Question

1. find the solution to the linear system using determinants.

\

$$\begin{cases} x - 2y = -8 \\\\ 3x + 2y = 0 \\end{cases}$$

\bigcirc\\ (2, 3)
\bigcirc\\ (2, -3)
\bigcirc\\ (-2, 3)
\bigcirc\\ (-2, -3)

1. find the result of applying the given elementary row operations.

\

$$\begin{bmatrix} 1 & 3 & \\bigm| & 1 & 0 \\\\ 0 & -13 & \\bigm| & -5 & 1 \\end{bmatrix}$$

• \frac{1}{13} r_2 \to r_2 \to

Explanation:

Step1: Recall Cramer's Rule

For a system \(

$$\begin{cases}a_1x + b_1y = c_1\\a_2x + b_2y = c_2\end{cases}$$

\), \(x=\frac{D_x}{D}\), \(y = \frac{D_y}{D}\), where \(D=

$$\begin{vmatrix}a_1&b_1\\a_2&b_2\end{vmatrix}$$

\), \(D_x=

$$\begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix}$$

\), \(D_y=

$$\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}$$

\), and \(D
eq0\).

For the system \(

$$\begin{cases}x - 2y=-8\\3x + 2y = 0\end{cases}$$

\), we have \(a_1 = 1\), \(b_1=-2\), \(c_1=-8\), \(a_2 = 3\), \(b_2 = 2\), \(c_2 = 0\).

Step2: Calculate \(D\)

\(D=

$$\begin{vmatrix}1&-2\\3&2\end{vmatrix}$$

=1\times2-(-2)\times3=2 + 6 = 8\)

Step3: Calculate \(D_x\)

\(D_x=

$$\begin{vmatrix}-8&-2\\0&2\end{vmatrix}$$

=(-8)\times2-(-2)\times0=-16-0=-16\)

Step4: Calculate \(D_y\)

\(D_y=

$$\begin{vmatrix}1&-8\\3&0\end{vmatrix}$$

=1\times0-(-8)\times3=0 + 24 = 24\)

Step5: Find \(x\) and \(y\)

\(x=\frac{D_x}{D}=\frac{-16}{8}=-2\)

\(y=\frac{D_y}{D}=\frac{24}{8}=3\)

Answer:

\((-2, 3)\) (Option: \((-2,3)\))