If A is any 2 `xx` matrix such that `[{:(1,2),(0,3):}]A=[{:(-1,0),(6,3):}]` then what is A equal to ?

Options

A. `[{:(-5,1),(-2,2):}]`
B. `[{:(-5,-2),(1,2):}]`
C. `[{:(-5,-2),(2,1):}]`
D. `[{:(5,2),(-2,-1):}]`

Explanation

To solve the problem, we need to find the matrix \( A \) given that:

\[ B \cdot A = C \]

where

\[ B = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} \quad \text{and} \quad C = \begin{pmatrix} -1 & 0 \\ 6 & 3 \end{pmatrix} \]

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

To find \( A \), we can rearrange the equation:

\[ A = B^{-1} \cdot C \]

First, we need to calculate the inverse of matrix \( B \).

The formula for the inverse of a \( 2 \times 2 \) 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 \( B \):

- \( a = 1 \)
- \( b = 2 \)
- \( c = 0 \)
- \( d = 3 \)

**Calculate the determinant of \( B \)**:

\[ \text{det}(B) = ad - bc = (1)(3) - (0)(2) = 3 \]

**Now, apply the inverse formula**:

\[ B^{-1} = \frac{1}{3} \begin{pmatrix} 3 & -2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{pmatrix} \]

### Step 2: Multiply \( B^{-1} \) with \( C \)

Now, we can find \( A \):

\[ A = B^{-1} \cdot C = \begin{pmatrix} 1 & -\frac{2}{3} \\ 0 & \frac{1}{3} \end{pmatrix} \cdot \begin{pmatrix} -1 & 0 \\ 6 & 3 \end{pmatrix} \]

**Perform the multiplication**:

1. First row, first column:

\[ 1 \cdot (-1) + \left(-\frac{2}{3}\right) \cdot 6 = -1 - 4 = -5 \]

2. First row, second column:

\[ 1 \cdot 0 + \left(-\frac{2}{3}\right) \cdot 3 = 0 - 2 = -2 \]

3. Second row, first column:

\[ 0 \cdot (-1) + \frac{1}{3} \cdot 6 = 0 + 2 = 2 \]

4. Second row, second column:

\[ 0 \cdot 0 + \frac{1}{3} \cdot 3 = 0 + 1 = 1 \]

Thus, we have:

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

### Final Answer

The matrix \( A \) is:

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