```Question 243802
If we look at the 2x2 matrix

a b
b a

we can clearly see that the determinant is {{{a^2-b^2}}} (just use the definition of the 2x2 determinant).

Moving onto the 3x3 matrix

a b 0
0 a b
b 0 a

If we compute this determinant, we get

{{{a*abs(matrix(2,2,a,b,0,a))+0+b*abs(matrix(2,2,b,0,a,b))=a*a^2+b*b^2=a^3+b^3}}}

Similarly, for the 4x4 matrix

a b 0 0
0 a b 0
0 0 a b
b 0 0 a

If we apply the laplace expansion along the first column, we get {{{a*abs(matrix(3,3,a,b,0,0,a,b,0,0,a))-b*abs(matrix(3,3,b,0,0,a,b,0,0,a,b))=a*a^3-b*b^3=a^4-b^4}}}. Also, for the 5x5 matrix

a b 0 0 0
0 a b 0 0
0 0 a b 0
0 0 0 a b
b 0 0 0 a

the determinant is {{{a^5+b^5}}}. Are you starting to see a pattern? If not, then try to find the determinant of the 6x6 matrix

a b 0 0 0 0
0 a b 0 0 0
0 0 a b 0 0
0 0 0 a b 0
0 0 0 0 a b
b 0 0 0 0 a

You'll find that the determinant is {{{a^6-b^6}}}

In general, the determinant of the n x n matrix

a b 0 0...0
0 a b 0...0
0 0 a b...0
...........
0 0 0 0...b
b 0 0 0...a

is {{{a^n+(-1)^(n+1)b^n}}}

I left some work out due to time constraints. Let me know if you need to see how to get any determinants.```