Question: > 
> Place the numbers 1 through 9 in the available input boxes so that the
>  results of performing the mathematical operations in any row and
>  column (from left to right and top to bottom) always equals 13.
> 
> 
> \begin{bmatrix}\square&+&\square&-&\square\\
>  -& &+& &\times\\ \square&\times&\square&+&\square\\
>  +&&-&&+\\ \square&\times&\square&-&\square\\ \end{bmatrix}
> 
> 
> 

It takes me an hour to solve this. Do you have an easy way to solve this? Let's explore.

[Original image here.](https://i.stack.imgur.com/eOKS7.png)

Answer: 
I tried it a bit, and my answer for this is $$8 \ \ 7 \ \ 2$$$$1 \ \ 9 \ \ 4$$ $$6 \ \ 3 \ \ 5$$
My approach was to notice that for $z$ in $x+y-z = 13$, $z$ can only be $1,2,3,4$, meaning the middle-left and bottom-middle ones could only be one of these four, and went on from there.
Label spaces A through I.

A+B, B+E,A+G are all at least 14. So A,B,E, and G are all greater than or equal to 5.

D x E must be less than 13 and E is 5 or more so D = 1 or 2.

If D=2 then E=5. If E = 5 then B+5-H=13 so B=9, H=1. So Gx1 -I =13 is impossible.

So D isn't 2. So D=1. So E+F = 13, so F is at least 4. C x F +I = 13 but F is at least 4 and I is at least 3 and C is at least 2. 

If C is at least 3 then CxD+I is at least 14 so C is at most 2 and at least 2. So C is 2 and I is at least 3.

2x F + I =13 and F is at least 4 so we have either 2x4 + 5. Or 2x5 +3.

If F=4, I=5, E=9. GxH = 18 and neither G or H is 9 so G and H are 3 and 6. And G is at least 5 so G = 6, H =3 and B+9-3=13 so B=7. That leaves A =8. And that works as A+7-2=13 and A-1+6=13.

That's a solution.

any other solution would have F=5 and I =3 and E=8.

So A+B=11, GxH= 16, A+G=12, B-H=5. 1,2,3 are assigned so G and H are at least 4, so G xH =16 is impossible.

So only solution is

A=8,B=7,C=2,D=1,E=9,F=4,G=6,H=3,I=5.