Supplement #1 Matrix Determinant & Ero

MATH HEMATICS S TUTOR RIAL INTER RNATIONAL UNDE ERGRADU UATE PRO OGRAMM ME By Gillang (gum milangaryo [email protected])

SUP PPLEM MENT #1 #

“Imaginatiion is more e importantt than know wledge” – Einstein E –

HOW TO T CALC CULATE DETERMI D INANT? 1. Sarrus s’ method d 2. Laplac ce expans sion CHAN NGE IN DE ETERMIN NANT’S VA ALUE – ROW R & CO OLUMN M MODIFICA ATION 1. If a ROW is multip plied with a scalar, the deterrminant va alue is multiplied d by the same s scallar. 2. If the ROW W is exch hanged, th he determ minant is multiplied m d by (-1)n. Where n is the number n off switches s made, th hat is, n is s 2 if you exchange e the ROW twic ce. plication of o a ROW, the 3. If a ROW is added by a scalar multip determina ants is UN NCHANGE ED. NOTE: The 3 ru ules above e also app ply to COL LUMN mo odification n.

Example: 1 2 1

2 1 5

1 3 0

1. Find the determinant of the matrix above! | |

0

6

10

| |

10

1

15)

2. Multiply the first row by 10, then find the value of the determinant! 10 2 1 | |

60

20 1 5 100

| |

10 3 0 10

150)

100

Note that the determinant is 10 times the previous determinant. The same scalar (10) multiplied to the first row. Rule 1 is PROVED! 3. Multiply the second column by 2, then find the value of the determinant!

| |

0

1 2 1

4 2 10

1 3 0

12

20

2

| | Note

that

the

determinant

30 )

20 is

twice

(2

times)

the

previous

determinant. The same scalar multiplied to the second column. Rule 1 is PROVED! 4. Exchange the first column with the third column, then find the value of the determinant! (Hint: see rule no. 2) 1 3 0 | |

1 | |

2 1 5 0

15 10

1 2 1 0

10

6)

The number of exchange is 1 (one i.e. the first column is exchanged with the third column, so the value of n is 1). The value of the new determinant is -10 that is (-1)1 times the old determinant. Rule 2 is PROVED! 5. Multiply the first column by 2 and add it to the third column, the find the value of the determinant! (Hint: see rule no. 3) 1 2 1

| | | |

2

14

2 1 5 30

| |

3 7 2 3

35

10

EXAMPLE Now use rule No.3 to solve the 4x4 matrix below. 1 0 1 5 0 7 2 4 -

2 4

3 1 3 6 5 1

We can use Laplace expansion along 1st column 0 5 7 4

-

0 5 7 4

2 6 3 1

3 4 6 5

R2-6R4 & R3+3R4 1 0 0 0

-

3 1 3 6 5 1

R2+R1 & R4-2R1 1 0 0 0

-

2 4

0 2 19 0 19 0 4 1

3 34 9 5

Now use we can calculate determinant easier 1

19 9

0

19 9 19 34

34 19 34 19 9

8)

19 25 475 -

Ain’t it simpler guys?

ELEMENTARY ROWS OPERATION (ERO) There are many way to solve a system of n-unknowns. For example of systems of n-unknowns: 1. System of 2-unknowns 2 2

3

2. System of 3 unknowns 4

3 2

3

8

5

4

2

4

9

There are many alternatives to solve systems above, here are some of the alternatives: 1. Elimination method 2. Matrix inversion 3. Cramer’s Rule I want to introduce a method, named elementary rows operation or ERO, that is able to solve systems above. There are some advantage to ERO, that is, ERO needs less time. But precision is needed and practicing ERO is essential. So the choice is yours to make. The steps to ERO will be discussed below. Before we discuss the steps to ERO, lets look at these examples of linear equations below. 1. 2

3

6

4

6

3 2

2. 2

3

5

3. 3 •

3

15

System No. 1 is INCONSISTENT. Why? The left hand side equation of the second equation is twice of the first equation. But the right hand side equation of the second equation is half of the first equation. Thus the system is inconsistent.

•

System No. 2 has A UNIQUE SOLUTION. Why?

•

System No. 3 has INFINITELY MANY SOLUTIONS. Why? The second equation is three times the first equation, thus the system has infinitely many solutions.

Here are the steps in ERO. 1. STEP #1 Define the linear equation, say equation No.2 above 2 2

3

or you can write the system above as 2 2. STEP #2 Form the augmented matrix of the equation above, which is 1 2

1 2 1 3

or you can write the augmented matrix above as 1 2

1 1

3. STEP #3 Then you can apply three ROW modifications (row ONLY). a. Switch one ROW with another b. Multiply a non-zero ROW with a scalar c. Add a ROW with a scalar (positive or negative) multiplication of another ROW. 4. STEPS #4 Apply steps #3 to obtain this form:

1 0 0 1 EXAMPLE OF ERO 1. Example #1 Let the system of linear equation be, 2 2

3

The augmented matrix of the system above is, 1 2

1 2 1 3

Now, we apply steps #3 -

R2-2R1 – add 2nd row with negative scalar -2 times 1st row (see steps #3 point c) 1 2

1 1 1 0

-

2 3

1 3

R2-2R1

2 1

R2 – multiply 2nd row by –(1/3) (see steps #3 point b) 1 0

1 3

2 1

(-1/3)R2

1 1 2 0 1 1/3 -

R1-1R2 – add 1st row with negative scalar -1 times 2nd row (see steps #3 point c) 1 1 2 0 1 1/3

R1-1R2

1 0 5/3 0 1 1/3 -

The solution is {x,y|x=5/3 and y=1/3}

2. Example #2

Let the system of linear equation be, 5 3

3

15

The augmented matrix of the system above is, 1 3

1 3

5 15

Now, we apply steps #3 -

R2-3R1 1 2

1 3 1 0

5 15

R2-3R1

1 5 0 0

No unique solution available! The example above yield infinitely many solutions! 3. Example #3 Let the system of linear equation be, 2

9

2

4

3

1

3

6

5

0

The augmented matrix of the system above is, 1 1 2 4 3 6

2 9 3 1 5 0

Now, we apply steps #3 -

R2-2R1 & R3-3R1 1 1 0 2 0 3

-

R3-(3/2)R2 & R1-(1/2)R2

2 7 11

9 17 27

1 0 0 -

0 2 0

2R1 & (-2)R3 2 0 0 2 0 0

-

11/2 35/2 7 17 1/2 3/2 11 7 1

35 17 3

R1-11R3 & R2+7R3 2 0 0 2 0 2 0 4 0 0 1 3

-

(1/2)R1 & (1/2)R2 1 0 0 1 0 1 0 2 0 0 1 3

-

The solution is {x,y,z|x=1; y=1/3; z=3}

NOTE: Advantage: -

It’s faster to solve system of 3-unknowns and above, ain’t it?!

-

It doesn’t take up that much space on your paper

But Remember: ERO is only an alternative way, it is not the ONLY WAY! Do lots of practiceS if you want to use ERO. It requires accuracy and correctness. Otherwise, use methods that you are confident with such as elimination, matrix inversion, or Cramer’s rule to solve problem of systems of n-unknowns eventhough it take will take more time.