Question Solve the following system of equations using an inverse matrix. You must also indicate the inverse matrix, $$ A^{-1} $$, that was used to solve the system. You may optionally write the inverse matrix with a scalar coefficient. $$ \begin{array}{r}-x-6 y=8 \ -3 x-10 y=6\end{array} $$

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To solve the system of equations using an inverse matrix, we first need to express the system in matrix form. The given equations are:

1. $$-x - 6y = 8$$
2. $$-3x - 10y = 6$$

We can rewrite this system in the form $$AX = B$$, where:

$$
A = \begin{pmatrix} -1 & -6 \ -3 & -10 \end{pmatrix}, \quad X = \begin{pmatrix} x \ y \end{pmatrix}, \quad B = \begin{pmatrix} 8 \ 6 \end{pmatrix}
$$

Next, we need to find the inverse of matrix $$A$$, denoted as $$A^{-1}$$. The formula for the inverse of a 2x2 matrix

$$
\begin{pmatrix} a & b \ c & d \end{pmatrix}
$$

is given by

$$
\frac{1}{ad - bc} \begin{pmatrix} d & -b \ -c & a \end{pmatrix}
$$

Now, let's calculate the determinant $$ad - bc$$ for our matrix $$A$$:

- $$a = -1$$, $$b = -6$$, $$c = -3$$, $$d = -10$$
- $$ad = (-1)(-10) = 10$$
- $$bc = (-6)(-3) = 18$$
- Therefore, $$ad - bc = 10 - 18 = -8$$

Now we can find the inverse matrix $$A^{-1}$$:

$$
A^{-1} = \frac{1}{-8} \begin{pmatrix} -10 & 6 \ 3 & -1 \end{pmatrix} = \begin{pmatrix} \frac{10}{8} & -\frac{6}{8} \ -\frac{3}{8} & \frac{1}{8} \end{pmatrix} = \begin{pmatrix} \frac{5}{4} & -\frac{3}{4} \ -\frac{3}{8} & \frac{1}{8} \end{pmatrix}
$$

Now, we can use the inverse matrix to solve for $$X$$:

$$
X = A^{-1}B
$$

Let's calculate $$X$$:

$$
X = \begin{pmatrix} \frac{5}{4} & -\frac{3}{4} \ -\frac{3}{8} & \frac{1}{8} \end{pmatrix} \begin{pmatrix} 8 \ 6 \end{pmatrix}
$$

Now, we will perform the matrix multiplication:

1. For $$x$$:
   $$
   x = \frac{5}{4} \cdot 8 + \left(-\frac{3}{4}\right) \cdot 6 = 10 - 4.5 = 5.5
   $$

2. For $$y$$:
   $$
   y = -\frac{3}{8} \cdot 8 + \frac{1}{8} \cdot 6 = -3 + 0.75 = -2.25
   $$

Thus, the solution to the system of equations is:

$$
x = 5.5, \quad y = -2.25
$$

In summary, the inverse matrix used is:

$$
A^{-1} = \begin{pmatrix} \frac{5}{4} & -\frac{3}{4} \ -\frac{3}{8} & \frac{1}{8} \end{pmatrix}
$$

And the solution to the system is:

$$
\begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} 5.5 \ -2.25 \end{pmatrix}
$$
To solve the system of equations using an inverse matrix, we first write the system in matrix form: $$ A = \begin{pmatrix} -1 & -6 \ -3 & -10 \end{pmatrix}, \quad X = \begin{pmatrix} x \ y \end{pmatrix}, \quad B = \begin{pmatrix} 8 \ 6 \end{pmatrix} $$ Next, we find the inverse of matrix $$A$$: $$ A^{-1} = \begin{pmatrix} \frac{5}{4} & -\frac{3}{4} \ -\frac{3}{8} & \frac{1}{8} \end{pmatrix} $$ Using $$A^{-1}$$, we solve for $$X$$: $$ X = A^{-1}B = \begin{pmatrix} 5.5 \ -2.25 \end{pmatrix} $$ So, the solution is $$x = 5.5$$ and $$y = -2.25$$.

Question Solve the following system of equations using an inverse matrix. You must also indicate the inverse matrix, \( A^{-1} \), that was used to solve the system. You may optionally write the inverse matrix with a scalar coefficient. \[ \begin{array}{r}-x-6 y=8 \\ -3 x-10 y=6\end{array} \]

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