Find the matrix `x` for which `[(3,2),(7,5)]x[(-1,1),(-2,1)]=[(2,-1),(0,4)]`

Answer

To solve the problem of finding the matrix \( X \) such that

\[ \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} X \begin{pmatrix} -1 & 1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ 0 & 4 \end{pmatrix}, \]

we can follow these steps:

### Step 1: Define the matrices
Let:
- \( A = \begin{pmatrix} 3 & 2 \\ 7 & 5 \end{pmatrix} \)
- \( B = \begin{pmatrix} -1 & 1 \\ -2 & 1 \end{pmatrix} \)
- \( P = \begin{pmatrix} 2 & -1 \\ 0 & 4 \end{pmatrix} \)

### Step 2: Rearranging the equation
We need to isolate \( X \). We can rearrange the equation as follows:
\[ AXB = P \implies X = A^{-1}PB^{-1} \]

### Step 3: Find the inverse of matrix \( A \)
To find \( A^{-1} \), we first compute the determinant of \( A \):
\[ \text{det}(A) = (3)(5) - (2)(7) = 15 - 14 = 1 \]
Since the determinant is 1, we can find the inverse using the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \]

### Step 4: Find the inverse of matrix \( B \)
Now, we compute the determinant of \( B \):
\[ \text{det}(B) = (-1)(1) - (1)(-2) = -1 + 2 = 1 \]
Thus, the inverse of \( B \) is:
\[ B^{-1} = \frac{1}{\text{det}(B)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & -1 \end{pmatrix} \]

### Step 5: Compute \( X \)
Now we can compute \( X \):
\[ X = A^{-1}PB^{-1} \]
First, calculate \( PB^{-1} \):
\[ PB^{-1} = \begin{pmatrix} 2 & -1 \\ 0 & 4 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 2 \cdot 1 + (-1) \cdot 2 & 2 \cdot 1 + (-1) \cdot (-1) \\ 0 \cdot 1 + 4 \cdot 2 & 0 \cdot 1 + 4 \cdot (-1) \end{pmatrix} = \begin{pmatrix} 0 & 3 \\ 8 & -4 \end{pmatrix} \]

Next, compute \( A^{-1} \) times \( PB^{-1} \):
\[ X = \begin{pmatrix} 5 & -2 \\ -7 & 3 \end{pmatrix} \begin{pmatrix} 0 & 3 \\ 8 & -4 \end{pmatrix} = \begin{pmatrix} 5 \cdot 0 + (-2) \cdot 8 & 5 \cdot 3 + (-2) \cdot (-4) \\ -7 \cdot 0 + 3 \cdot 8 & -7 \cdot 3 + 3 \cdot (-4) \end{pmatrix} = \begin{pmatrix} -16 & 27 \\ 24 & -21 \end{pmatrix} \]

### Final Result
Thus, the matrix \( X \) is:
\[ X = \begin{pmatrix} -16 & 27 \\ 24 & -21 \end{pmatrix} \]