Question

If the product of the matrices \[\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\] , \[\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right].......\]\[\left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]\] is equal to the matrix \[\left[ \begin{matrix}
   1 & 378 \\
   0 & 1 \\
\end{matrix} \right]\] then the value of n is equal to
a. 26
b. 27
c. 377
d. 378

Answer 1

Hint: These types of problems are pretty straight forward and are very easy to solve. For the problem of the given type, there are a total of ‘n’ terms, and we need to evaluate the product of all the matrix terms. We know that since ‘n’ is not defined here, we cannot find the value of the product of the matrices. In such cases, we try to find the pattern that the multiplication of the matrices follow. Thereby, we perform multiplication of the first few terms (say $3$ ) and then try to judge the pattern for the rest of the terms.

Complete step-by-step answer:
We now start off with the solution to the problem by evaluating the product of the first three terms. Multiplying the first two terms of the matrices, we get,
\[\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]\]
We now perform matrix multiplication,
\[\Rightarrow \left[ \begin{matrix}
   \left( 1\times 1 \right)+\left( 1\times 0 \right) & \left( 1\times 2 \right)+\left( 1\times 1 \right) \\
   \left( 0\times 1 \right)+\left( 1\times 0 \right) & \left( 0\times 2 \right)+\left( 1\times 1 \right) \\
\end{matrix} \right]\]
Evaluating, we get,
\[\Rightarrow \left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]\]
Now the third term of the matrix series is \[\left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]\] . Now, multiplying this third term with the matrix that we have formed by multiplying the first and the second, is,
\[\left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]\]
Performing matrix multiplication we get,
\[\Rightarrow \left[ \begin{matrix}
   \left( 1\times 1 \right)+\left( 3\times 0 \right) & \left( 1\times 3 \right)+\left( 3\times 1 \right) \\
   \left( 0\times 1 \right)+\left( 1\times 0 \right) & \left( 0\times 3 \right)+\left( 1\times 1 \right) \\
\end{matrix} \right]\]
Evaluating, we get,
\[\Rightarrow \left[ \begin{matrix}
   1 & 6 \\
   0 & 1 \\
\end{matrix} \right]\]
Now, here we observe a pattern. Whenever we multiply the terms of the matrices, all the values of the matrix remain unchanged, except the first row and second column value. We see that when the product of the first two terms was done the first row and second column value was equal to the sum of all its terms till that term. Similar result was found after multiplying the third term. So we can generalise the product of all the terms of the matrix series and find the net result as,
\[\left[ \begin{matrix}
   1 & \dfrac{n\left( n+1 \right)}{2} \\
   0 & 1 \\
\end{matrix} \right]\] , because it is the sum of all the terms, till the nth term.
Now, comparing this result with our given result we find that,
\[\left[ \begin{matrix}
   1 & \dfrac{n\left( n+1 \right)}{2} \\
   0 & 1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   1 & 378 \\
   0 & 1 \\
\end{matrix} \right]\]
Equating both sides we get,
\[\dfrac{n\left( n+1 \right)}{2}=378\]
Finding the value of ‘n’,
\[\begin{align}
  & n\left( n+1 \right)=756 \\
 & \Rightarrow {{n}^{2}}+n-756=0 \\
 & \Rightarrow {{n}^{2}}+28n-27n-756=0 \\
 & \Rightarrow n\left( n+28 \right)-27\left( n+28 \right)=0 \\
 & \Rightarrow \left( n+28 \right)\left( n-27 \right)=0 \\
\end{align}\]
Now, from here \[\left( n+28 \right)=0\] cannot be a solution, because ‘n’ cannot be a negative value, thus, we write,
\[\begin{align}
  & \left( n-27 \right)=0 \\
 & \Rightarrow n=27 \\
\end{align}\]
So, the correct answer is “Option (b)”.

Note: For these types of problems, we should not forget the matrix multiplication and follow each and every step. We need to perform the product of the first few terms, and then judge the pattern by analysing the previous product terms. Once we find the general solution, we can equate it to the given matrix of the question, to find our required answer.