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Consistency Condition for Linear Equations

CAT 2003 Slot 1 · Quantitative Ability · Hard · Algebra

This is a hard Quantitative Ability question from the CAT 2003 Slot 1 paper. It tests Algebra. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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Which one of the following conditions must p,qp, q and rr satisfy so that the following system of linear simultaneous equations has at least one solution, such that p+q+r≠0p + q + r \neq 0?

x+2y−3z=p2x+6y−11z=qx−2y+7z=r\begin{aligned} x + 2y - 3z &= p \\ 2x + 6y - 11z &= q \\ x - 2y + 7z &= r \end{aligned}
  1. A.

    5p - 2q - r = 0

  2. B.

    5p + 2q + r = 0

  3. C.

    5p + 2q - r = 0

  4. D.

    5p - 2q + r = 0

Answer

A

Explanation

We combine the equations to eliminate variables: Multiply Eq(1) by 5: 5x+10y−15z=5p5x + 10y - 15z = 5p Multiply Eq(2) by -2: −4x−12y+22z=−2q-4x - 12y + 22z = -2q Multiply Eq(3) by -1: −x+2y−7z=−r-x + 2y - 7z = -r Summing these coefficients: x(5−4−1)+y(10−12+2)+z(−15+22−7)=0x+0y+0z=5p−2q−rx(5 - 4 - 1) + y(10 - 12 + 2) + z(-15 + 22 - 7) = 0x + 0y + 0z = 5p - 2q - r. For the system to have a solution, we must have 5p−2q−r=05p - 2q - r = 0.

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