Question

If the product of n matrices $\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right],\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]......\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$ is equal to the matrix $\left[\begin{array}{cc}1& 378\\ 0& 1\end{array}\right]$ then the value of $n$ is equal to

• A. $26$
• B. $27$
• C. $377$
• D. $378$

Answer 1

Answer: (B)

$27$

$\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 1+2\\ 0& 1\end{array}\right]$

$\left[\begin{array}{cc}1& 1+2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 1+2+3\\ 0& 1\end{array}\right]$

On multiplying $n$ matrices, we get

$\left[\begin{array}{cc}1& 1+2+...+n\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 378\\ 0& 1\end{array}\right]$

$\Rightarrow n(n+1)=378\times 2$

$\Rightarrow n(n+1)=756=27\times 28$

$\Rightarrow n(n+1)=27(27+1)$

$\Rightarrow n=27$
Hence, option B.