MTH MTH Lecture 6. Yevgeniy Kovchegov Oregon State University

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MTH 306 0 MTH 306 - Lecture 6 Yevgeniy Kovchegov Oregon State University

MTH 306 1 Topics Lines and planes. Systems of linear equations. Systematic elimination of unknowns. Coe cient matrix. Augmented matrix. Determinants.

MTH 306 2 Lines. L P= 0 (x 0,y 0) P=(x,y) v= (x-x 0,y-y 0 ) (0,0) A line L is determined by two points, P 0 and P. Alternatively line L can be described by point P 0 and direction vector ~v.

MTH 306 3 Parametric equation. L r -v 0 (0,0) P 0 r 0= (x 0,y 0) v r +v 0 r +2v 0 Given a line going through point P 0 in the direction ~v it can be described as the following parametric equation ~r = ~r 0 + t~v, for 1t1, where ~r 0 is the vector from the origin to P 0.

MTH 306 4 Parametric equation. L r -v 0 (0,0) P 0 r 0= (x 0,y 0) v r +v 0 r +2v 0 Parametric equation ~r = ~r 0 + t~v Example 2D. Given ~r 0 = (x 0,y 0 ) and ~v =(a, b), then (x, y) =(x 0,y 0 )+t(a, b) =(x 0 + at, y 0 + bt)

MTH 306 5 Example 2D. Given ~r 0 = (x 0,y 0 ) and ~v =(a, b), then (x, y) =(x 0,y 0 )+t(a, b) =(x 0 + at, y 0 + bt) gives x = x 0 + at y = y 0 + bt for 1t1. These equations are called (scalar) parametric equations for the line L. The parametric equations for lines can be obtained similarly in 3-D, and higher dimensions. See the book.

MTH 306 6 Example Find parametric equations for the line L determined by the two points, (2, 5) and (1, 7). Solution Take ~r 0 = (2, 5) (as P 0 = (2, 5) is a point on L), and ~v = (1 2, 7 5) = ( 1, 2) The parametric equation ~r = ~r 0 + t~v reads x =2 t y =5+2t

MTH 306 7 Planes in 3-D. IR 3 n =(a,b,c) P=(x,y,z) P= 0 (x 0,y 0,z 0 ) v= (x-x 0,y-y 0,z-z 0) Π A plane in 3-D space R 3 is determined by a point P 0 =(x 0,y 0,z 0 ) on it and a vector ~n perpendicular (normal) to.

MTH 306 Planes in 3-D. IR 3 n =(a,b,c) P=(x,y,z) P= 0 (x 0,y 0,z 0 ) v= (x-x 0,y-y 0,z-z 0) Π Every point (x, y, z) on the plane has to satisfy (x x 0,y y 0,z z 0 )? ~n In other words, (x x 0,y y 0,z z 0 ) ~n = 0, which we write as a(x x 0 )+b(y y 0 )+c(z z 0 )=0

MTH 306 9 Equation a(x x 0 )+b(y y 0 )+c(z z 0 )=0 Example Find an equation for the plane in 3-D that contains point (1, 0, 2) and is perpendicular to vector ~n = (4, 2, 1). Answer We use (x x 0,y y 0,z z 0 ) ~n =0 with (x 0,y 0,z 0 ) = (1, 0, 2), obtaining 4(x 1) + 2(y 0) (z 2) = 0 which we simplify to 4x +2y z 2=0

MTH 306 10 Systems of linear equations. Consider a system of two linear equations x 2y =0 2x +3y = 14 To solve it means to find all (x, y) that satisfy both equations. Solution use the first equation to express x via y, obtaining x =2y. Then plug x =2y into the second equation 2(2y)+ 3y = 14. Thus 7y = 14. So, y = 2 and x =2y = 4. Answer x = 4, y =2

MTH 306 11 Systems of linear equations. x 2y =0 (L 1 ) 2x +3y = 14 (L 2 ) To solve it means to find all (x, y) that satisfy both equations. 6 5 4 3 2 1 L 1-3 -2-1 1 2 3 4 5 6 7 L 2

MTH 306 12 Systems of linear equations. ax + by = e (L 1 ) cx + dy = f (L 2 ) Case I. L 1 and L 2 intersect in a unique point. L 1 (0,0) L 2

MTH 306 13 Systems of linear equations. ax + by = e (L 1 ) cx + dy = f (L 2 ) Case II. L 1 and L 2 are distinct parallel lines. L 1 L 2 (0,0)

MTH 306 14 Systems of linear equations. ax + by = e (L 1 ) cx + dy = f (L 2 ) Case III. L 1 and L 2 coincide. L 1 L 2 (0,0)

