if ` 2A +B=[{:(5,-1),(3,2):}]and A-2B =[{:(1,-4),(0,5):}]` then find the matrices A and B .

Explanation

To solve for the matrices \( A \) and \( B \) given the equations:

1. \( 2A + B = \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} \) (Equation 1)
2. \( A - 2B = \begin{pmatrix} 1 & -4 \\ 0 & 5 \end{pmatrix} \) (Equation 2)

we will follow these steps:

### Step 1: Multiply Equation 1 by 2
We start by multiplying the entire Equation 1 by 2:

\[ 2(2A + B) = 2 \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} \]

This gives us:

\[ 4A + 2B = \begin{pmatrix} 10 & -2 \\ 6 & 4 \end{pmatrix} \tag{Equation 3} \]

### Step 2: Add Equation 2 to Equation 3
Next, we add Equation 2 to Equation 3:

\[ (4A + 2B) + (A - 2B) = \begin{pmatrix} 10 & -2 \\ 6 & 4 \end{pmatrix} + \begin{pmatrix} 1 & -4 \\ 0 & 5 \end{pmatrix} \]

This simplifies to:

\[ 5A + 0B = \begin{pmatrix} 11 & -6 \\ 6 & 9 \end{pmatrix} \]

### Step 3: Solve for Matrix A
From the equation \( 5A = \begin{pmatrix} 11 & -6 \\ 6 & 9 \end{pmatrix} \), we can solve for \( A \):

\[ A = \frac{1}{5} \begin{pmatrix} 11 & -6 \\ 6 & 9 \end{pmatrix} = \begin{pmatrix} \frac{11}{5} & -\frac{6}{5} \\ \frac{6}{5} & \frac{9}{5} \end{pmatrix} \]

### Step 4: Substitute A back into Equation 1 to find B
Now that we have \( A \), we substitute it back into Equation 1 to find \( B \):

\[ 2A + B = \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} \]

Calculating \( 2A \):

\[ 2A = 2 \begin{pmatrix} \frac{11}{5} & -\frac{6}{5} \\ \frac{6}{5} & \frac{9}{5} \end{pmatrix} = \begin{pmatrix} \frac{22}{5} & -\frac{12}{5} \\ \frac{12}{5} & \frac{18}{5} \end{pmatrix} \]

Now, substituting \( 2A \) into the equation:

\[ \begin{pmatrix} \frac{22}{5} & -\frac{12}{5} \\ \frac{12}{5} & \frac{18}{5} \end{pmatrix} + B = \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} \]

### Step 5: Solve for B
To find \( B \), we rearrange the equation:

\[ B = \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} - \begin{pmatrix} \frac{22}{5} & -\frac{12}{5} \\ \frac{12}{5} & \frac{18}{5} \end{pmatrix} \]

Calculating \( B \):

\[ B = \begin{pmatrix} 5 - \frac{22}{5} & -1 + \frac{12}{5} \\ 3 - \frac{12}{5} & 2 - \frac{18}{5} \end{pmatrix} = \begin{pmatrix} \frac{25}{5} - \frac{22}{5} & -\frac{5}{5} + \frac{12}{5} \\ \frac{15}{5} - \frac{12}{5} & \frac{10}{5} - \frac{18}{5} \end{pmatrix} \]

This simplifies to:

\[ B = \begin{pmatrix} \frac{3}{5} & \frac{7}{5} \\ \frac{3}{5} & -\frac{8}{5} \end{pmatrix} \]

### Final Result
Thus, the matrices \( A \) and \( B \) are:

\[ A = \begin{pmatrix} \frac{11}{5} & -\frac{6}{5} \\ \frac{6}{5} & \frac{9}{5} \end{pmatrix}, \quad B = \begin{pmatrix} \frac{3}{5} & \frac{7}{5} \\ \frac{3}{5} & -\frac{8}{5} \end{pmatrix} \]