Hi,

I have a system of equations with 3 unknowns, the coefficients are shown as a matrix below, and wondered if this would be a good way to solve for the variables?

 2270       -1800         0          |    15

-1800        3480       -680       |     0

    0           -680         1150      |    -8

I'd like to calculate all the steps manually, but all the matrix examples i've seen so far seem to use much simpler, single digit numbers, so what i'm trying to ask in a roundabout way is; do numbers this size lend themselves to a matrix, or is their a better way to solve?

Thanks for looking, and any help is much appreciated.

Matrix is probably the best way.  Use a calculator.

I agree with Mathematic.  But, if you don't use a matrix use substitution.   Solve the top equation for x as a function of y (assuming the variables are labeled x, y and z), the bottom for z as a function of y, then put those two results in the center equation and you will have only one variable (y).  The numbers look like they will be messy, but it would work.

"but all the matrix examples I've seen so far seem to use much simpler, single digit numbers".  So it is the arithmetic that is stopping you?

Using x, y, and z as unknowns, the equations are

2270x+ 1800y= 15. 

-1800x+ 3480y- 680z= 0

-680y- 1150z= -8

 Noticing that "y" is the only unknown to appear in all three equations, I would solve the first and third equation for x and z in terms of y:  x= (15- 1800y)/2270= (3- 160y)/454.

z= (8- 680y)/1150= (4- 340y)/575.

So the middle equation becomes -1800((3- 160y)/454)+ 3480y- 680((4- 340y)/575)= 0.  Use a calculator if those numbers are too much for you.