advertisement

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*

equation reads

*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