Math 252 Applied Linear Algebra Problem Notes on Take-Home Exams I to IX 1. 2 2 Linear System Find all solutions to x 5x =0, x +5x =0. Express the answer in one of 1 2 1 2 the three p ossible parametric forms: A point. This represents the intersection of two lines. A line. The parametric form is x = c + td , 1 1 y = c + td , 2 2 with 1 <t<1. A plane. In this case, all p oints x; y are answers. Solution to 1. To explain geometrically the exp ected kind of answer, observe that x +5x =0 is the 1 2 same as the rst equation x 5x = 0, therefore there are not two equations, but only one! The set of planar 1 2 points satisfying the two equations is exactly the set of points on the straight line x 5x =0 an in nite 1 2 number of points. The standard form of the solution is obtained by solving for x in terms of x , e.g., 1 2 x =5x , then write out the vector solution X as follows: 1 2 ! x 1 X = x 2 ! 5x 2 = x 2 ! 5 = x 2 1 Toeachvalue of x corresp onds a solution of the system of equations, i.e., there are in nitely many solutions, 2 represented go emetrically as a line. 1a. 2 4 Linear System Find all solutions to x x +7x x =0,2x +3x 8x + x =0. 1 2 3 4 1 2 3 4 Solution to 1a. Subtract two times the rst equation from the second to get 5x 22x +3x =0. Divide 2 3 4 3 22 x + x =0. Keep this as the replacement for the second equation. Add the new equation by5toget x 3 4 2 5 5 13 2 it to the rst equation to get its replacement x + x x =0. The replacement equations are therefore 1 3 4 5 5 2 13 x x =0; x + 3 4 1 5 5 22 3 x x + x =0; 2 3 4 5 5 which corresp ond exactly to the reduced row echelon form of the system. The general solution is 0 1 0 1 1 0 2 13 x 1 5 5 B C B C C B 22 3 x B C B C C B 2 5 5 + x : = x B C B C C B 4 3 @ A @ A A @ 1 0 x 3 0 1 x 4 2. 4 4 Linear System Find a fraction-free Gauss-elimination form and the reduced row-echelon form for the following equations: x 2x + x + x =2,3x +2x 2x = 8, 4x x x =1,5x +3x x =0. 1 2 3 4 1 3 4 2 3 4 1 3 4 1 Solution to 2. A fraction-free Gauss-elimination form can b e obtained from the reduced rowechelon form by multiplying each row by a suitable factor, to clear the fractions. In reality, there are in nitely many fraction-free forms, so there is no way to give an answer that everyone will arriveat. It turns out that the reduced rowechelon form is also fraction-free, so it can b e rep orted as the fraction-free answer! The reduced row-echelon form is obtained from the augmented matrix by row op erations, using the basic pivot algorithm. The answer for both questions: 1 0 0 1 0 0 4 C B 0 1 0 2 0 C B rref = C B A @ 0 0 0 1 7 0 0 0 0 1 3. 4 3 Linear System Find all solutions to the 4 3 system x + x x = 0, 4x x +5x = 0, 1 2 3 1 2 3 2x + x 2x =0,3x +2x 6x =0. 1 2 3 1 2 3 Solution to 3. The augmented matrix and its reduced rowechelon form are: 0 1 0 1 1 1 1 0 1 0 0 0 B C B C 4 1 5 0 0 1 0 0 B C B C aug = ; rref = B C B C @ A @ A 1 1 2 0 0 1 0 0 0 0 3 2 6 0 0 0 It follows that X = 0 is the only solution. 4. 3 4 Linear System Find all solutions to x x + x x = 2, 2x +3x x +2x = 5, 1 2 3 4 1 2 3 4 4x 2x +2x 3x =6. 