Q.
If ∣∣a1a2a3b1b2b3c1c2c3∣∣=5 , then the value of ∣∣b2c3−b3c2b3c1−b1c3b1c2−b2c1a3c2−a2c3a1c3−a3c1a2c1−a1c2a2b3−a3b2a3b1−a1b3a1b2−a2b1∣∣ is
2574
191
NTA AbhyasNTA Abhyas 2020Matrices
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Solution:
Let Δ=∣∣a1a2a3b1b2b3c1c2c3∣∣=5 and Δ′=∣∣b2c3−b3c2b3c1−b1c3b1c2−b2c1a3c2−a2c3a1c3−a3c1a2c1−a1c2a2b3−a3b2a3b1−a1b3a1b2−a2b1∣∣
here Δ′ is co-factor determinant of Δ
hence Δ′=Δn−1⇒Δ3−1=Δ2 =(5)2=25