We are given the following system of linear equations:

$2x - y + z = 5 \quad \text{(Equation 1)}$ $4x + 2y + 3z = 8 \quad \text{(Equation 2)}$ $3x - 4y - z = 3 \quad \text{(Equation 3)}$

We will solve this system using two methods: Matrices and Cramer's Rule.

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Method 1: Using Matrices

First, let's represent the system as a matrix equation: $A \cdot X = B$

Where:

X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \quad B = \begin{pmatrix} 5 \\ 8 \\ 3 \end{pmatrix}$$ We solve for $$X$$ by finding the inverse of matrix $$A$$ and using the relation: $$X = A^{-1} \cdot B$$ #### Step 1: Calculate the inverse of matrix $$A$$. First, we find the determinant of $$A$$ to ensure that the matrix is invertible. $$\text{det}(A) = \begin{vmatrix} 2 & -1 & 1 \\ 4 & 2 & 3 \\ 3 & -4 & -1 \end{vmatrix}$$ We'll calculate this determinant. #### Step 2: Apply the inverse matrix to the right-hand side (vector $$B$$). Let me calculate both steps.The determinant of matrix $$A$$ is $$\text{det}(A) = -15$$, which is non-zero, meaning that the matrix is invertible. By using the inverse of matrix $$A$$, we find the solution vector $$X$$: $$X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0.33 \\ -1.27 \\ 3.07 \end{pmatrix}$$ Thus, the solution is approximately: $$x = 0.33, \quad y = -1.27, \quad z = 3.07$$ --- ### **Method 2: Using Cramer's Rule** Cramer's Rule uses determinants to find each variable. Given the system $$A \cdot X = B$$, where $$A$$ is the coefficient matrix, we can find $$x$$, $$y$$, and $$z$$ as follows: - $$x = \frac{\text{det}(A_x)}{\text{det}(A)}$$ - $$y = \frac{\text{det}(A_y)}{\text{det}(A)}$$ - $$z = \frac{\text{det}(A_z)}{\text{det}(A)}$$ Where $$A_x$$, $$A_y$$, and $$A_z$$ are matrices formed by replacing the corresponding column of matrix $$A$$ with vector $$B$$. Let me now compute the determinants for $$A_x$$, $$A_y$$, and $$A_z$$.Using **Cramer's Rule**, we calculate the following determinants: - $$\text{det}(A_x) = -5$$ - $$\text{det}(A_y) = 19$$ - $$\text{det}(A_z) = -46$$ Thus, the values of $$x$$, $$y$$, and $$z$$ are: $$x = \frac{\text{det}(A_x)}{\text{det}(A)} = \frac{-5}{-15} = 0.33$$ $$y = \frac{\text{det}(A_y)}{\text{det}(A)} = \frac{19}{-15} = -1.27$$ $$z = \frac{\text{det}(A_z)}{\text{det}(A)} = \frac{-46}{-15} = 3.07$$ So, the solution from **Cramer's Rule** matches the solution from the matrix method: $$x = 0.33, \quad y = -1.27, \quad z = 3.07$$ --- Do you want more details or have any questions? Here are some related questions: 1. How do you calculate the determinant of a 3x3 matrix? 2. What are the properties of invertible matrices? 3. How does Cramer's Rule work for larger systems? 4. Can Cramer's Rule be used if the determinant of the matrix is zero? 5. How is matrix inversion used in solving systems of equations? **Tip:** Always check the determinant of a matrix before using methods like Cramer's Rule or inversion, as they require the matrix to be invertible.