Answer :
To solve the given system of linear equations, we will use the Gaussian elimination method to determine whether there are any solutions and, if so, what form they take.
The system of equations is:
[tex]\[ \left\{\begin{array}{r} x-y+2 z=-2 \\ 4 x+3 y+7 z=-9 \\ x+6 y+z=-3 \end{array}\right. \][/tex]
Step 1: Write the augmented matrix for the system:
[tex]\[ \begin{pmatrix} 1 & -1 & 2 & | & -2 \\ 4 & 3 & 7 & | & -9 \\ 1 & 6 & 1 & | & -3 \end{pmatrix} \][/tex]
Step 2: Use the first row to eliminate [tex]\( x \)[/tex] from the second and third rows.
1. For the second row:
[tex]\[ R2 \leftarrow R2 - 4R1 \][/tex]
Calculation:
[tex]\[ \begin{pmatrix} 4 & 3 & 7 & | & -9 \end{pmatrix} - 4 \cdot \begin{pmatrix} 1 & -1 & 2 & | & -2 \end{pmatrix} = \begin{pmatrix} 0 & 7 & -1 & | & -1 \end{pmatrix} \][/tex]
2. For the third row:
[tex]\[ R3 \leftarrow R3 - R1 \][/tex]
Calculation:
[tex]\[ \begin{pmatrix} 1 & 6 & 1 & | & -3 \end{pmatrix} - \begin{pmatrix} 1 & -1 & 2 & | & -2 \end{pmatrix} = \begin{pmatrix} 0 & 7 & -1 & | & -1 \end{pmatrix} \][/tex]
After these operations, the augmented matrix becomes:
[tex]\[ \begin{pmatrix} 1 & -1 & 2 & | & -2 \\ 0 & 7 & -1 & | & -1 \\ 0 & 7 & -1 & | & -1 \end{pmatrix} \][/tex]
Step 3: Make further simplifications to see if any inconsistencies appear.
Notice that the second and third rows are identical. So the system can be represented with just two equations:
[tex]\[ \left\{\begin{array}{r} x-y+2 z=-2 \\ 7 y - z=-1 \\ \end{array}\right. \][/tex]
Step 4: Solve for [tex]\( y \)[/tex] and [tex]\( z \)[/tex] from the second equation:
[tex]\[ 7y - z = -1 \rightarrow z = 7y + 1 \][/tex]
Step 5: Substitute [tex]\( z = 7y + 1 \)[/tex] back into the first equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x - y + 2(7y + 1) = -2 \\ x - y + 14y + 2 = -2 \\ x + 13y + 2 = -2 \\ x = -2 - 2 - 13y \\ x = -4 - 13y \][/tex]
Therefore, the solution set is:
[tex]\[ \left\{\begin{array}{r} x = -4 - 13y \\ y = y \\ z = 7y + 1 \\ \end{array}\right. \][/tex]
Since [tex]\( y \)[/tex] and [tex]\( z \)[/tex] can be any real values and they satisfy the equations, we have infinitely many solutions.
So, the correct choice is:
A. There are infinitely many solutions. If [tex]\( z \)[/tex] is allowed to be any real number, then
[tex]\[ x = -4 - 13y \quad \text{and} \quad y = y \][/tex]
In terms of [tex]\( z \)[/tex]:
[tex]\[ y = \frac{z - 1}{7} \][/tex]
Substituting [tex]\( y \)[/tex] back into the expression for [tex]\( x \)[/tex]:
[tex]\[ x = -4 - 13 \left(\frac{z - 1}{7}\right) \][/tex]
[tex]\[ x = -4 - \frac{13(z - 1)}{7} \][/tex]
[tex]\[ x = -4 - \frac{13z - 13}{7} \][/tex]
[tex]\[ x = -4 - \frac{13z}{7} + \frac{13}{7} \][/tex]
[tex]\[ x = -4 + \frac{13}{7} - \frac{13z}{7} \][/tex]
[tex]\[ x = -\frac{28}{7} + \frac{13}{7} - \frac{13z}{7} \][/tex]
[tex]\[ x = -\frac{15}{7} - \frac{13z}{7} \][/tex]
Thus,
[tex]\[ x = -\frac{15 + 13z}{7} \][/tex]
is the simplified form.
So the final answer in the choice is:
\[
A. There are infinitely many solutions. If [tex]\( z \)[/tex] is allowed to be any real number, then [tex]\( x = -\frac{15 + 13z}{7} \)[/tex] and [tex]\( y = \frac{z - 1}{7} \)[/tex].
