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Question
Without expanding determinants, show that
`|(1, 3, 6),(6, 1, 4),(3, 7, 12)| + |(2, 3, 3),(2, 1, 2),(1, 7, 6)| = 10|(1, 2, 1),(3, 1, 7),(3, 2, 6)|`
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Solution
L.H.S. = `|(1, 3, 6),(6, 1, 4),(3, 7, 12)| + 4 |(2, 3, 3),(2, 1, 2),(1, 7, 6)|`
In 1st determinant, taking 2 common from C3, we get
L.H.S. = `2|(1,3,3),(6,1,2),(3,7,6)| + 4|(2,3,3),(2,1,2),(1,7,6)|`
= `|(2, 3, 3),(12, 1, 2),(6, 7, 6)| + |(8, 3, 3),(8, 1, 2),(4, 7, 6)|`
= `|(2 + 8, 3, 3),(12 + 8, 1, 2),(6 + 4, 7, 6)|`
= `|(10, 3, 3),(20, 1, 2),(10, 7, 6)|`
Interchanging rows and columns, we get
L.H.S. = `|(10, 20, 10),(3, 1, 7),(3, 2, 6)|`
Taking 10 common from R1, we get
L.H.S. = `10|(1, 2, 1),(3, 1, 7),(3, 2, 6)|`
= R.H.S.
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