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Let $$A = \begin{pmatrix}1&a_1+\hat a_1 &a_2 + \hat a_2\\1& b_1 + \hat b_1 & b_2 + \hat b_2 \\ 1 & c_1 + \hat c_1 & c_2 + \hat c_2\end{pmatrix}$$ be a real $3\times3$ matrix. I need help proving $$\det A = \det\begin{pmatrix}1&a_1 &a_2\\1& b_1 & b_2 \\ 1 & c_1 & c_2 \end{pmatrix} + \det\begin{pmatrix}1&a_1 &a_2\\1& \hat b_1 & \hat b_2 \\ 1 & \hat c_1 & \hat c_2 \end{pmatrix} +\det\begin{pmatrix}1& \hat a_1 & \hat a_2\\1& b_1 & b_2 \\ 1 & \hat c_1 & \hat c_2 \end{pmatrix} +\det\begin{pmatrix}1&\hat a_1 &\hat a_2\\1& \hat b_1 & \hat b_2 \\ 1 & c_1 & c_2 \end{pmatrix}$$ I am only allowed to use "properties" of the determinant (i.e. multilinearity in each argument etc.), calculating both sides and comparing is obviously not allowed. I first used the multilinearity in each row to get $$\det A = \det\begin{pmatrix}1&a_1 &a_2\\1& b_1 & b_2 \\ 1 & c_1 & c_2 \end{pmatrix} + \det\begin{pmatrix}1&a_1 &\hat a_2\\1& b_1 & \hat b_2 \\ 1 & c_1 & \hat c_2 \end{pmatrix} +\det\begin{pmatrix}1& \hat a_1 & a_2\\1&\hat b_1 & b_2 \\ 1 & \hat c_1 & c_2 \end{pmatrix} +\det\begin{pmatrix}1&\hat a_1 &\hat a_2\\1& \hat b_1 & \hat b_2 \\ 1 &\hat c_1 &\hat c_2 \end{pmatrix}$$ but from here I don't know what properties I could use to get the result.

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HINT: Do the exact same thing you did but use multilinearity of the determinant as a function of column vectors instead of row vectors.

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  • $\begingroup$ Ahhh, I see! Thanks! $\endgroup$ – Staki42 Jan 20 '17 at 22:54

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