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

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 $|\Delta|=0$