If you were to use Gauss-Jordan elimination on the following linear system, what would the augmented matrix be

Question

? (please help iam stuck)

3w − x = 2y + z − 4
9x − y + z = 10
4w + 3y − z = 7
12x + 17 = 2y − z + 6

Answer 1

Rewriting the equations in a consistent form:
3w - x - 2y - z = -4
0w + 9x - y + z = 10
4w + 0x + 3y - z = 7
0w + 12x - 2y + z = -11

So the initial augmented matrix is
[ 3 -1 -2 -1 | -4]
[ 0 9 -1 1 | 10]
[ 4 0 3 -1 | 7]
[ 0 12 -2 1 | -11]

(sorry about the alignment!)

Answer 2

3w - x = 2y + z - 4
9x - y + z = 10
4w + 3y - z = 7
12x + 17 = 2y - z + 6

To obtain the augmented matrix, your first step would be to write every equation in the form aw + bx + cy + dz = k, for constants
a, b, c, d, and k. The result should be:

3w - x - 2y - z = -4
0w + 9x - y + z = 10
4w + 0x + 3y - z = 7
0w + 12x - 2y + z = -11

Note how I put 0w, or 0x, if the variable was nonexistent.

The augmented matrix would contain all of the coefficients, as well as all of the answers, of the above.

[3 -1 -2 -1 | -4]
[0 9 -1 1 | 10]
[4 0 3 -1 | 7]
[0 12 -2 1 | -11]

{But it would look much cleaner and aligned on paper}.

Answer 3

it extremely is b a million*x + 0*y -3*z = 6 0*x + a million*y +2*z = 7, so the proposed answer satisfies the equations. The 3x3 matrix (the left facet 3 columns of your matrix) is obviously singular (the final row is all zeros), meaning you have greater unknows (ranges of freedom) than equations (constraints). The rank of your 3x3 matrix is two, you have 3 columns, so which you're unfastened to set certainly one of your variables (e.g. x), although you like it, and nonetheless set the different 2 (y and z) to fulfill the equations. a 0.33 one (y)