Question

write the system of equations corresponding to the given augmented matrix using the variables x and y. then perform the row operation r₂ = -5r₁ + r₂ on the given augmented matrix.\
\\\

$$\begin{bmatrix}1 & 7 & \\bigm| & 1\\\\5 & -5 & \\bigm| & -1\\end{bmatrix}$$

\
\\\
which of the following is the system of equations corresponding to the augmented matrix?\
a. \\(\

$$\begin{cases}x + 7y = -1\\\\5x - 5y = 1\\end{cases}$$

\\)\
b. \\(\

$$\begin{cases}x + 5y = 1\\\\7x - 5y = -1\\end{cases}$$

\\)\
c. \\(\

$$\begin{cases}x + 7y = 1\\\\5x - 5y = -1\\end{cases}$$

\\)\
d. \\(\

$$\begin{cases}-5x + 7y = 1\\\\5x + y = -1\\end{cases}$$

\\)\
perform the row operation. what is the resultant matrix?\
\\\

$$\begin{bmatrix}\\square & \\square & \\bigm| & \\square\\\\\\square & \\square & \\bigm| & \\square\\end{bmatrix}$$

\
\\

Explanation:

Step1: Analyze the augmented matrix

The augmented matrix is \(

$$\begin{bmatrix}1 & 7 &\bigm| & 1\\5 & -5 &\bigm| & -1\end{bmatrix}$$

\). For a system of linear equations \(a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), the augmented matrix is \(

$$\begin{bmatrix}a_1 & b_1 &\bigm| & c_1\\a_2 & b_2 &\bigm| & c_2\end{bmatrix}$$

\). So the first equation is \(x + 7y=1\) and the second is \(5x - 5y=-1\), which matches option C.

Step2: Perform row operation \(R_2=-5r_1 + r_2\)

First, find \(-5r_1\): \(-5\times

$$\begin{bmatrix}1 & 7 &\bigm| & 1\end{bmatrix}$$

=

$$\begin{bmatrix}-5 & -35 &\bigm| & -5\end{bmatrix}$$

\)
Then add \(r_2\) (which is \(

$$\begin{bmatrix}5 & -5 &\bigm| & -1\end{bmatrix}$$

\)) to \(-5r_1\):
For the first element: \(-5 + 5 = 0\)
For the second element: \(-35+(-5)=-40\)
For the third element (the constant term): \(-5+(-1)=-6\)
Wait, no, wait. Wait, the row operation is \(R_2=-5r_1 + r_2\), so for the matrix \(

$$\begin{bmatrix}r_1\ _2\end{bmatrix}$$

\), new \(r_2=-5r_1 + r_2\).
So \(r_1=

$$\begin{bmatrix}1 & 7 &\bigm| & 1\end{bmatrix}$$

\), \(r_2=

$$\begin{bmatrix}5 & -5 &\bigm| & -1\end{bmatrix}$$

\)
New \(r_2\) elements:
First element: \(-5\times1 + 5=0\)
Second element: \(-5\times7+(-5)=-35 - 5=-40\)
Third element: \(-5\times1+(-1)=-5 - 1=-6\)? Wait, no, maybe I misread the original matrix. Wait the original matrix is \(

$$\begin{bmatrix}1 & 7 &\bigm| & 1\\5 & -5 &\bigm| & -1\end{bmatrix}$$

\)? Wait the user's matrix: first row [1,7,1], second row [5,-5,-1]. So row operation \(R_2=-5r_1 + r_2\):
\(R_2\) first element: \(-5(1)+5 = 0\)
\(R_2\) second element: \(-5(7)+(-5)=-35 - 5=-40\)
\(R_2\) third element: \(-5(1)+(-1)=-6\)? Wait, but maybe I made a mistake. Wait, no, let's check again. Wait the original matrix: first row 1,7,1; second row 5,-5,-1. So \(R_2=-5r_1 + r_2\) means:
For the coefficients of x: \(-5\times1 + 5 = 0\)
For coefficients of y: \(-5\times7+(-5)=-35 - 5=-40\)
For the constants: \(-5\times1+(-1)=-6\). So the new matrix is \(

$$\begin{bmatrix}1 & 7 &\bigm| & 1\\0 & -40 &\bigm| & -6\end{bmatrix}$$

\)
Wait, but maybe the original matrix was different? Wait the user's image: the augmented matrix is \(

$$\begin{bmatrix}1 & 7 &\bigm| & 1\\5 & -5 &\bigm| & -1\end{bmatrix}$$

\)? Wait the first part: "Write the system of equations corresponding to the given augmented matrix using the variables x and y. Then perform the row operation \(R_2 = -5r_1 + r_2\) on the given augmented matrix."
So first, the system: from the augmented matrix \(

$$\begin{bmatrix}1 & 7 &\bigm| & 1\\5 & -5 &\bigm| & -1\end{bmatrix}$$

\), the system is \(x + 7y = 1\) and \(5x - 5y=-1\), which is option C.
Then row operation \(R_2=-5r_1 + r_2\):
Compute each element of \(r_2\) as \(-5\times\) (element of \(r_1\)) + (element of \(r_2\))
First element (x coefficient): \(-5(1)+5 = 0\)
Second element (y coefficient): \(-5(7)+(-5)=-35 - 5=-40\)
Third element (constant): \(-5(1)+(-1)=-6\)
So the resultant matrix is \(

$$\begin{bmatrix}1 & 7 &\bigm| & 1\\0 & -40 &\bigm| & -6\end{bmatrix}$$

\)

Answer:

The system of equations corresponds to option C: \(

$$\begin{cases}x + 7y = 1\\5x - 5y=-1\end{cases}$$

\). The resultant matrix after \(R_2=-5r_1 + r_2\) is \(

$$\begin{bmatrix}1 & 7 &\bigm| & 1\\0 & -40 &\bigm| & -6\end{bmatrix}$$

\)