Linear combinations, transposes and symmetric matrices

The matrices $A$ and $B$ both have two rows and three columns, so the linear combination is defined and is again a $2\times 3$ matrix. Multiplication by a scalar multiplies every entry, therefore one first computes $3A$ and $-2B$. The sum is then performed entry by entry, adding the entries in the same position; for example the entry in position $(2,3)$ is $3\cdot (-2)+(-2)\cdot 5=-16$.

The matrix $A$ is $2\times 3$, so $A^T$ is $3\times 2$, like $B$; therefore the equation makes sense and the unknown $X$ is a $3\times 2$ matrix. By the properties of the sum and of multiplication by a scalar the equation is solved like a first-degree numerical equation: one adds $A^T$ to both sides and multiplies by $\frac{1}{2}$. In the transpose the first row $(1,0,-2)$ of $A$ becomes the first column of $A^T$ and the second row $(3,1,4)$ becomes the second column. The result is checked by substitution: $2X-A^T$ has first row $(4-1,6-3)=(3,3)$, which is the first row of $3B$, and likewise for the other rows.

By the properties of the transpose the matrix $A+A^T$ is symmetric, since $(A+A^T{)}^T=A^T+A$, and the matrix $A-A^T$ is skew-symmetric, since $(A-A^T{)}^T=A^T-A=-(A-A^T)$; multiplying by $\frac{1}{2}$ these properties remain true. Moreover $S+K=\frac{1}{2}A+\frac{1}{2}{A}^T+\frac{1}{2}A-\frac{1}{2}{A}^T=A$, therefore $S$ and $K$ are the required matrices. In the computation the transpose is obtained by writing the rows of $A$ as columns. The result is checked in two ways: $S$ has equal entries in mirror positions with respect to the diagonal and $K$ has zero diagonal and opposite entries in mirror positions; moreover the sum $S+K$ gives back $A$, for example $s_{23}+k_{23}=1+5=6=a_{23}$.