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To solve the system of equations using matrices and row operations, we start by writing the augmented matrix of the system of equations:
[tex]\[ \left[\begin{array}{ccc|c} 2 & -4 & -6 & 1 \\ 6 & 13 & 32 & 0 \\ -2 & 3 & 4 & 1 \end{array}\right] \][/tex]
### Step 1: Normalize the first row
First, we need to make the leading coefficient of the first row a 1. This is already achieved by dividing the first row by 2:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -2 & -3 & \frac{1}{2} \\ 6 & 13 & 32 & 0 \\ -2 & 3 & 4 & 1 \end{array}\right] \][/tex]
### Step 2: Eliminate the x-term from the second and third rows
Next, we use row operations to eliminate the [tex]\(x\)[/tex]-term from the second and third rows.
For the second row [tex]\((R2)\)[/tex]:
[tex]\[ R2 = R2 - 6 \cdot R1 \][/tex]
[tex]\[ \left[\begin{array}{ccc|c} 1 & -2 & -3 & \frac{1}{2} \\ 0 & 25 & 50 & -3 \\ -2 & 3 & 4 & 1 \end{array}\right] \][/tex]
For the third row [tex]\((R3)\)[/tex]:
[tex]\[ R3 = R3 + 2 \cdot R1 \][/tex]
[tex]\[ \left[\begin{array}{ccc|c} 1 & -2 & -3 & \frac{1}{2} \\ 0 & 25 & 50 & -3 \\ 0 & -1 & -2 & 2 \end{array}\right] \][/tex]
### Step 3: Normalize the second row
We now normalize the second row by multiplying by [tex]\(\frac{1}{25}\)[/tex]:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -2 & -3 & \frac{1}{2} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & -1 & -2 & 2 \end{array}\right] \][/tex]
### Step 4: Eliminate the y-term from the first and third rows
Now, use the second row to eliminate the [tex]\(y\)[/tex]-term from the first and third rows.
For the first row [tex]\((R1)\)[/tex]:
[tex]\[ R1 = R1 + 2 \cdot R2 \][/tex]
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 1 & \frac{1}{2} - \frac{6}{25} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & -1 & -2 & 2 \end{array}\right] \][/tex]
[tex]\(\frac{1}{2} - \frac{6}{25} = \frac{25}{50} - \frac{6}{25} = \frac{25}{50} - \frac{12}{50} = \frac{13}{50}\)[/tex]
Thus:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 1 & \frac{13}{50} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & -1 & -2 & 2 \end{array}\right] \][/tex]
For the third row [tex]\((R3)\)[/tex]:
[tex]\[ R3 = R3 + R2 \][/tex]
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 1 & \frac{13}{50} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & 0 & 0 & 2 - \frac{3}{25} \end{array}\right] \][/tex]
### Step 5: Check for consistency
To find out if the system is consistent, examine the third row.
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 1 & \frac{13}{50} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & 0 & 0 & 2 - \frac{3}{25} \end{array}\right] \][/tex]
Note:
[tex]\[ 2 - \frac{3}{25} = \frac{50}{25} - \frac{3}{25} = \frac{47}{25} \neq 0 \][/tex]
This results in a contradiction because [tex]\(0 = \frac{47}{25}\)[/tex]. Thus, the system has no solution.
Therefore, the system is inconsistent.
[tex]\[ \left[\begin{array}{ccc|c} 2 & -4 & -6 & 1 \\ 6 & 13 & 32 & 0 \\ -2 & 3 & 4 & 1 \end{array}\right] \][/tex]
### Step 1: Normalize the first row
First, we need to make the leading coefficient of the first row a 1. This is already achieved by dividing the first row by 2:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -2 & -3 & \frac{1}{2} \\ 6 & 13 & 32 & 0 \\ -2 & 3 & 4 & 1 \end{array}\right] \][/tex]
### Step 2: Eliminate the x-term from the second and third rows
Next, we use row operations to eliminate the [tex]\(x\)[/tex]-term from the second and third rows.
For the second row [tex]\((R2)\)[/tex]:
[tex]\[ R2 = R2 - 6 \cdot R1 \][/tex]
[tex]\[ \left[\begin{array}{ccc|c} 1 & -2 & -3 & \frac{1}{2} \\ 0 & 25 & 50 & -3 \\ -2 & 3 & 4 & 1 \end{array}\right] \][/tex]
For the third row [tex]\((R3)\)[/tex]:
[tex]\[ R3 = R3 + 2 \cdot R1 \][/tex]
[tex]\[ \left[\begin{array}{ccc|c} 1 & -2 & -3 & \frac{1}{2} \\ 0 & 25 & 50 & -3 \\ 0 & -1 & -2 & 2 \end{array}\right] \][/tex]
### Step 3: Normalize the second row
We now normalize the second row by multiplying by [tex]\(\frac{1}{25}\)[/tex]:
[tex]\[ \left[\begin{array}{ccc|c} 1 & -2 & -3 & \frac{1}{2} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & -1 & -2 & 2 \end{array}\right] \][/tex]
### Step 4: Eliminate the y-term from the first and third rows
Now, use the second row to eliminate the [tex]\(y\)[/tex]-term from the first and third rows.
For the first row [tex]\((R1)\)[/tex]:
[tex]\[ R1 = R1 + 2 \cdot R2 \][/tex]
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 1 & \frac{1}{2} - \frac{6}{25} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & -1 & -2 & 2 \end{array}\right] \][/tex]
[tex]\(\frac{1}{2} - \frac{6}{25} = \frac{25}{50} - \frac{6}{25} = \frac{25}{50} - \frac{12}{50} = \frac{13}{50}\)[/tex]
Thus:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 1 & \frac{13}{50} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & -1 & -2 & 2 \end{array}\right] \][/tex]
For the third row [tex]\((R3)\)[/tex]:
[tex]\[ R3 = R3 + R2 \][/tex]
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 1 & \frac{13}{50} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & 0 & 0 & 2 - \frac{3}{25} \end{array}\right] \][/tex]
### Step 5: Check for consistency
To find out if the system is consistent, examine the third row.
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 1 & \frac{13}{50} \\ 0 & 1 & 2 & -\frac{3}{25} \\ 0 & 0 & 0 & 2 - \frac{3}{25} \end{array}\right] \][/tex]
Note:
[tex]\[ 2 - \frac{3}{25} = \frac{50}{25} - \frac{3}{25} = \frac{47}{25} \neq 0 \][/tex]
This results in a contradiction because [tex]\(0 = \frac{47}{25}\)[/tex]. Thus, the system has no solution.
Therefore, the system is inconsistent.
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