Download Ch. 6 – Systems of Equations

Ch. 7 – Matrices and Systems of
Equations
7.3 – Multivariable Linear Systems
Systems of 3 equations
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Here is an example of a system of 3 linear equations:
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Can do elimination to solve, but must do it a lot
x  2 y  3z  9
 x  3 y  4
2 x  5 y  5 z  17
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Here is an example of the same system of 3 linear equations in
Row-Echelon Form (REF):
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

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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!
x  2 y  3z  9
y  3z  5
z2
REF it!
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Ex 1: Solve this system of equations.




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x  2 y  3z  9
 x  3 y  4
2 x  5 y  5 z  17
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…
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Now 2nd equation is led by y!
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x  2 y  3z  9
y  3z  5
2 x  5 y  5 z  17
x  2 y  3z  9
y  3z  5
REF it!
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(cont’d): Solve this system of equations:

2 x  5 y  5 z  17
Step 3: Third equation must be led by z

Must get rid of x, so add -2•(first equation) and second equation
2 x  4 y  6 z  18
2 x  5 y  5 z  17
 y  z  1

x  2 y  3z  9
y  3z  5
 y  z  1
Must get rid of y, so add 2nd and 3rd equations
y  3z  5
 y  z  1
z2
2z  4
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Back-substitute to get y = -1 and x = 1, so we get…

…(1, -1, 2)
x  2 y  3z  9
y  3z  5
z2
REF it!

Ex 2: Solve this system of equations.
2 x  4 y  z  4
2 x  4 y  6 z  13
x  2 y  0.5 z  2
2 x  4 y  6 z  13
4x  2 y  z  6
4x  2 y  z  6
Divide 1st equation by 2…
4 x  8 y  12 z  26
4x  2 y  z  6
6 y  11z  20
Add -2(E2) and E3 and replace for E2…
x  2 y  0.5 z  2
6 y  11z  20
4x  2 y  z  6
Add -4(E1) and E3 and replace for E3…
REF it!
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Ex 2: Cont’d.
4 x  8 y  2 z  8
4x  2 y  z  6
10 y  z  14
x  2 y  0.5 z  2
6 y  11z  20
 10 y  z  14
Add 5(E2) and 3(E3) and replace for E3…
30 y  55 z  100
30 y  3 z  42
58z  58
x  2 y  0.5 z  2
6 y  11z  20
 58 z  58
Divide E2 and E3 down to REF…
REF it!
x  2 y  0.5 z  2
11
10
y z 
6
3
z 1
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Ex 2: Cont’d.
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Now back-substitute…
11
10
y 
6
3

3
y
2
Answer: ( ½ , -3/2 , 1 )
 3
x  2     0.5(1)  2
 2
1
x
2
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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!
x  2 y  3z  5
 x  3 y  5z  4
2 x  3z  0
Write as a 3x4 matrix…
1
2
3
1
3
5 4
5
0 1 2 9
1
2
3
1
2
3
0
5 4
3 0
5
Add E1 and E2 and replace for E2…
1 2
3
0
2
2 9
3 0
1
0
5
Add -2(E1) and E3 and replace for E3…
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Ex 3: cont’d.
2 4 6 10
0 3
2
0
0 4 9 10
1 2
3
0
0
2 9
9 10
1
4
5
Add -4(E2) and E3 and replace for E3…
0 4
8
0
9 10
4
36
0 0 1 46
1 2
3
0
0
2 9
1 46
1
0
5
Divide E3 by -1 to get…
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Ex 3: Cont’d.
3
0
0
2 9
1 46
1
0
5
x  2 y  3z  5
y  2z  9
z  46
Now back-substitute…
y  2(46)  9

1 2
y  101
Answer: ( 69 , 101 , 46 )
x  2(101)  3(46)  5
x  69
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Ex 4: Solve this system of equations.
2 x  y  3z  4
4 x  2 z  10
2 x  3 y  13z  8
Write as a 3x4 array…
4 2 6  8
4
0
2 10
0 2 8 2
2
4
1
0
3
2
4
10
2 3 13 8
Add -2(E1) and E2 and replace for E2…
2
0
1
2
3
8
2
3
13 8
4
2
Add E1 and E3 and replace for E3…
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Ex 4: Solve this system of equations.
2 1 3 4
2 3 13 8
0 4 16 4
2
3
1
4
0 2 8
2
0 4 16 4
Add 2(E2) and E3 and replace for E3…
0 4
16
0
16 4
0
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4
0
0
4
0
2
1
0 2
0 0
3 4
8
0
2
0
We’re left with 0 = 0 in the 3rd equation. Since this equation is true,
we have an infinite number of solutions…
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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
za
2 y  8(a)  2
2 x  y  3z  4
 2 y  8z  2
y  4a  1
2 x  (4a  1)  3(a)  4

Answer: ( 5/2 – ½a , 4a – 1 , a )
5 1
x  a
2 2
0z  0