First slide

Operations of matrix

Question

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

Moderate

A

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

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

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

$[\begin{matrix} 1 & - 12 \\ 13 & 1 \end{matrix}]$

Solution

Multiplying we get
$\begin{matrix} [\begin{matrix} 1 & 1 + 2 + 3 + \dots + n \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 78 \\ 0 & 1 \end{matrix}] \end{matrix} = [\begin{matrix} 1 & \frac{n (n - 1)}{2} \\ 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 78 \\ 0 & 1 \end{matrix}] (g i v e n) \Rightarrow \frac{n (n - 1)}{2} = 78$
$\Rightarrow n = 13$
$∴$ Inverse of $[\begin{matrix} 1 & 13 \\ 0 & 1 \end{matrix}]$ $=$ $[\begin{matrix} 1 & - 13 \\ 0 & 1 \end{matrix}]$

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