QuestionJuly 14, 2025

Solve the system of equations below. x+y+z=2 2x-y+3z=13 -x+2y-z=-8 Select one: a. (1,-2,3) b. (1,-2,1) C. (1,-6,1) d. (1,-2,0)

Solve the system of equations below. x+y+z=2 2x-y+3z=13 -x+2y-z=-8 Select one: a. (1,-2,3) b. (1,-2,1) C. (1,-6,1) d. (1,-2,0)
Solve the system of equations below. x+y+z=2 2x-y+3z=13 -x+2y-z=-8 Select one: a. (1,-2,3) b. (1,-2,1) C. (1,-6,1) d. (1,-2,0)

Solution
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Answer

(3, -2, 1) Explanation 1. Write the system in matrix form The system can be written as: \begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ -1 & 2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 13 \\ -8 \end{bmatrix} 2. Use Gaussian elimination Transform the matrix to row-echelon form. Start by eliminating x from the second and third rows. 1. Row 2: R_2 - 2R_1 \rightarrow R_2 \begin{bmatrix} 1 & 1 & 1 \\ 0 & -3 & 1 \\ -1 & 2 & -1 \end{bmatrix} 2. Row 3: R_3 + R_1 \rightarrow R_3 \begin{bmatrix} 1 & 1 & 1 \\ 0 & -3 & 1 \\ 0 & 3 & 0 \end{bmatrix} 3. Simplify further Eliminate y from the third row. 1. Row 3: R_3 + R_2 \rightarrow R_3 \begin{bmatrix} 1 & 1 & 1 \\ 0 & -3 & 1 \\ 0 & 0 & 1 \end{bmatrix} 4. Back substitution Solve for z, then y, and finally x. 1. From Row 3: z = 1 2. Substitute z = 1 into Row 2: -3y + 1 = 13 \Rightarrow y = -2 3. Substitute y = -2 and z = 1 into Row 1: x - 2 + 1 = 2 \Rightarrow x = 3

Explanation

1. Write the system in matrix form<br /> The system can be written as:<br />$$<br />\begin{bmatrix}<br />1 & 1 & 1 \\<br />2 & -1 & 3 \\<br />-1 & 2 & -1<br />\end{bmatrix}<br />\begin{bmatrix}<br />x \\<br />y \\<br />z<br />\end{bmatrix}<br />=<br />\begin{bmatrix}<br />2 \\<br />13 \\<br />-8<br />\end{bmatrix}<br />$$<br /><br />2. Use Gaussian elimination<br /> Transform the matrix to row-echelon form. Start by eliminating $x$ from the second and third rows.<br /><br />1. Row 2: $R_2 - 2R_1 \rightarrow R_2$<br /> $$ <br /> \begin{bmatrix}<br /> 1 & 1 & 1 \\<br /> 0 & -3 & 1 \\<br /> -1 & 2 & -1<br /> \end{bmatrix}<br /> $$<br /><br />2. Row 3: $R_3 + R_1 \rightarrow R_3$<br /> $$ <br /> \begin{bmatrix}<br /> 1 & 1 & 1 \\<br /> 0 & -3 & 1 \\<br /> 0 & 3 & 0<br /> \end{bmatrix}<br /> $$<br /><br />3. Simplify further<br /> Eliminate $y$ from the third row.<br /><br />1. Row 3: $R_3 + R_2 \rightarrow R_3$<br /> $$ <br /> \begin{bmatrix}<br /> 1 & 1 & 1 \\<br /> 0 & -3 & 1 \\<br /> 0 & 0 & 1<br /> \end{bmatrix}<br /> $$<br /><br />4. Back substitution<br /> Solve for $z$, then $y$, and finally $x$.<br /><br />1. From Row 3: $z = 1$<br />2. Substitute $z = 1$ into Row 2: $-3y + 1 = 13 \Rightarrow y = -2$<br />3. Substitute $y = -2$ and $z = 1$ into Row 1: $x - 2 + 1 = 2 \Rightarrow x = 3$
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