Q.If $A = \begin{bmatrix}0 & -x\\ x & 0\end{bmatrix}$, $B = \begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$ and $x^2 = -1$, then show that $(A + B)^2 = A^2 + B^2$.

Concept: Matrix Multiplication Compatibility — both $A$ and $B$ are $2 \times 2$, so addition and multiplication are defined.

Step 1: Compute $A^2$ and $B^2$ separately.

$A^2 = \begin{bmatrix}0 & -x\\ x & 0\end{bmatrix} \begin{bmatrix}0 & -x\\ x & 0\end{bmatrix} = \begin{bmatrix}-x^2 & 0\\ 0 & -x^2\end{bmatrix}$.

Since $x^2 = -1$, we get $-x^2 = 1$, so $A^2 = \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} = I$.

$B^2 = \begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix} \begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix} = \begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} = I$.

Step 2: Compute $A+B = \begin{bmatrix}0 & -x+1\\ x+1 & 0\end{bmatrix}$. …

We need to show that $(A+B)^2 = A^2 + B^2$ for the given matrices, where $x^2 = -1$. The natural instinct is to expand $(A+B)^2 = A^2 + AB + BA + B^2$. For this to equal $A^2 + B^2$, we require $AB + BA = 0$, i.e., $AB = -BA$. So the problem reduces to checking whether $A$ and $B$ anti-commute.

Let’s verify this step by step.

1. Write down the matrices clearly.

   $A = \begin{bmatrix}0 & -x \\ x & 0\end{bmatrix}$, $B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$, and we are given $x^2 = -1$. Note that $x$ is not a real number — it behaves like the imaginary unit $i$, but we treat it algebraically.

2. Compute $AB$.

   Multiply $A$ and $B$:

3. Compute $BA$. Multiply in the reverse order:

4. Observe the anti-commutation.

   From steps 2 and 3, $AB = \begin{bmatrix} -x & 0 \\ 0 & x \end{bmatrix}$ and $BA = \begin{bmatrix} x & 0 \\ 0 & -x \end{bmatrix}$. Clearly $AB = -BA$, so $AB + BA = 0$.

   Tip

   This anti-commutation property is the entire reason the cross terms vanish. In general, $(A+B)^2 = A^2 + B^2$ if and only if $AB = -BA$. Always check this first.

5. Now compute $A^2$ and $B^2$ individually.

   First, $A^2$:

Since $x^2 = -1$, we have $-x^2 = -(-1) = 1$, so …

Method: Checking $(A+B)^2=A^2+B^2$ — when do the cross terms vanish?

For matrices, $(A+B)^2$ equals $A^2+B^2$ only when the cross terms cancel. The general method is to expand, isolate $AB+BA$, and use the problem's data (here a scalar condition like $x^2=-1$) to show that sum is the zero matrix.

Steps

Step 1: Expand keeping order.

So $(A+B)^2=A^2+B^2$ holds precisely when $AB+BA=O$ (the matrices anti-commute).

Step 2: Compute the individual squares using the given constraint.

Evaluate $A^2$ and $B^2$ by multiplication, then substitute the given scalar relation (e.g. $x^2=-1$) to simplify entries. …

Common Mistakes

Mistake 1: Assuming $(A+B)^2=A^2+B^2$ automatically, as if $2AB$ never appears.

Why it's wrong: the true expansion is $A^2+AB+BA+B^2$; the identity only holds because here $AB+BA=O$. Correct approach: expand first and justify why the cross terms disappear.

Mistake 2: Forgetting to substitute $x^2=-1$.

Why it's wrong: $A^2=\begin{bmatrix}-x^2 & 0\\0 & -x^2\end{bmatrix}$ simplifies to $I$ only after using $-x^2=1$; leaving it as $-x^2$ gives an unfinished, wrong-looking answer. Correct approach: apply the given scalar relation to every affected entry. …