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CAT 2020 Slot 3 — QA Question 26

2 Variable EquationsEasy

Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if

Answer & solution

  • A

    |k| = 2

  • B

    k = 2

  • |k| ≠ 2

  • D

    k ≠ 2

Solution

Easy

A pair of linear equations a1x+b1y=c1a_1x+b_1y=c_1 and a2x+b2y=c2a_2x+b_2y=c_2 has a unique solution exactly when the coefficient lines are not parallel, i.e. a1a2≠b1b2\dfrac{a_1}{a_2}\ne\dfrac{b_1}{b_2} (equivalently the determinant a1b2−a2b1≠0a_1b_2-a_2b_1\ne 0). Find which kk violate this, then negate.

1

Write the no-unique-solution condition. Identify the coefficients: a1=k, b1=1, a2=4, b2=ka_1=k,\ b_1=1,\ a_2=4,\ b_2=k. A unique solution fails when the ratios are equal.

a1a2=b1b2⇒ k4=1k(substitute coefficients)\begin{aligned} &\frac{a_1}{a_2}=\frac{b_1}{b_2}\\ &\Rightarrow\ \frac{k}{4} = \frac{1}{k} \quad\text{(substitute coefficients)} \end{aligned}
2

Solve for the bad values. Cross-multiply.

k⋅k=4⋅1⇒ k2=4⇒ k=±2\begin{aligned} &k\cdot k = 4\cdot 1\\ &\Rightarrow\ k^2 = 4\\ &\Rightarrow\ k = \pm 2 \end{aligned}
3

Negate to get the unique-solution condition. The system has a unique solution for all other kk, i.e. whenever k≠2k\ne 2 and k≠−2k\ne -2.

unique  ⟺  k≠2 and k≠−2⇒ ∣k∣≠2\begin{aligned} &\text{unique} \iff k\ne 2 \text{ and } k\ne -2\\ &\Rightarrow\ |k|\ne 2 \end{aligned}
∣k∣≠2|k|\ne 2
CAT 2020 Slot 3 QA Q26: Let k be a constant. The equations kx + y = 3 and 4x + ky = 4 have a unique solution if and only if — Solution | TheCATExam