How do you solve this system of equations: #8x - y + 2z = 21, x + 2y - z = 1, and 2x + 2y - 3z = - 3#?

1 Answer
Nov 22, 2017

I like to use matrix algebra - it is much easier than elimination and other methods. See below.

Explanation:

Step 1: Put the coefficients of the equations into a matrix we'll call A:

#((8,-1,2),(1,2,-1),(2,2,-3))#

The first horizontal row of the matrix is the coefficients in front of the variables in each equation (i.e. #8x-y+2z# using negatives for subtraction) The second row is the same except for the 2nd equation and so on.

Step 2: Put the answers to the equations into another matrix we'll call B:

#((21),(1),(-3))#

Again, each row here is the answer for the 1st, 2nd, and 3rd equation.

Step 3: Use the equation:

#((8,-1,2),(1,2,-1),(2,2,-3))*((x),(y),(z))=((21),(1),(-3))#

To solve this equation, we have to divide by matrix A on both sides in order to isolate our variables x, y, and z.

Step 4: Take the inverse of the first matrix and multiply it by the second matrix (taking the inverse is the same as dividing because we can't directly divide by a matrix)

#((8,-1,2),(1,2,-1),(2,2,-3))^-1*((21),(1),(-3))=((x),(y),(z))#

Simplify...

#((2),(1),(3))=((x),(y),(z))#

And there's our answer for x, y, and z without doing the long, strenuous elimination method.

If you need clarification on any of my steps or better clarification on what matrices are and how they work, feel free to leave a comment.

I hope this helps!