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Question 150173: Please help me solve this using matrices:
x + 2y - z = 2
2x - 2y + z = -1
6x +4y +3z = 5
Thank you so much, Lennie: Please help me solve this using matrices:
x + 2y - z = 2
2x - 2y + z = -1
6x +4y +3z = 5
Thank you so much, Lennie
Answer by Fombitz(1744) About Me  (Show Source):
You can put this solution on YOUR website!
In Matrix notation, the problem is,
(<BR> matrix( 3, 3, <BR> 1, 2, -1,<BR> 2, -2, 1,<BR> 6, 4, 3<BR> ))*<BR> (<BR> matrix( 3, 1, <BR> x, <BR> y, <BR> z<BR> )<BR> )=<BR> (<BR> matrix( 3, 1, <BR> 2, <BR> -1, <BR> 5<BR> )<BR> )
or simplified
[A][x]=[b]
[Ainv][A][x]=[Ainv][b]
[x]=[Ainv][b]
Find the inverse of matrix A, multiply by the right hand side to find the solution.
In this case,
[A]=(<BR> matrix( 3, 3, <BR> 1, 2, -1,<BR> 2, -2, 1,<BR> 6, 4, 3<BR> ))
The inverse of [A] is,
[Ainv]=(1/30)*(<BR> matrix( 3, 3, <BR> 10, 10, 0,<BR> 0, -9, 3,<BR> -20, -8, 6<BR> ))
Then the matrix multiplication is,
[x]=(1/14)*(<BR> matrix( 3, 3, <BR> 10, 10, 0,<BR> 0, -9, 3,<BR> -20, -8, 6<BR> ))*(<BR> matrix( 3, 1, <BR> 2, <BR> -1, <BR> 5<BR> )<BR> )
x=(1/30)(10*2+(10)(-1)+0(5))
x=(1/30)(20-10)
x=(1/30)(10)
x=1/3
y=(1/30)(0*2+(-9)(-1)+(3)(5))
y=(1/30)(9+15)
y=(1/30)(24)
y=4/5
z=(1/30)(20*2+(8)(-1)+(-6)(5))
z=(1/30)(40-8-30)
z=(1/30)(2)
z=-1/15
Your solution matrix is then
[x]=(<BR> matrix( 3, 1, <BR> 1/3, <BR> 4/5, <BR> -1/15<BR> )<BR> )