JEE Main 2019 — Mathematics Algebra

Given $[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}][\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}][\begin{matrix}1 & 3 \\ 0 & 1\end{matrix}]....[\begin{matrix}1 & n-1 \\ 0 & 1\end{matrix}]=[\begin{matrix}1 & 78 \\ 0 & 1\end{matrix}]$[10​11​][10​21​][10​31​]....[10​n−11​]=[10​781​]

And, we have $[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}][\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}]=[\begin{matrix}1 & 3 \\ 0 & 1\end{matrix}]=[\begin{matrix}1 & 1+2 \\ 0 & 1\end{matrix}]$[10​11​][10​21​]=[10​31​]=[10​1+21​]

Also, $[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}][\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}][\begin{matrix}1 & 3 \\ 0 & 1\end{matrix}]=[\begin{matrix}1 & 3 \\ 0 & 1\end{matrix}][\begin{matrix}1 & 3 \\ 0 & 1\end{matrix}]$[10​11​][10​21​][10​31​]=[10​31​][10​31​]

$$=[\begin{matrix}1 & 6 \\ 0 & 1\end{matrix}]=[\begin{matrix}1 & 1+2+3 \\ 0 & 1\end{matrix}]$$=[10​61​]=[10​1+2+31​]

$$\Rightarrow [\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}][\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}][\begin{matrix}1 & 3 \\ 0 & 1\end{matrix}]....[\begin{matrix}1 & n-1 \\ 0 & 1\end{matrix}]=[\begin{matrix}1 & 1+2+3+....+n-1 \\ 0 & 1\end{matrix}]$$⇒[10​11​][10​21​][10​31​]....[10​n−11​]=[10​1+2+3+....+n−11​]

$$\Rightarrow [\begin{matrix}1 & 1+2+3+....+n-1 \\ 0 & 1\end{matrix}]=[\begin{matrix}1 & 78 \\ 0 & 1\end{matrix}]$$⇒[10​1+2+3+....+n−11​]=[10​781​]

Using the sum of first $n$n natural numbers i.e. $1+2+3+...+n=\frac{n(n+1)}{2},$1+2+3+...+n=2n(n+1)​, we get

$$\frac{n(n-1)}{2}=78$$2n(n−1)​=78

$$\Rightarrow {n}^{2}-n-156=0$$⇒n2−n−156=0

$$\Rightarrow (n-13)(n+12)=0$$⇒(n−13)(n+12)=0

$\Rightarrow n=13$⇒n=13 or $n=-12$n=−12 (reject as $n$n is a natural number)

$\therefore$∴ We have to find inverse of $[\begin{matrix}1 & 13 \\ 0 & 1\end{matrix}]$[10​131​]

Which can be find by using ${A}^{-1}=\frac{adjA}{|A|}$A−1=∣A∣adjA​

Now, $|A|=|\begin{matrix}1 & 13 \\ 0 & 1\end{matrix}|=1-0=1$∣A∣=∣10​131​∣=1−0=1

$${A}_{11}=1,{A}_{12}=-13,{A}_{21}=0,{A}_{22}=1$$A11​=1,A12​=−13,A21​=0,A22​=1

Hence, $adjA=[\begin{matrix}1 & -13 \\ 0 & 1\end{matrix}]$adjA=[10​−131​]

$\therefore$∴${A}^{-1}=[\begin{matrix}1 & -13 \\ 0 & 1\end{matrix}].$A−1=[10​−131​].