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Mathematics Question on Determinants

If [11 01]\begin {bmatrix} 1 & 1 \\\ 0 & 1 \end {bmatrix} . [12 01]\begin {bmatrix} 1 & 2 \\\ 0 & 1 \end {bmatrix} . [13 01]\begin {bmatrix} 1 & 3 \\\ 0 & 1 \end {bmatrix} ..... [1n−1 11]\begin {bmatrix} 1 & n-1 \\\ 1 & 1 \end {bmatrix} = [178 01]\begin {bmatrix} 1 & 78 \\\ 0 & 1 \end {bmatrix}, then the inverse of [1n 01]\begin {bmatrix} 1 & n \\\ 0 & 1 \end {bmatrix} is

A

[1−13 01]\begin {bmatrix} 1 & -13 \\\ 0 & 1 \end {bmatrix}

B

[10 121]\begin {bmatrix} 1 & 0 \\\ 12 & 1 \end {bmatrix}

C

[1−12 01]\begin {bmatrix} 1 & -12 \\\ 0 & 1 \end {bmatrix}

D

[10 131]\begin {bmatrix} 1 & 0 \\\ 13 & 1 \end {bmatrix}

Answer

[1−13 01]\begin {bmatrix} 1 & -13 \\\ 0 & 1 \end {bmatrix}

Explanation

Solution

The correct answer is A:[1−13\01]\begin{bmatrix}1&-13\\\0&1\end{bmatrix}
Given that;
[11\01][12\01][13\01]......[1(n−1)\01]=[178\01]\begin{bmatrix}1&1\\\0&1\end{bmatrix}\begin{bmatrix}1&2\\\0&1\end{bmatrix}\begin{bmatrix}1&3\\\0&1\end{bmatrix}......\begin{bmatrix}1&(n-1)\\\0&1\end{bmatrix}=\begin{bmatrix}1&78\\\0&1\end{bmatrix}
⇒[11+2+3+....+n−1\01]=[178\01]\begin{bmatrix}1&1+2+3+....+n-1\\\0&1\end{bmatrix}=\begin{bmatrix}1&78\\\0&1\end{bmatrix}
⇒1+2+3+.....(n−1)=781+2+3+.....(n-1)=78
⇒n(n−2)2=78\frac{n(n-2)}{2}=78
⇒n2−2n−156=0n^2-2n-156=0
∴n=13,12\therefore n=13,12
Now inverse of [1n\01]\begin{bmatrix}1&n\\\0&1\end{bmatrix} is;
[113\01]−1=[1−13\01]\begin{bmatrix}1&13\\\0&1\end{bmatrix}^{-1}=\begin{bmatrix}1&-13\\\0&1\end{bmatrix}
[\bigg[if A=[ac\bd]  A−1=1ad−ac[d−b−ca]]A=\begin{bmatrix}a&c\\\b&d\end{bmatrix}\space A^{-1}=\frac{1}{ad-ac}\begin{bmatrix}d&-b\\\\-c&a\end{bmatrix}\bigg]
and [1−13\01]\begin{bmatrix}1&-13\\\0&1\end{bmatrix} also
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