Answer :
Sure, let's solve this system of linear equations step by step.
We have three equations:
1. \( x + y + 2z = 3 \)
2. \( 2x + y - 3z = 4 \)
3. \( -x - 2y = -5 \)
Step 1: Expressing the system of equations in matrix form:
[tex]\[ \begin{pmatrix} 1 & 1 & 2 \\ 2 & 1 & -3 \\ -1 & -2 & 0 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ -5 \\ \end{pmatrix} \][/tex]
Step 2: Using Gaussian elimination or any other method to solve the system. Let's detail the steps we would typically follow:
Row operations to transform the matrix to the row echelon form:
1. First, we use the first row to eliminate the \( x \) term in the subsequent rows.
- Row 2: Subtract 2 times Row 1 from Row 2:
[tex]\[ R2 \rightarrow R2 - 2R1 = [2 - 2(1), 1 - 2(1), -3 - 2(2)] = [0, -1, -7] \][/tex]
- Row 3: Add Row 1 to Row 3:
[tex]\[ R3 \rightarrow R3 + R1 = [-1 + 1, -2 + 1, 0 + 2] = [0, -1, 2] \][/tex]
2. Now the system looks like:
[tex]\[ \begin{pmatrix} 1 & 1 & 2 \\ 0 & -1 & -7 \\ 0 & -1 & 2 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ -2 \\ \end{pmatrix} \][/tex]
3. We then use the second row to eliminate the \( y \) term in the third row. Subtract Row 2 from Row 3:
[tex]\[ R3 \rightarrow R3 - R2 = [0 - 0, -1 + 1, 2 + 7] = [0, 0, 9] \][/tex]
4. We now have an upper triangular matrix:
[tex]\[ \begin{pmatrix} 1 & 1 & 2 \\ 0 & -1 & -7 \\ 0 & 0 & 9 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ -2 \\ \end{pmatrix} \][/tex]
Step 3: Back-substitution to find \( x \), \( y \), and \( z \):
1. From the third row, solve for \( z \):
[tex]\[ 9z = -2 \implies z = -2 / 9 \implies z = 0 \][/tex]
2. Substitute \( z \) into the second row to find \( y \):
[tex]\[ -y - 7(0) = -2 \implies -y = -2 \implies y = 2 \][/tex]
3. Substitute \( y \) and \( z \) into the first row to find \( x \):
[tex]\[ x + 2 + 2(0) = 3 \implies x + 2 = 3 \implies x = 1 \][/tex]
Hence, the solution to the system of equations is:
[tex]\[ x = 1, \quad y = 2, \quad z = 0 \][/tex]
So, the final answer is:
[tex]\[ (x, y, z) = (1, 2, 0) \][/tex]
We have three equations:
1. \( x + y + 2z = 3 \)
2. \( 2x + y - 3z = 4 \)
3. \( -x - 2y = -5 \)
Step 1: Expressing the system of equations in matrix form:
[tex]\[ \begin{pmatrix} 1 & 1 & 2 \\ 2 & 1 & -3 \\ -1 & -2 & 0 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ -5 \\ \end{pmatrix} \][/tex]
Step 2: Using Gaussian elimination or any other method to solve the system. Let's detail the steps we would typically follow:
Row operations to transform the matrix to the row echelon form:
1. First, we use the first row to eliminate the \( x \) term in the subsequent rows.
- Row 2: Subtract 2 times Row 1 from Row 2:
[tex]\[ R2 \rightarrow R2 - 2R1 = [2 - 2(1), 1 - 2(1), -3 - 2(2)] = [0, -1, -7] \][/tex]
- Row 3: Add Row 1 to Row 3:
[tex]\[ R3 \rightarrow R3 + R1 = [-1 + 1, -2 + 1, 0 + 2] = [0, -1, 2] \][/tex]
2. Now the system looks like:
[tex]\[ \begin{pmatrix} 1 & 1 & 2 \\ 0 & -1 & -7 \\ 0 & -1 & 2 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ -2 \\ \end{pmatrix} \][/tex]
3. We then use the second row to eliminate the \( y \) term in the third row. Subtract Row 2 from Row 3:
[tex]\[ R3 \rightarrow R3 - R2 = [0 - 0, -1 + 1, 2 + 7] = [0, 0, 9] \][/tex]
4. We now have an upper triangular matrix:
[tex]\[ \begin{pmatrix} 1 & 1 & 2 \\ 0 & -1 & -7 \\ 0 & 0 & 9 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 3 \\ -2 \\ -2 \\ \end{pmatrix} \][/tex]
Step 3: Back-substitution to find \( x \), \( y \), and \( z \):
1. From the third row, solve for \( z \):
[tex]\[ 9z = -2 \implies z = -2 / 9 \implies z = 0 \][/tex]
2. Substitute \( z \) into the second row to find \( y \):
[tex]\[ -y - 7(0) = -2 \implies -y = -2 \implies y = 2 \][/tex]
3. Substitute \( y \) and \( z \) into the first row to find \( x \):
[tex]\[ x + 2 + 2(0) = 3 \implies x + 2 = 3 \implies x = 1 \][/tex]
Hence, the solution to the system of equations is:
[tex]\[ x = 1, \quad y = 2, \quad z = 0 \][/tex]
So, the final answer is:
[tex]\[ (x, y, z) = (1, 2, 0) \][/tex]
