Question

If $A=\left\{\begin{array}{cc}2& 1\\ 0& 1\end{array}\right\},AB=I$, then $B=$

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

Answer 1

Answer: (B)

$\frac{1}{2}\left\{\begin{array}{cc}1& -1\\ 0& 2\end{array}\right\}$

If $A=\left[\begin{array}{cc}2& 1\\ 0& 1\end{array}\right]$
and$AB=I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
$So,A^{-1}AB=A^{-1}I$
$B=A^{-1}$
So we have to find inverse of A,
$\therefore \left[\begin{array}{cc}2& 1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
by transformation on A, we convert it into I
by $r_1\to \frac{r_1}{2}$
$\Rightarrow \left[\begin{array}{cc}1& 1/2\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1/2& 0\\ 0& 1\end{array}\right]$
$r_1\to r_1-\frac{r_1}{2}$
So,$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1/2& -1/2\\ 0& 1\end{array}\right]$
So, $A^{-1}=\left[\begin{array}{cc}1/2& -1/2\\ 0& 1\end{array}\right]$
$A^{-1}=\frac{1}{2}\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]$