Perform the indicated operation.



40−4
010
001
−8
−1
3





⇒ 40
⇒ 0
010
001
⇒ 4
−1
3


Multiply row 1 by the reciprocal of 4.



400
010
001
4
−1
3





100
010
001
−1
3

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]

"A set of numbers arranged in rows and columns so as to form a rectangular array."

"A matrix formed by combining the columns of two matrices to form a new matrix."

"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:

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