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

Options

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

Explanation

To solve the problem of finding \( 2A^{-1} \) for the matrix \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of Matrix A
The determinant of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by the formula:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
- \( a = 1 \)
- \( b = 2 \)
- \( c = 3 \)
- \( d = 4 \)

Calculating the determinant:
\[ \text{det}(A) = (1)(4) - (2)(3) = 4 - 6 = -2 \]

### Step 2: Calculate the Adjoint of Matrix A
The adjoint of a 2x2 matrix is obtained by swapping the elements on the main diagonal and changing the signs of the off-diagonal elements. For matrix \( A \):
\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} \]

### Step 3: Calculate the Inverse of Matrix A
The inverse of a matrix \( A \) is given by the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \]
Substituting the values we calculated:
\[ A^{-1} = \frac{1}{-2} \cdot \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} \frac{4}{-2} & \frac{-2}{-2} \\ \frac{-3}{-2} & \frac{1}{-2} \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} \]

### Step 4: Calculate \( 2A^{-1} \)
Now, we multiply the inverse of matrix \( A \) by 2:
\[ 2A^{-1} = 2 \cdot \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} 2 \cdot -2 & 2 \cdot 1 \\ 2 \cdot \frac{3}{2} & 2 \cdot -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} \]

### Final Result
Thus, the final answer is:
\[ 2A^{-1} = \begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} \]

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