Question

If $\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]\cdots \left[ \begin{matrix}
   1 & n-1 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 78 \\
   0 & 1 \\
\end{matrix} \right]$, then the inverse of $\left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]$ is
[a] $\left[ \begin{matrix}
   1 & -13 \\
   0 & 1 \\
\end{matrix} \right]$
[b] $\left[ \begin{matrix}
   1 & 0 \\
   12 & 1 \\
\end{matrix} \right]$
[c] $\left[ \begin{matrix}
   1 & -12 \\
   0 & 1 \\
\end{matrix} \right]$
[d] $\left[ \begin{matrix}
   1 & 0 \\
   13 & 1 \\
\end{matrix} \right]$

Answer 1

Hint: Assume that $\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]\cdots \left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & f\left( n \right) \\
   0 & 1 \\
\end{matrix} \right]$. Form a recursive relation for f(n) and solve for f(n) by considering the product $\left[ \begin{matrix}
   1 & f\left( n-1 \right) \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & f\left( n \right) \\
   0 & 1 \\
\end{matrix} \right]$. Hence find the value of n satisfying $f\left( n-1 \right)=78$ and hence find the inverse of $\left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]$.

Complete step-by-step solution:
Let us take $\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]\cdots \left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & f\left( n \right) \\
   0 & 1 \\
\end{matrix} \right]$
Replacing n by n-1, we get
$\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]\cdots \left[ \begin{matrix}
   1 & n-1 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & f\left( n-1 \right) \\
   0 & 1 \\
\end{matrix} \right]$
Hence, we have
$\left[ \begin{matrix}
   1 & f\left( n-1 \right) \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & f\left( n \right) \\
   0 & 1 \\
\end{matrix} \right]$
Now, performing matrix multiplication, we have
$\left[ \begin{matrix}
   1 & n+f\left( n-1 \right) \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & f\left( n \right) \\
   0 & 1 \\
\end{matrix} \right]$
Equating elements of both matrices, we get
$\begin{align}
  & n+f\left( n-1 \right)=f\left( n \right) \\
 & \Rightarrow f\left( n \right)-f\left( n-1 \right)=n \\
\end{align}$
Replacing n by n-1, we get
$f\left( n-1 \right)-f\left( n-2 \right)=n-1$
Replacing n by n-1, we get
$\begin{align}
  & f\left( n-2 \right)-f\left( n-3 \right)=n-2 \\
 & \vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \ \ \ \ \ \ \ \ \ \ \ \ \ \vdots \\
\end{align}$
Continuing this way, we get
$f\left( 2 \right)-f\left( 1 \right)=2$
Adding all these equations formed, we get
$\begin{align}
  & f\left( n \right)-f\left( n-1 \right)+f\left( n-1 \right)-f\left( n-2 \right)+\cdots +f\left( 2 \right)-f\left( 1 \right)=n+n-1+n-2+\cdots +2 \\
 & \Rightarrow f\left( n \right)-f\left( 1 \right)=2+3+4+\cdots +n \\
\end{align}$
Also, we have
$\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & f\left( 1 \right) \\
   0 & 1 \\
\end{matrix} \right]$
Hence, we have
f(1) = 1
Hence, we have
$f\left( n \right)=1+2+\cdots +n$
We know that the sum of the first n natural numbers is $\dfrac{n\left( n+1 \right)}{2}$
Hence, we have
$f\left( n \right)=\dfrac{n\left( n+1 \right)}{2}$
From the given equation, we have
$\begin{align}
  & f\left( n-1 \right)=78 \\
 & \Rightarrow \dfrac{n\left( n-1 \right)}{2}=78 \\
 & \Rightarrow {{n}^{2}}-n=156 \\
 & \Rightarrow {{n}^{2}}-n-156=0 \\
\end{align}$
We solve this quadratic equation using splitting the middle term method
We have $13-12=1$ and $13\times 12=156$
Hence, we have
${{n}^{2}}-13n+12n-156=0$
Taking n common from the first two terms and 12 common from the last two terms, we get
$n\left( n-13 \right)+12\left( n-13 \right)=0$
Taking n-13 common from the two terms, we get
$\left( n+12 \right)\left( n-13 \right)=0$
Since n is natural number n > 0 and hence n + 12 > 0
Hence, we have
$\begin{align}
  & n-13=0 \\
 & \Rightarrow n=13 \\
\end{align}$
Hence, the matrix $M=\left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]$ is given by $M=\left[ \begin{matrix}
   1 & 13 \\
   0 & 1 \\
\end{matrix} \right]$
Here det(M) = 1
We know that \[A=\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\Rightarrow {{A}^{-1}}=\dfrac{1}{\det A}\left[ \begin{matrix}
   d & -b \\
   -c & a \\
\end{matrix} \right]\]
Hence, we have
${{M}^{-1}}=\left[ \begin{matrix}
   1 & -13 \\
   0 & 1 \\
\end{matrix} \right]$
Hence option [a] is correct.

Note: The equations of the form ${{T}_{n}}-{{T}_{n-1}}=g\left( n \right)$ form a telescopic series on writing down of all terms and hence ${{T}_{n}}={{T}_{1}}+\sum\limits_{i=2}^{n}{g\left( i \right)}$. Although we have shown how the total sum comes, the student is advised to remember the result