Question

For a given $2\times 2$ matrix $A$, it is observed that
$A\left|\begin{array}{c}1\\ -1\end{array}\right|=-1\left|\begin{array}{c}1\\ -1\end{array}\right|$ and $A\left|\begin{array}{c}1\\ -2\end{array}\right|=-2\left|\begin{array}{c}1\\ -2\end{array}\right|$ then the matrix $A$ is

• A. $A=\left|\begin{array}{cc}2& 1\\ -1& -1\end{array}\right|\left|\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right|\left|\begin{array}{cc}1& 1\\ -1& -2\end{array}\right|$
• B. $A=\left|\begin{array}{cc}1& 1\\ -1& -2\end{array}\right|\left|\begin{array}{cc}1& 0\\ 1& 2\end{array}\right|\left|\begin{array}{cc}2& 1\\ -1& -1\end{array}\right|$
• C. $A=\left|\begin{array}{cc}1& 1\\ -1& -2\end{array}\right|\left|\begin{array}{cc}-1& 0\\ 1& -2\end{array}\right|\left|\begin{array}{cc}2& 1\\ -1& -1\end{array}\right|$
• D. $A=\left|\begin{array}{cc}1& -2\\ 1& -3\end{array}\right|$

Answer 1

The correct option is C $A=\left|\begin{array}{cc}1& 1\\ -1& -2\end{array}\right|\left|\begin{array}{cc}-1& 0\\ 1& -2\end{array}\right|\left|\begin{array}{cc}2& 1\\ -1& -1\end{array}\right|$

Given $A\left[\begin{array}{c}1\\ -1\end{array}\right]=(-1)\left[\begin{array}{c}1\\ -1\end{array}\right]$ and $A\left[\begin{array}{c}1\\ -2\end{array}\right]=(-2)\left[\begin{array}{c}1\\ -2\end{array}\right]$
$\Rightarrow$, Now, we know that square matrix $A_{n\times n}$ is said to be diagonalizable if there exists a non-singular matrix P such that $P^{-1}AP=D$ (diagonal matrix) whose diagonal elements are eigen values of $A$ and columns of $P$ are eigen vectors of $A$ and $P$ is known as MODAL matrix
i.e., $P=\left[x_{1}x2\right],D=\left[\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right]$
So, $P^{-1}AP=D\Rightarrow A=PD{P}^{-1}$
$\therefore A=\left|\begin{array}{cc}1& 1\\ -1& -2\end{array}\right|\left|\begin{array}{cc}-1& 0\\ 1& -2\end{array}\right|\left|\begin{array}{cc}2& 1\\ -1& -1\end{array}\right|$
Hence option (c) is correct