Question

If the product of the matrices
$\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]$ $\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]$ $\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right]$ .......$\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]$ = $\left[\begin{array}{cc}1& 378\\ 0& 1\end{array}\right]$, then $n$ is equal to

• A. $27$
• B. $26$
• C. $376$
• D. $378$

Answer 1

The correct option is A $27$

$\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right].....\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 378\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 1+2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 0& 1\end{array}\right].....\left[\begin{array}{cc}1& n\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 378\\ 0& 1\end{array}\right]$

Performing multiplications of matrices in this order...

$\left[\begin{array}{cc}1& 1+2+3+.....+n\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 378\\ 0& 1\end{array}\right]$

By the rule of equality

$1+2+3+.....+n=378$

$\frac{n(n+1)}{2}=378n^2+n=756n^2+n-756=0n^2+28n-27n-756=0n(n+28)-27(n+28)=0(n-27)(n+28)=0n=27orn=-28$

but in the equation it is $n$ cannot be $-28$

So $n=27$