Question

2 + my +2 1. Consider the linear system (LS): 2 + y - 2 m is a constant. 2x + 3y + m2 = -1, (a) The LS can be written as Az = b. Use Maple to generate the matrix A and the vector b. (b) Construct the augmented matrix (Ab) and compute its reduced echelon form. (C) Using (b), compute the values of m for which the system has a unique solution.

Answer 1

sos: a b 2 3 Augmented montix <ajb> = rom a -- - 2 3 m odiagonal consist all element above augmented matrix is in reduced now echelon form if the main of is om ons and and below the main diagonal are oz. So, we need to reduced the above matrix given <Alb) - 1 1 1 2 3 Apply: (R2 = R2 -Rp) AND (R3 = R₃ - Q R1); 2 1 m o 5. 1-m 0 3-2m m-2 Apply: (R2 R2 -m 2 m ma 0 3-2mm-2 - 5

Aer & B - me): 2. - m4 15 - У Чm- 3-2 м М-2- 1m-1 АРР'Я (R = Ra - 3 - 2 m), ): - (m+1) - 2 М - 1 m-1 О О m2+m- 2--0m. m) m - Apply (R3 M2+ m-4 m2 o - (m+1) 11 2 1 m- -- 1 2 - 2 h-m-ч АРР'Я R, TR, + (mt) ] О m2--3m - 10 m-t m-ч 11 m-1 о m- o o 1 а – 3 м, m2 + may

Apply [R2 = R₂ - ( 2 m . 1) R3]: O o m2-3m-10 m² m-4 m+8 metm-4 O 2-Bm matm-u the given aug this is the reduced echelon form of mented mafreix <Alby. since, in the above augmented matteix, each Columns contain leading ones and hence it has no free variables, therefore the given augmented malorix will always produce unique solutions irrespective of the value of m. but 70 So, in cand take real value any bat mafm -4 to So, m+124 -4-1/4 (m+12)2 = . 17 4 m2m + m & 17-1 So, m will take all possible values exception