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I have the following question which I know I should use the determinant to solve. Here it is:

Determine if there exists an invertible $3\times3$ matrix $A$ such that $$\begin{align*} a_{11}+a_{12}+a_{13}&=0 \\ a_{21}+a_{22}+a_{23}&=0 \\ a_{31}+a_{32}+a_{33}&=0 \end{align*}$$

I know that the answer is no, that there is no such matrix, but I’d like to know how to solve this question with determinants.

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You say you should use the determinant to solve this, but I would like to add that I think you should not do it this way. – JLA Aug 9 '14 at 21:08
up vote 3 down vote accepted

The determinant is invariant under adding single multiples of one column to another. So using this we can add Column $2$ to Column $1$. Then add Column $3$ to Column $1$. This results in a column of zeroes. Hence, the determinant of $A$ is $0$ so it cannot be invertible.

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but I don't understand a a column of zeros results, we didn't subtract columns, so how did we receive zeros? – Alan Aug 9 '14 at 20:45
Perhaps I'm mistaken but you seem to impose a condition on $A$ that says if you add the columns together the resulting vector is the zero vector. – user71352 Aug 9 '14 at 20:48
Wow I'm an idiot. Thank you!! – Alan Aug 9 '14 at 20:50
Not at all. You're welcome!! – user71352 Aug 9 '14 at 20:53

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