Consistency Condition for Linear Equations

CAT 2003 Slot 1 · QA · Hard · Algebra

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+r0p + q + r \neq 0?

x+2y3z=p2x+6y11z=qx2y+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+10y15z=5p5x + 10y - 15z = 5p Multiply Eq(2) by -2: 4x12y+22z=2q-4x - 12y + 22z = -2q Multiply Eq(3) by -1: x+2y7z=r-x + 2y - 7z = -r Summing these coefficients: x(541)+y(1012+2)+z(15+227)=0x+0y+0z=5p2qrx(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 5p2qr=05p - 2q - r = 0.

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