LINEAR ALGEBRA YULVI ZAIKA WHAT IS LINEAR ALGEBRA

LINEAR ALGEBRA YULVI ZAIKA

WHAT IS LINEAR ALGEBRA? • DEVELOPS FROM THE IDEA OF TRYING TO SOLVE AND ANALYZE SYSTEMS OF LINEAR EQUATIONS. • THEORY OF MATRICES AND DETERMINANTS ARISE FROM THIS EFFORT • WE WILL THEN GENERALIZE THESE IDEAS TO THE ABSTRACT CONCEPT OF A VECTOR SPACE • LINEAR TRANSFORMATIONS, EIGENVALUES, INNER PRODUCTS. . .

RELATIONSHIPS BETWEEN CONSEPT • ONE OF THE IMPORTANT GOALS OF A COURSE IN LINEAR ALGEBRA IS TO ESTABLISH THE INTRICATE THREAD OF RELATIONSHIPS BETWEEN SYSTEMS OF LINEAR EQUATIONS, MATRICES, DETERMINANTS, VECTORS, LINEAR TRANSFORMATIONS, AND EIGENVALUES

WHY IS LINEAR ALGEBRA INTERESTING? • IT HAS MANY APPLICATIONS IN MANY DIVERSE FIELDS. (COMPUTER GRAPHICS, CHEMISTRY, BIOLOGY, DIFFERENTIAL EQUATIONS, ECONOMICS, BUSINESS, ENGINEERING. ESPECIALLY TRAFFIC MANAGEMENT. . . ) • IT STRIKES A NICE BALANCE BETWEEN COMPUTATION AND THEORY. • GREAT AREA IN WHICH TO USE TECHNOLOGY.

WHAT IS A LINEAR EQUATION? • A LINEAR EQUATION IS AN EQUATION OF THE FORM, • ANXN+ AN-1 XN-1+. . . + A 1 X 1 = B. • A SOLUTION TO A LINEAR EQUATION IS AN ASSIGNMENT OF VALUES TO THE VARIABLES (XI’S) THAT MAKE THE EQUATION TRUE.

WHAT IS A SYSTEM OF LINEAR EQUATIONS? • A SYSTEM OF LINEAR EQUATIONS IS SIMPLY A SET OF LINEAR EQUATIONS. I. E. A 1, 1 X 1+ A 1, 2 X 2+. . . + A 1, NXN = B 1 A 2, 1 X 1+ A 2, 2 X 2+. . . + A 2, NXN = B 2. . . AM, 1 X 1+ AM, 2 X 2+. . . + AM, NXN = BM

• A SOLUTION TO A SYSTEM OF EQUATIONS IS SIMPLY AN ASSIGNMENT OF VALUES TO THE VARIABLES THAT SATISFIES (IS A SOLUTION TO) ALL OF THE EQUATIONS IN THE SYSTEM. • IF A SYSTEM OF EQUATIONS HAS AT LEAST ONE SOLUTION, WE SAY IT IS CONSISTENT. • IF A SYSTEM DOES NOT HAVE ANY SOLUTIONS WE SAY THAT IT IS INCONSISTENT.

EXAMPLES This is a consistent system as x =1, y =1 is a solution. This is an inconsistent system. Why? ?

HOW MANY SOLUTIONS DOES THIS SYSTEM HAVE?

HOW MANY SOLUTIONS COULD A SYSTEM HAVE? • AS WE SAW, A SYSTEM OF LINEAR EQUATIONS CAN HAVE NO SOLUTIONS, ONE SOLUTION, OR INFINITELY MANY SOLUTIONS. • WE ALSO SAW THAT GRAPHING THINGS IS ONE WAY TO FIND THE SOLUTIONS. • HOWEVER, THIS IS DIFFICULT IN MORE THAN THREE VARIABLES.

AUGMENTED MATRICS

OPERATIONS THAT LEAD TO EQUIVALENT SYSTEMS 1. INTERCHANGE TWO EQUATIONS. 2. MULTIPLY AN EQUATION BY A NONZERO CONSTANT. 3. ADD A MULTIPLE OF AN EQUATION TO ANOTHER EQUATION.

