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Without expanding, show that : $ \begin{vmatrix} 42 & 1 & 6 \\ 28 & 7 & 4 \\ 14 & 3 & 2 \end{vmatrix} = 0 $

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Toolbox:
  • If two rows or columns of a determinant are identical, then its value is zero
  • If a row or columns of a determinant is mutiplied by k, then the value of the determinant is also multiplied by k
Let $\Delta= \begin{vmatrix} 42 & 1 & 6 \\ 28 & 7 & 4 \\ 14 & 3 & 2 \end{vmatrix} $
Let us take 7 as the common factor from $C_1$
Therefore $\Delta =7 \begin{vmatrix} 6 & 1 & 6 \\ 4 & 7 & 4 \\ 2 & 3 & 2 \end{vmatrix} $
Now we can see that $C_1$ and $C_3$ are identical
we know that any two rows or columns are identical, then the determinant value is 0
Therefore $|\Delta|=0$
answered Apr 5, 2013 by meena.p
 

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