Certainly! To solve the linear system using determinants, we will utilize Cramer's rule. Let's consider the given system of equations:
[tex]\[
\begin{array}{l}
2 x + 3 y = 6 \\
-8 x - 3 y = 12
\end{array}
\][/tex]
We can represent this system in matrix form as [tex]\(AX = B\)[/tex].
Here, the coefficient matrix [tex]\(A\)[/tex] is:
[tex]\[
A = \begin{pmatrix}
2 & 3 \\
-8 & -3
\end{pmatrix}
\][/tex]
The variables matrix [tex]\(X\)[/tex] is:
[tex]\[
X = \begin{pmatrix}
x \\
y
\end{pmatrix}
\][/tex]
The constants matrix [tex]\(B\)[/tex] is:
[tex]\[
B = \begin{pmatrix}
6 \\
12
\end{pmatrix}
\][/tex]
### Step 1: Calculate the Determinant of Matrix [tex]\(A\)[/tex], denoted as [tex]\(|A|\)[/tex]:
[tex]\[
|A| = \begin{vmatrix}
2 & 3 \\
-8 & -3
\end{vmatrix} = (2 \times -3) - (3 \times -8) = -6 + 24 = 18
\][/tex]
Thus, [tex]\(|A| = 18\)[/tex].
### Step 2: Calculate the Determinant of Matrix [tex]\(A_x\)[/tex], denoted as [tex]\(|A_x|\)[/tex]:
Matrix [tex]\(A_x\)[/tex] is formed by replacing the first column of matrix [tex]\(A\)[/tex] with the constants from matrix [tex]\(B\)[/tex]:
[tex]\[
A_x = \begin{pmatrix}
6 & 3 \\
12 & -3
\end{pmatrix}
\][/tex]
[tex]\[
|A_x| = \begin{vmatrix}
6 & 3 \\
12 & -3
\end{vmatrix} = (6 \times -3) - (3 \times 12) = -18 - 36 = -54
\][/tex]
Therefore, [tex]\(|A_x| = -54\)[/tex].
### Step 3: Calculate the Determinant of Matrix [tex]\(A_y\)[/tex], denoted as [tex]\(|A_y|\)[/tex]:
Matrix [tex]\(A_y\)[/tex] is formed by replacing the second column of matrix [tex]\(A\)[/tex] with the constants from matrix [tex]\(B\)[/tex]:
[tex]\[
A_y = \begin{pmatrix}
2 & 6 \\
-8 & 12
\end{pmatrix}
\][/tex]
[tex]\[
|A_y| = \begin{vmatrix}
2 & 6 \\
-8 & 12
\end{vmatrix} = (2 \times 12) - (6 \times -8) = 24 + 48 = 72
\][/tex]
So, [tex]\(|A_y| = 72\)[/tex].
### Summary of Determinants:
[tex]\[
|A| = 18
\][/tex]
[tex]\[
|A_x| = -54
\][/tex]
[tex]\[
|A_y| = 72
\][/tex]
