If `A = {:((1,-1),(1,2)):}`and `I={:((1,0),(0,1)):}` then `3A^(-1)` is equal to :

Options

A. `3I-A`
B. `4I-A`
C. `6I-A`
D. `A - 6I`

Explanation

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

### Step 1: Find the inverse of matrix \( A \)

The formula for the inverse of a 2x2 matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:

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

For our matrix \( A \):

- \( a = 1 \)
- \( b = -1 \)
- \( c = 1 \)
- \( d = 2 \)

First, we calculate the determinant \( |A| = ad - bc \):

\[ |A| = (1)(2) - (-1)(1) = 2 + 1 = 3 \]

Now, we can find \( A^{-1} \):

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

### Step 2: Calculate \( 3A^{-1} \)

Now we multiply \( A^{-1} \) by 3:

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

### Step 3: Express \( 3A^{-1} \) in terms of the identity matrix \( I \)

The identity matrix \( I \) is:

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

We can express \( 3A^{-1} \) in terms of \( I \) and \( A \):

We know that:

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

Thus, we have:

\[ 3A^{-1} = 3I - A \]

### Final Answer

Therefore, the final answer is:

\[ 3A^{-1} = 3I - A \]