`if[{:(2,1),(3,2):}]A[{:(-3,2),(5,-3):}]=[{:(1,0),(0,1):}],"then" A=?`

Options

A. `[{:(0,1),(1,1):}]`
B. `[{:(1,1),(1,0):}]`
C. `[{:(1,0),(0,1):}]`
D. None of these

Explanation

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

\[ B A C = I \]

where

\[ B = \begin{pmatrix} 2 & 1 \\ 3 & 2 \end{pmatrix}, \quad C = \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix}, \quad I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

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

To find the inverse of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \), we use the formula:

\[ \text{If } \text{det}(B) \neq 0, \quad B^{-1} = \frac{1}{\text{det}(B)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

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

\[ \text{det}(B) = (2)(2) - (3)(1) = 4 - 3 = 1 \]

**Now, compute \( B^{-1} \):**

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

### Step 2: Find the inverse of matrix \( C \)

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

\[ \text{det}(C) = (-3)(-3) - (5)(2) = 9 - 10 = -1 \]

**Now, compute \( C^{-1} \):**

\[ C^{-1} = \frac{1}{-1} \begin{pmatrix} -3 & -2 \\ -5 & -3 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 5 & 3 \end{pmatrix} \]

### Step 3: Use the relationship \( A = B^{-1} C^{-1} \)

Now we can find \( A \):

\[ A = B^{-1} C^{-1} = \begin{pmatrix} 2 & -1 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} 3 & 2 \\ 5 & 3 \end{pmatrix} \]

### Step 4: Perform the matrix multiplication

\[ A = \begin{pmatrix} (2)(3) + (-1)(5) & (2)(2) + (-1)(3) \\ (-3)(3) + (2)(5) & (-3)(2) + (2)(3) \end{pmatrix} \]

Calculating each element:

- First row, first column: \( 6 - 5 = 1 \)
- First row, second column: \( 4 - 3 = 1 \)
- Second row, first column: \( -9 + 10 = 1 \)
- Second row, second column: \( -6 + 6 = 0 \)

Thus, we have:

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

### Final Answer

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