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 $$2\times3$$ matrix.

Hide solution and development

Development:

$$\left( \begin{array}{cc} 2 & 6 \\ 1 & 3 \end{array} \right) \cdot \left( \begin{array}{cc} 4 & 1 \\ 0 & 5 \end{array} \right)=\left( \begin{array}{cc} 2\cdot4+6\cdot0 & 2\cdot1+6\cdot5 \\ 1\cdot4+3\cdot0 & 1\cdot1+3\cdot5 \end{array} \right) = \left( \begin{array}{cc} 8 & 32 \\ 4 & 16 \end{array} \right)$$

Solution:

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

Hide solution and development

Development:

$$\left( \begin{array}{ccc} 2 & 0 & 1 \end{array} \right) \cdot \left( \begin{array}{cc} -1 & 2 \\ 3 & 1 \\ 4 & 2 \end{array} \right) = \left( \begin{array}{cc} 2\cdot(-1)+0\cdot3+1\cdot4 & 2\cdot2+0\cdot1+1\cdot2 \end{array} \right)= \left( \begin{array}{cc} 2 & 6 \end{array} \right)$$

Solution:

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

Hide solution and development

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

Hide solution and development