Question

Problem 2 We have seen that, if a matrix A is invertible, then we can express the unique solution of Az = b as I = A-15. Soon, we will introduce ideas that help us understand At = b better when A is not invertible. This problem is preparation for that! 11 301 1-3 -9 2 Let A = 2 6 0 1-2 -6 -5 (a) Solve the system A7 = 5. (b) What does the solution set of Az = b look like graphically? (Is it a line, circle, etc.?) Does it pass through the origin? Use Mathematica to check your answer (see HW2 #6(b) for the relevant Mathematica commands). (c) Solve the system Az = 0. (Can you re-use your work from #2(a)? How does your final answer compare with your answer to #2(a)?) (d) What does the solution of Az = 0 look like graphically? How does it relate (graph- ically) to the solution set of At=b? Mathematica tip: To graph two or more things together, you can use the Show command. Here is an example: Show [ ParametricPlot 3D[{1,2,0}+t*{3,-2,1}, {t,-3,3}], ParametricPlot 3D (s*{4,-3,7}+t{0,1,2},{s, -3,3} , {t,-3,3}] (e) If C is any vector in R4, what can you say about the number of solutions of the system At = c? (Must there be a solution? Could the system have exactly one solution? Could it have infinitely many solutions?)

Answer 1

moyo A = 1-3 -9 2 and b e 2 6 0 L-? 6.-5 . / 31 To solue Ax = I we make an augmente- matrix [A: b] and find the rank of A and rank of [A:6]. 9 rank CAD - rand A!6] then solution exist If dank (A) # rank [A! b] then solution will not exist. [A:b] - 3027 -9 2 :-20 2 6 0 : 4 1-2 -6 -5:31] R2 R₂ + 3 R1, R₂ + R2 - 2 R1 Ry Ryt 2R, 3 0 :27 N lo o 2 :141 Ry G Rz -ņa - OOO 0 0 -5 : 35] 2 ſi 30:27 O 02 !-14 - CR2² R2/21 loo-5:35 R₂ + R3/5) Loo 0:

11 3 0 : 27 " lo o I :-) I (R2 R₃ - R₂] lo 01:-1 LO 0 Oio. si 30 ! 27 No :-) LO 0 0; o) - rank(A) = 2 = rank [A!b], Solution will exist. 3=-7 x + 3y =2 X = 2-3y Here y is free ariable , by choosing the value of g we can find the value of x. Put y = I x = -1 The line solution with set all of 3 Ax = 6 shaus a as a fix value. .

system Az = o Chomog. eneous we solue the system). A. ſi 3 0 1 R2 + R2+3R,, 1-3-92 . Rat R3-2R, 2 6 0 L-2-6-5] Ry> Rutari, [ 3 0 0 o 07 2 1 o o N ó -5l 11 3 of 0 0 1 o o.s. Loo of ſi 3 01 OBO- ooow O-oo 2=0 , x + 3y = 0 X=-34 x = -3 Put y = 1 shows also a line but with AX=0 ? coordinate zero.