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- truster
- 2005-12-11 00:59:18
well ,we are here to learn, now im curious about the shorter 'matrix' way.. thanks 
- mathsyperson
- 2005-12-11 00:56:38
I don't know about matrices, but this just looks like a simultaneous equation.
There might be a better way of solving this, but I'd do it like this:
Rearrange equation 3: B = 29 - 3A - 2C
Substitute into equation 2:
5A + 2(29 - 3A - 2C) - 4C = 37
A + 8C = 21
A = 21 - 8C
Substitute into equation 1:
5(21 - 8C) + 3(29 - 3(21 - 8C) - 2C) - 2C = 51
105 - 40C + 87 - 189 + 72C - 6C - 2C = 51
24C = 48
C = 2
Use C to find A:
A = 21 - 8*2 = 5
Use A and C to find B:
B = 29 - 3*5 - 2*2 = 10
So, A = 5, B = 10, C = 2.
As I said, there's probably an shorter way that I don't know about.
- truster
- 2005-12-10 21:42:17
I need help with solving the matrix, I googled but I just don't understand. Found a good site (http://www.purplemath.com/modules/mtrxinvr.htm) but my English isn't that great.
I really want to know how to solve the matrix, please can anyone solve this with the details attached; so I can see how you did it.
The matrix I want to solve goes like this:
5A+3B-2C=51
5A+2B-4C=37
3A+ B+2C=29
I don't know if this is die-hard for 'the math is fun forum', but I'm just a curious beginner, please help me out.