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Question
Answer the following question:
Evaluate determinant along second column
`|(1, -1, 2),(3, 2, -2),(0, 1, -2)|`
Solution
Let D = `|(1, -1, 2),(3, 2, -2),(0, 1, -2)|`
The expansion of the determinant D along second column is
D = – 1 (cofactor of – 1) + 2 (cofactor of 2) + 1 (cofactor of 1)
= `-1 xx (-1)^(1 + 2) |(3, -2),(0, -2)| + 2 xx (-1)^(2 + 2)|(1, 2),(0, -2)| + 1 xx (-1)^(2 + 3)|(1, 2),(3, -2)|`
= –1( –1) ( – 6 + 0) + 2 (1)( – 2 – 0) +1 ( –1)( –2 – 6)
= 1 ( – 6) + 2( – 2) – 1( – 8)
= –6 – 4 + 8
= – 2.
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