Let $X, Y, Z$ be $2 \times 2$ matrices with real entries. Define $\odot$ as follows $X \odot Y = \frac{1}{2}(XY + YX)$ then:
Step-by-Step Solution
Key Concept: The operation $\odot$ is the symmetric part of matrix multiplication, which is automatically commutative and satisfies properties like $X \odot X = X^2$ due to the averaging of $XY$ and $YX$.
We verify each option systematically. For option 1: $X \odot Y = \frac{1}{2}(XY + YX) = \frac{1}{2}(YX + XY) = Y \odot X$ by commutativity of addition, so this is TRUE. For option 2: $X \odot I = \frac{1}{2}(XI + IX) = \frac{1}{2}(X + X) = X$ since $XI = IX = X$, so this is TRUE. For option 3: $X \odot X = \frac{1}{2}(XX + XX) = \frac{1}{2}(2X^2) = X^2$, so this is TRUE. For option 4: $X \odot (Y \odot Z) = X \odot \frac{1}{2}(YZ + ZY) = \frac{1}{2}(X \cdot \frac{1}{2}(YZ + ZY) + \frac{1}{2}(YZ + ZY) \cdot X)$, while $X \odot Y + X \odot Z$ gives different terms involving $XY$, $YX$, $XZ$, $ZX$. These are not equal in general due to non-commutativity of matrix multiplication, so this is FALSE.
Correct Answer: 1,2,3