If `A=[{:(,1,4),(,1,-3):}] and B=[{:(,1,2),(,-1,-1):}]`, find:
`(A+B)^2`

Answer

To solve the problem, we need to find \((A + B)^2\), where:

\[ A = \begin{pmatrix} 1 & 4 \\ 1 & -3 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 2 \\ -1 & -1 \end{pmatrix} \]

### Step 1: Calculate \(A + B\)

To add matrices \(A\) and \(B\), we add their corresponding elements:

\[ A + B = \begin{pmatrix} 1 + 1 & 4 + 2 \\ 1 + (-1) & -3 + (-1) \end{pmatrix} \]

Calculating each element:

- First row, first column: \(1 + 1 = 2\)
- First row, second column: \(4 + 2 = 6\)
- Second row, first column: \(1 - 1 = 0\)
- Second row, second column: \(-3 - 1 = -4\)

Thus, we have:

\[ A + B = \begin{pmatrix} 2 & 6 \\ 0 & -4 \end{pmatrix} \]

### Step 2: Calculate \((A + B)^2\)

Now we need to find \((A + B)^2\), which is equal to \((A + B) \times (A + B)\):

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

To multiply these matrices, we use the matrix multiplication rule:

\[ \text{Result}(i,j) = \sum_{k} \text{Matrix1}(i,k) \times \text{Matrix2}(k,j) \]

Calculating each element of the resulting matrix:

- First row, first column:
\[ 2 \times 2 + 6 \times 0 = 4 + 0 = 4 \]

- First row, second column:
\[ 2 \times 6 + 6 \times (-4) = 12 - 24 = -12 \]

- Second row, first column:
\[ 0 \times 2 + (-4) \times 0 = 0 + 0 = 0 \]

- Second row, second column:
\[ 0 \times 6 + (-4) \times (-4) = 0 + 16 = 16 \]

Thus, we have:

\[ (A + B)^2 = \begin{pmatrix} 4 & -12 \\ 0 & 16 \end{pmatrix} \]

### Final Result

The final result is:

\[ (A + B)^2 = \begin{pmatrix} 4 & -12 \\ 0 & 16 \end{pmatrix} \]

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