If $A=\left(\begin{array}{cc}2& 2\\ 9& 4\end{array}\right)$ and $I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, then $10A^{-1}$ is equal to:

1. Calculate the determinant $|A|$:
$$|A|=(2)(4)-(2)(9)=8-18=-10$$
2. Calculate the Adjoint of $A$ (swap diagonal, negate anti-diagonal):
$$\text{adj }A=\left(\begin{array}{cc}4& -2\\ -9& 2\end{array}\right)$$
3. Find $A^{-1}$:
$$A^{-1}=\frac{1}{|A|}\text{adj }A=\frac{1}{-10}\left(\begin{array}{cc}4& -2\\ -9& 2\end{array}\right)$$
4. Multiply by 10:
$$10A^{-1}=10\cdot \frac{1}{-10}\left(\begin{array}{cc}4& -2\\ -9& 2\end{array}\right)=-1\left(\begin{array}{cc}4& -2\\ -9& 2\end{array}\right)=\left(\begin{array}{cc}-4& 2\\ 9& -2\end{array}\right)$$
5. Check options:Option B: $A-6I=\left(\begin{array}{cc}2& 2\\ 9& 4\end{array}\right)-\left(\begin{array}{cc}6& 0\\ 0& 6\end{array}\right)=\left(\begin{array}{cc}-4& 2\\ 9& -2\end{array}\right)$.Matches.