Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$, and a system of linear equations $x+y+z=5$ $x+2y+3z=\mu$ $x+3y+\lambda z=1$ is constructed. If $\rho$ is the probability that the system has a unique solution $q$ is the probability that the system has no solution, then

(A) $\mathrm{p}=\frac{5}{6}$ and $\mathrm{q}=\frac{5}{36}$

(B) $p=\frac{1}{6}$ and $q=\frac{1}{36}$

(C) $\mathrm{p}=\frac{1}{6}$ and $\mathrm{q}=\frac{5}{36}$

(D) $\mathrm{p}=\frac{5}{6}$ and $\mathrm{q}=\frac{1}{36}$

Matrices — JEE Main · 2021 · 26 Aug 2021 · Shift 2 Question with Solution

Two fair dice are thrown. The numbers on them are taken as

λ

and

μ

, and a system of linear equations

x + y + z = 5

x + 2 y + 3 z = μ

x + 3 y + λ z = 1

is constructed. If

ρ

is the probability that the system has a unique solution

q

is the probability that the system has no solution, then

Answer: (A)

The coefficient determinant is

D = 111123 13 λ = (2 λ - 9) - (λ - 3) + (3 - 2) = λ - 5

The system has a unique solution exactly when

D \neq = 0

, that is

λ \neq = 5

. Five of the six faces give this, so

p = \frac{5}{6}

.
When

λ = 5

the coefficient rows satisfy

(row 3) = 2 (row 2) - (row 1)

, since

2 (1, 2, 3) - (1, 1, 1) = (1, 3, 5)

. Consistency then forces the same relation on the constants,

1 = 2 μ - 5

, that is

μ = 3

. The system therefore has no solution precisely when

λ = 5

and

μ \neq = 3

, whose probability is

q = \frac{1}{6} \times \frac{5}{6} = \frac{5}{36}

p = \frac{5}{6}

and

q = \frac{5}{36}

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option