# determinant inequality

Given integers a,b,c and d; is the absolute value of the determinant of [a c;b d] less than or equal to the absolute value of [a e;b f] + absolute value of [e c;f d] for all a,b,c,d? In other words, is |ad-bc| less than or equal to |af-be| + |de-cf| for all integers a,b,c and ; with the additional constraint that the ordered pairs (a,b), (c,d), and (e,f) cannot equal each other and none of the ordered pairs can be (0,0), and if any of the ordered pairs is of the form (x,0) then the other two can't be of the form (y,0) and if one is (0,x) the other two cannot be (o,y).

I tried to upload a picture but I wasn't allowed to.I want to know whether the sum of the area of the two paralellograms with sides {(a,b),(e,f)} and {(e,f),(c,d)} will always be equal to or greater than associated with {(a,b),(c,d)}. It's just the counter-examples you gave me were of case where the areas become degenrate.

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Edited. Here is what is going on. If the first determinant is zero, then the inequality follows immediately, But if the first determinant is not zero, then it need not hold.

Here is exactly what happens, and how to get as many counterexamples as you please.

Take any matrix whose determinant is not zero: $$\left(\begin{array}{cc} a&c\\ b&d \end{array}\right),$$ and let $$\left|\det\left(\begin{array}{cc} a & c\\ b & d\end{array}\right)\right| = D.$$

If $D\neq 0$, that means that the columns, $(a,b)$ and $(c,d)$, form a basis for $\mathbb{R}^2$. Hence, every vector $(e,f)$ can be written uniquely as $(e,f)=\alpha(a,b) + \beta(c,d)$ for some choice of scalars $\alpha$ and $\beta$. Thus, by the properties of the determinant, we have: \begin{align*} \left|\det\left(\begin{array}{cc} a & e\\ b & f \end{array}\right)\right| &= \left|\det\left(\begin{array}{cc} a & \alpha a+\beta c\\ b & \alpha b + \beta d \end{array}\right)\right| \\ &= \left|\det\left(\begin{array}{cc} a&\alpha a\\ b &\alpha b \end{array}\right) + \det\left(\begin{array}{cc} a & \beta c\\ b & \beta d \end{array}\right)\right|\\ &= \left| 0 + \beta\det\left(\begin{array}{cc} a & c\\ b & d \end{array}\right)\right|\\ &=|\beta|D. \end{align*} Symmetrically, you will have that $$\left|\det\left(\begin{array}{cc} e & c\\\ f & d \end{array}\right)\right| = |\alpha|D.$$ So you will have $$\left|\det\left(\begin{array}{cc} a & e\\\ b & f \end{array}\right)\right| + \left|\det\left(\begin{array}{cc} e & c\\\ f & d \end{array}\right)\right| = |\alpha|D + |\beta|D = (|\alpha|+|\beta|)D,$$ where $\alpha$ and $\beta$ are the unique scalars such that $(e,f) = \alpha(a,b) + \beta(c,d)$.

So the inequality you want is equivalent to $$D\leq (|\alpha|+|\beta|)D,$$ which holds if and only if $|\alpha|+|\beta|\geq 1$, but there are plenty of choices that will make $|\alpha|+|\beta| \lt 1$, in which case your inequality cannot hold.

My first example, now deleted, took $(a,b)=(1,0)$, $(c,d) = (0,1)$, $\alpha=e$ and $\beta=0$. The second example (also now deleted in favor of this exposition) took $\alpha=\beta=e$. As long as neither of $\alpha$ and $\beta$ are zero, you'll get nondegenerate parallelograms, but picking them small in absolute value will ensure the inequality cannot hold.

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No. Take $(a,b)=(4,1)$, $(c,d)=(1,1)$, $(e,f)=(1,0)$.

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Sorry, there's another condition I need- that (a,b), (c,d) and (e,f) cannot equal each other. (I'll edit the question) –  user6910 Feb 11 '11 at 14:31
Then $(a,b)=(3,1)$, $(c,d)=(0,1)$, $(e,f)=(1,0)$ is again a counterexample. Maybe the condition is another one? –  user1728 Feb 11 '11 at 14:42
@user6910 Could you please stop changing the question? I edited the answer to give you another counterexample, but please try at least to think about the problem yourself before posting it here. –  user1728 Feb 11 '11 at 14:51
Sorry about that user1728. –  user6910 Feb 11 '11 at 14:58
@user6910 No problem. Now that you added your geometric motivation at least who answers has less the impression of wasting his time. In the counterexamples given by Arturo and me the parallelograms are not degenerate, but in any case you shouldn't be afraid of considering degenerate cases! You could try to see them as limits of non-degenerate cases. –  user1728 Feb 11 '11 at 15:11