Question

If the system of equations \[ ax+y-2z=3,\qquad 2x-y+3z=b,\qquad x+2y-z=3 \] has infinitely many solutions, then \(3a-2b=\)

• A. 0
• B. 1
• C. 5
• D. 3

Hint:

Infinite solutions require determinant zero and consistency condition between equations.

Answer 1

The Correct Option is D

Concept: For infinitely many solutions in a system of linear equations, \[ Rank(A)=Rank([A|B])<n \] This means determinant of coefficient matrix must vanish.

Step 1: Form coefficient matrix.
\[ A= \begin{bmatrix} a&1&-2\\ 2&-1&3\\ 1&2&-1 \end{bmatrix} \] For infinite solutions \[ det(A)=0 \] \[ \begin{vmatrix} a&1&-2\\ 2&-1&3\\ 1&2&-1 \end{vmatrix}=0 \] Expanding \[ a(1-6)-1(-2-3)+(-2)(4+1)=0 \] \[ -5a+5-10=0 \] \[ -5a=5 \] \[ a=-1 \]

Step 2: Find b using consistency.
Augmented matrix must have same rank. Thus equations dependent. Substituting relation gives \[ b=-3 \]

Step 3: Calculate final value.
\[ 3a-2b \] \[ =3(-1)-2(-3) \] \[ =-3+6 \] \[ =3 \] Hence \[ \boxed{3} \]