To find $A^2$ for the given matrix $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$, we need to multiply $A$ by itself:

$A^2 = A \cdot A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$

Calculating each entry:

• Top left: $1 \cdot 1 + 2 \cdot 3 = 1 + 6 = 7$
• Top right: $1 \cdot 2 + 2 \cdot 4 = 2 + 8 = 10$
• Bottom left: $3 \cdot 1 + 4 \cdot 3 = 3 + 12 = 15$
• Bottom right: $3 \cdot 2 + 4 \cdot 4 = 6 + 16 = 22$

Thus,

$A^2 = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix}$

Would you like further details on matrix multiplication or any additional questions?

Here are some related questions:

1. What is the trace of matrix $A^2$?
2. How do you find the determinant of $A^2$?
3. What are the eigenvalues of the matrix $A^2$?
4. How does the rank of $A$ compare to the rank of $A^2$?
5. Can we find $A^3$ by multiplying $A^2$ with $A$?

Tip: Remember that matrix multiplication is not commutative, so $AB \neq BA$ in general.