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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. [[1, 0], [12, 1]]
2. [[1, -13], [0, 1]]
3. [[1, -12], [0, 1]]
4. [[1, 0], [13, 1]]

Correct answer: [[1, -13], [0, 1]]

Solution

Multiplying the upper-triangular matrices adds the off-diagonal entries: 1+2+...+(n-1)=n(n-1)/2=78 -> n(n-1)=156 -> n=13. The inverse of [[1,13],[0,1]] is [[1,-13],[0,1]].

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