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Exercises · Q12

Q.Evaluate ∣246135012∣\begin{vmatrix}2&4&6\\1&3&5\\0&1&2\end{vmatrix} by first taking out a common factor from the first row.

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Row 1 is (2,4,6)=2(1,2,3)(2,4,6)=2(1,2,3), so by the common factor property (Section 5):

∣246135012∣=2∣123135012∣\begin{vmatrix}2&4&6\\1&3&5\\0&1&2\end{vmatrix}=2\begin{vmatrix}1&2&3\\1&3&5\\0&1&2\end{vmatrix}

Expand the simplified determinant along row 1:

∣123135012∣=1∣3512∣−2∣1502∣+3∣1301∣\begin{vmatrix}1&2&3\\1&3&5\\0&1&2\end{vmatrix}=1\begin{vmatrix}3&5\\1&2\end{vmatrix}-2\begin{vmatrix}1&5\\0&2\end{vmatrix}+3\begin{vmatrix}1&3\\0&1\end{vmatrix}

=1(6−5)−2(2−0)+3(1−0)=1(1)−2(2)+3(1)=1−4+3=0=1(6-5)-2(2-0)+3(1-0)=1(1)-2(2)+3(1)=1-4+3=0

So the original determinant is 2×0=02\times0=0.

Independent check — expand the original determinant directly without factoring: …

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