If the entries in a 3×3 determinant are either 0 or 1, then the greatest value of their determinant is
A
1
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B
2
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C
3
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D
9
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Solution
The correct option is B2 Given : △=∣∣
∣∣a1a2a3b1b2b3c1c2c3∣∣
∣∣
Using 'Sarrus' rule : △=[a1b2c3+b1c2a3+c1a2b3]−[a3b2c1+b3c2a1+c3a2b1]
For Δ to be maximum, [a3b2c1+b3c2a1+c3a2b1]=0 ⇒a1=b2=c3=0 ∴Δ=b1c2a3+c1a2b3
It will be maximum, when a2=a3=b3=b1=c1=c2=1 ∴Δmax=2