bmatrix}a\:&\:b\:\\ c\:&\:d\:\end{bmatrix}}\begin{bmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{bmatrix}`

`A^(-1)=\frac{1}{\det \begin{bmatrix}2&6\\ 5&-1\end{bmatrix}}\begin{bmatrix}-1&-6\\ -5&2\end{bmatrix}`

`A^(-1)=(1)/(-32)\begin{bmatrix}-1&-6\\ -5&2\end{bmatrix}`

`X=(1)/(-32)\begin{bmatrix}-1&-6\\ -5&2\end{bmatrix}\begin{bmatrix}38\\ \:15\end{bmatrix}`

`X=(1)/(-32)\begin{bmatrix}\left(-1\right)\cdot \:38+\left(-6\right)\cdot \:15\\ \left(-5\right)\cdot \:38+2\cdot \:15\end{bmatrix}`

`X=(1)/(-32)\begin{bmatrix}-128\\ -160\end{bmatrix}`

`\begin{bmatrix}x\\ \:y\end{bmatrix}=\begin{bmatrix}4\\ 5\end{bmatrix}`

Therefore, the value of x is 4 and value of y is 5.