If `{:A+B=[(1,0),(1,1)]andA-2B=[(-1,1),(0,-1)]:}," then "A=`

Options

A. `(1)/(3){:[(1,1),(2,1)]:}`
B. `(1)/(3){:[(2,1),(1,2)]:}`
C. `{:[(1,1),(2,1)]:}`
D. `{:[(1,1),(1,1)]:}`

Answer

To solve the problem, we need to find the matrix \( A \) given the 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 1: Multiply Equation 1 by 2

To eliminate \( B \) when we add the two equations, we can multiply the first equation by 2:

\[ 2(A + B) = 2 \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \]

This gives us:

\[ 2A + 2B = \begin{pmatrix} 2 & 0 \\ 2 & 2 \end{pmatrix} \quad (Equation 3) \]

### Step 2: Write down Equation 2

We can keep Equation 2 as it is:

\[ A - 2B = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \quad (Equation 2) \]

### Step 3: Add Equation 3 and Equation 2

Now we will add Equation 3 and Equation 2 together:

\[ (2A + 2B) + (A - 2B) = \begin{pmatrix} 2 & 0 \\ 2 & 2 \end{pmatrix} + \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \]

This simplifies to:

\[ 2A + 2B + A - 2B = 3A \]

And on the right side, we calculate:

\[ \begin{pmatrix} 2 - 1 & 0 + 1 \\ 2 + 0 & 2 - 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \]

### Step 4: Set up the equation for A

Now we have:

\[ 3A = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} \]

### Step 5: Solve for A

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 Result

Thus, the matrix \( A \) is:

\[ A = \begin{pmatrix} \frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{1}{3} \end{pmatrix} \]

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