9
$\begingroup$

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?

$\endgroup$
17
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ Wow...Well spotted... $\endgroup$ – sktsasus Feb 7 '18 at 22:11
  • 2
    $\begingroup$ bangs head against wall $\endgroup$ – Mehrdad Feb 8 '18 at 8:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.