If `A=[(1,-1),(2,-1)],B = [(x,1),(4,-1)] and A^2+B^2=(A+B)^2` find the value of x.
State, whether `A^2 + B^2 and (A + B)^2` are always equal or not.

Answer

To solve the problem, we need to find the value of \( x \) such that the equation \( A^2 + B^2 = (A + B)^2 \) holds true, where:

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

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we perform matrix multiplication of \( A \) with itself:

\[ A^2 = A \cdot A = \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} \]

Calculating each element:

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

Thus,

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

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

Next, we calculate \( B^2 \):

\[ B^2 = B \cdot B = \begin{pmatrix} x & 1 \\ 4 & -1 \end{pmatrix} \cdot \begin{pmatrix} x & 1 \\ 4 & -1 \end{pmatrix} \]

Calculating each element:

- First row, first column: \( x \cdot x + 1 \cdot 4 = x^2 + 4 \)
- First row, second column: \( x \cdot 1 + 1 \cdot (-1) = x - 1 \)
- Second row, first column: \( 4 \cdot x + (-1) \cdot 4 = 4x - 4 \)
- Second row, second column: \( 4 \cdot 1 + (-1) \cdot (-1) = 4 + 1 = 5 \)

Thus,

\[ B^2 = \begin{pmatrix} x^2 + 4 & x - 1 \\ 4x - 4 & 5 \end{pmatrix} \]

### Step 3: Calculate \( A + B \)

Now we calculate \( A + B \):

\[ A + B = \begin{pmatrix} 1 & -1 \\ 2 & -1 \end{pmatrix} + \begin{pmatrix} x & 1 \\ 4 & -1 \end{pmatrix} = \begin{pmatrix} 1 + x & -1 + 1 \\ 2 + 4 & -1 - 1 \end{pmatrix} = \begin{pmatrix} x + 1 & 0 \\ 6 & -2 \end{pmatrix} \]

### Step 4: Calculate \( (A + B)^2 \)

Next, we calculate \( (A + B)^2 \):

\[ (A + B)^2 = (A + B) \cdot (A + B) = \begin{pmatrix} x + 1 & 0 \\ 6 & -2 \end{pmatrix} \cdot \begin{pmatrix} x + 1 & 0 \\ 6 & -2 \end{pmatrix} \]

Calculating each element:

- First row, first column: \( (x + 1)(x + 1) + 0 \cdot 6 = (x + 1)^2 \)
- First row, second column: \( (x + 1) \cdot 0 + 0 \cdot (-2) = 0 \)
- Second row, first column: \( 6(x + 1) + (-2) \cdot 6 = 6x + 6 - 12 = 6x - 6 \)
- Second row, second column: \( 6 \cdot 0 + (-2)(-2) = 4 \)

Thus,

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

### Step 5: Set up the equation \( A^2 + B^2 = (A + B)^2 \)

Now we set \( A^2 + B^2 \) equal to \( (A + B)^2 \):

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

Calculating \( A^2 + B^2 \):

\[ A^2 + B^2 = \begin{pmatrix} -1 + (x^2 + 4) & 0 + (x - 1) \\ 0 + (4x - 4) & -1 + 5 \end{pmatrix} = \begin{pmatrix} x^2 + 3 & x - 1 \\ 4x - 4 & 4 \end{pmatrix} \]

### Step 6: Equate the matrices

Now we equate the corresponding elements:

1. From the first row, first column:
\[ x^2 + 3 = (x + 1)^2 \]
Expanding the right side:
\[ x^2 + 3 = x^2 + 2x + 1 \implies 3 = 2x + 1 \implies 2 = 2x \implies x = 1 \]

2. From the first row, second column:
\[ x - 1 = 0 \implies x = 1 \]

3. From the second row, first column:
\[ 4x - 4 = 6x - 6 \]
Rearranging gives:
\[ 2 = 2x \implies x = 1 \]

4. From the second row, second column:
\[ 4 = 4 \quad \text{(always true)} \]

### Conclusion

The value of \( x \) is \( \boxed{1} \).