2w-3x+4y-z=07w-x-8y+9z=02w+8x+y-z=0

To solve the system of equations:
$2w-3x+4y-z=0$
$7w-x-8y+9z=0$
$2w+8x+y-z=0$
We can use the method of elimination to solve the system. The goal is to eliminate one variable at a time by adding or subtracting equations. Here's how we can proceed:
Step 1: Let's eliminate the variable $w$ by multiplying the first equation by $-7$ and the second equation by $2$. This will allow us to add the equations and eliminate $w$.
$-7(2w-3x+4y-z)=-7(0)$
$2(7w-x-8y+9z)=2(0)$
Simplifying these equations gives us:
$-14w+21x-28y+7z=0$
$14w-2x-16y+18z=0$
Adding the above two equations:
$(-14w+21x-28y+7z)+(14w-2x-16y+18z)=0+0$
Simplifying this equation yields:
$19x-44y+25z=0$
Step 2: Now, let's eliminate the variable $x$ by multiplying the first equation by $2$ and the third equation by $3$. This will allow us to add the equations and eliminate $x$.
$2(2w-3x+4y-z)=2(0)$
$3(2w+8x+y-z)=3(0)$
Simplifying these equations gives us:
$4w-6x+8y-2z=0$
$6w+24x+3y-3z=0$
Adding the above two equations:
$(4w-6x+8y-2z)+(6w+24x+3y-3z)=0+0$
Simplifying this equation yields:
$10w+32y-5z=0$
Step 3: Now, let's eliminate the variable $y$ by multiplying the first equation by $22$ and the third equation by $11$. This will allow us to add the equations and eliminate $y$.
$22(2w-3x+4y-z)=22(0)$
$11(2w+8x+y-z)=11(0)$
Simplifying these equations gives us:
$44w-66x+88y-22z=0$
$22w+88x+11y-11z=0$
Adding the above two equations:
$(44w-66x+88y-22z)+(22w+88x+11y-11z)=0+0$
Simplifying this equation yields:
$66w+22x+77z=0$
Now, we have three equations:
$19x-44y+25z=0$
$10w+32y-5z=0$
$66w+22x+77z=0$
We can solve this system of equations using any method of our choice. One common method is Gaussian elimination. Let's proceed with Gaussian elimination to solve the system.
To solve the system using Gaussian elimination, we'll create an augmented matrix with the coefficients of the variables:
$\left[\begin{array}{ccc}19& -44& 25\\ 10& 32& -5\\ 66& 22& 77\end{array}\right]$
Now, we'll perform row operations to transform the matrix into row-echelon form. Our goal is to obtain zeros below the main diagonal.
First, let's interchange the first and second rows:
$\left[\begin{array}{ccc}10& 32& -5\\ 19& -44& 25\\ 66& 22& 77\end{array}\right]$
Next, we'll multiply the first row by $19$ and the second row by $10$, and then subtract the second row from the first row:
$\left[\begin{array}{ccc}190& 608& -95\\ 0& -1160& 480\\ 66& 22& 77\end{array}\right]$
Now, let's multiply the first row by $66$ and the third row by $190$, and then subtract the third row from the first row:
$\left[\begin{array}{ccc}0& 208& -4085\\ 0& -1160& 480\\ 66& 22& 77\end{array}\right]$
Finally, let's divide the second row by $-1160$ to simplify the matrix:
$\left[\begin{array}{ccc}0& 1& \frac{17}{116}\\ 0& 1& -\frac{6}{29}\\ 66& 22& 77\end{array}\right]$
Now, the matrix is in row-echelon form. We can use back substitution to find the values of the variables.
From the first row, we have:
$0x+1y+\frac{17}{116}z=0$
This simplifies to:
$y+\frac{17}{116}z=0$
From the second row, we have:
$0x+1y-\frac{6}{29}z=0$
This simplifies to:
$y-\frac{6}{29}z=0$
From the third row, we have:
$66w+22x+77z=0$
We can solve the above equations simultaneously. From the first two equations, we can express $y$ in terms of $z$:
$y=-\frac{17}{116}z$
$y=\frac{6}{29}z$
Setting these two expressions equal to each other, we have:
$-\frac{17}{116}z=\frac{6}{29}z$
To solve for $z$, we can cross-multiply:
$-\frac{17}{116}z·29=\frac{6}{29}z·116$
$-493z=696z$
$-493z-696z=0$
$-1189z=0$
Dividing both sides by $-1189$, we find:
$z=0$
Substituting this value of $z$ back into the equation $y=\frac{6}{29}z$, we get:
$y=\frac{6}{29}·0$
$y=0$
Finally, substituting $z=0$ and $y=0$ into the equation $66w+22x+77z=0$, we have:
$66w+22x+77·0=0$
$66w+22x=0$
Dividing both sides by $22$, we get:
$3w+x=0$
We have obtained the values of $w$, $x$, $y$, and $z$:
$w$ can take any value,
$x=-3w$,
$y=0$,
$z=0$.
Therefore, the solution to the system of equations is:
$w=w$,
$x=-3w$,
$y=0$,
$z=0$.