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EEJ MAIN Mathematics QUESTION #6943
Question 1

If $\begin{pmatrix}1&1\\0&1\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}1&3\\0&1\end{pmatrix}\cdots\begin{pmatrix}1&n-1\\0&1\end{pmatrix} = \begin{pmatrix}1&78\\0&1\end{pmatrix}$, find the inverse of $\begin{pmatrix}1&n\\0&1\end{pmatrix}$.

  • $\begin{pmatrix}1&0\\12&1\end{pmatrix}$
  • $\begin{pmatrix}1&-13\\0&1\end{pmatrix}$
  • $\begin{pmatrix}1&-12\\0&1\end{pmatrix}$✔️
  • $\begin{pmatrix}1&0\\13&1\end{pmatrix}$
Correct Answer Explanation

The product of upper-triangular matrices $\begin{pmatrix}1&k\\0&1\end{pmatrix}$ gives $\begin{pmatrix}1&\sum k\\0&1\end{pmatrix}$.

$\sum_{k=1}^{n-1}k = \dfrac{(n-1)n}{2} = 78 \Rightarrow n(n-1)=156 \Rightarrow n=13$

The inverse of $\begin{pmatrix}1&n\\0&1\end{pmatrix} = \begin{pmatrix}1&13\\0&1\end{pmatrix}$ is $\begin{pmatrix}1&-13\\0&1\end{pmatrix}$.