If \left[\right. 1 & 1 \\ 0 & 1 \left]\right. \cdot \left[\right. 1 & 2 \\ 0 & 1 \left]\right. \cdot \left[\right. 1 & 3 \\ 0 & 1 \left]\right. \cdot . .......... \cdot \left[\right. 1 & n - 1 \\ 0 & 1 \left]\right. = \left[\right. 1 & 78 \\ 0 & 1 \left]\right. , then the inverse of \left[\right. 1 & n \\ 0 & 1 \left]\right. is

Engineering

Mathematics

Multiplication of Matrices

Question

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

$[\begin{matrix} 1 & 0 \\ 12 & 1 \end{matrix}]$

$[\begin{matrix} 1 & - 12 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 1 & - 13 \\ 0 & 1 \end{matrix}]$

$[\begin{matrix} 1 & 0 \\ 13 & 1 \end{matrix}]$

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$[\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}] \cdot [\begin{matrix} 1 & 3 \\ 0 & 1 \end{matrix}] \cdot . .......... \cdot [\begin{matrix} 1 & n - 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 78 \\ 0 & 1 \end{matrix}]$

$\Rightarrow [\begin{matrix} 1 & 1 + 2 + 3 + . ..... + n - 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 78 \\ 0 & 1 \end{matrix}]$

$\Rightarrow \frac{n (n - 1)}{2} = 78 \Rightarrow n = 13, - 12 (reject)$

$∴ we have to find inverse of [\begin{matrix} 1 & 13 \\ 0 & 1 \end{matrix}]$

$∴ [\begin{matrix} 1 & - 13 \\ 0 & 1 \end{matrix}]$