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 and F are real numbers. The absolute value of the difference of these two solutions, is

Options

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

Explanation

To solve the matrix equation

\[ \begin{pmatrix} 2A + 1 & -5 \\ -4 & A \end{pmatrix}^{-1} \begin{pmatrix} A - 5 & B \\ 2A - 2 & C \end{pmatrix} = \begin{pmatrix} 14 & D \\ E & F \end{pmatrix} \]

we will follow these steps:

### Step 1: Find the inverse of the first matrix

The inverse of a 2x2 matrix

\[ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]

is given by

\[ \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

For our matrix

\[ \begin{pmatrix} 2A + 1 & -5 \\ -4 & A \end{pmatrix} \]

we have:

- \(a = 2A + 1\)
- \(b = -5\)
- \(c = -4\)
- \(d = A\)

The determinant \(D\) is

\[ D = (2A + 1)A - (-5)(-4) = 2A^2 + A - 20 \]

Thus, the inverse is

\[ \frac{1}{2A^2 + A - 20} \begin{pmatrix} A & 5 \\ 4 & 2A + 1 \end{pmatrix} \]

### Step 2: Multiply the inverse with the second matrix

Now we need to multiply this inverse with the second matrix

\[ \begin{pmatrix} A - 5 & B \\ 2A - 2 & C \end{pmatrix} \]

The product is

\[ \frac{1}{2A^2 + A - 20} \begin{pmatrix} A & 5 \\ 4 & 2A + 1 \end{pmatrix} \begin{pmatrix} A - 5 & B \\ 2A - 2 & C \end{pmatrix} \]

Calculating the first element of the product:

\[ A(A - 5) + 5(2A - 2) = A^2 - 5A + 10A - 10 = A^2 + 5A - 10 \]

Calculating the second element of the first row:

\[ A \cdot B + 5 \cdot C \]

Calculating the first element of the second row:

\[ 4(A - 5) + (2A + 1)(2A - 2) = 4A - 20 + (4A^2 - 4A + 2A - 2) = 4A^2 + 4A - 22 \]

Calculating the second element of the second row:

\[ 4B + (2A + 1)C \]

So we have:

\[ \frac{1}{2A^2 + A - 20} \begin{pmatrix} A^2 + 5A - 10 & AB + 5C \\ 4A^2 + 4A - 22 & 4B + (2A + 1)C \end{pmatrix} = \begin{pmatrix} 14 & D \\ E & F \end{pmatrix} \]

### Step 3: Set up equations

From the equality of matrices, we can set up the following equations:

1. \( \frac{A^2 + 5A - 10}{2A^2 + A - 20} = 14 \)
2. \( \frac{AB + 5C}{2A^2 + A - 20} = D \)
3. \( \frac{4A^2 + 4A - 22}{2A^2 + A - 20} = E \)
4. \( \frac{4B + (2A + 1)C}{2A^2 + A - 20} = F \)

### Step 4: Solve the first equation

From the first equation:

\[ A^2 + 5A - 10 = 14(2A^2 + A - 20) \]

Expanding gives:

\[ A^2 + 5A - 10 = 28A^2 + 14A - 280 \]

Rearranging gives:

\[ -27A^2 - 9A + 270 = 0 \]

Dividing by -9:

\[ 3A^2 + A - 30 = 0 \]

### Step 5: Factor or use the quadratic formula

Using the quadratic formula:

\[ A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 + 360}}{6} = \frac{-1 \pm 19}{6} \]

Thus, we have:

1. \( A = 3 \)
2. \( A = -\frac{10}{3} \)

### Step 6: Find the absolute difference

The absolute difference is:

\[ |3 - (-\frac{10}{3})| = |3 + \frac{10}{3}| = | \frac{9}{3} + \frac{10}{3} | = | \frac{19}{3} | = \frac{19}{3} \]

### Final Answer

The absolute value of the difference of these two solutions is

\[ \frac{19}{3} \]