# Solve the system of linear equations and interpret your solution geometrically: [1] 2x + y + 2z...

## Question:

Solve the system of linear equations and interpret your solution geometrically:

[1] 2x + y + 2z - 4 = 0

[2] x - y - z - 2 = 0

[3] x + 2y - 6z - 12 = 0

## Systems of Linear Equations in multiple variables

Systems of linear equations can have many variables. A solution to a system of linear equations in multiple variables is a solution to each equation simultaneously. Systems of linear equations can be solved using standard techniques of substitution or elimination, which are commonly learned in the case of two equations and two variables, or by using a matrix and performing Gaussian elimination or Gauss-Jordan elimination. Although tedious, using a matrix to solve a system can be easier than using substitution or elimination, in particular when more equations and more variables are involved. A system of linear equations can have one solution, no solution, or infinitely many solutions

Putting the system into an augmented matrix, but in the order Equation 2, Equation 1, Equation 3, we have

{eq}\begin{bmatrix} 1&-1&-1&|&2\\ 2&1&2&|&4\\ 1&2&-6&|&12 \end{bmatrix} {/eq}

Performing the row operations R2-> R2-2R1 and R3->R3-R1,

{eq}\begin{bmatrix} 1&-1&-1&|&2\\ 0&3&4&|&0\\ 0&3&-5&|&10 \end{bmatrix} {/eq}

Then performing the row operation R3->R3-R2,

{eq}\begin{bmatrix} 1&-1&-1&|&2\\ 0&3&4&|&0\\ 0&0&-9&|&10 \end{bmatrix} {/eq}

Translating back into a system of equations, we have

{eq}x-y-z=2\\ 3y+4z=0\\ -9z=10 {/eq}

First solving equation 3 for z, then substituting into equation 2 to find y, then substituting both into equation 1 to find x, we have

{eq}x=\frac{64}{27}\\ y=\frac{40}{27}\\ z=\frac{-10}{9} {/eq}

Geometrically, this means that the three planes defined by the three original equations intersect each other at exactly one point, the point {eq}\left(\frac{64}{27} , \frac{40}{27} , \frac{-10}{9}\right) {/eq}