Question

Math

Posted 4 months ago

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Problem 1. (10 points)
Linear System - Three Variables
Solve the following system of equations using Gaussian elimination method. If there are no solutions, type "N" for both $x, y$ and $z$. If there are infinitely many solutions, type " $z$ " for $z$, and expressions in terms of $z$ for $x$ and $y$.
$\begin{aligned}
4 x-3 y-5 z & =-4 \\
3 x+1 y-8 z & =-1 \\
-5 x+1 y+6 z & =-3
\end{aligned}$
$x=$ $\qquad$ $y=$ $\qquad$ $z=$ $\qquad$
Answer(s) submitted:
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

Write the augmented matrix for the given system of equations:
$\begin{pmatrix}
4 & -3 & -5 & | & -4 \\
3 & 1 & -8 & | & -1 \\
-5 & 1 & 6 & | & -3
\end{pmatrix}$

step 2

Perform row operations to get a leading 1 in the first row, first column. Divide the first row by 4:
$\begin{pmatrix}
1 & -\frac{3}{4} & -\frac{5}{4} & | & -1 \\
3 & 1 & -8 & | & -1 \\
-5 & 1 & 6 & | & -3
\end{pmatrix}$

step 3

Eliminate the first column entries below the leading 1 by performing row operations:
$R2 = R2 - 3R1 \quad \text{and} \quad R3 = R3 + 5R1$
$\begin{pmatrix}
1 & -\frac{3}{4} & -\frac{5}{4} & | & -1 \\
0 & \frac{13}{4} & -\frac{17}{4} & | & 2 \\
0 & -\frac{11}{4} & \frac{11}{4} & | & -8
\end{pmatrix}$

step 4

Normalize the second row by dividing by $\frac{13}{4}$:
$R2 = \frac{4}{13} R2$
$\begin{pmatrix}
1 & -\frac{3}{4} & -\frac{5}{4} & | & -1 \\
0 & 1 & -\frac{17}{13} & | & \frac{8}{13} \\
0 & -\frac{11}{4} & \frac{11}{4} & | & -8
\end{pmatrix}$

step 5

Eliminate the second column entries below and above the leading 1 in the second row:
$R3 = R3 + \frac{11}{4} R2 \quad \text{and} \quad R1 = R1 + \frac{3}{4} R2$
$\begin{pmatrix}
1 & 0 & -\frac{29}{13} & | & -\frac{5}{13} \\
0 & 1 & -\frac{17}{13} & | & \frac{8}{13} \\
0 & 0 & 0 & | & -\frac{10}{13}
\end{pmatrix}$

step 6

Since the third row results in $0 = -\frac{10}{13}$, which is a contradiction, the system has no solutions

Answer

$x = N$, $y = N$, $z = N$

Key Concept

Gaussian Elimination

Explanation

Gaussian elimination is used to solve systems of linear equations by transforming the system's augmented matrix into row-echelon form. If a row results in a contradiction, the system has no solutions.

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