Question

If A = $\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$ then ${{A}^{n}}$ is equal to $[n\in N]$
A . $\left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]$
B . $\left[ \begin{matrix}
   n & n \\
   0 & n \\
\end{matrix} \right]$
C . $\left[ \begin{matrix}
   n & 1 \\
   0 & 1 \\
\end{matrix} \right]$
D . none of these

Answer 1

Hint: We are given a matrix and we have to find the n terms of that matrix. We know a square matrix is a matrix which has same number of rows and columns. First we find the square of matrix. To Find the square of A we multiply A with A and after simplifying it , we get the value of square of matrix A and similary we find the cube of A by multiplying square of A with A then by comparing the values, we find the ${{A}^{n}}$.
Complete Step- by- Step Solution:
We have given the matrix A = $\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$
First we find the matrix ${{A}^{2}}$
We will now perform matrix multiplication
${{A}^{2}}$= $\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$ $\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$
${{A}^{2}}$= $\left[ \begin{matrix}
   [1\times 1]+[1\times 0] & [1\times 1]+[1\times 1] \\
   [0\times 1]+[1\times 0] & [0\times 1]+[1\times 1] \\
\end{matrix} \right]$
Simplifying the above equation, we get
${{A}^{2}}$= $\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]$
Now we find the ${{A}^{3}}$
For this we multiply ${{A}^{2}}$with A
${{A}^{3}}$= $\left[ \begin{matrix}
   1 & 2 \\
   0 & 1 \\
\end{matrix} \right]$$\left[ \begin{matrix}
   1 & 1 \\
   0 & 1 \\
\end{matrix} \right]$
${{A}^{3}}$= $\left[ \begin{matrix}
   [1\times 1]+[2\times 0] & [1\times 1]+[2\times 1] \\
   [0\times 1]+[1\times 0] & [0\times 1]+[1\times 1] \\
\end{matrix} \right]$
Simplifying the above equation, we get
${{A}^{3}}$= $\left[ \begin{matrix}
   1 & 3 \\
   0 & 1 \\
\end{matrix} \right]$
Here we observe a pattern. Whenever we multiply the terms, all the values remains same except first row and second column value. So there we generalize the value of ${{A}^{n}}$
Now we find the value of ${{A}^{n}}$
${{A}^{n}}$= $\left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]$
Hence the value of ${{A}^{n}}$= $\left[ \begin{matrix}
   1 & n \\
   0 & 1 \\
\end{matrix} \right]$
Thus, Option (A) is correct.
Note: Students must take care while multiplying the two matrices. In Multiplication of matrices, remember that the number of column of first matrix match the number of rows of second matrix. When we want to multiply the matrices, then the parts of the rows in first matrix are multiplied with the columns in the second matrix.