Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