ELEMENTARY ROW OPERATION (OPERASI BARIS ELEMENTER)

CONTOH DIKETAHUI …………(i) …………(iii) kalikan pers (i) dengan (-2), kemudian tambahkan ke pers (ii). kalikan baris (i) dengan (-2), lalu tambahkan ke baris (ii). kalikan pers (i) dengan (-3), kemudian tambahkan ke pers (iii). kalikan baris (i) dengan (-3), lalu tambahkan ke baris (iii). kalikan pers (ii) dengan (1/2). kalikan baris (ii) dengan (1/2).

LANJUTAN CONTOH kalikan pers (ii) dengan (1/2). kalikan baris (ii) dengan (1/2). kalikan pers (ii) dengan (-3), lalu tambahkan ke pers (iii). kalikan brs (ii) dengan (-3), lalu tambahkan ke brs (iii). kalikan pers (iii) dengan (-2). kalikan brs (iii) dengan (-2). kalikan pers (ii) dengan (-1), lalu tambahkan ke pers (i). kalikan brs (ii) dengan (-1), lalu tambahkan ke brs (i).

LANJUTAN CONTOH kalikan pers (ii) dengan (-1), lalu tambahkan ke pers (i). kalikan pers (iii) dengan (-11/2), lalu tambahkan ke pers (i) dan kalikan pers (ii) dg (7/2), lalu tambahkan ke pers (ii) kalikan brs (ii) dengan (-1), lalu tambahkan ke brs (i). kalikan brs (iii) dengan (-11/2), lalu tambahkan ke brs (i) dan kalikan brs (ii) dg (7/2), lalu tambahkan ke brs (ii) Diperoleh penyelesaian x = 1, y = 2, z = 3. Terdapat kaitan menarik antara bentuk SPL dan representasi matriksnya. Metoda ini berikutnya disebut dengan METODA ELIMINASI GAUSS.

• WE SAY THAT A SYSTEM IN THIS FORM IS IN ROW -ECHELON FORM. • WE CAN FIND THE SOLUTIONS TO A SYSTEM IN ROW-ECHELON FORM USING BACK SUBSTITUTION.

SOLVING A SYSTEM NOT IN R-E FORM. • WE SAY THAT TWO SYSTEMS OF EQUATIONS (OR EQUATIONS) ARE EQUIVALENT, IF THEY HAVE EXACTLY THE SAME SOLUTIONS. • ONE METHOD OF SOLVING A SYSTEM OF EQUATIONS IS TO TRANSFORM IT TO AN EQUIVALENT SYSTEM IN R-E FORM.

• USING THESE THREE OPERATIONS WE CAN SOLVE A SYSTEM OF LINEAR EQUATIONS IN THE FOLLOWING MANNER. 1. SAVE THE X TERM IN THE FIRST EQUATION AND USE IT TO ELIMINATE ALL OTHER X TERMS. 2. IGNORE THE FIRST EQUATION AND USE THE SECOND TERM IN THE SECOND EQUATION TO ELIMINATE ALL OTHER SECOND TERMS IN THE REMAINING EQUATIONS. 3. CONTINUE IN THIS MANNER UNTIL THE SYSTEM IS IN R-E FORM.

BENTUK ECHELON-BARIS MISALKAN SPL DISAJIKAN DALAM BENTUK MATRIKS BERIKUT: MAKA SPL INI MEMPUNYAI PENYELESAIAN X = 1, Y = 2, Z = 3. MATRIKS INI DISEBUT BENTUK ECHELON-BARIS TEREDUKSI. UNTUK DAPAT MENCAPAI BENTUK INI MAKA SYARATNYA ADALAH SBB: 1. JIKA SUATU BRS MATRIKS TIDAK NOL SEMUA MAKA ELEMEN TAK NOL PERTAMA ADALAH 1. BRS INI DISEBUT MEMPUNYAI LEADING 1. 2. SEMUA BRS YG TERDIRI DARI NOL SEMUA DIKUMPULKAN DI BAGIAN BAWAH. 3. LEADING 1 PADA BARIS LEBIH ATAS POSISINYA LEBIH KIRI DARIPADA LEADING 1 BARIS BERIKUT. 4. SETIAP KOLOM YANG MEMUAT LEADING 1, ELEMEN LAIN SEMUANYA 0.

