https://file.aiquickdraw.com/mathbot/file/1725340971101om7ea6f4.jpg

To solve the system of linear equations using Cramer's rule, we'll go through each step carefully. Here are the equations extracted from the image:

\[
\begin{aligned}
6x + y - 3z &= 5 \quad \text{(Equation 1)} \\
x + 3y - 2z &= 5 \quad \text{(Equation 2)} \\
2x + y + 4z &= 8 \quad \text{(Equation 3)}
\end{aligned}
\]

### Step 1: Write the system in matrix form
The system of equations can be written as:

\[
\mathbf{A} \cdot \mathbf{x} = \mathbf{B}
\]

Where:

\[
\mathbf{A} = \begin{pmatrix} 6 & 1 & -3 \\ 1 & 3 & -2 \\ 2 & 1 & 4 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 5 \\ 5 \\ 8 \end{pmatrix}
\]

### Step 2: Calculate the determinant of matrix \(\mathbf{A}\)

\[
\text{det}(\mathbf{A}) = \begin{vmatrix} 6 & 1 & -3 \\ 1 & 3 & -2 \\ 2 & 1 & 4 \end{vmatrix}
\]

This determinant can be expanded as:

\[
\text{det}(\mathbf{A}) = 6 \begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} - 1 \begin{vmatrix} 1 & -2 \\ 2 & 4 \end{vmatrix} + (-3) \begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix}
\]

Calculating each 2x2 determinant:

\[
\begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} = (3)(4) - (-2)(1) = 12 + 2 = 14
\]
\[
\begin{vmatrix} 1 & -2 \\ 2 & 4 \end{vmatrix} = (1)(4) - (-2)(2) = 4 + 4 = 8
\]
\[
\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} = (1)(1) - (3)(2) = 1 - 6 = -5
\]

Now, substitute back:

\[
\text{det}(\mathbf{A}) = 6(14) - 1(8) - 3(-5) = 84 - 8 + 15 = 91
\]

### Step 3: Calculate determinants of matrices \(\mathbf{A}_x\), \(\mathbf{A}_y\), and \(\mathbf{A}_z\)

#### Determinant of \(\mathbf{A}_x\)
Replace the first column of \(\mathbf{A}\) with \(\mathbf{B}\):

\[
\mathbf{A}_x = \begin{pmatrix} 5 & 1 & -3 \\ 5 & 3 & -2 \\ 8 & 1 & 4 \end{pmatrix}
\]

\[
\text{det}(\mathbf{A}_x) = 5 \begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} - 1 \begin{vmatrix} 5 & -2 \\ 8 & 4 \end{vmatrix} + (-3) \begin{vmatrix} 5 & 3 \\ 8 & 1 \end{vmatrix}
\]

Calculate the 2x2 determinants:

\[
\begin{vmatrix} 3 & -2 \\ 1 & 4 \end{vmatrix} = 14 \quad \text{(as calculated earlier)}
\]
\[
\begin{vmatrix} 5 & -2 \\ 8 & 4 \end{vmatrix} = (5)(4) - (-2)(8) = 20 + 16 = 36
\]
\[
\begin{vmatrix} 5 & 3 \\ 8 & 1 \end{vmatrix} = (5)(1) - (3)(8) = 5 - 24 = -19
\]

Substitute back:

\[
\text{det}(\mathbf{A}_x) = 5(14) - 1(36) - 3(-19) = 70 - 36 +

Learn how to solve a system of linear equations using Cramer's rule with detailed calculation steps. This problem involves matrix operations and determinants, suitable for advanced high school or college-level students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation