Notes 7.3 – Multivariate Linear Systems and Row Operations.

Notes 7.3 – Multivariate Linear Systems and Row Operations

I. Solving SystemsA.) Ex. 1 – Solve the system:

2 7

2 7

3

x y z

y z

z

2

1

3

x

y

z

B.) Ex. 2 – Now try this one:

2 7

3 5 14

2 2 3

x y z

x y z

x y z

3 6 3 21

3 5 14

x y z

x y z

3 2 7

3 5 14

2 2 3

x y z

x y z

x y z

2 7y z

2 2 7

2 2 3

x y z

x y z

2 4 2 14

2 2 3

x y z

x y z

2 3 11y z

Furthermore...

2 7

2 3 11

y z

y z

2 4 14

2 3 11

y z

y z

2 2 7

2 3 11

y z

y z

3z

2(3) 7

1

y

y

2( 1) 3 7

2

x

x

2, 1,3

II. Matrix Algebra

1 2 1

3 5 1

2 2 1

x

y

z

7

14

3

A.)

2 7

3 5 14

2 2 3

x y z

x y z

x y z

Triangular Form

Matr

ix For

m

A B = C

x

y

z

2

1

3

B.) On the Calculator…

A-1CB =

1 2 1 7

3 5 1 14

2 2 1 3

1 # # #

0 1 # #

0 0 1 #

Called the Augmented Matrix

The idea is to get a matrix that will look like this

1 2 1 7

3 5 1 14

2 2 1 3

Top row is ok – Starts with 1

This row needs to start with a 0

We can multiply Row 1 by -3 and then add it to Row 2. If we replace Row 2 with the new sum, we have an equivalent matrix.

1 2 1 7

3 5 1 14

2 2 1 3

1 2 1 7

0 1 2 7

0 2 3 11

Notation

-3R1 + R2

1 2 1 7

0 1 2 7

2 2 1 3

1 2 1 7

0 1 2 7

2 2 1 3

-2R1 + R3

1 2 1 7

0 1 2 7

0 2 3 11

-2R2 + R3

1 2 1 7

0 1 2 7

0 0 1 3

Now, this matrix is in row echelon form. What does this mean?

0 1 2 7

2(3) 7

1

x y z

y

y

1 2 1 7

2( 1) 3 7

2

x y z

x

x

0 0 1 3

3

x y z

z

C.) Def. - A matrix is in row echelon form if the following conditions are satisfied:

1.) Rows of all 0’s (if there are any) occur at the bottom.

2.) The first entry in any row with nonzero entries is 1.

3.) The column subscript of the leading 1 entries increases as the row subscript increases.

D.) Ex. 3 – Put the following matrix in row echelon form.

2 1 1 2

1 2 3 0

3 1 1 2

R211 2 3 0

2 1 1 2

3 1 1 2

2R1+R2

1 2 3 0

0 3 5 2

0 7 10 2

-3R1+R3

-1/3R2

1 2 3 0

5 20 1

3 30 7 10 2

1 2 3 0

5 20 1

3 30 7 10 2

-7R2+R3

1 2 3 0

5 20 1

3 35 20

0 03 3

1 2 3 0

5 20 1

3 30 0 1 4

3/5R3

Solving we get

0

6

4

x

y

z

III. Reduced Row Echelon Form

1 0 0 ...

0 1 0 ...

0 0 1 ...

0 0 ... 1

a

b

c

n

A.) A matrix is in reduced row echelon form if its main diagonal consists of all 1’s and zeros everywhere but the last column.

B.) Ex. 5 - Place the matrix in reduced row echelon form.

1 2 1 7

0 1 0 1

0 0 1 3

1 2 1 7

0 1 2 7

0 0 1 3

2R3 + R2

1 0 1 5

0 1 0 1

0 0 1 3

2R2 + R1-R3 + R1 1 0 0 2

0 1 0 1

0 0 1 3

C.) Ex. 6 – Solve the following system using your calculator.

1 1 1 3

2 1 4 8

1 2 1 1

3

2 4 8

2 1

x y z

x y z

x y z

1 0 3 5

0 1 2 2

0 0 0 0

Rref(A) Infinite Solutions !!!

3 5,2 2,z z z

1 3 5

2 2

x z

y z

1 0 3 5

0 1 2 2

0 0 0 0