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If $[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ $[\begin{array}{l} 1 & 2 \\ 0 & 1 \end{array}] [\begin{array}{l} 1 & 3 \\ 0 & 1 \end{array}] \dots \dots \cdot [\begin{matrix} 1 & n - 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 78 \\ 0 & 1 \end{matrix}] then the inverse of [\begin{array}{l} 1 & n \\ 0 & 1 \end{array}] is$

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10121

1−1301

10131

1−12131

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detailed solution

Correct option is B

Multiplying we get 11+2+3+⋯+n01=17801 =1n(n-1)201 =17801 given ⇒n(n-1)2=78⇒n=13∴Inverse of 11301 =1−1301

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If 110112011301⋯⋯·1n-101=17801 then the inverse of 1n01 is

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