If `[(2,1),(7,4)]A[(-3,2),(5,-3)]=[(1,0),(0,1)]` then matrix A equals

Options

A. `[(7,5),(-11,-8)]`
B. `[(2,1),(5,3)]`
C. `[(7,1),(34,5)]`
D. `[(5,3),(13,8)]`

Answer

To solve the equation \([(2,1),(7,4)]A[(-3,2),(5,-3)]=[(1,0),(0,1)]\), we need to find the matrix \(A\).

### Step 1: Identify the matrices
Let:
- \(B = \begin{pmatrix} 2 & 1 \\ 7 & 4 \end{pmatrix}\)
- \(C = \begin{pmatrix} -3 & 2 \\ 5 & -3 \end{pmatrix}\)
- \(I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) (the identity matrix)

### Step 2: Rewrite the equation
The equation can be rewritten as:
\[ B A C = I \]

### Step 3: Multiply both sides by \(C^{-1}\)
To isolate \(A\), we multiply both sides by \(C^{-1}\) from the right:
\[ B A C C^{-1} = I C^{-1} \]
Since \(C C^{-1} = I\), this simplifies to:
\[ B A = C^{-1} \]

### Step 4: Multiply both sides by \(B^{-1}\)
Now, we multiply both sides by \(B^{-1}\) from the left:
\[ B^{-1} B A = B^{-1} C^{-1} \]
This simplifies to:
\[ A = B^{-1} C^{-1} \]

### Step 5: Calculate \(B^{-1}\)
To find \(B^{-1}\), we first calculate the determinant of \(B\):
\[ \text{det}(B) = (2)(4) - (1)(7) = 8 - 7 = 1 \]
Now, the inverse of \(B\) is given by:
\[ B^{-1} = \frac{1}{\text{det}(B)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = \begin{pmatrix} 4 & -1 \\ -7 & 2 \end{pmatrix} \]

### Step 6: Calculate \(C^{-1}\)
Next, we calculate the determinant of \(C\):
\[ \text{det}(C) = (-3)(-3) - (2)(5) = 9 - 10 = -1 \]
Now, the inverse of \(C\) is given by:
\[ C^{-1} = \frac{1}{\text{det}(C)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} = -1 \begin{pmatrix} -3 & -2 \\ -5 & -3 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 5 & 3 \end{pmatrix} \]

### Step 7: Calculate \(A\)
Now we can calculate \(A\):
\[ A = B^{-1} C^{-1} = \begin{pmatrix} 4 & -1 \\ -7 & 2 \end{pmatrix} \begin{pmatrix} 3 & 2 \\ 5 & 3 \end{pmatrix} \]
Calculating the product:
- First row, first column: \(4 \cdot 3 + (-1) \cdot 5 = 12 - 5 = 7\)
- First row, second column: \(4 \cdot 2 + (-1) \cdot 3 = 8 - 3 = 5\)
- Second row, first column: \(-7 \cdot 3 + 2 \cdot 5 = -21 + 10 = -11\)
- Second row, second column: \(-7 \cdot 2 + 2 \cdot 3 = -14 + 6 = -8\)

Thus, we have:
\[ A = \begin{pmatrix} 7 & 5 \\ -11 & -8 \end{pmatrix} \]

### Final Answer
The matrix \(A\) is:
\[ A = \begin{pmatrix} 7 & 5 \\ -11 & -8 \end{pmatrix} \]