If the matrix A is such that `({:(1,3),(0,1):})A=({:(1,1),(0,-1):})`, then what is A equal to ?

Options

A. `({:(1,4),(0,-1):})`
B. `({:(1,4),(0,1):})`
C. `({:(-1,4),(0,-1):})`
D. `({:(1,-4),(0,-1):})`

Explanation

To find the matrix \( A \) such that

\[ \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} A = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}, \]

we can follow these steps:

### Step 1: Identify the matrices involved
We have the matrix \( B = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \) and the resulting matrix \( C = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} \). We need to find the matrix \( A \).

### Step 2: Multiply both sides by the inverse of matrix \( B \)
To isolate \( A \), we can multiply both sides of the equation by the inverse of \( B \) from the left:

\[ B^{-1} B A = B^{-1} C \]

This simplifies to:

\[ A = B^{-1} C \]

### Step 3: Calculate the inverse of matrix \( B \)
The formula for the inverse of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:

\[ \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]

For our matrix \( B = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \):

- \( a = 1 \)
- \( b = 3 \)
- \( c = 0 \)
- \( d = 1 \)

The determinant \( \text{det}(B) = ad - bc = 1 \cdot 1 - 3 \cdot 0 = 1 \).

Thus, the inverse \( B^{-1} \) is:

\[ B^{-1} = \frac{1}{1} \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix} \]

### Step 4: Multiply \( B^{-1} \) by \( C \)
Now we need to calculate \( A = B^{-1} C \):

\[ A = \begin{pmatrix} 1 & -3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix} \]

### Step 5: Perform the matrix multiplication
To multiply the matrices, we calculate each element:

1. First row, first column: \( 1 \cdot 1 + (-3) \cdot 0 = 1 \)
2. First row, second column: \( 1 \cdot 1 + (-3) \cdot (-1) = 1 + 3 = 4 \)
3. Second row, first column: \( 0 \cdot 1 + 1 \cdot 0 = 0 \)
4. Second row, second column: \( 0 \cdot 1 + 1 \cdot (-1) = 0 - 1 = -1 \)

Thus, we have:

\[ A = \begin{pmatrix} 1 & 4 \\ 0 & -1 \end{pmatrix} \]

### Final Answer
Therefore, the matrix \( A \) is:

\[ A = \begin{pmatrix} 1 & 4 \\ 0 & -1 \end{pmatrix} \]