If A = `{:[(1, 2), (3,4)]:}` , then `2A^(-1)` =

Options

A. 3I - A
B. 5 I - A
C. A - 5I
D. 3A + I

Explanation

To solve the problem, we need to find the value of \( 2A^{-1} \) given the matrix \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \).

### Step-by-Step Solution:

1. **Find the Characteristic Polynomial**:
We start by finding the characteristic polynomial of the matrix \( A \). The characteristic polynomial is given by the determinant of \( A - \lambda I \), where \( I \) is the identity matrix.

\[ A - \lambda I = \begin{pmatrix} 1 - \lambda & 2 \\ 3 & 4 - \lambda \end{pmatrix} \]

The determinant is calculated as follows:

\[ \text{det}(A - \lambda I) = (1 - \lambda)(4 - \lambda) - (2)(3) \]

Expanding this:

\[ = (1 - \lambda)(4 - \lambda) - 6 = 4 - 5\lambda + \lambda^2 - 6 = \lambda^2 - 5\lambda - 2 \]

Thus, the characteristic polynomial is:

\[ \lambda^2 - 5\lambda - 2 = 0 \]

2. **Use the Characteristic Polynomial to Find \( A^{-1} \)**:
From the characteristic polynomial, we can express \( A \) in terms of \( I \):

\[ A^2 - 5A - 2I = 0 \]

Rearranging gives:

\[ A^2 - 5A = 2I \]

Now, multiplying both sides by \( A^{-1} \):

\[ A - 5I = 2A^{-1} \]

3. **Solve for \( 2A^{-1} \)**:
Rearranging the equation above gives:

\[ 2A^{-1} = A - 5I \]

4. **Calculate \( A - 5I \)**:
Now we need to compute \( A - 5I \):

\[ 5I = 5 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} \]

Therefore:

\[ A - 5I = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} - \begin{pmatrix} 5 & 0 \\ 0 & 5 \end{pmatrix} = \begin{pmatrix} 1 - 5 & 2 - 0 \\ 3 - 0 & 4 - 5 \end{pmatrix} = \begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} \]

5. **Final Result**:
Thus, we find:

\[ 2A^{-1} = A - 5I = \begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} \]

### Conclusion:
The final answer is:

\[ 2A^{-1} = \begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} \]