Solve the system by finding the reduced row-echelon form of the augmented matrix

Question

Solve the system by finding the reduced row-echelon form of the augmented matrix.
x-4y-2z=2
2x-11y-10z=13
-x+6y+6z=-8
reduced row-echelon form:
I ANSWERED
1 -4   -2    2
0 1   -10   -3
0   0   0      0
(It's wrong, and I don't know why ? )


If there is one solution, give its coordinates in the answer spaces below.
If there are infinitely many solutions, enter z in the answer blank for , enter a formula for in terms of in the answer blank for and enter a formula for in terms of in the answer blank for .

If there are no solutions, leave the answer blanks for , and empty.

How can I find my X, Y ,Z ? Is it going to be Zero ?!

Explanation / Answer

Solve the system by finding the reduced row-echelon form of the augmented matrix.
x-4y-2z=2
2x-11y-10z=13
-x+6y+6z=-8
reduced row-echelon form:
I ANSWERED
1 -4   -2    2
0 1   -10   -3
0   0   0      0
(It's wrong, and I don't know why ? )

SEE CORRECT SOLUTION BELOW

If there is one solution, give its coordinates in the answer spaces below.

THERE ARE INFINITE SOLUTIONS ...

If there are infinitely many solutions, enter z in the answer blank for , enter a formula for in terms of in the answer blank for and enter a formula for in terms of in the answer blank for .

Z = ANY REAL NUMBER

Y=-3-2Z

X=-10-6Z

EXAMPLES ARE

ETC...

How can I find my X, Y ,Z ? Is it going to be Zero ?!

AUGMENTED MATRIX 1 -4 -2 2 2 -11 -10 13 -1 6 6 -8 NR2=R2-2R1….NR3=R3+R1 1 -4 -2 2 0 -3 -6 9 0 2 4 -6 NR1=R1-4R2/3….NR2=-R2/3…..NR3=R3+2R2/3 1 0 6 -10 0 1 2 -3 0 0 0 0 SINCE LAST ROW IS ALL ZEROS , THE EQUATIONS ARE DEPENDENT AND CONSISTENT. HENCE WE HAVE 2 INDEPENDENT EQNS.WITH 3 VARIABLES. SO ONE VARIABLE IS FREE…LET IT BE Z . SO FIRST 2 EQNS. GIVE US X+6Z=-10….X=-10-6Z Y+2Z=-3…..Y=-3-2Z HENCE THE SOLUTION SET IS X=-10-6Z Y=-3-2Z Z=ANY REAL NUMBER IF WE TAKE Z=0 , THEN THE SOLUTION SET IS X=-10……Y=-3…..Z=0 IF WE TAKE Z=1 THEN THE SOLUTION SET IS X=-16 …….Y=-5….Z=1

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