Q.If $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$, then $A^2$ is equal to
(A) $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
(B) $\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$
(C) $\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$
(D) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Concept: Idempotent Matrix Property — we compute $A^2$ directly by matrix multiplication.

Step 1: Write $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. Then

$A^2 = A \cdot A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.

Step 2: Multiply:

• First row, first column: $(0)(0) + (1)(1) = 1$
• First row, second column: $(0)(1) + (1)(0) = 0$
• Second row, first column: $(1)(0) + (0)(1) = 0$ …

The problem asks for $A^2$ where $A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$. This matrix has a special property: it swaps the two coordinates when multiplied with a vector. Squaring a swap operation brings everything back to where it started — that’s the intuition behind an idempotent or, more precisely, an involutory matrix.

A matrix $M$ is called involutory if $M^2 = I$, the identity matrix. Here, $A$ is a classic example: it’s the permutation matrix that exchanges the first and second rows (or columns). Applying it twice undoes itself.

Let’s verify by direct multiplication.

1. Set up the multiplication

   $A^2 = A \cdot A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$.

2. Compute the (1,1) entry

   Row 1 of first matrix: $[0, 1]$. Column 1 of second matrix: $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$.

   Dot product: $0 \cdot 0 + 1 \cdot 1 = 1$.

3. Compute the (1,2) entry

   Row 1: $[0, 1]$. Column 2: $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$.

   Dot product: $0 \cdot 1 + 1 \cdot 0 = 0$.

4. Compute the (2,1) entry

   Row 2: $[1, 0]$. Column 1: $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$.

   Dot product: $1 \cdot 0 + 0 \cdot 1 = 0$.

5. Compute the (2,2) entry …

Method: Squaring a matrix with the row-by-column rule

Use this to compute $A^2 = A\cdot A$ for any square matrix — never guess from the entries alone.

Steps

Step 1: Write $A^2$ as the product $A \cdot A$.

$A^2$ means $A$ multiplied by itself, so the ordinary matrix-multiplication rule applies.

Step 2: Compute each entry as a row-times-column dot product.

For a $2\times 2$ matrix, the $(i,j)$ entry of the product is

i.e. row $i$ of the first copy paired against column $j$ of the second copy, multiply corresponding terms and add. …

Common Mistakes

Mistake 1: Squaring each entry separately instead of doing matrix multiplication.

Why it's wrong: entrywise squaring of $\begin{bmatrix}0&1\\1&0\end{bmatrix}$ would return the same matrix, which is a wrong answer. Correct approach: use the row-by-column rule; here $A^2 = \begin{bmatrix}1&0\\0&1\end{bmatrix} = I$, not $A$.

Mistake 2: Confusing "involutory" ($A^2=I$) with "idempotent" ($A^2=A$). …