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{Comments Off on Game of Thrones Season 6 Full Episode 7 Download}Q: Matrix multiplication of non-square matrices (that are square up-to the $n$th power) I have two matrices $A$ and $B$, that are $n$x$n$-square, i.e. $A, B \in \mathbb{R}^{n \times n}$. How can I multiply them to get a matrix $C$ such that $C \in \mathbb{R}^{n \times n}$? I guess it would be somehow a simple $C=A \cdot B$, but if so, why does it not work in general? If it would work for $A, B \in \mathbb{R}^{n \times n}$, then it would be the same as multiplying $A$ and $B$ columnwise, the same for both the $A$ and $B$ as $A, B \in \mathbb{R}^{n \times n}$ we would be multiplying the $n$ matrices in each column (including $n$x$n$ matrices, respectively, columnwise and the result would be the $n$x$n$ result. A more mathematical way to formulate this question is as follows: Let $A, B \in \mathbb{R}^{n \times n}$. How do you «formally» multiply them to get $C \in \mathbb{R}^{n \times n}$, i.e. $C = AB$? A: The approach below works for any choice of $A$, $B$ and $C$, as long as $C$ is a square matrix. The result is called Kronecker product and has the form $$C=\left(A\otimes B\right)$$ where $\otimes$ means the Kronecker product. Of course, you can recover $A$ and $B$ from $C$ by applying the appropriate matrices on the appropriate rows or columns. A: You need to write $C$ as a matrix with two different dimensions, one for $A$ and one for $B$. $C$ is a $n*n$ matrix, so you need a \$(n*n)*(