CK-12-Pre-Calculus Concepts

http://www.ck12.org Chapter 8. Systems and Matrices

The following system of equations has the solution (1, 3, 7). You can verify this by substituting 1 forx, 3 fory, and
7 forzinto each equation.

x+ 2 y−z= 0

7 x− 0 y+z= 14

0 x+y+z= 10

One thing to be mindful of when given a system of equations is whether or not the equations are linearly indepen-
dent. Three equations are linearly independent if each equation cannot be produced by a linear combination of the
other two.
When solving a system of three equations and three variables, there are a few general guidelines that can be helpful:

• Start by trying to eliminate the first variable in the second row.


• Next eliminate the first and second variables in the third row. This will create zero coefficients in the lower
  right hand corner.


• Repeat this process for the upper right hand corner and you should end up with a very nice diagonal indicating
  whatx,yandzequal.

Example A
Even though you know the solution, solve the system of equations below:

x+ 2 y−z= 0

7 x− 0 y+z= 14

0 x+y+z= 10

Solution: There are a number of ways to solve this system. Common techniques involve swapping rows, dividing
and multiplying a row by a constant and adding or subtracting a multiple of one row to another.
Step 1: Swap rows 2 and 3. Change-0 to +0.

x+ 2 y−z= 0

0 x+y+z= 10

7 x+ 0 y+z= 14

Step 2: Subtract 7 times row 1 to row 3.

x+ 2 y−z= 0

0 x+y+z= 10

0 x− 14 y+ 8 z= 14

Step 3: Add 14 times row 2 to row 3