QuestionJuly 15, 2025

Solve the following system of equations using Gaussian elimination. -5x+4y+15z=-23 3x-2y-20z=-33 -4x-4y+5z=-78 ((996)/(41),-(241)/(41),(1667)/(205)) (21,4,(22)/(5)) (-(241)/(41),(996)/(41),-(241)/(41)) ((22)/(5),4,(22)/(5))

Solve the following system of equations using Gaussian elimination. -5x+4y+15z=-23 3x-2y-20z=-33 -4x-4y+5z=-78 ((996)/(41),-(241)/(41),(1667)/(205)) (21,4,(22)/(5)) (-(241)/(41),(996)/(41),-(241)/(41)) ((22)/(5),4,(22)/(5))
Solve the following system of equations using Gaussian elimination.
-5x+4y+15z=-23
3x-2y-20z=-33
-4x-4y+5z=-78
((996)/(41),-(241)/(41),(1667)/(205))
(21,4,(22)/(5))
(-(241)/(41),(996)/(41),-(241)/(41))
((22)/(5),4,(22)/(5))

Solution
4.3(263 votes)

Answer

(\frac {22}{5},4,\frac {22}{5}) Explanation 1. Write the augmented matrix Convert the system into an augmented matrix: \begin{bmatrix} -5 & 4 & 15 & | & -23 \\ 3 & -2 & -20 & | & -33 \\ -4 & -4 & 5 & | & -78 \end{bmatrix} 2. Perform row operations to get upper triangular form Use row operations to eliminate x from rows 2 and 3. Multiply Row 1 by \frac{3}{-5} and add to Row 2, and multiply Row 1 by \frac{-4}{-5} and add to Row 3. 3. Simplify the matrix After performing the row operations, the matrix becomes: \begin{bmatrix} -5 & 4 & 15 & | & -23 \\ 0 & \frac{2}{5} & -11 & | & -46.8 \\ 0 & \frac{12}{5} & 17 & | & -96.4 \end{bmatrix} 4. Eliminate y from Row 3 Multiply Row 2 by \frac{12}{2} and subtract from Row 3 to eliminate y in Row 3. 5. Back substitution Solve for z using the last row, then substitute back to find y, and finally x.

Explanation

1. Write the augmented matrix<br /> Convert the system into an augmented matrix: <br />$\begin{bmatrix} -5 & 4 & 15 & | & -23 \\ 3 & -2 & -20 & | & -33 \\ -4 & -4 & 5 & | & -78 \end{bmatrix}$<br /><br />2. Perform row operations to get upper triangular form<br /> Use row operations to eliminate $x$ from rows 2 and 3. Multiply Row 1 by $\frac{3}{-5}$ and add to Row 2, and multiply Row 1 by $\frac{-4}{-5}$ and add to Row 3.<br /><br />3. Simplify the matrix<br /> After performing the row operations, the matrix becomes:<br />$\begin{bmatrix} -5 & 4 & 15 & | & -23 \\ 0 & \frac{2}{5} & -11 & | & -46.8 \\ 0 & \frac{12}{5} & 17 & | & -96.4 \end{bmatrix}$<br /><br />4. Eliminate $y$ from Row 3<br /> Multiply Row 2 by $\frac{12}{2}$ and subtract from Row 3 to eliminate $y$ in Row 3.<br /><br />5. Back substitution<br /> Solve for $z$ using the last row, then substitute back to find $y$, and finally $x$.
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