The matrix A satisfying the equation `[(1,3),(0,1)]A=[(1,1),(0,-1)]` is

Options

A. `[(1,4),(-1,0)]`
B. `[(1,-4),(1,0)]`
C. `[(1,4),(0,-1)]`
D. None

Explanation

To find the matrix \( A \) that satisfies the equation

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

we will denote the matrix \( A \) as

\[ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}. \]

### Step 1: Set up the equation

We start by multiplying the two matrices on the left side:

\[ \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & -1 \end{pmatrix}. \]

### Step 2: Perform the matrix multiplication

Now, we perform the multiplication:

\[ \begin{pmatrix} 1 \cdot a + 3 \cdot c & 1 \cdot b + 3 \cdot d \\ 0 \cdot a + 1 \cdot c & 0 \cdot b + 1 \cdot d \end{pmatrix} = \begin{pmatrix} a + 3c & b + 3d \\ c & d \end{pmatrix}. \]

### Step 3: Set up the system of equations

Now we equate the resulting matrix to the right-hand side:

1. \( a + 3c = 1 \) (from the first row, first column)
2. \( b + 3d = 1 \) (from the first row, second column)
3. \( c = 0 \) (from the second row, first column)
4. \( d = -1 \) (from the second row, second column)

### Step 4: Solve the equations

From equation (3), we have:

\[ c = 0. \]

Substituting \( c = 0 \) into equation (1):

\[ a + 3(0) = 1 \implies a = 1. \]

From equation (4), we have:

\[ d = -1. \]

Substituting \( d = -1 \) into equation (2):

\[ b + 3(-1) = 1 \implies b - 3 = 1 \implies b = 4. \]

### Step 5: Write the final matrix

Now we have all the values:

\[ a = 1, \quad b = 4, \quad c = 0, \quad d = -1. \]

Thus, the matrix \( A \) is:

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

### Final Answer

The matrix \( A \) satisfying the equation is

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

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