QuestionJuly 15, 2025

Solve the following system of equations using Gaussian elimination. -3x+3y-2z=-24 3x-9y+3z=-3 4x+9y-3z=45 (5,3,(16)/(3)) ((20)/(3),13,6) (6,(20)/(3),13) (3,(16)/(3),5)

Solve the following system of equations using Gaussian elimination. -3x+3y-2z=-24 3x-9y+3z=-3 4x+9y-3z=45 (5,3,(16)/(3)) ((20)/(3),13,6) (6,(20)/(3),13) (3,(16)/(3),5)
Solve the following system of equations using Gaussian elimination.
-3x+3y-2z=-24
3x-9y+3z=-3
4x+9y-3z=45
(5,3,(16)/(3))
((20)/(3),13,6)
(6,(20)/(3),13)
(3,(16)/(3),5)

Solution
4.3(122 votes)

Answer

(6, \frac{20}{3}, \frac{16}{3}) Explanation 1. Write the augmented matrix Convert the system into an augmented matrix: \begin{bmatrix} -3 & 3 & -2 & | & -24 \\ 3 & -9 & 3 & | & -3 \\ 4 & 9 & -3 & | & 45 \end{bmatrix} 2. Perform row operations to get upper triangular form Add Row 1 to Row 2, and replace Row 2: \begin{bmatrix} -3 & 3 & -2 & | & -24 \\ 0 & -6 & 1 & | & -27 \\ 4 & 9 & -3 & | & 45 \end{bmatrix} Multiply Row 1 by \frac{4}{-3} and add to Row 3: \begin{bmatrix} -3 & 3 & -2 & | & -24 \\ 0 & -6 & 1 & | & -27 \\ 0 & 5 & -\frac{11}{3} & | & 13 \end{bmatrix} 3. Continue row operations to achieve row-echelon form Multiply Row 2 by \frac{5}{-6} and add to Row 3: \begin{bmatrix} -3 & 3 & -2 & | & -24 \\ 0 & -6 & 1 & | & -27 \\ 0 & 0 & -\frac{1}{3} & | & -\frac{16}{3} \end{bmatrix} 4. Back substitution Solve for z from Row 3: z = \frac{16}{3} Substitute z into Row 2 to solve for y: -6y + 1 \cdot \frac{16}{3} = -27 y = \frac{20}{3} Substitute y and z into Row 1 to solve for x: -3x + 3 \cdot \frac{20}{3} - 2 \cdot \frac{16}{3} = -24 x = 6

Explanation

1. Write the augmented matrix<br /> Convert the system into an augmented matrix:<br />$$<br />\begin{bmatrix}<br />-3 & 3 & -2 & | & -24 \\<br />3 & -9 & 3 & | & -3 \\<br />4 & 9 & -3 & | & 45<br />\end{bmatrix}<br />$$<br /><br />2. Perform row operations to get upper triangular form<br /> Add Row 1 to Row 2, and replace Row 2:<br />$$<br />\begin{bmatrix}<br />-3 & 3 & -2 & | & -24 \\<br />0 & -6 & 1 & | & -27 \\<br />4 & 9 & -3 & | & 45<br />\end{bmatrix}<br />$$<br /><br /> Multiply Row 1 by $\frac{4}{-3}$ and add to Row 3:<br />$$<br />\begin{bmatrix}<br />-3 & 3 & -2 & | & -24 \\<br />0 & -6 & 1 & | & -27 \\<br />0 & 5 & -\frac{11}{3} & | & 13<br />\end{bmatrix}<br />$$<br /><br />3. Continue row operations to achieve row-echelon form<br /> Multiply Row 2 by $\frac{5}{-6}$ and add to Row 3:<br />$$<br />\begin{bmatrix}<br />-3 & 3 & -2 & | & -24 \\<br />0 & -6 & 1 & | & -27 \\<br />0 & 0 & -\frac{1}{3} & | & -\frac{16}{3}<br />\end{bmatrix}<br />$$<br /><br />4. Back substitution<br /> Solve for $z$ from Row 3:<br />$$ z = \frac{16}{3} $$<br /><br /> Substitute $z$ into Row 2 to solve for $y$:<br />$$ -6y + 1 \cdot \frac{16}{3} = -27 $$<br />$$ y = \frac{20}{3} $$<br /><br /> Substitute $y$ and $z$ into Row 1 to solve for $x$:<br />$$ -3x + 3 \cdot \frac{20}{3} - 2 \cdot \frac{16}{3} = -24 $$<br />$$ x = 6 $$
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