Explanation
Step 1: Matrix A ke determinant ∣A∣ ki calculation
Sabse pehle hum determinant ∣A∣ ki value nikalenge. Matrix ke elements bade factorials hain, isliye hum rows se common factors bahar nikalenge:
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R1 se 5! common lene par.
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R2 se 6! common lene par.
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R3 se 7! common lene par.
∣A∣=5!6!7!1×(5!⋅6!⋅7!)111amp;6amp;7amp;8amp;42amp;56amp;72
∣A∣=111amp;6amp;7amp;8amp;42amp;56amp;72
Ab row operations R2→R2−R1 aur R3→R3−R2 apply karne par:
∣A∣=100amp;6amp;1amp;1amp;42amp;14amp;16
Expanding along C1:
Step 2: ∣2A∣ ki value find karna
Property: ∣kA∣=kn∣A∣ (Yahan n=3 hai kyunki matrix 3×3 hai)
Step 3: ∣adj(adj(M))∣ ki property apply karna
Property: ∣adj(adj(M))∣=∣M∣(n−1)2
Yahan humein ∣adj(adj(2A))∣ nikalna hai, toh M=2A aur n=3:
∣adj(adj(2A))∣=∣2A∣(3−1)2
Step 4: Final Calculation
∣2A∣=24 ki value upar waale formula mein rakhne par:
Sahi Answer: (1) 216