MTH 306 15 Systematic elimination of unknowns. x 2y =0 Eq.1 2x +3y = 14 + x 2y =0 Eq.1 Eq.2 7y = 14 Eq.2 2 Eq.1! Eq.2 + x 2y =0 Eq.1 y =2 x =4 y =2 + 1 7Eq.2! Eq.2 Eq.1 + 2 Eq.2 Eq.2

MTH 306 16 System of Equations x 2y =0 2x +3y = 14 + + x 2y =0 7y = 14 + + x 2y =0 Augmented Matrix " # 1 2 0 R1 2 3 14 R2 " 1 2 0 0 7 14 " 1 2 0 0 1 2 y =2 + + x =4 y =2 " 1 0 4 0 1 2 # # R1 R2 2R1 R1 1 7 R2 # R1+2R2 R2

MTH 306 17 Coe cient matrix. Augmented matrix. Consider a system of two linear equations ax + by = e with two unknowns cx + dy = f Group the coe cients of the unknowns into a rectangular " array # called the coe cient matrix A = a b c d The system is fully described by its augmented matrix " a b e c d f #

MTH 306 1 Theorem. An n n linear system of algebraic equations either has a unique solution, no solution, or infinitely many solutions. L 1 L 1 L 1 L2 L2 > > (0,0) L (0,0) (0,0) 2 a 1,1 x 1 + a 1,2 x 2 + + a 1,n x n = b 1 a 2,1 x 1 + a 2,2 x 2 + + a 2,n x n = b 2... a n,1 x 1 + a n,2 x 2 + + a n,n x n = b n

MTH 306 19 Example. x +2y =1 " 1 2 1 2 1 7 2x y =7 + + x +2y =1 " 1 2 1 0 5 5 5y =5 + + x +2y =1 " 1 2 1 0 1 1 y = 1 + + x =3 y = 1 " 1 0 3 0 1 1 # R1 R2 # # R1 R2 2R1 R1 1 5 R2 # R1 2R2 R2

MTH 306 20 x +2y =1 2x y =7 ) x =3 y = 1 Here x +2y = 1 and 2x a unique point (3, 1). y = 7 intersect in 6 5 4 3 2 1 2x-y=7-4 -3-2 -1-2 -3-4 -5-6 -7 1 2 3 4 5 6 7 x+2y=1

MTH 306 21 Example. x +2y =1 " 1 2 1 3 6 3 3x +6y =3 + + x +2y =1 0=0 " 1 2 1 0 0 0 # R1 R2 # R1 R2 3R1 Here x +2y = 1 and 3x +6y =3coincide. Thus there are infinitely many solutions. The solutions line x +2y = 1 can be expresses as a parametric equation (x, y) = (1 2t, t) = (1, 0)+( 2, 1)t, where y = t

MTH 306 22 x +2y =1 3x +6y =3 ) x +2y =1 Here x +2y = 1 and 3x +6y =3coincide. 6 5 4 3 2 1-4 -3-2 -1-2 -3-4 1 2 4 5 6 7 x+2y=1 3x+6y=3 (x, y) = (1 2t, t) = (1, 0)+( 2, 1)t, where y = t

MTH 306 23 Example. x +2y =1 " 1 2 1 1 2 4 x +2y =4 + + x +2y =1 0=3 " 1 2 1 0 0 3 # R1 R2 # R1 R2 R1 Here x +2y = 1 and x +2y =4aredistinct parallel lines. The system has no solutions. As otherwise 0 would be equal to 3.

MTH 306 24 x +2y =1 x +2y =4 ) x +2y =1 0=3 Here x +2y = 1 and x +2y =4aredistinct parallel lines. The system has no solutions. 6 5 4 3 2 1 x+2y=4-4 -3-2 -1-2 -3-4 1 2 3 4 5 6 7 x+2y=1

MTH 306 25 Determinant of a 2 2 coe cient matrix. " # a b If A = is a coe cient matrix of a linear system, then its determinant is defined c d as Other notations a b c d det = ad bc a c b d!, A, or det(a)

MTH 306 26 Determinant of a 2 2 coe cient matrix. Consider a system of two linear equations ax + by = e with two unknowns cx + dy = f a b c d 6=0, there is a unique solution. x 2y =0 Example. was shown to 2x +3y = 14 have a unique solution (x =4,y = 2), and 1 2 2 3 =3+4=76= 0

MTH 306 27 a b c d 6=0, there is a unique solution. x +2y =1 Example. was shown to 2x y =7 have a unique solution (x =3,y = 1), and 1 2 2 1 = 1 4= 5 6= 0 x +2y =1 Example. was shown to 3x +6y =3 have infinitely many solutions, and 1 2 3 6 =6 6=0

MTH 306 2 a b c d 6=0, there is a unique solution. Example. x +2y =1 x +2y =4 was shown to have no solutions, and 1 2 1 2 =2 2=0