Question

If the system of equations \[ \begin{cases} 2x + y + pz = -1 \\ 3x - 2y + z = q \\ 5x - 8y + 9z = 5 \end{cases} \] has more than one solution, then \( q - p \) is equal to:

• A. \(2\)
• B. \(-2\)
• C. \(4\)
• D. \(-4\)

Hint:

For infinitely many solutions, ensure both: \begin{itemize} \item determinant of coefficient matrix is zero, \item constants follow the same linear dependence as the equations. \end{itemize}

Answer 1

The Correct Option is C

For a system of three linear equations to have more than one solution, the determinant of the coefficient matrix must be zero and the system must be consistent. 
Step 1: Determinant of coefficient matrix \[ \begin{vmatrix} 2 & 1 & p\\ 3 & -2 & 1\\ 5 & -8 & 9 \end{vmatrix} =2((-2)\cdot 9-1\cdot(-8)) -1(3\cdot 9-1\cdot 5) +p(3\cdot(-8)-(-2)\cdot 5) \] \[ =2(-18+8)-1(27-5)+p(-24+10) =-20-22-14p \] \[ \Rightarrow -42-14p=0 \Rightarrow p=-3 \] Step 2: Consistency condition With \(p=-3\), observe that the rows satisfy: \[ -2R_1+3R_2=R_3 \] For consistency, the constants must satisfy the same relation: \[ -2(-1)+3q=5 \Rightarrow 2+3q=5 \Rightarrow q=1 \] Step 3: Compute \(q-p\) \[ q-p=1-(-3)=4 \] \[ \boxed{4} \]