Question

Using elementary transformations, find the inverse of matrix
$\left[\begin{array}{cc}6& 5\\ 5& 4\end{array}\right]$

• A. $\left[\begin{array}{cc}4& 5\\ -5& 6\end{array}\right]$
• B. $\left[\begin{array}{cc}4& 5\\ 5& -6\end{array}\right]$
• C. $\left[\begin{array}{cc}4& -5\\ -5& 6\end{array}\right]$
• D. $\left[\begin{array}{cc}-4& 5\\ 5& -6\end{array}\right]$

Answer 1

Answer: (D)

Given matrix$A=\left[\begin{array}{cc}6& 5\\ 5& 4\end{array}\right],thenA=IA$

$\Rightarrow \left[\begin{array}{cc}6& 5\\ 5& 4\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]A$

Applying $R_1\to R_1-R_2$, we get

$\left[\begin{array}{cc}1& 1\\ 5& 4\end{array}\right]=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]A$

Applying $R_2\to R_2-5R_1$, we get

$\left[\begin{array}{cc}1& 1\\ 0& -1\end{array}\right]=\left[\begin{array}{cc}-4& 5\\ -5& 6\end{array}\right]A$

Applying $R_2\to \left(-1\right)R_2$, we get

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}-4& 5\\ 5& -6\end{array}\right]A$

$A^{-1}=\left[\begin{array}{cc}-4& 5\\ 5& -6\end{array}\right]$