BENTUK ECHELON-BARIS DAN ECHELON-BARIS TEREDUKSI MATRIKS YANG MEMENUHI KONDISI (1), (2), (3) DISEBUT BENTUK ECHELON-BARIS. CONTOH BENTUK ECHELON-BARIS TEREDUKSI: CONTOH BENTUK ECHELON-BARIS:

BENTUK UMUM ECHELON-BARIS dimana lambang ∗ dapat diisi bilanangan real sebarang.

BENTUK UMUM ECHELON-BARIS TEREDUKSI dimana lambang ∗ dapat diisi bilanangna real sebarang.

PENYELESAIAN SPL MELALUI BENTUK ECHELON-BARIS MISAL DIBERIKAN BENTUK MATRIKS SPL SBB: Tentukan penyelesaian masing-masing SPL di atas.

PROBLEM 1. CONSIDER THE SYSTEM OF EQUATIONS INDICATE WHAT WE CAN SAY ABOUT THE RELATIVE POSITIONS OF THE LINES WHEN (a) THE SYSTEM HAS NO SOLUTIONS (b) THE SYSTEM HAS EXACTLY ONE SOLUTION (c) THE SYSTEM HAS INFINITELY MANY SOLUTIONS • IF THE SYSTEM OF EQUATIONS IS CONSISTENT, EXPLAIN WHY AT LEAST ONE EQUATION CAN BE DISCARDED FROM THE SYSTEM WITHOUT ALTERING THE SOLUTION SET

HOMOGENEOUS LINEAR SYSTEMS • A SYSTEM OF LINEAR EQUATIONS IS SAID TO BE HOMOGENEOUS IF THE CONSTANT TERMS ARE ALL ZERO; THAT IS, THE SYSTEM HAS THE FORM • EVERY HOMOGENEOUS SYSTEM OF LINEAR EQUATIONS IS CONSISTENT, SINCE ALL SUCH SYSTEMS HAVE X 1=0 , X 2=0 , …XN AS A SOLUTION. • THIS SOLUTION IS CALLED THE TRIVIAL SOLUTION; IF THERE ARE

• BECAUSE A HOMOGENEOUS LINEAR SYSTEM ALWAYS HAS THE TRIVIAL SOLUTION, THERE ARE ONLY TWO POSSIBILITIES FOR ITS SOLUTIONS: • THE SYSTEM HAS ONLY THE TRIVIAL SOLUTION • THE SYSTEM HAS INFINITELY MANY SOLUTIONS IN ADDITION TO THE TRIVIAL SOLUTION • IN THE SPECIAL CASE OF A HOMOGENEOUS LINEAR SYSTEM OF TWO EQUATIONS IN TWO UNKNOWNS, SAY • THE GRAPHS OF THE EQUATIONS ARE LINES THROUGH THE ORIGIN, AND THE TRIVIAL SOLUTION CORRESPONDS TO THE POINT OF INTERSECTION AT THE ORIGIN

CONTOH Reducing this matrix to reduced row echelon The augmented matrix for the system is The general solution

E 1. 2 : 31 -32 Indicate whether the statement is always true or sometimes false. Justify your answer by giving a logical argument or a counterexample (a) If a matrix is reduced to reduced row-echelon form by two different sequences of elementary row operations, the resulting matrices will be different (b) If a matrix is reduced to row-echelon form by two different sequences of elementary row operations, the resulting matrices might be different. (c)If the reduced row-echelon form of the augmented matrix for a linear system has a row of zeros, then the system must have infinitely many solutions. (d) If three lines in the -plane xy are sides of a triangle, then the system of equations formed from their equations has three solutions, one corresponding to each vertex. (d) A linear system of three equations in five unknowns must be consistent (e)A linear system of five equations in three unknowns cannot be consistent (f) If a linear system of n equations in n unknowns has n leading 1's in the reduced rowechelon form of its augmented matrix, then the system has exactly one solution (g)If a linear system of n equations in n unknowns has two equations that are multiples of one another, then the system is inconsistent