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CONVERT A MATRIX FROM QUESTION 5 TO UPPER TRIANGULAR MATRIX

To convert the matrix from Question 5 into an upper triangular matrix, we first extract the linear system:

\[
\begin{aligned}
    3x + 2y - z &= 1, \\
    x + 2y + 2z &= 0, \\
    2x + y + 3z &= -1.
\end{aligned}
\]

This can be written in matrix form as \( AX = B \), where:

\[
A = \begin{bmatrix} 3 & 2 & -1 \\ 1 & 2 & 2 \\ 2 & 1 & 3 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}.
\]

### Steps to Convert to Upper Triangular Matrix

1. **Initial Matrix:** Form the augmented matrix \([A|B]\):

\[
\left[ \begin{array}{ccc|c}
3 & 2 & -1 & 1 \\
1 & 2 & 2 & 0 \\
2 & 1 & 3 & -1
\end{array} \right].
\]

2. **Row Operations:** We will use row reduction (Gaussian elimination) to convert this matrix to upper triangular form.

   - **Step 1:** Make the element at the first row, first column (pivot) equal to 1. Swap \( R_1 \) with \( R_2 \):

\[
\left[ \begin{array}{ccc|c}
1 & 2 & 2 & 0 \\
3 & 2 & -1 & 1 \\
2 & 1 & 3 & -1
\end{array} \right].
\]

   - **Step 2:** Eliminate the elements below the pivot in the first column:
     - For \( R_2 \): \( R_2 = R_2 - 3R_1 \):

\[
\left[ \begin{array}{ccc|c}
1 & 2 & 2 & 0 \\
0 & -4 & -7 & 1 \\
2 & 1 & 3 & -1
\end{array} \right].
\]

     - For \( R_3 \): \( R_3 = R_3 - 2R_1 \):

\[
\left[ \begin{array}{ccc|c}
1 & 2 & 2 & 0 \\
0 & -4 & -7 & 1 \\
0 & -3 & -1 & -1
\end{array} \right].
\]

   - **Step 3:** Move to the second pivot (second row, second column) and make it 1. Multiply \( R_2 \) by \( -\frac{1}{4} \):

\[
\left[ \begin{array}{ccc|c}
1 & 2 & 2 & 0 \\
0 & 1 & \frac{7}{4} & -\frac{1}{4} \\
0 & -3 & -1 & -1
\end{array} \right].
\]

   - **Step 4:** Eliminate the element below the pivot in the second column:
     - For \( R_3 \): \( R_3 = R_3 + 3R_2 \):

\[
\left[ \begin{array}{ccc|c}
1 & 2 & 2 & 0 \\
0 & 1 & \frac{7}{4} & -\frac{1}{4} \\
0 & 0 & \frac{19}{4} & -\frac{7}{4}
\end{array} \right].
\]

3. **Final Upper Triangular Matrix:** The augmented matrix is now in upper triangular form:

\[
\left[ \begin{array}{ccc|c}
1 & 2 & 2 & 0 \\
0 & 1 & \frac{7}{4} & -\frac{1}{4} \\
0 & 0 & \frac{19}{4} & -\frac{7}{4}
\end{array} \right].
\]

### Summary

The matrix has been successfully converted into an upper triangular form using Gaussian elimination.

Would you like a step-by-step explanation of how to solve this system using back-substitution or any further details?

### 5 Related Questions
1. What is the next step after obtaining the upper triangular form of a matrix?
2. How can you find the determinant of a matrix using its triangular form?
3. What is the difference between row echelon form and reduced row echelon form?
4. How do row operations affect the solutions of a linear system?
5. Can Gaussian elimination be used on non-square matrices?

### Tip
Always start Gaussian elimination by choosing the row with the largest absolute value in the pivot column to minimize numerical errors during computations.

Learn how to convert a linear system of equations into an upper triangular matrix using Gaussian elimination. This guide provides a detailed, step-by-step explanation, suitable for first-year undergraduate students studying matrix algebra.

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