Math  /  Algebra

QuestionConsider the following. {x15x2+2x3=93x1+x2x3=82x2+5x3=7\left\{\begin{array}{rr} x_{1}-5 x_{2}+2 x_{3}= & -9 \\ -3 x_{1}+x_{2}-x_{3}= & 8 \\ -2 x_{2}+5 x_{3}= & -7 \end{array}\right. (a) Write the system of linear equations as a matrix equation, AX=BA X=B. (b) Use Gauss-Jordan elimination on [AB][A \vdots B] to solve for the matrix XX. X=[x1x2x3]=[]X=\left[\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right]=\left[\begin{array}{c} \square \\ \square \\ \square \end{array}\right] \Rightarrow

Studdy Solution

STEP 1

1. We are given a system of linear equations and need to express it in matrix form.
2. We will use Gauss-Jordan elimination to solve the system.
3. The solution will be expressed in terms of the matrix X X .

STEP 2

1. Write the system of linear equations as a matrix equation AX=B A X = B .
2. Form the augmented matrix [AB][A \vdots B].
3. Use Gauss-Jordan elimination to solve for X X .

STEP 3

Write the system of equations in matrix form AX=B A X = B .
The system of equations is:
\begin{align*} x_{1} - 5x_{2} + 2x_{3} &= -9 \\ -3x_{1} + x_{2} - x_{3} &= 8 \\ -2x_{2} + 5x_{3} &= -7 \end{align*}
This can be written as:
A=[152311025],X=[x1x2x3],B=[987]A = \begin{bmatrix} 1 & -5 & 2 \\ -3 & 1 & -1 \\ 0 & -2 & 5 \end{bmatrix}, \quad X = \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix}, \quad B = \begin{bmatrix} -9 \\ 8 \\ -7 \end{bmatrix}
Thus, the matrix equation is:
AX=BA X = B

STEP 4

Form the augmented matrix [AB][A \vdots B].
[AB]=[152931180257][A \vdots B] = \begin{bmatrix} 1 & -5 & 2 & \vdots & -9 \\ -3 & 1 & -1 & \vdots & 8 \\ 0 & -2 & 5 & \vdots & -7 \end{bmatrix}

STEP 5

Use Gauss-Jordan elimination to solve for X X .
First, we will perform row operations to transform the augmented matrix into reduced row-echelon form.
1. Start with the first row as the pivot row.
2. Eliminate the x1 x_1 term from the second row by adding 3 times the first row to the second row:
R2=R2+3R1R_2 = R_2 + 3R_1
[15290145190257]\begin{bmatrix} 1 & -5 & 2 & \vdots & -9 \\ 0 & -14 & 5 & \vdots & -19 \\ 0 & -2 & 5 & \vdots & -7 \end{bmatrix}
3. Eliminate the x2 x_2 term from the third row by adding 17\frac{1}{7} times the second row to the third row:
R3=R3+17R2R_3 = R_3 + \frac{1}{7}R_2
[152901451900257707]\begin{bmatrix} 1 & -5 & 2 & \vdots & -9 \\ 0 & -14 & 5 & \vdots & -19 \\ 0 & 0 & \frac{25}{7} & \vdots & -\frac{70}{7} \end{bmatrix}
4. Simplify the third row:
R3=725R3R_3 = \frac{7}{25}R_3
[15290145190012]\begin{bmatrix} 1 & -5 & 2 & \vdots & -9 \\ 0 & -14 & 5 & \vdots & -19 \\ 0 & 0 & 1 & \vdots & -2 \end{bmatrix}
5. Back substitute to eliminate x3 x_3 from the first and second rows:
R1=R12R3R_1 = R_1 - 2R_3 R2=R25R3R_2 = R_2 - 5R_3
[1505014090012]\begin{bmatrix} 1 & -5 & 0 & \vdots & -5 \\ 0 & -14 & 0 & \vdots & -9 \\ 0 & 0 & 1 & \vdots & -2 \end{bmatrix}
6. Simplify the second row:
R2=114R2R_2 = -\frac{1}{14}R_2
[15050109140012]\begin{bmatrix} 1 & -5 & 0 & \vdots & -5 \\ 0 & 1 & 0 & \vdots & \frac{9}{14} \\ 0 & 0 & 1 & \vdots & -2 \end{bmatrix}
7. Back substitute to eliminate x2 x_2 from the first row:
R1=R1+5R2R_1 = R_1 + 5R_2
[1005140109140012]\begin{bmatrix} 1 & 0 & 0 & \vdots & \frac{5}{14} \\ 0 & 1 & 0 & \vdots & \frac{9}{14} \\ 0 & 0 & 1 & \vdots & -2 \end{bmatrix}
The solution is:
X=[5149142]X = \begin{bmatrix} \frac{5}{14} \\ \frac{9}{14} \\ -2 \end{bmatrix}

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