If `A+B=[(1,0),(1,1)]`and `A-2B=[(-1,1),(0,-1)]` then A is equal to

Options

A. `[(1,1),(2,1)]`
B. `[(2//3,1//3),(1//3,2//3)]`
C. `[(1//3,1//3),(2//3,1//3)]`
D. None of these

Answer

To find the matrix \( A \) given the equations \( A + B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \) and \( A - 2B = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \), we can follow these steps:

### Step 1: Set up the equations
We have two equations:
1. \( A + B = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \) (Equation 1)
2. \( A - 2B = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \) (Equation 2)

### Step 2: Multiply Equation 1 by 2
To eliminate \( B \), we can manipulate the equations. First, we multiply Equation 1 by 2:
\[ 2(A + B) = 2 \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 2 & 2 \end{pmatrix} \]
This gives us:
\[ 2A + 2B = \begin{pmatrix} 2 & 0 \\ 2 & 2 \end{pmatrix} \quad \text{(Equation 3)} \]

### Step 3: Add Equation 2 to Equation 3
Now we add Equation 2 to Equation 3:
\[ (2A + 2B) + (A - 2B) = \begin{pmatrix} 2 & 0 \\ 2 & 2 \end{pmatrix} + \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \]
This simplifies to:
\[ 2A + 2B + A - 2B = 3A = \begin{pmatrix} 2 - 1 & 0 + 1 \\ 2 + 0 & 2 - 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \]

### Step 4: Solve for \( A \)
Now, we can isolate \( A \):
\[ 3A = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \]
To find \( A \), we divide both sides by 3:
\[ A = \frac{1}{3} \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} \end{pmatrix} \]

### Final Answer
Thus, the matrix \( A \) is:
\[ A = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} \end{pmatrix} \]