Explanation
1. Total Outcomes n(S)
Matrix M=[acamp;bamp;d] ke 4 positions hain, aur har position ke liye 3 choices {0,1,2} hain.
2. Non-Invertible Matrices n(Ac) nikalna
Invertible hone ke liye ∣M∣=0 hona chahiye. Hum pehle wo cases nikalenge jahan ∣M∣=ad−bc=0, yani ad=bc.
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Case 1: ad=bc=0
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ad=0 ke liye 5 pairs hain: (0,0),(0,1),(0,2),(1,0),(2,0).
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bc=0 ke liye bhi 5 pairs hain.
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Total cases =5×5=25.
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Case 2: ad=bc=1
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ad=1 tabhi hoga jab (1,1) ho (sirf 1 pair).
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bc=1 ke liye bhi sirf 1 pair (1,1).
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Total cases =1×1=1.
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Case 3: ad=bc=2
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ad=2 ke liye 2 pairs: (1,2),(2,1).
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bc=2 ke liye bhi 2 pairs: (1,2),(2,1).
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Total cases =2×2=4.
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Case 4: ad=bc=4
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ad=4 ke liye sirf 1 pair: (2,2).
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bc=4 ke liye sirf 1 pair: (2,2).
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Total cases =1×1=1.
3. Probability Calculation
Non-invertible matrices ki total sankhya n(Ac)=25+1+4+1=31.
Invertible matrices ki sankhya n(A)=n(S)−n(Ac)=81−31=50.
Ab probability P(A) hogi: