Question 40913
22. Use a determinant to find an equation of the line that passes through the center of the circle (x+3)^2+(y-2)^2=7 and the point (1,4).
COMPARING WITH STANDARD EQN. OF A CIRCLE
(X-H)^2+(Y-K)^2=R^2
,CENTRE IS (H,K)....THAT IS (-3,2)...(X1,Y1)
LINE JOINING THIS WITH (1,4).....(X2,Y2) IS..(Y-Y1)/(Y2-Y1)=(X-X1)/(X2-X1)
(Y-4)/(4-2)=(X-1)/(1+3)
(Y-4)4=2(X-1)
2Y-8=X-1
2Y-X-7=0

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15. Use the inverse matrix method to solve the system of linear equations.
6x-6y-5z=11
3x+6y-z=6
9x-3y+z=0
SEE THE FOLLOWING EXAMPLE AND TRY .IF STILL IN DIFFICULTY PLEASE COME BACK
This is a systems of equations problem and I need to
solve for x,y,z. I do not how to get any of the
variables or what steps to take to get the answers.
The problem:
2x-5y+ ___=-22
__+y+3z=10
x+___+8z=15
The underscores represents blanks or O
If this had to be an augmented matrix, how would I do
that and get those variables?
1 solutions
Answer 20202 by venugopalramana(1462) About Me on
2006-04-16 22:01:33 (Show Source):
2x-5y+ ___=-22...........................I
__+y+3z=10.....................II
x+___+8z=15...........................III
2*EQN.III-EQN.I
2X+16Z-2X+5Y=30+22=52
5Y+16Z=52...........................IV
EQN.IV-5*EQN.II
5Y+16Z-5Y-15Z=52-50=2
Z=2....SUBSTITUTING IN EQN.III
X+8*2=15
X=15-16=-1
SUBSTITUTING FOR Z IN EQN.II
Y+3*2=10
Y=10-6=4
USING MATRIX METHOD...AUGMENTED MATRIX IS
2 -5 0 -22 1 0 0 ?
0 1 3 10 0 1 0 ?
1 0 8 15 0 0 1 ?
NR1=R1/2
1 -2.5 0 -11
0 1 3 10
1 0 8 15
NR3=R3-R1
1 -2.5 0 -11
0 1 3 10
0 2.5 8 26
NR3=R3-2.5*R2
1 -2.5 0 -11
0 1 3 10
0 0 0.5 1
NR3=R3/0.5
1 -2.5 0 -11
0 1 3 10
0 0 1 2
NR2=R2-3R3
1 -2.5 0 -11
0 1 0 4
0 0 1 2
NR1=R1+2.5R2
1 0 0 -1
0 1 0 4
0 0 1 2
HENCE
X=-1
Y=4
Z=2