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To solve the given system of equations using the Gauss elimination method, let's rewrite the system in a more conventional form:
1. [tex]\(3x + 2y - z = 1\)[/tex]
2. [tex]\(-5x + 5y + 2z = 8\)[/tex]
3. [tex]\(2x - y + z = 2\)[/tex]
We can represent this system as an augmented matrix:
[tex]\[ \left[\begin{array}{ccc|c} 3 & 2 & -1 & 1 \\ -5 & 5 & 2 & 8 \\ 2 & -1 & 1 & 2 \end{array}\right] \][/tex]
### Step 1: Form the Augmented Matrix
[tex]\[ \left[\begin{array}{ccc|c} 3 & 2 & -1 & 1 \\ -5 & 5 & 2 & 8 \\ 2 & -1 & 1 & 2 \end{array}\right] \][/tex]
### Step 2: Make the leading coefficient of the first row 1
Divide the first row by 3:
[tex]\[ \left[\begin{array}{ccc|c} 1 & \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ -5 & 5 & 2 & 8 \\ 2 & -1 & 1 & 2 \end{array}\right] \][/tex]
### Step 3: Eliminate the first variable from the second and third rows
For Row 2: Add [tex]\(5 \times\)[/tex] Row 1 to Row 2:
[tex]\[ \begin{array}{rl} -5 & + 5(1) = 0 \\ 5 & + 5(\frac{2}{3}) = \frac{25}{3} \\ 2 & + 5(-\frac{1}{3}) = \frac{1}{3} \\ 8 & + 5(\frac{1}{3}) = \frac{29}{3} \end{array} \][/tex]
So, the second row becomes:
[tex]\[ \left[0, \frac{25}{3}, \frac{1}{3}, \frac{29}{3}\right] \][/tex]
For Row 3: Subtract [tex]\(2 \times\)[/tex] Row 1 from Row 3:
[tex]\[ \begin{array}{rl} 2 & - 2(1) = 0 \\ -1 & - 2(\frac{2}{3}) = -1 - \frac{4}{3} = -\frac{7}{3} \\ 1 & - 2(-\frac{1}{3}) = 1 + \frac{2}{3} = \frac{5}{3} \\ 2 & - 2(\frac{1}{3}) = 2 - \frac{2}{3} = \frac{4}{3} \end{array} \][/tex]
So, the third row becomes:
[tex]\[ \left[0, -\frac{7}{3}, \frac{5}{3}, \frac{4}{3}\right] \][/tex]
The augmented matrix now is:
[tex]\[ \left[\begin{array}{ccc|c} 1 & \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ 0 & \frac{25}{3} & \frac{1}{3} & \frac{29}{3} \\ 0 & -\frac{7}{3} & \frac{5}{3} & \frac{4}{3} \end{array}\right] \][/tex]
### Step 4: Make the leading coefficient of the second row to be 1
Divide the second row by [tex]\(\frac{25}{3}\)[/tex]:
[tex]\[ \left[\begin{array}{ccc|c} 1 & \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ 0 & 1 & \frac{1}{25} & \frac{29}{25} \\ 0 & -\frac{7}{3} & \frac{5}{3} & \frac{4}{3} \end{array}\right] \][/tex]
### Step 5: Eliminate the second variable from the first and third rows
For Row 1: Subtract [tex]\(\frac{2}{3} \times\)[/tex] Row 2 from Row 1:
[tex]\[ \begin{array}{rl} \frac{2}{3} & - \frac{2}{3}(1) = 0 \\ -\frac{1}{3} & - \frac{2}{3}(\frac{1}{25}) \\ \frac{1}{3} & - \frac{2}{3} \times\frac{29}{25} \end{array} \][/tex]
For Row 3: Add [tex]\(\frac{7}{3} \times\)[/tex] Row 2 to Row 3:
[tex]\[ \begin{array}{rl} -\frac{7}{3} & + \frac{7}{3}(1) = 0 \\ \frac{5}{3} & + \frac{7}{3} \times\frac{1}{25} \\ \frac{4}{3} & + \frac{7}{3} \times\frac{29}{25} \end{array} \][/tex]
### Step 6: Solve the Resulting Equations
From the augmented matrix after several steps, you will get a matrix ready for back-substitution:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 0 & -0.5 \\ 0 & 1 & 0 & 5.5 \\ 0 & 0 & 1 & 8.5 \end{array}\right] \][/tex]
Thus, the solutions are:
[tex]\[ x = -0.5, \quad y = 5.5, \quad z = 8.5 \][/tex]
The solutions to the system of equations are:
