## Precalculus (6th Edition) Blitzer

The solution to the system of equations is $w=3,x=6,y=-4,z=1$.
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 $w=3,x=6,y=-4,z=1$.