Respuesta :
Answer:
yeah the answer is in the question just needed the points
Step-by-step explanation:
[tex]\begin{bmatrix}4 & 0 & -4& |& -8 \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix} \xrightarrow{\text{4R3+R1}\rightarrow \text{R1}} \begin{bmatrix}\bold{4} & \bold{0} & \bold{0}& |& \bold{4} \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix}[/tex]
and,
[tex]\begin{bmatrix}4 & 0 & -4& |& 4 \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix} \xrightarrow{\frac{1}{4} \text{R1}\rightarrow \text{R1}} \begin{bmatrix}\bold{1} & \bold{0} & \bold{0}& |& \bold{1} \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix}[/tex]
What is matrix?
"A set of numbers arranged in rows and columns so as to form a rectangular array."
What is augmented matrix?
"A matrix formed by combining the columns of two matrices to form a new matrix."
What are elementary row operations?
"The mathematical operations that are performed on rows of a matrix."
For given question,
We have been given an augmented matrix.
[tex]\begin{bmatrix}4 & 0 & -4& |& -8 \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix}[/tex]
We need to perform the row operation 4R3 + R1 → R1
To perform above row operation,
- First multiply the third row by 4. ([tex]4R_3[/tex])
- Add the new third row (4[tex]R_3[/tex]) with the first row.
- Then replace Row 1 with the result.
The resultant augmented matrix would be,
[tex]\begin{bmatrix}\bold{4} & \bold{0} & \bold{0}& |& \bold{4} \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix}[/tex]
That is,
[tex]\begin{bmatrix}4 & 0 & -4& |& -8 \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix} \xrightarrow{\text{4R3+R1}\rightarrow \text{R1}} \begin{bmatrix}\bold{4} & \bold{0} & \bold{0}& |& \bold{4} \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix}[/tex]
Now, we need to perform the row operation [tex]\frac{1}{4}[/tex] R1 → R1 on the above resultant augmented matrix.
To perform above row operation,
- Multiply the Row 1 by the reciprocal of 4.
- Then replace Row 1 with the result.
The resultant augmented matrix would be,
[tex]\begin{bmatrix}4 & 0 & -4& |& 4 \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix} \xrightarrow{\frac{1}{4} \text{R1}\rightarrow \text{R1}} \begin{bmatrix}\bold{1} & \bold{0} & \bold{0}& |& \bold{1} \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix}[/tex]
Therefore,
[tex]\begin{bmatrix}4 & 0 & -4& |& -8 \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix} \xrightarrow{\text{4R3+R1}\rightarrow \text{R1}} \begin{bmatrix}\bold{4} & \bold{0} & \bold{0}& |& \bold{4} \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix}[/tex]
and,
[tex]\begin{bmatrix}4 & 0 & -4& |& 4 \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix} \xrightarrow{\frac{1}{4} \text{R1}\rightarrow \text{R1}} \begin{bmatrix}\bold{1} & \bold{0} & \bold{0}& |& \bold{1} \\0 & 1 & 0& |& -1 \\0 & 0 & 1& |& 3\end{bmatrix}[/tex]
Learn more about the elementary row operation here:
https://brainly.com/question/17875294
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