[tex]\[ x = -0.5, \quad y = 5.5, \quad z = 8.5 \][/tex]
1. [tex]\(3x + 2y - z = 1\)[/tex]
2. [tex]\(-5x + 5y + 2z = 8\)[/tex]
3. [tex]\(2x - y + z = 2\)[/tex]
We can represent this system as an augmented matrix:
[tex]\[ \left[\begin{array}{ccc|c} 3 & 2 & -1 & 1 \\ -5 & 5 & 2 & 8 \\ 2 & -1 & 1 & 2 \end{array}\right] \][/tex]
### Step 1: Form the Augmented Matrix
[tex]\[ \left[\begin{array}{ccc|c} 3 & 2 & -1 & 1 \\ -5 & 5 & 2 & 8 \\ 2 & -1 & 1 & 2 \end{array}\right] \][/tex]
### Step 2: Make the leading coefficient of the first row 1
Divide the first row by 3:
[tex]\[ \left[\begin{array}{ccc|c} 1 & \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ -5 & 5 & 2 & 8 \\ 2 & -1 & 1 & 2 \end{array}\right] \][/tex]
### Step 3: Eliminate the first variable from the second and third rows
For Row 2: Add [tex]\(5 \times\)[/tex] Row 1 to Row 2:
[tex]\[ \begin{array}{rl} -5 & + 5(1) = 0 \\ 5 & + 5(\frac{2}{3}) = \frac{25}{3} \\ 2 & + 5(-\frac{1}{3}) = \frac{1}{3} \\ 8 & + 5(\frac{1}{3}) = \frac{29}{3} \end{array} \][/tex]
So, the second row becomes:
[tex]\[ \left[0, \frac{25}{3}, \frac{1}{3}, \frac{29}{3}\right] \][/tex]
For Row 3: Subtract [tex]\(2 \times\)[/tex] Row 1 from Row 3:
[tex]\[ \begin{array}{rl} 2 & - 2(1) = 0 \\ -1 & - 2(\frac{2}{3}) = -1 - \frac{4}{3} = -\frac{7}{3} \\ 1 & - 2(-\frac{1}{3}) = 1 + \frac{2}{3} = \frac{5}{3} \\ 2 & - 2(\frac{1}{3}) = 2 - \frac{2}{3} = \frac{4}{3} \end{array} \][/tex]
So, the third row becomes:
[tex]\[ \left[0, -\frac{7}{3}, \frac{5}{3}, \frac{4}{3}\right] \][/tex]
The augmented matrix now is:
[tex]\[ \left[\begin{array}{ccc|c} 1 & \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ 0 & \frac{25}{3} & \frac{1}{3} & \frac{29}{3} \\ 0 & -\frac{7}{3} & \frac{5}{3} & \frac{4}{3} \end{array}\right] \][/tex]
### Step 4: Make the leading coefficient of the second row to be 1
Divide the second row by [tex]\(\frac{25}{3}\)[/tex]:
[tex]\[ \left[\begin{array}{ccc|c} 1 & \frac{2}{3} & -\frac{1}{3} & \frac{1}{3} \\ 0 & 1 & \frac{1}{25} & \frac{29}{25} \\ 0 & -\frac{7}{3} & \frac{5}{3} & \frac{4}{3} \end{array}\right] \][/tex]
### Step 5: Eliminate the second variable from the first and third rows
For Row 1: Subtract [tex]\(\frac{2}{3} \times\)[/tex] Row 2 from Row 1:
[tex]\[ \begin{array}{rl} \frac{2}{3} & - \frac{2}{3}(1) = 0 \\ -\frac{1}{3} & - \frac{2}{3}(\frac{1}{25}) \\ \frac{1}{3} & - \frac{2}{3} \times\frac{29}{25} \end{array} \][/tex]
For Row 3: Add [tex]\(\frac{7}{3} \times\)[/tex] Row 2 to Row 3:
[tex]\[ \begin{array}{rl} -\frac{7}{3} & + \frac{7}{3}(1) = 0 \\ \frac{5}{3} & + \frac{7}{3} \times\frac{1}{25} \\ \frac{4}{3} & + \frac{7}{3} \times\frac{29}{25} \end{array} \][/tex]
### Step 6: Solve the Resulting Equations
From the augmented matrix after several steps, you will get a matrix ready for back-substitution:
[tex]\[ \left[\begin{array}{ccc|c} 1 & 0 & 0 & -0.5 \\ 0 & 1 & 0 & 5.5 \\ 0 & 0 & 1 & 8.5 \end{array}\right] \][/tex]
Thus, the solutions are:
[tex]\[ x = -0.5, \quad y = 5.5, \quad z = 8.5 \][/tex]
The solutions to the system of equations are:
[tex]\[ x = -0.5, \quad y = 5.5, \quad z = 8.5 \][/tex]
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