Answer :
Let's solve the system of equations using the Gauss-Jordan elimination method step-by-step.
We start with the augmented matrix representation of the system:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 & -5 & -1 & 2 \\ 2 & -7 & 9 & \infty \end{array} \][/tex]
### Step 1: Make the element in the first row, first column equal to 1 (Already 1).
The first row already has a 1 as the first element, so we don't need to change anything:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 & -5 & -1 & 2 \\ 2 & -7 & 9 & \infty \end{array} \][/tex]
### Step 2: Make the elements below the leading 1 (in the first column) equal to 0.
We can make the elements below the 1 in the first column (the second and third rows) 0 by subtracting twice the first row from the second row, and subtracting twice the first row from the third row:
- For the second row: [tex]\( R2 \leftarrow R2 - 2 \times R1 \)[/tex]
- For the third row: [tex]\( R3 \leftarrow R3 - 2 \times R1 \)[/tex]
Let's perform these operations:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 - 2 \times 1 & -5 - 2 \times (-3) & -1 - 2 \times 2 & 2 - 2 \times (-1) \\ 2 - 2 \times 1 & -7 - 2 \times (-3) & 9 - 2 \times 2 & \infty - 2 \times (-1) \end{array} \][/tex]
Which simplifies to:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & \infty + 2 \end{array} \][/tex]
### Step 3: Make the element in the second row, second column equal to 1.
The second row already has a 1 in the second column, so we don't need to change anything:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & \infty + 2 \end{array} \][/tex]
### Step 4: Make the elements above and below the leading 1 in the second column equal to 0.
We can make the elements above and below the 1 in the second column (the first and third rows) 0 by adding 3 times the second row to the first row, and adding the second row to the third row:
- For the first row: [tex]\( R1 \leftarrow R1 + 3 \times R2 \)[/tex]
- For the third row: [tex]\( R3 \leftarrow R3 + R2 \)[/tex]
Let's perform these operations:
[tex]\[ \begin{array}{ccc|c} 1 + 3 \times 0 & -3 + 3 \times 1 & 2 + 3 \times (-5) & -1 + 3 \times 4 \\ 0 & 1 & -5 & 4 \\ 0 + 0 & -1 + 1 & 5 - 5 & (\infty + 2) + 4 \end{array} \][/tex]
Which simplifies to:
[tex]\[ \begin{array}{ccc|c} 1 & 0 & -13 & 11 \\ 0 & 1 & -5 & 4 \\ 0 & 0 & 0 & \infty + 6 \end{array} \][/tex]
### Step 5: Make the element in the third row, third column equal to 1.
Here, we encounter an issue: the third row is [tex]\(0 = \infty + 6\)[/tex], which is a contradiction and implies that the system of equations is inconsistent.
### Conclusion:
The system of equations has no solution due to the inconsistency introduced by the [tex]\(\infty\)[/tex] term in the third row. Thus, the equations cannot be solved.
We start with the augmented matrix representation of the system:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 & -5 & -1 & 2 \\ 2 & -7 & 9 & \infty \end{array} \][/tex]
### Step 1: Make the element in the first row, first column equal to 1 (Already 1).
The first row already has a 1 as the first element, so we don't need to change anything:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 & -5 & -1 & 2 \\ 2 & -7 & 9 & \infty \end{array} \][/tex]
### Step 2: Make the elements below the leading 1 (in the first column) equal to 0.
We can make the elements below the 1 in the first column (the second and third rows) 0 by subtracting twice the first row from the second row, and subtracting twice the first row from the third row:
- For the second row: [tex]\( R2 \leftarrow R2 - 2 \times R1 \)[/tex]
- For the third row: [tex]\( R3 \leftarrow R3 - 2 \times R1 \)[/tex]
Let's perform these operations:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 2 - 2 \times 1 & -5 - 2 \times (-3) & -1 - 2 \times 2 & 2 - 2 \times (-1) \\ 2 - 2 \times 1 & -7 - 2 \times (-3) & 9 - 2 \times 2 & \infty - 2 \times (-1) \end{array} \][/tex]
Which simplifies to:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & \infty + 2 \end{array} \][/tex]
### Step 3: Make the element in the second row, second column equal to 1.
The second row already has a 1 in the second column, so we don't need to change anything:
[tex]\[ \begin{array}{ccc|c} 1 & -3 & 2 & -1 \\ 0 & 1 & -5 & 4 \\ 0 & -1 & 5 & \infty + 2 \end{array} \][/tex]
### Step 4: Make the elements above and below the leading 1 in the second column equal to 0.
We can make the elements above and below the 1 in the second column (the first and third rows) 0 by adding 3 times the second row to the first row, and adding the second row to the third row:
- For the first row: [tex]\( R1 \leftarrow R1 + 3 \times R2 \)[/tex]
- For the third row: [tex]\( R3 \leftarrow R3 + R2 \)[/tex]
Let's perform these operations:
[tex]\[ \begin{array}{ccc|c} 1 + 3 \times 0 & -3 + 3 \times 1 & 2 + 3 \times (-5) & -1 + 3 \times 4 \\ 0 & 1 & -5 & 4 \\ 0 + 0 & -1 + 1 & 5 - 5 & (\infty + 2) + 4 \end{array} \][/tex]
Which simplifies to:
[tex]\[ \begin{array}{ccc|c} 1 & 0 & -13 & 11 \\ 0 & 1 & -5 & 4 \\ 0 & 0 & 0 & \infty + 6 \end{array} \][/tex]
### Step 5: Make the element in the third row, third column equal to 1.
Here, we encounter an issue: the third row is [tex]\(0 = \infty + 6\)[/tex], which is a contradiction and implies that the system of equations is inconsistent.
### Conclusion:
The system of equations has no solution due to the inconsistency introduced by the [tex]\(\infty\)[/tex] term in the third row. Thus, the equations cannot be solved.
