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I have a determinant to expand which is

$\triangle = \begin{bmatrix} p& 1 & \frac{-q}{2}{}\\ 1& 2 &-q \\ 2& 2 & 3 \end{bmatrix} = 0 $

But when I am expanding the determinant along the first row such as $ p(6+2q) - (3+2q) = 0 $ but when I am trying to expand along first column I am getting $p(6+2q) - (3+q) = 0$ but I have been told by my teacher that we can expand the determinant of $3\times 3$ matrix in along any row and any column giving the same result. where lies the error ?

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    use$\LaTeX$to get $\LaTeX$2011-01-24

1 Answers 1

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Seems you like you forgot the third column ($\frac{-q}{2}$) when expanding using the first row.

In the first row expansion, just using the first and second columns gives you $p(6+q) - (3 +2q)$. The third column gives an extra $q$, after adding which it matches the second expression you got, using the first column.

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    Aha ! Thanks Mr Moron but in this case I acted like a moron :P2011-01-24