If AX=B, where `A={:[(3,1),(-1,2)],:}B={:[(7,3),(0,6)],:}` then X=

Options

A. `{:[(1,0),(2,3)]:}`
B. `{:[(0,3),(1,2)]:}`
C. `{:[(3,2),(0,1)]:}`
D. `{:[(2,0),(1,3)]:}`

Answer

To solve the equation \( AX = B \) where

\[ A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 7 & 3 \\ 0 & 6 \end{pmatrix} \]

we need to find the matrix \( X \).

### Step 1: Find the Determinant of Matrix A

First, we calculate the determinant of matrix \( A \):

\[ \text{det}(A) = (3)(2) - (1)(-1) = 6 + 1 = 7 \]

### Step 2: Check if the Inverse Exists

Since \( \text{det}(A) = 7 \) (which is not zero), the inverse of \( A \) exists.

### Step 3: Find the Inverse of Matrix A

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

\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

For our matrix \( A \):

\[ A^{-1} = \frac{1}{7} \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} \]

### Step 4: Multiply Both Sides by \( A^{-1} \)

Now, we can find \( X \) by multiplying both sides of the equation \( AX = B \) by \( A^{-1} \):

\[ X = A^{-1}B \]

### Step 5: Calculate \( A^{-1}B \)

Substituting the values we have:

\[ X = \frac{1}{7} \begin{pmatrix} 2 & -1 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 7 & 3 \\ 0 & 6 \end{pmatrix} \]

Now, we perform the matrix multiplication:

1. First row, first column:
\[ (2)(7) + (-1)(0) = 14 + 0 = 14 \]

2. First row, second column:
\[ (2)(3) + (-1)(6) = 6 - 6 = 0 \]

3. Second row, first column:
\[ (1)(7) + (3)(0) = 7 + 0 = 7 \]

4. Second row, second column:
\[ (1)(3) + (3)(6) = 3 + 18 = 21 \]

Thus, we have:

\[ X = \frac{1}{7} \begin{pmatrix} 14 & 0 \\ 7 & 21 \end{pmatrix} \]

### Step 6: Simplify the Result

Now, we simplify \( X \):

\[ X = \begin{pmatrix} \frac{14}{7} & \frac{0}{7} \\ \frac{7}{7} & \frac{21}{7} \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} \]

### Final Result

Thus, the matrix \( X \) is:

\[ X = \begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} \]

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