∣11112313λ∣=λ−5
Δ1=∣12313λ5μ1∣=10λ+3μ−λμ−44
Δ2=∣11113λ5μ1∣=5λ+μ−λμ−13
Δ3=∣1111235μ1∣=−2μ+6
For unique solution
λ=5,μ∈1,2,3,4,5,6
⇒p=65×1=65
For No Solution
λ=5,μ=3
⇒q=61×65=365
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+2y+3z=μ
x+3y+λz=1
is constructed. If p is the probability that the system has a unique solution and q is the probability that the system has no solution, then:
Held on 26 Aug 2021 · Verified 6 Jul 2026.
p=61 and q=365
p=65 and q=361
p=61 and q=361
p=65 and q=365
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