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If $\det \begin{pmatrix}a&1&d\\ b&1&e\\ c&1&f\end{pmatrix}=1$ and $\det \begin{pmatrix}a&1&d\\ b&2&e\\ c&3&f\end{pmatrix}=1$, what is $\det \begin{pmatrix}a&-4&d\\ b&-5&e\\ c&-6&f\end{pmatrix}$?

So I am aware about all the different operations and what changes they bring to the value of the determinant, but I am not exactly sure which one of them is being applied here. There is no constant multiplication or an addition or subtraction by a row. I have tried to add and subtract various multiples of the matrices from each other as well but not to any avail. I believe I am just not spotting something.

Any help?

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1 Answer 1

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By the multilinearity of determinant, $$\begin{aligned} & \begin{vmatrix}a&-4&d\\ b&-5&e\\ c&-6&f\end{vmatrix} \\ &= \begin{vmatrix}a&-1&d\\ b&-2&e\\ c&-3&f\end{vmatrix}+\begin{vmatrix}a&-3&d\\ b&-3&e\\ c&-3&f\end{vmatrix} \\ &= -\begin{vmatrix}a&1&d\\ b&2&e\\ c&3&f\end{vmatrix}-3\begin{vmatrix}a&1&d\\ b&1&e\\ c&1&f\end{vmatrix} \\ &= -1 -3(1) =-4. \end{aligned} $$

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  • 2
    $\begingroup$ Wow...Well spotted... $\endgroup$
    – sktsasus
    Commented Feb 7, 2018 at 22:11
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    $\begingroup$ bangs head against wall $\endgroup$
    – user541686
    Commented Feb 8, 2018 at 8:57

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