If `X=[{:(3,-4),(1,-1):}],B=[{:(5,2),(-2,1):}]and A=[{:(p,q),(r,s):}]` satisfy the equation AX = B, then the matrix A is equal to

Options

A. `[{:(-7,26),(1,-5):}]`
B. `[{:(7,26),(4,17):}]`
C. `[{:(-7,-4),(26,13):}]`
D. `[{:(-7,26),(-6,23):}]`

Explanation

To solve the equation \( AX = B \) where \( X = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \), \( B = \begin{pmatrix} 5 & 2 \\ -2 & 1 \end{pmatrix} \), and \( A = \begin{pmatrix} p & q \\ r & s \end{pmatrix} \), we need to perform matrix multiplication and then compare the resulting matrix to matrix \( B \).

### Step 1: Write the equation for matrix multiplication
We start with the equation:
\[ AX = B \]
Substituting the matrices, we have:
\[ \begin{pmatrix} p & q \\ r & s \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 5 & 2 \\ -2 & 1 \end{pmatrix} \]

### Step 2: Perform the multiplication
Calculating the left-hand side:
\[ AX = \begin{pmatrix} p \cdot 3 + q \cdot 1 & p \cdot -4 + q \cdot -1 \\ r \cdot 3 + s \cdot 1 & r \cdot -4 + s \cdot -1 \end{pmatrix} \]
This simplifies to:
\[ AX = \begin{pmatrix} 3p + q & -4p - q \\ 3r + s & -4r - s \end{pmatrix} \]

### Step 3: Set up equations by comparing elements
Now we compare the elements of the resulting matrix with matrix \( B \):
1. \( 3p + q = 5 \)
2. \( -4p - q = 2 \)
3. \( 3r + s = -2 \)
4. \( -4r - s = 1 \)

### Step 4: Solve for \( p \) and \( q \)
From the first equation:
\[ q = 5 - 3p \]
Substituting \( q \) into the second equation:
\[ -4p - (5 - 3p) = 2 \]
This simplifies to:
\[ -4p - 5 + 3p = 2 \implies -p - 5 = 2 \implies -p = 7 \implies p = -7 \]
Now substituting \( p \) back to find \( q \):
\[ q = 5 - 3(-7) = 5 + 21 = 26 \]

### Step 5: Solve for \( r \) and \( s \)
From the third equation:
\[ s = -2 - 3r \]
Substituting \( s \) into the fourth equation:
\[ -4r - (-2 - 3r) = 1 \]
This simplifies to:
\[ -4r + 2 + 3r = 1 \implies -r + 2 = 1 \implies -r = -1 \implies r = 1 \]
Now substituting \( r \) back to find \( s \):
\[ s = -2 - 3(1) = -2 - 3 = -5 \]

### Step 6: Form the matrix \( A \)
Now we have:
\[ p = -7, \quad q = 26, \quad r = 1, \quad s = -5 \]
Thus, the matrix \( A \) is:
\[ A = \begin{pmatrix} -7 & 26 \\ 1 & -5 \end{pmatrix} \]

### Final Answer
The matrix \( A \) is:
\[ \begin{pmatrix} -7 & 26 \\ 1 & -5 \end{pmatrix} \]