Question

Find $P^{-1}$,if exists and given $p=\left[\begin{array}{cc}10& -2\\ -5& 1\end{array}\right]$

• A. $\left[\begin{array}{cc}2& 2\\ 5& 10\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 2\\ 5& 10\end{array}\right]$
• C. $\left[\begin{array}{cc}1& 2\\ 4& 10\end{array}\right]$
• D. $P^{-1}$ does not exist

Answer 1

Answer: (D)

We have, $P-IP$
$\therefore \left[\begin{array}{cc}10& -2\\ -5& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]P$
Applying $R_1\to \frac{1}{10}{R}_{{}^{}1}$,we get
$\left[\begin{array}{cc}1& -\frac{1}{5}\\ -5& 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{10}& 0\\ 0& 1\end{array}\right]P$
Applying$R_2\to R_2+5R_1$, we get
$\left[\begin{array}{cc}1& -\frac{1}{5}\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}\frac{1}{10}& 0\\ \frac{1}{2}& 1\end{array}\right]P$
Since, all the elements in the second row of the matrix on $LHS$ are zero, therefore the given matrix is non- invertible. Hence, $P^{-1}$ does not exist.