There are two possible values of A in the solution of the
matrix equation `[[2A+1,-5],[-4,A]]^(-1) [[A-5,B],[2A-2,C]]= [[14,D],[E,F]]`,
where A, B, C, D, E, F are real numbers. The absolute
value of the difference of these two solutions, is

Options

A. `8/3`
B. `11/3`
C. `1/3`
D. `19/3`

Explanation

To solve the matrix equation given in the problem, we will follow these steps:

### Step 1: Write down the matrix equation
The matrix equation is given as:
\[ \begin{bmatrix} 2A + 1 & -5 \\ -4 & A \end{bmatrix}^{-1} \begin{bmatrix} A - 5 & B \\ 2A - 2 & C \end{bmatrix} = \begin{bmatrix} 14 & D \\ E & F \end{bmatrix} \]

### Step 2: Find the inverse of the first matrix
For a 2x2 matrix
\[ \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \]
the inverse is given by:
\[ \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}. \]
Here, \( a = 2A + 1 \), \( b = -5 \), \( c = -4 \), and \( d = A \).

#### Calculate the determinant:
\[ \text{det} = (2A + 1)A - (-5)(-4) = 2A^2 + A - 20. \]

#### Calculate the adjoint:
\[ \text{adjoint} = \begin{bmatrix} A & 5 \\ 4 & 2A + 1 \end{bmatrix}. \]

#### Therefore, the inverse is:
\[ \begin{bmatrix} 2A + 1 & -5 \\ -4 & A \end{bmatrix}^{-1} = \frac{1}{2A^2 + A - 20} \begin{bmatrix} A & 5 \\ 4 & 2A + 1 \end{bmatrix}. \]

### Step 3: Multiply the inverse with the second matrix
Now we need to multiply the inverse with the second matrix:
\[ \frac{1}{2A^2 + A - 20} \begin{bmatrix} A & 5 \\ 4 & 2A + 1 \end{bmatrix} \begin{bmatrix} A - 5 & B \\ 2A - 2 & C \end{bmatrix}. \]

#### Performing the multiplication:
The resulting matrix will be:
\[ \begin{bmatrix} A(A - 5) + 5(2A - 2) & AB + 5C \\ 4(A - 5) + (2A + 1)(2A - 2) & 4B + (2A + 1)C \end{bmatrix}. \]

### Step 4: Set the resulting matrix equal to the right-hand side
This gives us:
\[ \begin{bmatrix} A^2 - 5A + 10A - 10 & AB + 5C \\ 4A - 20 + 4A^2 - 4A + 2A + 1 & 4B + (2A + 1)C \end{bmatrix} = \begin{bmatrix} 14 & D \\ E & F \end{bmatrix}. \]

### Step 5: Equate the elements
From the first element:
\[ A^2 + 5A - 10 = 14 \implies A^2 + 5A - 24 = 0. \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 5, c = -24 \):
\[ A = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} = \frac{-5 \pm \sqrt{25 + 96}}{2} = \frac{-5 \pm \sqrt{121}}{2} = \frac{-5 \pm 11}{2}. \]
Thus, we have two solutions:
1. \( A = \frac{6}{2} = 3 \)
2. \( A = \frac{-16}{2} = -8 \)

### Step 7: Find the absolute difference
The absolute value of the difference between the two solutions \( A = 3 \) and \( A = -8 \) is:
\[ |3 - (-8)| = |3 + 8| = 11. \]

### Final Answer
The absolute value of the difference of these two solutions is \( \boxed{11} \).