Solving linear system equation

Question

How would I solve this linear system equation?

$$\begin{cases} 2w+x-y=4\\ 3z-x=6\\ -2y-x+9z+4w=7 \end{cases}$$

First I arranged them so I could make the metric and then I was stuck and I don't realy know how to continue, please help me.

Answer 1

$$2w+x-y=4$$ $$3z-x=6$$ $$-2y-x+9z+4w=7$$ Can be rewritten to $$2w+x-y+0z=4$$ $$0w-x+0y+3z=6$$ $$4w-x-2y+9z=7$$ $$0w+0x+0y+0z=0$$ So now we have $$\left[\begin{array}{cccc|c} 2 & 1 & -1 & 0 & 4 \\ 0 & -1 & 0 & 3 & 6 \\ 4 & -1 & -2 & 9 & 7 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array}\right]$$ $$=\left[\begin{array}{cccc|c} 2 & 1 & -1 & 0 & 4 \\ 0 & -1 & 0 & 3 & 6 \\ 0 & 3 & 0 & -9 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array}\right]$$ $$=\left[\begin{array}{cccc|c} 2 & 1 & -1 & 0 & 4 \\ 0 & -1 & 0 & 3 & 6 \\ 0 & 0 & 0 & 0 & 19 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array}\right]$$ This system is inconsistent because we have $$0w+0x+0y+0z=19$$ Therefore this system has no solutions.

Answer 2

Try multiplying the first equation by $2$, and the second one by $3$: \begin{cases} 4w+2x-2y=8\\ 9z-3x=18\\ -2y-x+9z+4w=7 \end{cases} Then, \begin{cases} 4w-2y=8-2x\\ 9z=18+3x\\ 4w-2y+9z=7+x \end{cases}

And the rest is by positioning the elements from the first two equations in the third one

Answer 3

Write your system into normal form: \begin{cases} x-y+2w=4\\ x-3z=-6\\ x+2y-9z-4w=-7 \end{cases}

Now, depending on the tools you have available, there are several possibilities. The most efficient is Gaussian elimination: the matrix of the system is \begin{align} \left[\begin{array}{cccc|c} 1 & -1 & 0 & 2 & 4 \\ 1 & 0 & -3 & 0 & -6 \\ 1 & 2 & -9 & -4 & -7 \end{array}\right] &\to \left[\begin{array}{cccc|c} 1 & -1 & 0 & 2 & 4 \\ 0 & 1 & -3 & -2 & -10 \\ 1 & 2 & -9 & -4 & -7 \end{array}\right] && R_2\gets R_2-R_1 \\[6px] &\to \left[\begin{array}{cccc|c} 1 & -1 & 0 & 2 & 4 \\ 0 & 1 & -3 & -2 & -10 \\ 0 & 3 & -9 & -6 & -11 \end{array}\right] && R_3\gets R_3-R_1 \\[6px] &\to \left[\begin{array}{cccc|c} 1 & -1 & 0 & 2 & 4 \\ 0 & 1 & -3 & -2 & -10 \\ 0 & 0 & 0 & 0 & 19 \end{array}\right] && R_3\gets R_3-3R_2 \\[6px] \end{align} which shows the system has no solution.

Another possibility is to get $x$ from the second equation: $x=3z-6$. Substitute in the third equation to get $$ 3z-6+2y-9z-4w=-7 $$ or $$ 2y-6z-4w=-1 $$ Substitute also in the first equation to find $3z-6-y+2w=4$, so you get $$ y=3z+2w-10 $$ and now back in the other equation $$ 6z+4w-6z-4w-20=-1 $$ or $$ -20=-1 $$ that is false.