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If $c_1 \geq d_1, c_2 \geq d_2$ and $c_3 \geq d_3$ and we change the third coulumn of $ \det\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\a_3 & b_3 & c_3 \end{pmatrix}$ as $ \det\begin{pmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\a_3 & b_3 & d_3 \end{pmatrix}$, what will be the relation between these determinants. My intuition is: $$\det\begin{pmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\a_3 & b_3 & c_3 \end{pmatrix} \geq \det\begin{pmatrix} a_1 & b_1 & d_1 \\ a_2 & b_2 & d_2 \\a_3 & b_3 & d_3 \end{pmatrix}.$$

But I am not getting any reference for this.

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    $\begingroup$ No, the determinant can decrease a well (there are numerous counterexamples). $\endgroup$
    – Jean Marie
    Mar 31, 2021 at 17:01

4 Answers 4

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This is not true. Determinants are crazy, so changing some set of values results in weird changes overall. The determinant for the matrix in question is

$$D=c_1(a_2b_3-a_3b_2)-c_2(a_1b_3-a_3b_1)+c_3(a_1b_2-a_2b_1)$$

If we increase the $c_i$'s, this changes the determinant based on the determinants of the minors. In principle, the change in the determinant with respect to a particular element is equal in magnitude to the determinant of the element's corresponding minor, with a possible sign change based on the element's location.

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This is false. Let the first matrix be $\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} $ and the second $\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} $. Then $d_1=0$, $d_2=0$, $d_3=1$, $c_1=1$, $c_2=0$, and $c_3=1$. These obey your inequalities, but the first matrix has determinant 1 while the second has determinant 0.

Moreover, this isn't true even if you replace the $\leq$ with strict inequalities, as I can use the matrices $\begin{bmatrix} 1 & 2 & 0\\ 0 & 2 & 0 \\ 1 & 2 & 1 \end{bmatrix} $ and $\begin{bmatrix} 1 & 2 & 2\\ 0 & 2 & 2 \\ 1 & 2 & 2 \end{bmatrix} $, the former of which has determinant 2 while the latter has determinant 0.

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  • $\begingroup$ Based on your idea $\begin{vmatrix}1&0&0\\0&1&0\\1&0&1\end{vmatrix}=1,\ \begin{vmatrix}1&0&1+x\\0&1&0\\1&0&1\end{vmatrix}=-x,\ \begin{vmatrix}1&0&1\\0&1&0\\1&0&1+x\end{vmatrix}=x$ $\endgroup$
    – zwim
    Mar 31, 2021 at 17:24
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View the determinant as giving the signed volume of the parallelepiped defined by the three vectors $\bf a, \bf b, \bf c$.

Then if $\bf c$ is in the same "direction" as $\bf w = \bf a \times \bf b$, more rigorously if $0 < \bf c \cdot \bf w$, and untill the increase $\Delta \bf c = \bf d - \bf c$ is in the same direction, you obtain an increase of the (positive) volume.

Then it is easy to understand what happens in the four cases:
original volume = $\pm$, $\Delta \bf c \cdot \bf w = \pm$.

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Suppose this was true. Then $$det(a,b,c) \geq det(a,b,d)$$ $$-det(a,b,c) \leq -det(a,b,d)$$ $$det(-a,b,c) \leq det(-a,b,d)$$ provides a counterexample.

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