Question

How do you solve `[(1,-2,3), (2,-4,6), (3,-6,9)][(x), (y), (z)]=[(0),(0),(0)]`?

Answer 1

`x - 2y+3z = 0`

Explanation:

We can solve this system of linear equations by using Gaussian Elimination by setting up an augmented matrix of the equation coefficients.

`( (1, -2, 3, |, 0), (2, -4, 6, |, 0), (3, -6, 9, |, 0) )`

We can now perform elementary row operations:

`( (1, -2, 3, |, 0), (2, -4, 6, |, 0), (3, -6, 9, |, 0) ) stackrel(R_2-2R_1 rarr R_2)(rarr) ( (1, -2, 3, |, 0), (0, 0, 0, |, 0), (3, -6, 9, |, 0) )`

`( (1, -2, 3, |, 0), (0, 0, 0, |, 0), (3, -6, 9, |, 0) ) stackrel(R_3-3R_1 rarr R_3)(rarr) ( (1, -2, 3, |, 0), (0, 0, 0, |, 0), (0, 0, 0, |, 0) )`

We can now use back substitution to get the values of `x`, `y`, and `z`:

From Row `1` we have:

`x - 2y+3z = 0`

Thus we have infinitely many solutions.

To interpret this geometrically we have the insertion of three identical planes with equation:

`x - 2y+3z = 0`

whose points of intersection are the entire plane.