Solve the system of given equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. begin{cases}x+3y=0x+y+z=13x-y-z=11end{cases}

Solve the system of given equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. begin{cases}x+3y=0x+y+z=13x-y-z=11end{cases}

Question
Matrices
asked 2021-02-08
Solve the system of given equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination.
\(\begin{cases}x+3y=0\\x+y+z=1\\3x-y-z=11\end{cases}\)

Answers (1)

2021-02-09
Step 1
Given
x+3y=0
x+y+z=1
3x-y-z=11
Step 2
We write the given system of equation in the form
\(AX=B\)
where
\(A=\begin{bmatrix}1 & 3&0 \\1 & 1&1\\3&-1&-1 \end{bmatrix} , B=\begin{bmatrix}x \\y\\z \end{bmatrix} \text{ and } B=\begin{bmatrix}0 \\1\\11 \end{bmatrix}\)
The augmented matrix for the given system of the equation can be represented as
\([A|B]=\begin{bmatrix}1 & 3&0&0 \\1 & 1&1&1\\3&-1&-1&11 \end{bmatrix}\)
\(R_2 \rightarrow R_2-R_1 \text{ and } R_3 \rightarrow R_3-3R_1\)
\(\Rightarrow [A|B]=\begin{bmatrix}1 & 3&0&0 \\0 & -2&1&1\\0&-10&-1&11 \end{bmatrix}\)
\(R_2 \rightarrow -\frac{1}{2}R_2\)
\(\Rightarrow [A|B]=\begin{bmatrix}1 & 3&0&0 \\0 & 1&-\frac{1}{2}&-\frac{1}{2}\\0&-10&-1&11 \end{bmatrix}\)
\(R_3 \rightarrow R_3+10R_2\)
\(\Rightarrow [A|B]=\begin{bmatrix}1 & 3&0&0 \\0 & 1&-\frac{1}{2}&-\frac{1}{2}\\0&0&-6&6 \end{bmatrix}\)
\(R_3 \rightarrow -\frac{1}{6}R_3\)
\(\Rightarrow [A|B]=\begin{bmatrix}1 & 3&0&0 \\0 & 1&-\frac{1}{2}&-\frac{1}{2}\\0&0&1&-1 \end{bmatrix}\)
Step 3
We write the system of the linear equation as
\(x+3y=0\)
\(y-\frac{1}{2}z=-\frac{1}{2}\)
\(z=-1\)
Plug this in the above equation we get
\(y-\frac{1}{2}(-1)=-\frac{1}{2}\)
\(\Rightarrow y+\frac{1}{2}=-\frac{1}{2}\)
\(\Rightarrow y=-\frac{1}{2}-\frac{1}{2}\)
\(\Rightarrow y=-1\)
Plug this in the above equation
\(x+3(-1)=0\)
\(\Rightarrow x-3=0\)
\(\Rightarrow x=3\)
\(x=3 , y=-1 \text{ and } z=-1\)
0

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