Let be a natural number n≥2 and the n×n matrix whose entries at the i-th line and j-th column is min(i,j). Calculate:
a) its determinant
b) its inverse
As there was hardly any programming involved, this computer solution is more heavily dependent on "letting the computer do it" than most; such is the strength of MATLAB.
clc
clearvars
for sz=1:8
for i=1:sz
m(sz,i)=i;
m(i,sz)=i;
end
disp(m)
disp(det(m))
disp(inv(m))
disp(" ")
end
builds the matrices a layer at a time, reporting along the way:
In each set, first appears the matrix, then the determinant, then the inverse matrix. I started with n=1, and the pattern holds, though incomplete for n=1.
1
1
1
1 1
1 2
1
2 -1
-1 1
1 1 1
1 2 2
1 2 3
1
2 -1 0
-1 2 -1
0 -1 1
1 1 1 1
1 2 2 2
1 2 3 3
1 2 3 4
1
2 -1 0 0
-1 2 -1 0
0 -1 2 -1
0 0 -1 1
1 1 1 1 1
1 2 2 2 2
1 2 3 3 3
1 2 3 4 4
1 2 3 4 5
1
2 -1 0 0 0
-1 2 -1 0 0
0 -1 2 -1 0
0 0 -1 2 -1
0 0 0 -1 1
1 1 1 1 1 1
1 2 2 2 2 2
1 2 3 3 3 3
1 2 3 4 4 4
1 2 3 4 5 5
1 2 3 4 5 6
1
2 -1 0 0 0 0
-1 2 -1 0 0 0
0 -1 2 -1 0 0
0 0 -1 2 -1 0
0 0 0 -1 2 -1
0 0 0 0 -1 1
1 1 1 1 1 1 1
1 2 2 2 2 2 2
1 2 3 3 3 3 3
1 2 3 4 4 4 4
1 2 3 4 5 5 5
1 2 3 4 5 6 6
1 2 3 4 5 6 7
1
2 -1 0 0 0 0 0
-1 2 -1 0 0 0 0
0 -1 2 -1 0 0 0
0 0 -1 2 -1 0 0
0 0 0 -1 2 -1 0
0 0 0 0 -1 2 -1
0 0 0 0 0 -1 1
1 1 1 1 1 1 1 1
1 2 2 2 2 2 2 2
1 2 3 3 3 3 3 3
1 2 3 4 4 4 4 4
1 2 3 4 5 5 5 5
1 2 3 4 5 6 6 6
1 2 3 4 5 6 7 7
1 2 3 4 5 6 7 8
1
2 -1 0 0 0 0 0 0
-1 2 -1 0 0 0 0 0
0 -1 2 -1 0 0 0 0
0 0 -1 2 -1 0 0 0
0 0 0 -1 2 -1 0 0
0 0 0 0 -1 2 -1 0
0 0 0 0 0 -1 2 -1
0 0 0 0 0 0 -1 1
In all cases the determinant is 1, and the inverse matrix has element (n,n) = 1; all the other elements on the main diagonal are 2; any element immediately adjacent to the main diagonal is -1; all the rest of the elements are zero.
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Posted by Charlie
on 2021-01-29 18:04:44 |
Truly computer solution (spoiler)
