Answer:
[tex]x=-1223,y=-629,z=-31[/tex]
Step-by-step explanation:
The given systems of equations is
[tex]x-2y+z=4[/tex]
[tex]3x-5y-17z=3[/tex]
[tex]2x-6y+43z=-5[/tex]
The augmented matrix is
[tex]\left[\begin{array}{cccc}1&-2&1&|\:\:\:\:4\\3&-5&-17&|\:\:\:\:3\\2&-6&43&|-5\end{array}\right][/tex].
We perform the following row operations to reduce the matrix to reduced row echelon form using row 1 as our pivot row.
[tex]R_2-3R_1\rightarrow R_2[/tex]
[tex]R_3-2R_1\rightarrow R_3[/tex]
[tex]\left[\begin{array}{cccc}1&-2&1&|\:\:\:\:\:\:\:\:\:\:4\\0&1&-20&|\:\:\:\:-9\\0&-2&41&|-13\end{array}\right][/tex]
Next, we perform the following row operations using row 2 as our pivot row to obtain,
[tex]R_1+2R_2\rightarrow R_1[/tex]
[tex]R_3+2R_2\rightarrow R_3[/tex]
[tex]\left[\begin{array}{cccc}1&0&-39&|-14\\0&1&-20&|\:\:-9\\0&0&1&|-31\end{array}\right][/tex]
Next, we perform the following row operations using row 3 as our pivot row to get,
[tex]R_1+39R_3\rightarrow R_1[/tex]
[tex]R_2+20R_3\rightarrow R_2[/tex]
[tex]\left[\begin{array}{cccc}1&0&0&|\:-1223\\0&1&0&|\:\:\:\:-629\\0&0&1&|\:\:\:\:\:\:-31\end{array}\right][/tex]
The matrix is now in the reduced row echelon form,
Therefore the solution is,
[tex]x=-1223,y=-629,z=-31[/tex]
