Question

Below is the augmented matrix of a linear system. The unknown is a real number.

1 2 -1 0 4 2 3 -1 -1 5 0 1 -1 1 3 3 5 -2 0 10 | 1 0 2 d Several steps in finding the RREF have been done and now the matrix looks like: [1 1 0 1 3 31 0 1 -1 -3 -1 | -6 0 0 0 2 2 4 0 0 0 0 0 1 0 a) Write down the RREF of this matrix. b) Solve the system and write the solution as a linear combination of constant vectors. c) Find d. d) What is the rank of the coefficient matrix of the system?

Answer 1

Giren augmented matrix a the now reduced the (RREF) 1 2 -10 4111 | 2 3 -1 -1 5 10 od-] I 32 | 3 5 -2 0 1012] After several state in finding echelon form (RREF) then Tato 1313 7 R-R2 0 ] -1 -3 -1 Too o 22 1 4 1 R3 ooo 0 010] Ft 0 1 0 0 R-4R3 +7 L To 1 -1 0 210 R2 +3 Rz Too o 12 Loo 0 0010 makix looks like Ito'144197 To 1.-1 -3 -1/- lo o ott | o o o oo | 4 0 CRREF of the mabix o to b.) het hy l ERS is the soun ws then, at az=1 Kg - x3 +226=0 kytkg = 2 het ag=c, xz z C2 x4=2-61, 222-26 ; 2, ata ez TI-C27 Jól C-2411107 ez 21 LIJ TO7 s o teel I 아 C1, CZER) L Lace c) 34, +522-222 +2004 = d. 3) 3.61-62) +5 (€2-24)- 2.62 +104 = d. 2) 3 - 567 +562 - 102, +104 = d. i [223] Rank of the wetticient matrix - No matrix - Noi of nonzero rows in RREF =3