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Which determinants can be used to solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in the system of linear equations below?

[tex]\[
\begin{array}{l}
-3x + 2y = -9 \\
4x - 15y = -25
\end{array}
\][/tex]

A. [tex]\[
|A|=\left|\begin{array}{cc}
-3 & 2 \\
4 & -15
\end{array}\right|, \left|A_x\right|=\left|\begin{array}{cc}
-9 & 2 \\
-25 & -15
\end{array}\right|, \left|A_y\right|=\left|\begin{array}{cc}
-3 & -9 \\
4 & -25
\end{array}\right|
\][/tex]

B. [tex]\[
|A|=\left|\begin{array}{cc}
-3 & 2 \\
4 & -15
\end{array}\right|, \left|A_x\right|=\left|\begin{array}{cc}
-3 & -9 \\
4 & -25
\end{array}\right|, \left|A_y\right|=\left|\begin{array}{cc}
-9 & 2 \\
-25 & -15
\end{array}\right|
\][/tex]

C. [tex]\[
|A|=\left|\begin{array}{c}
-9 \\
-25
\end{array}\right|, \left|A_x\right|=\left|\begin{array}{cc}
-9 & 2 \\
-25 & -15
\end{array}\right|, \left|A_y\right|=\left|\begin{array}{cc}
-3 & -9 \\
4 & -25
\end{array}\right|
\][/tex]

D. [tex]\[
|A| = -9, |A| = \left|\begin{array}{ll}
-3 & -9
\end{array}\right|, |A| = \left|\begin{array}{ll}
-9 & 2
\end{array}\right|
\][/tex]

Sagot :

GinnyAnsweridn
To solve the given system of linear equations using determinants, we can apply Cramer's Rule. Here's the step-by-step process:

Given the system of linear equations:
[tex]\[ \begin{array}{l} -3x + 2y = -9 \\ 4x - 15y = -25 \end{array} \][/tex]

We can rewrite this system in matrix form [tex]\(AX = B\)[/tex], where:

[tex]\[ A = \begin{pmatrix} -3 & 2 \\ 4 & -15 \end{pmatrix}, \quad X = \begin{pmatrix} x \\ y \end{pmatrix}, \quad B = \begin{pmatrix} -9 \\ -25 \end{pmatrix} \][/tex]

According to Cramer's Rule, the solution for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] can be found using the determinants of matrices derived from [tex]\(A\)[/tex] by replacing the respective columns with the column matrix [tex]\(B\)[/tex]. The solution is given as:

[tex]\[ x = \frac{\text{det}(A_x)}{\text{det}(A)}, \quad y = \frac{\text{det}(A_y)}{\text{det}(A)} \][/tex]

Where:

- [tex]\(\text{det}(A)\)[/tex] is the determinant of the matrix [tex]\(A\)[/tex].
- [tex]\(\text{det}(A_x)\)[/tex] is the determinant of the matrix formed by replacing the first column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].
- [tex]\(\text{det}(A_y)\)[/tex] is the determinant of the matrix formed by replacing the second column of [tex]\(A\)[/tex] with [tex]\(B\)[/tex].

Let's identify each determinant:

1. The determinant of [tex]\(A\)[/tex]:
[tex]\[ A = \begin{pmatrix} -3 & 2 \\ 4 & -15 \end{pmatrix} \][/tex]

2. The determinant of [tex]\(A_x\)[/tex]:
[tex]\[ A_x = \begin{pmatrix} -9 & 2 \\ -25 & -15 \end{pmatrix} \][/tex]

3. The determinant of [tex]\(A_y\)[/tex]:
[tex]\[ A_y = \begin{pmatrix} -3 & -9 \\ 4 & -25 \end{pmatrix} \][/tex]

Given the determinants:
[tex]\[ \text{det}(A) = 37.000000000000014, \quad \text{det}(A_x) = 184.99999999999991, \quad \text{det}(A_y) = 110.99999999999997 \][/tex]

These determinants can be used to solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] using Cramer's Rule:
[tex]\[ x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{184.99999999999991}{37.000000000000014} \][/tex]

[tex]\[ y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{110.99999999999997}{37.000000000000014} \][/tex]

Thus, the determinants used to solve for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in the system of linear equations are:

- [tex]\(\text{det}(A) = 37.000000000000014\)[/tex]
- [tex]\(\text{det}(A_x) = 184.99999999999991\)[/tex]
- [tex]\(\text{det}(A_y) = 110.99999999999997\)[/tex]
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