SOLUTION: using matrices 2x-y-z=-3
x-2y+3z=6
x+y+z=6
thank you
Algebra.Com
Question 63397: using matrices 2x-y-z=-3
x-2y+3z=6
x+y+z=6
thank you
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLSolve the system of equations using matrices.
3x - 2y + z = -7
x + y - 2z = 12
3x + y - z = 10
METHOD:
1.WRITE THE GIVEN EQNS.AS A MATRIX,EACH EQN.IN ONE ROW AND
EACH COLUMN AN UNKOWN AND LASTLY THE CONSTANT TERM IN THE SAME ORDER
2.LITERALLY EACH ROW REPRESENTS ONE EQN. WITH MULTIPLICATION BY
X,Y,Z AS UNDERSTOOD AND OMITTED AND THE LAST CONSTANT HAVING
AN = SIGN IN FRONT.
3.HENCE IF WE MAKE IT IDENTITY MATRIX IN THE UNKNOWNS ZONE ,IT
WOULD MEAN THE CONSTANT TERMS GIVE VALUES OF UNKNOWNS IN THAT ORDER
4.ACHIEVE IT BY LINEAR TRANSFORMATIONS AS SHOWN BELOW.
GIVEN MATRIX EACH EQN.IN A ROW,EACH COLUMN FOR..X,Y,Z AND LAST COL. FOR CONSTANT.
3 -2 1 -7
1 1 -2 12
3 1 -1 10
R1=R1/3
1 -0.666666667 0.333333333 -2.333333333
1 1 -2 12
3 1 -1 10
R2=R2-R1
1 -0.666666667 0.333333333 -2.333333333
0 1.666666667 -2.333333333 14.33333333
3 1 -1 10
R3=R3-3*R1
1 -0.666666667 0.333333333 -2.333333333
0 1.666666667 -2.333333333 14.33333333
0 3 -2 17
R2=R2/(5/3)
1 -0.666666667 0.333333333 -2.333333333
0 1 -1.4 8.6
0 3 -2 17
R3=R3-3*R2
1 -0.666666667 0.333333333 -2.333333333
0 1 -1.4 8.6
0 0 2.2 -8.8
R3=R3/2.2
1 -0.666666667 0.333333333 -2.333333333
0 1 -1.4 8.6
0 0 1 -4
R2=R2+1.4*R3
1 -0.666666667 0.333333333 -2.333333333
0 1 0 3
0 0 1 -4
R1=R1-R3/3
1 -0.666666667 0 -1
0 1 0 3
0 0 1 -4
R1=R1+2R2/3
1 0 0 1
0 1 0 3
0 0 1 -4
HENCE X=1….Y=3…..Z=-4
LOWING EXAMPLE AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
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