The matrix A satisfying `A[[1, 5], [0, 1]]=[[3, -1], [6, 0]]` is

Options

A. `[[6, -6], [12, 4]]`
B. `[[3, -16], [6, -30]]`
C. `[[9, -16], [36, -45]]`
D. `[[9, 4], [36, 4]]`

Answer

To find the matrix \( A \) that satisfies the equation \( A \begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -1 \\ 6 & 0 \end{bmatrix} \), we can follow these steps:

### Step 1: Define the matrices
Let:
- \( B = \begin{bmatrix} 1 & 5 \\ 0 & 1 \end{bmatrix} \)
- \( C = \begin{bmatrix} 3 & -1 \\ 6 & 0 \end{bmatrix} \)

The equation can be rewritten as:
\[ A B = C \]

### Step 2: Find the inverse of matrix \( B \)
To isolate \( A \), we need to multiply both sides of the equation by \( B^{-1} \) (the inverse of \( B \)). Thus, we have:
\[ A = C B^{-1} \]

First, we need to calculate \( B^{-1} \). The formula for the inverse of a \( 2 \times 2 \) matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:
\[ B^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \]

For matrix \( B \):
- \( a = 1, b = 5, c = 0, d = 1 \)

Calculate the determinant:
\[ \text{det}(B) = ad - bc = (1)(1) - (5)(0) = 1 \]

Since the determinant is not zero, \( B^{-1} \) exists. Now, we can find \( B^{-1} \):
\[ B^{-1} = \frac{1}{1} \begin{bmatrix} 1 & -5 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -5 \\ 0 & 1 \end{bmatrix} \]

### Step 3: Compute \( A \)
Now substitute \( C \) and \( B^{-1} \) back into the equation:
\[ A = C B^{-1} = \begin{bmatrix} 3 & -1 \\ 6 & 0 \end{bmatrix} \begin{bmatrix} 1 & -5 \\ 0 & 1 \end{bmatrix} \]

### Step 4: Perform the matrix multiplication
Now, we perform the multiplication:
1. First row, first column: \( 3 \cdot 1 + (-1) \cdot 0 = 3 \)
2. First row, second column: \( 3 \cdot (-5) + (-1) \cdot 1 = -15 - 1 = -16 \)
3. Second row, first column: \( 6 \cdot 1 + 0 \cdot 0 = 6 \)
4. Second row, second column: \( 6 \cdot (-5) + 0 \cdot 1 = -30 + 0 = -30 \)

Thus, we find:
\[ A = \begin{bmatrix} 3 & -16 \\ 6 & -30 \end{bmatrix} \]

### Conclusion
The matrix \( A \) that satisfies the equation is:
\[ A = \begin{bmatrix} 3 & -16 \\ 6 & -30 \end{bmatrix} \]