Consider 24 matrices of order 2×2 Considering 24 matrices of the type 2 x 2 which can be obtained by some arrangement of 4 non-negative integers a, b, c and d. For certain assignment of non-negative integers, four of these matrices have ∆ = 16, four have ∆ = - 16 and sixteen matrices are singular.
I need to find the total number of possible values of a+b+c+d.
I tried assuming one matrix to give ∆ = 16 , i.e. ab - cd = 16. Consequently, ac - bd = 0 and ad - bc = 0. However, after simplifying it's coming to a=d and b=c.
Any way to solve it differently?
 A: Let's choose that
$$
\left|\begin{matrix}a&c\\d&b\end{matrix}\right|=ab-cd=16
$$
Then we are given that
$$
\left|\begin{matrix}a&d\\b&c\end{matrix}\right|=\left|\begin{matrix}a&c\\b&d\end{matrix}\right|=0
$$
Which is to say, $[a,b]^T$ and $[d,c]^T$ are parallel, as well as $[a,b]^T$ and $[c,d]^T$. This gives two possibilities:

*

*$[c,d]^T$ and $[d,c]^T$ are parallel, so $c=d$

*$a=b=0$
However, since $ab-cd$ is positive, the second option is invalid. Thus we must necessarily have $c=d$.
A very similar argument with a different pair of singular matrices shows we do now have two possibilities. Either:

*

*$a=b$

*$c=d=0$
In this case, however, we can't rule out the second possibility.
If we are in the top case, $a=b$, we are left with solving $a^2-c^2=16$. We know that $a^2-c^2=(a+c)(a-c)$, so all that's left is to factor $16$ and try to match it up with these two factors.
Since $a,c$ are both nonnegative integers, there are only three options:
$$
\cases{a+c=16\\a-c=1}\\
\cases{a+c=8\\a-c=2}\\
\cases{a+c=4\\a-c=4}
$$
Solving these, we get
$$
\cases{a=\frac{17}2\\c=\frac{15}2}\\
\cases{a=5\\c=3}\\
\cases{a=4\\c=0}
$$
The first one is invalid. The two others yield $a+b+c+d=16$ and $8$ respectively.
Now for the $c=d=0$ case. Then we just get $ab=16$, with the possibilities $a+b=17$, $10$ or $8$ (the last of which is a duplicate from above, with $a=b=4, c=d=0$).
So in conclusion, $a+b+c+d$ can be $17, 16, 10$ or $8$.
