If $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, B = \begin{bmatrix} 0 & 0 \\ 3 & 0 \end{bmatrix}$ then

Given:

$A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

$B = \begin{bmatrix} 0 & 0 \\ 3 & 0 \end{bmatrix}$

Matrix multiplication is not commutative, meaning $AB \neq BA$ in general.

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$AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 3 & 0 \end{bmatrix}$

Position (1,1): $(1)(0) + (0)(3) = 0$

Position (1,2): $(1)(0) + (0)(0) = 0$

Position (2,1): $(0)(0) + (0)(3) = 0$

Position (2,2): $(0)(0) + (0)(0) = 0$

$AB = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$

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$BA = \begin{bmatrix} 0 & 0 \\ 3 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

Position (1,1): $(0)(1) + (0)(0) = 0$

Position (1,2): $(0)(0) + (0)(0) = 0$

Position (2,1): $(3)(1) + (0)(0) = 3$

Position (2,2): $(3)(0) + (0)(0) = 0$

$BA = \begin{bmatrix} 0 & 0 \\ 3 & 0 \end{bmatrix} \neq 0$

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For Option 3:

$A^2 = A \times A$

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

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

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For Option 4:

$B^2 = \begin{bmatrix} 0 & 0 \\ 3 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 3 & 0 \end{bmatrix}$

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

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Therefore, $AB = 0$ and $BA \neq 0$.