If $$A = \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix}$$ and $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$, then $$10 A^{-1}$$ is equal to.

We are given the matrix $$A = \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix}$$ and the identity matrix $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$. Our task is to find $$10A^{-1}$$ and then compare it with the four expressions provided in the options.

First, we recall the standard formula for the inverse of a $$2 \times 2$$ matrix. For a general matrix $$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$, the inverse exists (provided the determinant is non-zero) and is given by

$$\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)^{-1} \;=\; \dfrac{1}{ad-bc}\,\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.$$

Here $$a = 2,\; b = 2,\; c = 9,\; d = 4$$. So we first compute the determinant:

$$\det(A) \;=\; ad - bc \;=\; (2)(4) - (2)(9) \;=\; 8 - 18 \;=\; -10.$$

Since the determinant is $$-10 \neq 0$$, the inverse exists. Using the same formula, the adjugate (or adjoint) matrix of $$A$$ is

$$\operatorname{adj}(A) \;=\; \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \;=\; \begin{pmatrix} 4 & -2 \\ -9 & 2 \end{pmatrix}.$$

Therefore, the inverse of $$A$$ is

$$A^{-1} \;=\; \dfrac{1}{\det(A)}\,\operatorname{adj}(A) \;=\; \dfrac{1}{-10}\,\begin{pmatrix} 4 & -2 \\ -9 & 2 \end{pmatrix}.$$

Carrying the scalar $$\dfrac{1}{-10}$$ inside, we get

$$A^{-1} \;=\; \begin{pmatrix} -\dfrac{4}{10} & \dfrac{2}{10} \\[4pt] \dfrac{9}{10} & -\dfrac{2}{10} \end{pmatrix}.$$ By simplifying the individual fractions we may keep them as they are, or proceed directly to the next requirement, which is to multiply the whole inverse by $$10$$.

Multiplying each entry of $$A^{-1}$$ by $$10$$ gives us

$$10A^{-1} \;=\; 10 \times \begin{pmatrix} -\dfrac{4}{10} & \dfrac{2}{10} \\[4pt] \dfrac{9}{10} & -\dfrac{2}{10} \end{pmatrix} \;=\; \begin{pmatrix} -4 & 2 \\ 9 & -2 \end{pmatrix}.$$

We now compare this matrix with the four candidate expressions. Let us explicitly evaluate each option.

We first compute $$A - 6I$$ because the numeric difference $$6$$ directly appears in two of the options and will likely match the magnitude of the numbers we obtained.

Since $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$, we have $$6I = \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix}.$$ Therefore,

$$A - 6I \;=\; \begin{pmatrix} 2 & 2 \\ 9 & 4 \end{pmatrix} \;-\; \begin{pmatrix} 6 & 0 \\ 0 & 6 \end{pmatrix} \;=\; \begin{pmatrix} 2-6 & 2-0 \\ 9-0 & 4-6 \end{pmatrix} \;=\; \begin{pmatrix} -4 & 2 \\ 9 & -2 \end{pmatrix}.$$

We see that

$$A - 6I \;=\; \begin{pmatrix} -4 & 2 \\ 9 & -2 \end{pmatrix} \;=\; 10A^{-1}.$$

Hence, the matrix $$10A^{-1}$$ is exactly equal to $$A - 6I$$. This corresponds to Option C in the list.

Hence, the correct answer is Option C.