# Math 2250-1 Fri Sept 14 Solutions to linear equations in 3 unknowns:

Math 2250-1 Fri Sept 14 Details of two worked exercises, originally from Wednesday's notes.

Solutions to linear equations in 3 unknowns: What is the geometric question you're answering?

Exercise 4) Consider the system x

C

2 y

C

z = 4 3 x

C

8 y

C

7 z = 20 called Gaussian elimination.

2 x

C

7 y

C

9 z = 23 .

Use elementary equation operations (or if you prefer, elementary row operations in the synthetic version) to find the solution set to this system. There's a systematic way to do this, which we'll talk about. It's Hint: The solution set is a single point, x, y, z = 5,

K

2, 3 .

, Do your work synthetically, just keeping track of coefficients and right-hand sides. Start with the augmented matrix 1 2 1 4 3 8 7 20 . 2 7 9 23 , Use elementary row operations (which correspond to elementary equation operations) to reduce the system to an equivalent one that is simpler. First, eliminate the

x

variable from the second two equations, by adding multiples of the first equation: K 3

R

1 C

R

2 /

R

2 ; K 2

R

1 C

R

3 /

R

3 : 1 2 1 4 0 2 4 8 .

1 0 3 7 2 1 15 , Now focus on the sub-matrix below the first row and to the right of the first column. Get a (row) "leading 1" in the

a

22 spot, so that you can eliminate the

y

variable in the 3

rd

equation: K

R

3 C

R

2 /

R

2 : 4 0 0 K 1 3 K 3 7 K 7 15 K

R

2 /

R

2 : 1 2 1 0 1 3 4 7 .

0 3 7 15 , Eliminate the

y

K coefficient in the 3

rd

row: K 3

R

2 C

R

3 /

R

3

1 2 0 1 1 3 4 7 .

0 0 K 2 K 6 , Get a leading 1 in the 3

rd

row:

R

3 K 2 /

R

3 1 2 1 4 0 1 3 7 .

0 0 1 3 At this stage we could backsolve to find the solution: The third equation is

z

y

C 3 \$ 3 = 7 so

y

= K 2 . Thus the first equation reads

x

C 2 K = 3 . Thus the second 2 C 1 \$ 3 = 4 so

x

= 5 . Thus the solution is

x

,

y

,

z

= 5, K 2, 3 . However, it is usually easier to continue this process to get an even easier system. Except now we work from the bottom right back up to the top left.

, Use the

z

K coefficient in the 3

rd

equation to remove the

z

K variable from the top 2 equations: K 3

R

3 C

R

2 /

R

2 ; K

R

3 C

R

1 /

R

1 : 1 2 0 1 0 1 0 K 2 .

0 0 1 1 0 0 3 , Move left and up; use the

y

-coefficient in the 2

nd

row to remove the

y

K variable in the first row: K 2

R

2 C

R

1 /

R

1 5 0 1 0 0 0 1 K 2 3 .

Thus

x

,

y

,

z

= 5, K 2, 3 .

Exercise 5 There are other possibilities. In the two systems below we kept all of the coeffients the same as in Exercise 4, except for a 33 , and we changed the right side in the third equation, for 5a. Work out what happens in each case. 5a) x

C

2 y

C

z = 4 3 x

C

8 y

C

7 z = 20 2 x

C

7 y

C

8 z = 20 .

As in the previous exercise we work from the top left to the lower right, and then reverse course:

1 2 1 3 8 7 4 20 2 7 8 20 K 3

R

1 C

R

2 /

R

2 ; K 2

R

1 C

R

3 /

R

3 : 1 2 1 0 2 4 4 8 .

0 3 6 12

R

2 2 /

R

2 : 1 2 1 0 1 2 4 4 .

0 3 6 12 K 3

R

2 C

R

3 /

R

3 1 2 1 4 0 1 2 4 .

0 0 0 1 0 K 3 0 , Work with the "leading 1" in the second row, to eliminate the

y

K variable in the first row.

K 2

R

1 C

R

2 /

R

1 K 4 0 1 2 4 .

0 0 0 0 The two equations corresponding to this augmented matrix read

x

K 3

z

= K 4

y

C 2

z

= 4 .

Thus we may choose vector form this reads

z

arbitrarily, and solve for

y

,

x

. Let

z

=

t

2 = . Then

y

= 4 K 2

t

,

x

= K 4 C 3

t

. In

x

K 4 C 3

t y

= 4

z

which is an entire line of solutions.

K 2

t t

= K 4 4 0 C 3

t

K 2

t t

= K 4 4 0 C

t

3 K 2 1