Question

If $\left[\begin{array}{cc}1& 2\\ 3& -5\end{array}\right]$, then $A^{-1}$ is equal to

• A. $\left[\begin{array}{cc}\frac{5}{11}& \frac{2}{11}\\ \frac{3}{11}& -\frac{1}{11}\end{array}\right]$
• B. $\left[\begin{array}{cc}-\frac{5}{11}& -\frac{2}{11}\\ -\frac{3}{11}& -\frac{1}{11}\end{array}\right]$
• C. $\left[\begin{array}{cc}\frac{5}{11}& \frac{2}{11}\\ \frac{3}{11}& \frac{1}{11}\end{array}\right]$
• D. $\left[\begin{array}{cc}5& 2\\ 3& -1\end{array}\right]$

Answer 1

Answer: (A)

$\left[\begin{array}{cc}\frac{5}{11}& \frac{2}{11}\\ \frac{3}{11}& -\frac{1}{11}\end{array}\right]$

Since $A=\left[\begin{array}{cc}1& 2\\ 3& -5\end{array}\right]$

$\therefore \left|A\right|=\left[\begin{array}{cc}1& 2\\ 3& -5\end{array}\right]=-5-6=-11$

and $adj\left(A\right)=\left[\begin{array}{cc}-5& -2\\ -3& 1\end{array}\right]$

$\therefore A^{-1}=\frac{1}{\left|A\right|}adj\left(A\right)$

$=-\frac{1}{11}\left[\begin{array}{cc}-5& -2\\ -3& 1\end{array}\right]=\frac{1}{11}\left[\begin{array}{cc}5& 2\\ 3& -1\end{array}\right]$

$=\left[\begin{array}{cc}\frac{5}{11}& \frac{2}{11}\\ \frac{3}{11}& -\frac{1}{11}\end{array}\right]$