Question

Matrix$\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$is similar to$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$.

Short answer

The above statement is true.

Matrix $\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$is similar to$\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ .

Step-by-step solution

Definition of similar matrix

Two square matrices A and B are said to be similar if there exists an invertible matrix P such that

$B=P^{-1}AP$

Finding an invertible matrix

Let us suppose,

$A=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$and$B=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$

are similar matrices then there exists an invertible matrix $P=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$such that

BP = PA

data-custom-editor="chemistry" $$\begin{gathered} \Rightarrow \left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}c& d\\ a& b\end{array}\right]=\left[\begin{array}{cc}a& -b\\ c& -d\end{array}\right] \\ \Rightarrow a=c \\ d=-b \end{gathered}$$

Therefore we get,

$P=\left[\begin{array}{cc}a& b\\ c& -b\end{array}\right]$

Thus, the matrix$P=\left[\begin{array}{cc}a& b\\ c& -b\end{array}\right]$ is an invertible matrix.

Hence,

Matrix $\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$is similar to $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$.