Answer
The solution to the system of equations is $ w=3,x=6,y=-4,z=1$.
Work Step by Step
Consider the following system of equations:
$\begin{align}
& w+x+y+z=6 \\
& w-x+3y+z=-14 \\
& w+2x-3z=12 \\
& 2w+3x+6y+z=1
\end{align}$
Express the above system in the form of a matrix as follows:
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 6 \\
1 & -1 & 3 & 1 & -14 \\
1 & 2 & 0 & -3 & 12 \\
2 & 3 & 6 & 1 & 1 \\
\end{matrix} \right]$
Using the elementary row operations we will obtain the echelon form of a matrix:
${{R}_{2}}\to {{R}_{2}}+\left( -1 \right){{R}_{1}},{{R}_{3}}\to {{R}_{3}}+\left( -1 \right){{R}_{1}},{{R}_{4}}\to {{R}_{4}}+\left( -2 \right){{R}_{1}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 6 \\
0 & -2 & 2 & 0 & -20 \\
0 & 1 & -1 & -4 & 6 \\
0 & 1 & 4 & -1 & -11 \\
\end{matrix} \right]$
${{R}_{2}}\to -\frac{1}{2}{{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 6 \\
0 & 1 & -1 & 0 & 10 \\
0 & 1 & -1 & -4 & 6 \\
0 & 1 & 4 & -1 & -11 \\
\end{matrix} \right]$
${{R}_{3}}\to {{R}_{3}}+\left( -1 \right){{R}_{2}},{{R}_{4}}\to {{R}_{4}}+\left( -1 \right){{R}_{2}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 6 \\
0 & 1 & -1 & 0 & 10 \\
0 & 0 & 0 & -4 & -4 \\
0 & 0 & 5 & -1 & -21 \\
\end{matrix} \right]$
${{R}_{3}}\leftrightarrow {{R}_{4}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 6 \\
0 & 1 & -1 & 0 & 10 \\
0 & 0 & 5 & -1 & -21 \\
0 & 0 & 0 & -4 & -4 \\
\end{matrix} \right]$
${{R}_{4}}\to -\frac{1}{4}{{R}_{4}}$ gives,
$\left[ \begin{matrix}
1 & 1 & 1 & 1 & 6 \\
0 & 1 & -1 & 0 & 10 \\
0 & 0 & 5 & -1 & -21 \\
0 & 0 & 0 & 1 & 1 \\
\end{matrix} \right]$
Rewrite the system corresponding to echelon form of the matrix:
$ w+x+y+z=6$ (I)
$ x-y=10$ (II)
$5y-z=-21$ (III)
$ z=1$ (IV)
Apply the back-substitution method:
Equation (IV) gives,
$ z=1$.
Substitute the value of z in equation (III) as follows:
$5y-\left( 1 \right)=-21$
It can be further simplified as:
$ y=-4$.
Substitute the values of y and z in equation (II) as follows:
$ x-\left( -4 \right)=10$
Simplify:
$ x=6$
Substitute the values of x, y, and z in equation (I) as follows:
$ w+\left( 6 \right)+\left( -4 \right)+\left( 1 \right)=6$
It can be further simplified as:
$ w=3$
Hence, the solution to the system is $ w=3,x=6,y=-4,z=1$.
