Problems from Product of matrices

Development:

As the first matrix has $3$ rows and $2$ columns, the second must have $2$ rows and $3$ columns. The result of the product will be a matrix with $3$ rows and $3$ columns.

In the second case, where $M$ is in front and it is the first factor, it must also have $2$ rows and $3$ columns. Nevertheless, the result of the multiplication will be a matrix with $2$ rows and $2$ columns.

Solution:

In both cases $M$ should be a $2\times 3$ matrix.

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Development:

$\left(\begin{array}{cc}2& 6\\ 1& 3\end{array}\right)\cdot \left(\begin{array}{cc}4& 1\\ 0& 5\end{array}\right)=\left(\begin{array}{cc}2\cdot 4+6\cdot 0& 2\cdot 1+6\cdot 5\\ 1\cdot 4+3\cdot 0& 1\cdot 1+3\cdot 5\end{array}\right)=\left(\begin{array}{cc}8& 32\\ 4& 16\end{array}\right)$

Solution:

$\left(\begin{array}{cc}8& 32\\ 4& 16\end{array}\right)$

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Development:

$\left(\begin{array}{ccc}2& 0& 1\end{array}\right)\cdot \left(\begin{array}{cc}-1& 2\\ 3& 1\\ 4& 2\end{array}\right)=\left(\begin{array}{cc}2\cdot (-1)+0\cdot 3+1\cdot 4& 2\cdot 2+0\cdot 1+1\cdot 2\end{array}\right)=\left(\begin{array}{cc}2& 6\end{array}\right)$

Solution:

$\left(\begin{array}{cc}2& 6\end{array}\right)$

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Development:

A simple example would be $$\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)\cdot \left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)$$

$$\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)\cdot \left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)$$

Solution:

Commute, for example, the matrix $\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)$ and $\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)$

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