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 \( 2A^{-1} \) where \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \), we will follow these steps:

### Step 1: Find the determinant of matrix A
The determinant of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
\[ \text{det}(A) = (1)(4) - (2)(3) = 4 - 6 = -2 \]

### Step 2: Find the inverse of matrix A
The inverse of a 2x2 matrix \( A \) is given by:
\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
Substituting the values from our matrix \( A \):
\[ A^{-1} = \frac{1}{-2} \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} \]

### Step 3: Multiply the inverse by 2
Now we need to calculate \( 2A^{-1} \):
\[ 2A^{-1} = 2 \begin{pmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} 2 \cdot -2 & 2 \cdot 1 \\ 2 \cdot \frac{3}{2} & 2 \cdot -\frac{1}{2} \end{pmatrix} = \begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix} \]

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

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