Question

What is the solution to the system (1)x-y+a=-1(2)x+y+3z=-3(3)2x-y+2z=0

Answer 1

Convert first the given into an augmented matrix.

`\mleft[\begin{array}c1 & -1 & 1 & 1 \\ 1 & 1 & 3 & 3 \\ 2 & -1 & 2 & 0\end{array}\mright]`

Next, perform Gauss-Jordan elimination or row reduced echelon form to find each values for x, y, and z

`\begin{gathered} R_2=R_2-R_1\mleft[\begin{array}ccc1 & -1 & 1 & 1 \\ 0 & 2 & 2 & 2 \\ 2 & -1 & 2 & 0\end{array}\mright] \\ R_3=R_3-2R_1\mleft[\begin{array}ccc1 & -1 & 1 & 1 \\ 0 & 2 & 2 & 2 \\ 0 & 1 & 0 & -2\end{array}\mright] \\ R_2=(R_(2))/(2)\mleft[\begin{array}c1 & -1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & -2\end{array}\mright] \end{gathered}`
`\begin{gathered} R_1=R_1+R_2\mleft[\begin{array}ccc1 & 0 & 2 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & -2\end{array}\mright] \\ R_3=R_3-R_2\mleft[\begin{array}c1 & 0 & 2 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & -1 & -3\end{array}\mright] \end{gathered}`
`\begin{gathered} R_3=-R_3\mleft[\begin{array}ccc1 & 0 & 2 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 3\end{array}\mright] \\ R_1=R_1-2R_3\mleft[\begin{array}ccc1 & 0 & 0 & -4 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 3\end{array}\mright] \\ R_2=R_2-R_3\mleft[\begin{array}ccc1 & 0 & 0 & -4 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 3\end{array}\mright] \end{gathered}`

At the last solution of the row reduced echelon form, we determine that the solution to the system is (x,y,z) = (-4,-2,3)