QJEEAdv08Maa84b: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Maa84b Solution: Consider the system of equations a x + b y = 0 ; c x + d y = 0…

Question

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Consider the system of equations

a x + b y = 0; c x + d y = 0,

where

a, b, c, d

\in {0, 1}

STATEMENT - 1 : The probability that the system of equations has a unique solution is

\frac{3}{8} .

and

STATEMENT - 2 : The probability that the system of equations has a solution is

1.

A STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT - 1
B STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1
C STATEMENT - 1 is True, STATEMENT - 2 is False.
D STATEMENT - 1 is False, STATEMENT - 2 is True.

Accepted Answer

We have,

$a x + b y = 0$

$c x + d y = 0$

since, the system of homogeneous equation is always consistent and has a solution.

Therefore, statement 2 is true.

Now, $Δ = | \begin{matrix} a & b \\ c & d \end{matrix} |$ and $a, b, c, d \in {0, 1}$

$= a d - b c$

No. of ways of selecting $a, b, c, d$ from the set {0, 1} is 2 $\times$ 2 $\times$ 2 $\times$ 2 = 16

If the system has unique solution,

Then $Δ \neq 0$

$\Rightarrow$ either $a d = 1, b c = 0$ or $a d = 0, b c = 1$ favourable case = 6

Therefore, probability that system of equation has unique solution is $\frac{6}{16} = \frac{3}{8}$ . Statement 1 is True.

Hence, the Statement 2 is True but is not a correct explanation of statement 1.

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