91影视

A square matrix 鈥楤鈥� is said to be similar to a matrix A if there exists an invertible matrix P such that

$B=P^{-1}AP$

Let us take n = 2. Then,

$I_2=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right){\text{and2I}}_2=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)$

Let us suppose that matrix $I_2$is similar to matrix$2I_2$ then there exists an invertible matrix $P=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$such that $2I_{2}P=P{I}_2.$

$$\begin{gathered} 鈬�\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ 鈬�\left(\begin{array}{cc}2a& 2b\\ 2c& 2d\end{array}\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) \end{gathered}$$

Thus, matrix$P=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$ is not invertible and so there doesn鈥檛 exist any invertible matrix P such that matrix $I_2$is similar to matrix $2I_2$.

Since this holds for n = 2, therefore, it will hold for every value of n.

Hence, matrix $I_n$is not similar to $2I_n$ where is$I_n$identity matrix of order .

Since there doesn鈥檛 exist any invertible matrix P such that matrix $I_2$is similar to matrix $2I_2$for n = 2, therefore, there doesn鈥檛 exist any invertible matrix for any value of n.

Hence, matrix $I_n$is not similar to$2I_n$ where is$I_n$ identity matrix of order .