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Question:

If [1 1; 0 1][1 2; 0 1][1 3; 0 1]...[1 n-1; 0 1] = [1 78; 0 1], then the inverse of [1 n; 0 1] is?

[1 0; 1 21]

[1 -1; 2 0 1]

[1 -1; 3 0 1]

[1 0; 1 31]

Solution:

[1101][1201][1201][1301][1n−1;01]=[17801]⇒[11+2+3+...+n−1;01]=[17801]⇒n(n−2;)2=78⇒n=13,−1;2(reject)∴We have to find inverse of[11301]∴[1−1;301]