hello mr. anderson… we’ve been waiting for you.. hello mr. anderson… we’ve been waiting for...

Hello Mr. Anderson…

We’ve been waiting for you.

Hello Mr. Anderson…

We’ve been waiting for you.

To beat the matrix, one must first obtain access to it’s core by understanding where the matrix comes from.

Mr. Anderson, what is a matrix?

A matrix is a rectangular array of terms usually taken from a system of equations.

For example, given the following system of equations:

I’ll use the coefficients to create a matrix.€

2x + 2y = 4

2x − y = 1

€

2x + 2y = 4

2x − y = 1

€

2 2 4

2 −1 1

⎡

⎣ ⎢

⎤

⎦ ⎥

Augmented matrix

€

x

€

y

€

constant

4 basic operations:

You can replace any row in the matrix by:

1. The product of that row and a constant.

2. The sum of that row and another.

3. The sum of that row and a product of another and a constant.

4. Another row (swap rows).

The overall goal is to get the matrix into what is known as row reduced echelon form.

€

1 0 k1

0 1 k2

⎡

⎣ ⎢

⎤

⎦ ⎥

€

1 0 0 k1

0 1 0 k2

0 0 1 k3

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

2 2 42 −1 1

⎡

⎣⎢

⎤

⎦⎥

1. Replace row 1 by multiplying it by one half.

1 1 22 −1 1

⎡

⎣⎢

⎤

⎦⎥

1 1 22 −1 1

⎡

⎣⎢

⎤

⎦⎥

2. Replace row 2 by multiplying row 1 by (-2) and adding it to row 2.

1 1 20 −3 −3

⎡

⎣⎢

⎤

⎦⎥

1 1 20 −3 −3

⎡

⎣⎢

⎤

⎦⎥

3. Replace row 2 by multiplying it by -1/3.

1 1 20 1 1

⎡

⎣⎢

⎤

⎦⎥

1 1 20 1 1

⎡

⎣⎢

⎤

⎦⎥

4. Replace row 1 by multiplying row 2 by -1 and adding it to row 1.

1 0 10 1 1

⎡

⎣⎢

⎤

⎦⎥

€

1 0 1

0 1 1

⎡

⎣ ⎢

⎤

⎦ ⎥

€

x

€

y

€

=

1 0 10 1 1

⎡

⎣⎢

⎤

⎦⎥

€

x

€

y

€

=

€

1x + 0y = 1

0x + 1y = 1or

€

x =1

y =1

€

⇒ solution is (1,1).

Why?

€

2x − 3y + z = 2

3x + y − z = −2

x − 5y + 7z = 5

System of equations.

2 −3 1 23 1 −1 −21 −5 7 5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Matrix form.

1 ? ? ?? ? ? ?? ? ? ?

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Most efficient method:

Start by getting the first entry to be a 1.

1 ? ? ?0 ? ? ?0 ? ? ?

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Work your way down the column to fill in the zeros below the 1 by using the first row.

1 ? ? ?0 1 ? ?0 ? ? ?

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Once the first column is completed, go to the next column and find the term on the major diagonal. This term needs to be a 1.

1 ? ? ?0 1 ? ?0 0 1 ?

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

This matrix is called “upper triangular” because it only contains entries in the upper triangular part of the matrix.

1 0 0 ?0 1 0 ?0 0 1 ?

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

Using the same idea that we did before, use the 3rd row to get the 3rd column to all be zeros.

2 −3 1 23 1 −1 −21 −5 7 5

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1. Swap rows 1 and 3 (to get the “1” in the upper left hand corner of the matrix).

1 −5 7 53 1 −1 −22 −3 1 2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 −5 7 53 1 −1 −22 −3 1 2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

2. Replace row 2 by multiplying row 1 by (-3) and add to row 2. (This is to get a zero directly below the 1 in the upper left hand corner).

1 −5 7 50 16 −22 −172 −3 1 2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 −5 7 50 16 −22 −172 −3 1 2

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

3. Replace row 3 by multiplying row 1 by (-2) and adding to row 3. (This is to finish the first column with the last zero.)

1 −5 7 50 16 −22 −170 7 −13 −8

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 −5 7 50 16 −22 −170 7 −13 −8

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

4. Multiply row 2 by 1/16 in order to get the next “1” where it has to be on the major diagonal.

1 −5 7 5

0 1 −118 −17160 7 −13 −8

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 −5 7 5

0 1 −118 −17160 7 −13 −8

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

5. Replace row 3 by multiplying row 2 by -7 and adding to row 3.

1 −5 7 5

0 1 −118 −1716

0 0 −27 8 −916

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1 −5 7 5

0 1 −118 −17160 0 −27 8 −916

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

6. Replace row 3 by multiplying it by -8/27. (This is to get the final “1” in the major diagonal.)

1 −5 7 5

0 1 −118 −17160 0 1 1

6

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1 −5 7 5

0 1 −118 −17160 0 1 1

6

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

7. Replace row 2 by multiplying row 3 by 11/8 and adding to row 2. (This is to get the zero directly above the 1 in the bottom row.)

1 −5 7 5

0 1 0 −5 60 0 1 1

6

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

8. Replace row 1 by multiplying row 3 by -7 and adding to row 1.

1 −5 7 5

0 1 0 −5 60 0 1 1

6

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1 −5 0 236

0 1 0 −5 60 0 1 1

6

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1 −5 0 236

0 1 0 −5 60 0 1 1

6

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

9. Replace row 1 by multiplying row 2 by 5 and adding to row 1.

1 0 0 −13

0 1 0 −5 60 0 1 1

6

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

1 0 0 −13

0 1 0 −5 60 0 1 1

6

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

Therefore, the solution is:

x =−13

y=−56z=1

6

or

−13,−5

6, 16( )

Get Me Outta Here!!