The given matrix is:
[tex]\begin{bmatrix}{1} & -{6} & {5}|-3 \\ 6 & {-4} & {8|-5} \\ -4{} & {-5} & {3|5}\text{ }\end{bmatrix}[/tex]Each term in the matrix is the argument of a system of equations in the form:
ax+by+cz=d
Where a, b, c and d, are the elements in each row of the matrix. Then, the system of equations is:
[tex]\begin{gathered} 1x-6y+5z=-3 \\ 6x-4y+8z=-5 \\ -4x-5y+3z=5 \end{gathered}[/tex]And by simplifying it we obtain:
[tex]\begin{gathered} x-6y+5z=-3 \\ 6x-4y+8z=-5 \\ -4x-5y+3z=5 \end{gathered}[/tex]The answer is option D.
b. The row operations are:
[tex]\begin{gathered} R_2=-6r_1+r_2 \\ R_3=4r_1+r_3 \end{gathered}[/tex]Let's start with the first operation to find R2:
[tex]\begin{gathered} -6r1+r2=R2 \\ -6(1)+6=-6+6=0\text{ (first term of the R2)} \\ -6(-6)+(-4)=36-4=32\text{ (second term of R2)} \\ -6(5)+8=-30+8=-22\text{ (third term of R2)} \\ -6(-3)+(-5)=18-5=13\text{ (fourth term of R2)} \end{gathered}[/tex]Now, for R3 we have:
[tex]\begin{gathered} 4r1+r3=R3 \\ 4(1)+(-4)=4-4=0\text{ (first term of the R3)} \\ 4(-6)+(-5)=-24-5=-29\text{ (second term of the R3)} \\ 4(5)+(3)=20+3=23\text{ (third term of the R3)} \\ 4(-3)+(5)=-12+5=-7\text{ (fourth term of the R3)} \end{gathered}[/tex]Then, the resulting matrix is:
[tex]\begin{bmatrix}{1} & -{6} & {5}|-3 \\ 0 & {32} & {-22|13} \\ 0{} & {-29} & {23|-7}\text{ }\end{bmatrix}[/tex]