If A and B are two matrices of order 2 × 2 such that A is a symmetric matrix and B is a skew-symmetric matrix, then:

A symmetric matrix satisfies $A = A^T$ (the matrix equals its transpose).

For example, if $A = \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}$, then $A^T = \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix} = A$

Elements are mirrored across the main diagonal.

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A skew-symmetric matrix satisfies $B = -B^T$ (the matrix equals negative of its transpose).

For example, if $B = \begin{bmatrix} 0 & 5 \\ -5 & 0 \end{bmatrix}$, then $B^T = \begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix} = -B$

Diagonal elements are always 0, and elements are negative mirror images.

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To check if $A^2$ is symmetric, we need to verify $(A^2)^T = A^2$.

When taking the transpose of a product: $(AB)^T = B^T \cdot A^T$ (the order reverses).

$(A^2)^T$

$= (A \cdot A)^T$

$= A^T \cdot A^T$

$= A \cdot A$

$= A^2$

Therefore, $A^2$ is symmetric.

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To check if $B^2$ is skew-symmetric, we need to verify $(B^2)^T = -B^2$.

$(B^2)^T$

$= (B \cdot B)^T$

$= B^T \cdot B^T$

$= (-B) \cdot (-B)$

$= B^2$

This means $(B^2)^T = B^2$, which makes $B^2$ symmetric, not skew-symmetric.

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For $A + B =$ Identity matrix, this would only be true for specific matrices, not all symmetric and skew-symmetric matrices.

For $A - B =$ Null matrix, this would mean $A = B$, which is not true in general.

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$A^2$ is also a symmetric matrix because when a symmetric matrix is multiplied by itself, the result remains symmetric due to the transpose properties.

Symmetric $×$ Symmetric $=$ Symmetric

Skew-Symmetric $×$ Skew-Symmetric $=$ Symmetric