Answer :
Certainly! Let's solve the system of linear equations using the Gauss-Jordan elimination method step-by-step.
The system of equations is:
[tex]\[ \begin{align*} 1. & \quad x - 3y + 2z = -1 \\ 2. & \quad 2x - 5y - z = 2 \\ 3. & \quad 2x - 7y + 9z = -6 \end{align*} \][/tex]
First, let's represent this system as an augmented matrix:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 & -5 & -1 & 2 \\ 2 & -7 & 9 & -6 \end{array}\right] \][/tex]
### Step 1: Make the leading coefficient of the first row 1
The first row is already in this form, so we can proceed.
### Step 2: Eliminate [tex]\( x \)[/tex] from the second and third rows
To eliminate the x-term from the second row, subtract 2 times the first row from the second row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 & -5 & -1 & 2 \\ 2 & -7 & 9 & -6 \end{array}\right] - 2 \times \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 2 & -7 & 9 & -6 \end{array}\right] \][/tex]
To eliminate the x-term from the third row, subtract 2 times the first row from the third row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 2 & -7 & 9 & -6 \end{array}\right] - 2 \times \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & -4 \end{array}\right] \][/tex]
### Step 3: Make the second leading coefficient 1
To make the second leading coefficient (1 in the second row) 1, we can proceed as it is already 1.
### Step 4: Eliminate the [tex]\( y \)[/tex] terms from the first and third rows
To eliminate the [tex]\( y \)[/tex]-term from the first row, add 3 times the second row to the first row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & -4 \end{array}\right] + 3 \times \left[\begin{array}{ccc|c} 0 & 1 & -5 & 4 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -13 & 11 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & -4 \end{array}\right] \][/tex]
To eliminate the [tex]\( y \)[/tex]-term from the third row, add the second row to the third row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -13 & 11 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & -4 \end{array}\right] + \left[\begin{array}{ccc|c} 0 & 1 & -5 & 4 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -13 & 11 \\ 0 & 1 & -5 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right] \][/tex]
### Step 5: Make the third leading coefficient 1
Now the third row is all zeros, which means it doesn't affect our variables anymore.
### Step 6: Back-substitution
Now, we back substitute to solve for the variables. From the second row, we solve for [tex]\( y \)[/tex]:
[tex]\[ y = 4z + 4 \implies y - 5z = 4 \][/tex]
From the first row, we solve for [tex]\( z \)[/tex]:
[tex]\[ x - 13z = 11 \implies x = 13z + 11 \][/tex]
Putting together the results, we have:
- [tex]\( x = 13z + 11 \)[/tex]
- [tex]\( y = 4z + 4 \)[/tex]
- [tex]\( z \)[/tex] remains as a free parameter.
Thus, the solution to the system in terms of [tex]\( z \)[/tex] is:
[tex]\[ (x, y, z) = (13z + 11, 4z + 4, z) \][/tex]
The system of equations is:
[tex]\[ \begin{align*} 1. & \quad x - 3y + 2z = -1 \\ 2. & \quad 2x - 5y - z = 2 \\ 3. & \quad 2x - 7y + 9z = -6 \end{align*} \][/tex]
First, let's represent this system as an augmented matrix:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 & -5 & -1 & 2 \\ 2 & -7 & 9 & -6 \end{array}\right] \][/tex]
### Step 1: Make the leading coefficient of the first row 1
The first row is already in this form, so we can proceed.
### Step 2: Eliminate [tex]\( x \)[/tex] from the second and third rows
To eliminate the x-term from the second row, subtract 2 times the first row from the second row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 & -5 & -1 & 2 \\ 2 & -7 & 9 & -6 \end{array}\right] - 2 \times \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 2 & -7 & 9 & -6 \end{array}\right] \][/tex]
To eliminate the x-term from the third row, subtract 2 times the first row from the third row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 2 & -7 & 9 & -6 \end{array}\right] - 2 \times \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & -4 \end{array}\right] \][/tex]
### Step 3: Make the second leading coefficient 1
To make the second leading coefficient (1 in the second row) 1, we can proceed as it is already 1.
### Step 4: Eliminate the [tex]\( y \)[/tex] terms from the first and third rows
To eliminate the [tex]\( y \)[/tex]-term from the first row, add 3 times the second row to the first row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & -4 \end{array}\right] + 3 \times \left[\begin{array}{ccc|c} 0 & 1 & -5 & 4 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -13 & 11 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & -4 \end{array}\right] \][/tex]
To eliminate the [tex]\( y \)[/tex]-term from the third row, add the second row to the third row:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & -13 & 11 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & -4 \end{array}\right] + \left[\begin{array}{ccc|c} 0 & 1 & -5 & 4 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -13 & 11 \\ 0 & 1 & -5 & 4 \\ 0 & 0 & 0 & 0 \end{array}\right] \][/tex]
### Step 5: Make the third leading coefficient 1
Now the third row is all zeros, which means it doesn't affect our variables anymore.
### Step 6: Back-substitution
Now, we back substitute to solve for the variables. From the second row, we solve for [tex]\( y \)[/tex]:
[tex]\[ y = 4z + 4 \implies y - 5z = 4 \][/tex]
From the first row, we solve for [tex]\( z \)[/tex]:
[tex]\[ x - 13z = 11 \implies x = 13z + 11 \][/tex]
Putting together the results, we have:
- [tex]\( x = 13z + 11 \)[/tex]
- [tex]\( y = 4z + 4 \)[/tex]
- [tex]\( z \)[/tex] remains as a free parameter.
Thus, the solution to the system in terms of [tex]\( z \)[/tex] is:
[tex]\[ (x, y, z) = (13z + 11, 4z + 4, z) \][/tex]
