If A1,B1,C1,.. are, respectively, the cofactors of the elements a1,b1,c1,.. of the determinant Δ=∣∣
∣∣a1b1c1a2b2c2a3b3c3∣∣
∣∣, Δ≠0, then the value of ∣∣∣B2C2B3C3∣∣∣ is equal to
A
a12Δ
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B
a1Δ
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C
a1Δ2
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D
a12Δ2
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Solution
The correct option is Ba1Δ Δ=∣∣
∣∣a1b1c1a2b2c2a3b3c3∣∣
∣∣ ⇒Δ=a1(b2c3−b3c2)+b1(a3c2−a2c3)+c1(a2b3−a3b2) We have to find ∣∣∣B2C2B3C3∣∣∣ Here, B2 is cofactor of element b2 So, B2=∣∣∣a1c1a3c3∣∣∣ ⇒B2=a1c3−a3c1 C2 is cofactor of element c2 So, C2=−∣∣∣a1b1a3b3∣∣∣ ⇒C2=a3b1−a1b3 B3 is cofactor of element b3 So, B3=−∣∣∣a1c1a2c2∣∣∣ ⇒B3=a2c1−a1c2 C3 is cofactor of element c3 So, C3=∣∣∣a1b1a2b2∣∣∣ ⇒C3=a1b2−a2b1 Now, ∣∣∣B2C2B3C3∣∣∣ =∣∣∣a1c3−a3c1a3b1−a1b3a2c1−a1c2a1b2−a2b1∣∣∣ =a12b2c3−a1a2b1c3−a1a3b2c1−a12b3c2+a1a2b3c1+a1a3b1c2 =a1[a1(b2c3−b3c2)+b1(a3c2−a2c3)+c1(a2b3−a3b2)] =a1Δ