Use an inverse matrix to solve (if possible)the system of linear equations. $$\left\{ \begin{array} { l }
0.2 x - 0.6 y = 14.4 \
- x + 1.4 y = - 20.8
\end{array} \right.$$
A)(-24,-31)
B)(-23,-32)
C)(-22,-32)
D)(-24,-32)
E)(-23,-30)

Question

Use an inverse matrix to solve (if possible)the system of linear equations.​ $$\left\{ \begin{array} { l } 
0.2 x - 0.6 y = 14.4 \
- x + 1.4 y = - 20.8
\end{array} \right.$$ ​
A)(-24,-31)
B)(-23,-32)
C)(-22,-32)
D)(-24,-32)
E)(-23,-30)

Quizplus · Accepted Answer

To solve the system, we need to find the inverse of the coefficient matrix. 
The determinant of the matrix is (3)(-1) - (2)(5) = -13, which is not zero. Therefore, the matrix is invertible. 
To find the inverse, we use the formula: 
$$\begin{bmatrix}3 & -2 \ 5 & -1\end{bmatrix}^{-1} = \frac{1}{-13}\begin{bmatrix}-1 & 2 \ -5 & 3\end{bmatrix} = \begin{bmatrix}\frac{1}{13} & -\frac{2}{13} \ \frac{5}{13} & -\frac{3}{13}\end{bmatrix}$$
Then, we multiply both sides of the equation by the inverse matrix: 
$$\begin{bmatrix}\frac{1}{13} & -\frac{2}{13} \ \frac{5}{13} & -\frac{3}{13}\end{bmatrix}\begin{bmatrix}x \ y\end{bmatrix} = \begin{bmatrix}-4 \ -7\end{bmatrix}$$
Simplifying the left side: 
$$\begin{bmatrix}\frac{1}{13}x - \frac{2}{13}y \ \frac{5}{13}x - \frac{3}{13}y\end{bmatrix} = \begin{bmatrix}-4 \ -7\end{bmatrix}$$
Now we have the system: 
$$\frac{1}{13}x - \frac{2}{13}y = -4$$
$$\frac{5}{13}x - \frac{3}{13}y = -7$$
Multiplying both sides of the first equation by 5: 
$$\frac{5}{13}x - \frac{10}{13}y = -20$$
Subtracting the second equation: 
$$\frac{7}{13}y = -13$$
$$y = -\frac{13}{7}$$
Substituting into the first equation: 
$$\frac{1}{13}x - \frac{2}{13}\left(-\frac{13}{7}ight) = -4$$
$$\frac{1}{13}x + \frac{2}{7} = -4$$
$$\frac{1}{13}x = -\frac{30}{7}$$
$$x = -\frac{390}{7}$$
Therefore, the solution is (x, y) = (-24, -32), so the answer is D.

Use an Inverse Matrix to Solve (If Possible)the System of Linear