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Question 6 Solve using Matrices 2x – 3y + 2z = 13 3x – 4y – 3z = 1 3x +y – z = 2 infinite number of solutions No Solution O (-1,2,1) (2,-1,3)

Answer 1

Writing given system of equations in romatrix from 2 2 13 3 -4 -3 у 3 -1 2 2 -3 2 13 So the awemented matrix is: 3 -4 -3 1 3 1 -1 2 Applying row transformations, on the augmented matrix, for echelon form, we have 3 -4 -3 1 = 2 -3 2 13 Swap matrix rows: R1 R2 3 1 -1 2 -4 -3 37 Cancel leading coefficient in row R2 by performing R2 –R2 – Ž 3 3 co 1 = 0 R1 متن ها و راحت 3 do = 10 - 4 37 3 Cancel leading coefficient in row Rz by performing R3-R3-1. Ri 5 0 3 0 2 -3 2 1 1 -4 5 = 1 37 Swap matrix rows: R2 R3 0 = R2 5 3 -4 -3 0 5 2 Cancel leading coefficient in row R3 by performing R3 — R3 + 1 0 0 62 62 15 Hence, the system of equations, reduces to: 37-44-32 = 1 sy +22=1 *2=3, y=-1, and 62 62 5 (On back substitution) and x=2 Za Therefore, the marked answer is perfectly correct. i.e. X = 2, y=-1 and z=3 OR (2, -1, 3)