1 2 3 4 Solution to 4. The augmented matrix and its reduced rowechelon form are: 0 1 0 1 1 0 0 1=2 5 1 1 1 1 2 B C B C aug = 2 3 1 2 5 ; rref = 0 1 0 1=4 4 @ A @ A 4 2 2 3 6 0 0 1 1=4 3 The standard form of the solution is obtained by identifying the lead variables and the arbitrary variables: Variables x , x and x are the lead variables, b ecause they corresp ond to a leading 1 in the RREF. 1 2 3 See the b oxed 1's ab ove. Variable x is the arbitrary variable, b ecause arbitrary variables are the variables left over after 4 removal of the lead variables. The standard form of the solution X is obtained by replacing each lead variable with its corresp onding equation, obtained from the RREF. The arbitrary variables are left untouched. Then: 1 0 1 0 1 0 0 1 1 1 x +5 5 x 4 1 2 2 C B C B C B B C 1 1 x +4 4 x C B C B C B B C 4 2 4 4 = x = + X = C B C B C B B C 4 1 A @ A @ A @ @ A x 3 3 3 x 4 3 4 x 1 0 x 4 4 2 4a. 3 4 Linear System Use Gauss-Jordan elimination to nd the general solution: x + 2x + 4x x = 3 1 2 3 4 3x + 4x + 5x x = 7 1 2 3 4 x + 3x + 4x + 5x = 4 1 2 3 4 Solution to 4a. The augmented matrix and its reduced rowechelon form are: 1 0 1 0 1 0 0 5 1 3 1 2 4 1 C B C B 3 4 5 1 7 ; rref = aug = 0 1 0 6 1 A @ A @ 1 3 4 5 4 0 0 1 2 0 Variables x , x and x are the lead variables, b ecause they corresp ond to a leading 1 in the RREF. 1 2 3 See the b oxed 1's ab ove. Variable x is the arbitrary variable, b ecause arbitrary variables are the variables left over after 4 removal of the lead variables. The standard form of the solution is obtained by replacing the lead variables x , x , x by their equations 1 2 3 x =5x +1, x = 6x +1, x =2x , but the arbitrary variable x is untouched. Then: 1 4 2 4 3 4 4 0 1 0 1 0 1 0 1 x 5x +1 5 1 1 4 B C B C B C B C x 6x +1 6 1 B C B C B C B C 2 4 X = = = x + B C B C B C B C 4 @ A @ A @ A @ A x 2x 2 0 3 4 x x 1 0 4 4 Imp ortant: This method for writing out X applies only in case the equations are in reduced echelon form. A matrix C is in reduced row echelon form provided each nonzero row starts with a leading 1, and ab ove and b elow that leading 1 app ear only zeros. 5. Linear Combinations Compute the result of the linear combination 2u + v 3w where 1 0 1 0 1 0 9 0 1 C B C B C B : ; w = 1 ; v = 0 u = 1 A @ A @ A @ 4 2 2 Solution to 5. The result of the linear combination is 1 0 1 0 1 0 1 0 25 9 0 1 C B C B C B C B : = 1 3 1 + 0 2u + v 3w =2 1 A @ A @ A @ A @ 10 4 2 2 5a. EqualityofVectors Let 0 1 0 1 0 1 1 x +1 9 B C B C B C u = x ; v = 0 ; w = 1 @ A @ A @ A 2 2 4x The linear combination p =2u + v 3w dep ends up on x. Is there a value of x such that 0 1 22:9 B C 0:8 p = ? @ A 11:2 3 Solution to 5a. The linear combination p is given by 0 1 0 1 0 1 0 1 1 x +1 9 x 24 B C B C B C B C p =2 x + 0 3 1 = 2x 3 : @ A @ A @ A @ A 2 2 4x 12x 2 There is a value of x such that 1 0 22:9 C B p = 0:8 A @ 11:2 exactly when the comp onents agree, i.e., x 24 = 22:9, 2x 3=0:8, 12x 3=11:2. This happ ens for x =1:1check all three equations!. 6. Largest Linearly Indep endent Set Extract from the list 1 1 0 1 0 1 0 0 0 1 2 1 C C B C B C B B ; 1 ; 0 ; 2 1 A A @ A @ A @ @ 1 1 0 0 a largest set of linearly indep endentvectors. Solution to 6. To extract a largest linearly indep endent set of vectors from a list v , v , v , v 4 requires an 1 2 3 algorithm b e followed.