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$$ \det \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} = 0 $$

Is it possible to find the determinant of the matrix

$$ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} $$

by inspection? By finding the determinant of the matrix by "inspection" I mean to find it without putting it into cofactor form or using diagonals.

I have already tried interpreting the determinant geometrically as the parallelepiped formed by the the three column vectors in the matrix, but fail to see that they would result in a parallelepiped with a volume of 0. I have not tried manipulating the determinant by adding multiples of one row/column to another row/column as I feel that there should be a simpler solution.

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In this particular case the middle column is equal to the average of the first and 3rd, which automatically means, that there is a column of zero after a couple of equivalent transformations, which makes it zero.

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The second row is the element by element average of the other rows, thus a linear combination. The determinant of a square matrix where any row is a linear combination of the others is zero.

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You can use the fact that the third line is twice the second one minus the first one.

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Alternatively, notice that $$ \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix}\begin{pmatrix}1\\-2\\1\end{pmatrix} = 0\begin{pmatrix}1\\-2\\1\end{pmatrix}, $$ so that $0$ is an eigenvalue of your matrix.

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Your matrix is $\begin{pmatrix} 1&1&1\\4&4&4\\7&7&7\end{pmatrix}+\begin{pmatrix} 0&1&2\\0&1&2\\0&1&2\end{pmatrix}\quad$ So if you set $u=\begin{pmatrix} 1\\4\\7\end{pmatrix}$ and $v=\begin{pmatrix} 1\\1\\1\end{pmatrix}$

You can see that $A(e_1)=u+0v,\ A(e_2)=u+1v,\ A(e_3)=u+2v$

Meaning $\operatorname{rank}(A)$ is at most $2$ since the image is in $\operatorname{span}(u,v)$ of dimension $2$.

Thus $\det(A)=0$

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0
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The middle column is the average of the first and the third which makes the determinant to become zero.

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