The system of equations is:
[tex]\[ \left\{\begin{array}{r} x-y+2 z=-2 \\ 4 x+3 y+7 z=-9 \\ x+6 y+z=-3 \end{array}\right. \][/tex]
Step 1: Write the augmented matrix for the system:
[tex]\[ \begin{pmatrix} 1 & -1 & 2 & | & -2 \\ 4 & 3 & 7 & | & -9 \\ 1 & 6 & 1 & | & -3 \end{pmatrix} \][/tex]
Step 2: Use the first row to eliminate [tex]\( x \)[/tex] from the second and third rows.
1. For the second row:
[tex]\[ R2 \leftarrow R2 - 4R1 \][/tex]
Calculation:
[tex]\[ \begin{pmatrix} 4 & 3 & 7 & | & -9 \end{pmatrix} - 4 \cdot \begin{pmatrix} 1 & -1 & 2 & | & -2 \end{pmatrix} = \begin{pmatrix} 0 & 7 & -1 & | & -1 \end{pmatrix} \][/tex]
2. For the third row:
[tex]\[ R3 \leftarrow R3 - R1 \][/tex]
Calculation:
[tex]\[ \begin{pmatrix} 1 & 6 & 1 & | & -3 \end{pmatrix} - \begin{pmatrix} 1 & -1 & 2 & | & -2 \end{pmatrix} = \begin{pmatrix} 0 & 7 & -1 & | & -1 \end{pmatrix} \][/tex]
After these operations, the augmented matrix becomes:
[tex]\[ \begin{pmatrix} 1 & -1 & 2 & | & -2 \\ 0 & 7 & -1 & | & -1 \\ 0 & 7 & -1 & | & -1 \end{pmatrix} \][/tex]
Step 3: Make further simplifications to see if any inconsistencies appear.
Notice that the second and third rows are identical. So the system can be represented with just two equations:
[tex]\[ \left\{\begin{array}{r} x-y+2 z=-2 \\ 7 y - z=-1 \\ \end{array}\right. \][/tex]
Step 4: Solve for [tex]\( y \)[/tex] and [tex]\( z \)[/tex] from the second equation:
[tex]\[ 7y - z = -1 \rightarrow z = 7y + 1 \][/tex]
Step 5: Substitute [tex]\( z = 7y + 1 \)[/tex] back into the first equation to solve for [tex]\( x \)[/tex]:
[tex]\[ x - y + 2(7y + 1) = -2 \\ x - y + 14y + 2 = -2 \\ x + 13y + 2 = -2 \\ x = -2 - 2 - 13y \\ x = -4 - 13y \][/tex]
Therefore, the solution set is:
[tex]\[ \left\{\begin{array}{r} x = -4 - 13y \\ y = y \\ z = 7y + 1 \\ \end{array}\right. \][/tex]
Since [tex]\( y \)[/tex] and [tex]\( z \)[/tex] can be any real values and they satisfy the equations, we have infinitely many solutions.
So, the correct choice is:
A. There are infinitely many solutions. If [tex]\( z \)[/tex] is allowed to be any real number, then
[tex]\[ x = -4 - 13y \quad \text{and} \quad y = y \][/tex]
In terms of [tex]\( z \)[/tex]:
[tex]\[ y = \frac{z - 1}{7} \][/tex]
Substituting [tex]\( y \)[/tex] back into the expression for [tex]\( x \)[/tex]:
[tex]\[ x = -4 - 13 \left(\frac{z - 1}{7}\right) \][/tex]
[tex]\[ x = -4 - \frac{13(z - 1)}{7} \][/tex]
[tex]\[ x = -4 - \frac{13z - 13}{7} \][/tex]
[tex]\[ x = -4 - \frac{13z}{7} + \frac{13}{7} \][/tex]
[tex]\[ x = -4 + \frac{13}{7} - \frac{13z}{7} \][/tex]
[tex]\[ x = -\frac{28}{7} + \frac{13}{7} - \frac{13z}{7} \][/tex]
[tex]\[ x = -\frac{15}{7} - \frac{13z}{7} \][/tex]
Thus,
[tex]\[ x = -\frac{15 + 13z}{7} \][/tex]
is the simplified form.
So the final answer in the choice is:
\[
A. There are infinitely many solutions. If [tex]\( z \)[/tex] is allowed to be any real number, then [tex]\( x = -\frac{15 + 13z}{7} \)[/tex] and [tex]\( y = \frac{z - 1}{7} \)[/tex].
