2007-02-03 16:57:34 · 4 answers · asked by Rosy 3 in Science & Mathematics ➔ Mathematics
If the augmented matrix is all zeros, then that means the system is equivalent to:
0x + 0y + 0z = 0
0x + 0y + 0z = 0
0x + 0y + 0z = 0
(if there's 3 variables).
In this case, there are infinitely many solutions (in fact, any ordered triple). If you pick any values for x, y, and z, all three linear equations are true!
2007-02-03 17:26:28 · answer #1 · answered by Anonymous · 1⤊ 0⤋
The same way you would if it were not all zeroes.
x+2y + 3z =0
2x -4y +6z =0
5x-9y -12z =0
is an example that the augmented portion of the matrix is all zroes. Do you think this system is not solvable?
2007-02-03 17:33:13 · answer #2 · answered by ironduke8159 7 · 0⤊ 1⤋
ask a maths teaher
2007-02-03 17:01:21 · answer #3 · answered by Anonymous · 0⤊ 1⤋
it's undefined. Trick question.
2007-02-03 16:59:57 · answer #4 · answered by Name Unknown 2 · 0⤊ 1⤋