1. Ch. 7 – Matrices and Systems of Equations
7.3 – Multivariable Linear Systems

2. Systems of 3 equations
Here is an example of a system of 3 linear equations:
Can do elimination to solve, but must do it a lot
Here is an example of the same system of 3 linear equations in Row-Echelon Form (REF):
REF  equations are in a stair-step pattern, leading coefficient is 1
It’s much easier to solve!
Just plug in 2 for z to find y, then plug that in to find x!
We want multivariable equations to look like this!

3. REF it! Ex 1: Solve this system of equations.
To solve, we will do several eliminations to get the equations in REF!
To do that, we need our stair-step pattern and leading coefficients of 1!
ADD and REPLACE!
Step 1: First equation must be led by x - CHECK!
Step 2: Second equation must be led by y
Must get rid of x, so add equations 1 and 2
The new equation is y + 3z = 5, so replace it for the second equation to get…
Now 2nd equation is led by y!

4. REF it! (cont’d): Solve this system of equations:
Step 3: Third equation must be led by z
Must get rid of x, so add -2•(first equation) and second equation
Must get rid of y, so add 2nd and 3rd equations
Back-substitute to get y = -1 and x = 1, so we get…
…(1, -1, 2)

5. REF it! Ex 2: Solve this system of equations.
Add -2(E2) and E3 and replace for E2…
Divide 1st equation by 2…
Add -4(E1) and E3 and replace for E3…

6. REF it! Ex 2: Cont’d. Add 5(E2) and 3(E3) and replace for E3…
Divide E2 and E3 down to REF…

7. REF it!
Ex 2: Cont’d.
Now back-substitute…
Answer: ( ½ , -3/2 , 1 )

8. Ex 3: Solve this system of equations.
You can always rewrite the system as an augmented matrix (3x4) of coefficients – it will save you lead!
Just get 1’s in the main diagonal!
Add E1 and E2 and replace for E2…
Write as a 3x4 matrix…
Add -2(E1) and E3 and replace for E3…

9. Ex 3: cont’d. Add -4(E2) and E3 and replace for E3…
Divide E3 by -1 to get…

10. Ex 3: Cont’d.
Now back-substitute…
Answer: ( 69 , 101 , 46 )

11. Ex 4: Solve this system of equations.
Add -2(E1) and E2 and replace for E2…
Write as a 3x4 array…
Add E1 and E3 and replace for E3…

12. Ex 4: Solve this system of equations.
We’re left with 0 = 0 in the 3rd equation. Since this equation is true, we have an infinite number of solutions…
Add 2(E2) and E3 and replace for E3…

13. Ex (cont’d):
If the answer is infinite solutions, we must find a generic triple that solves the system:
Step 1: Let z = a.
Step 2: Back-substitute to get expressions for x and y in terms of a
Answer: ( 5/2 – ½a , 4a – 1 , a )