Answer :
Let's solve the given system of equations using Gaussian elimination to find the solution.
The system of equations is:
[tex]\[ \left\{\begin{array}{rrr} -x + y + z &= -2 \\ -x + 4y - 5z &= -14 \\ 4x - 2y - 8z &= 0 \end{array}\right. \][/tex]
1. Write the augmented matrix:
[tex]\[ \left[ \begin{array}{ccc|c} -1 & 1 & 1 & -2 \\ -1 & 4 & -5 & -14 \\ 4 & -2 & -8 & 0 \end{array} \right] \][/tex]
2. Perform row operations to obtain row echelon form:
i. Row 1 operations:
[tex]\[ R1 = [-1\ 1\ 1\ |-2] \][/tex]
ii. Row 2 operations:
Subtract Row 1 from Row 2:
[tex]\[ R2 = [-1 + 1\ 4 - 1\ -5 - 1\ -14 + 2] = [0\ 3\ -6\ -12] \][/tex]
iii. Row 3 operations:
Multiply Row 1 by 4 and add to Row 3:
[tex]\[ R3 = [4 \cdot -1 + 4\ 4 \cdot 1 - 2\ 4 \cdot 1 - 8\ 4 \cdot -2 + 0] = [0\ 2\ -4\ -8] \][/tex]
The resulting matrix is:
[tex]\[ \left[ \begin{array}{ccc|c} -1 & 1 & 1 & -2 \\ 0 & 3 & -6 & -12 \\ 0 & 2 & -4 & -8 \end{array} \right] \][/tex]
3. Further simplification to obtain the row echelon form:
i. Simplify Row 3:
Multiply Row 2 by 2/3 and subtract from Row 3:
[tex]\[ R3 = R3 - \frac{2}{3}R2 = [0\ 2\ -4\ -8] - \frac{2}{3}[0\ 3\ -6\ -12] \][/tex]
Simplifying this:
[tex]\[ R3 = [0\ 2 - 2\ 3\ -4 + 4\ 0 -8 + 8\ 0] \][/tex]
Hence,
[tex]\[ R3 = [0\ 0\ 0\ 0] \][/tex]
The resulting augmented matrix in row echelon form is:
[tex]\[ \left[ \begin{array}{ccc|c} -1 & 1 & 1 & -2 \\ 0 & 3 & -6 & -12 \\ 0 & 0 & 0 & 0 \end{array} \right] \][/tex]
4. Back substitution:
From the second row equation:
[tex]\[ 3y - 6z = -12 \][/tex]
Simplifying:
[tex]\[ y - 2z = -4 \][/tex]
[tex]\[ y = 2z - 4 \][/tex]
From the first row equation:
[tex]\[ -x + y + z = -2 \][/tex]
Substitute [tex]\( y = 2z - 4 \)[/tex]:
[tex]\[ -x + (2z - 4) + z = -2 \][/tex]
[tex]\[ -x + 3z - 4 = -2 \][/tex]
[tex]\[ -x = -3z + 2 \][/tex]
[tex]\[ x = 3z - 2 \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ \left\{ (x, y, z) \mid x = 3z - 2, y = 2z - 4, z \in \mathbb{R} \right\} \][/tex]
So, the solution set is:
[tex]\[ \{((3z - 2, 2z - 4, z) \mid z \text{ is any real number})\} \][/tex]
The system of equations is:
[tex]\[ \left\{\begin{array}{rrr} -x + y + z &= -2 \\ -x + 4y - 5z &= -14 \\ 4x - 2y - 8z &= 0 \end{array}\right. \][/tex]
1. Write the augmented matrix:
[tex]\[ \left[ \begin{array}{ccc|c} -1 & 1 & 1 & -2 \\ -1 & 4 & -5 & -14 \\ 4 & -2 & -8 & 0 \end{array} \right] \][/tex]
2. Perform row operations to obtain row echelon form:
i. Row 1 operations:
[tex]\[ R1 = [-1\ 1\ 1\ |-2] \][/tex]
ii. Row 2 operations:
Subtract Row 1 from Row 2:
[tex]\[ R2 = [-1 + 1\ 4 - 1\ -5 - 1\ -14 + 2] = [0\ 3\ -6\ -12] \][/tex]
iii. Row 3 operations:
Multiply Row 1 by 4 and add to Row 3:
[tex]\[ R3 = [4 \cdot -1 + 4\ 4 \cdot 1 - 2\ 4 \cdot 1 - 8\ 4 \cdot -2 + 0] = [0\ 2\ -4\ -8] \][/tex]
The resulting matrix is:
[tex]\[ \left[ \begin{array}{ccc|c} -1 & 1 & 1 & -2 \\ 0 & 3 & -6 & -12 \\ 0 & 2 & -4 & -8 \end{array} \right] \][/tex]
3. Further simplification to obtain the row echelon form:
i. Simplify Row 3:
Multiply Row 2 by 2/3 and subtract from Row 3:
[tex]\[ R3 = R3 - \frac{2}{3}R2 = [0\ 2\ -4\ -8] - \frac{2}{3}[0\ 3\ -6\ -12] \][/tex]
Simplifying this:
[tex]\[ R3 = [0\ 2 - 2\ 3\ -4 + 4\ 0 -8 + 8\ 0] \][/tex]
Hence,
[tex]\[ R3 = [0\ 0\ 0\ 0] \][/tex]
The resulting augmented matrix in row echelon form is:
[tex]\[ \left[ \begin{array}{ccc|c} -1 & 1 & 1 & -2 \\ 0 & 3 & -6 & -12 \\ 0 & 0 & 0 & 0 \end{array} \right] \][/tex]
4. Back substitution:
From the second row equation:
[tex]\[ 3y - 6z = -12 \][/tex]
Simplifying:
[tex]\[ y - 2z = -4 \][/tex]
[tex]\[ y = 2z - 4 \][/tex]
From the first row equation:
[tex]\[ -x + y + z = -2 \][/tex]
Substitute [tex]\( y = 2z - 4 \)[/tex]:
[tex]\[ -x + (2z - 4) + z = -2 \][/tex]
[tex]\[ -x + 3z - 4 = -2 \][/tex]
[tex]\[ -x = -3z + 2 \][/tex]
[tex]\[ x = 3z - 2 \][/tex]
Thus, the solution to the system of equations is:
[tex]\[ \left\{ (x, y, z) \mid x = 3z - 2, y = 2z - 4, z \in \mathbb{R} \right\} \][/tex]
So, the solution set is:
[tex]\[ \{((3z - 2, 2z - 4, z) \mid z \text{ is any real number})\} \][/tex]
