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Multiplying matrices (pre-defined sizes in answers)

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Zero matrices AND AB = 0 does not imply that either A = 0 or B = 0.

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An $n \\times p$ matrix $A$ can be multiplied by a $p \\times n$ matrix $B$ to form an $n \\times m$ matrix $AB=C$.

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The number of columns of $A$ must match the number of rows of $B$.

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The element in the $i^{th}$ row and $j^{th}$ column of $C$ is obtained by multiplying the $i^{th}$ row of $A$ with the $j^{th}$ column of $B$.

", "advice": "

We are asked to carry out matix multiplications that have some (perhaps) surprising answers.

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The matrix $\\var{Z2}$ is called the zerot matrix of order $2$, and is usually denoted by the symbol $\\underline{0}$.

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And:

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$ \\Large A \\times \\underline{0}=\\underline{0} \\times A=\\underline{0}$    for any matrix $A$

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This explains why, for the first four multiplications, we get fairly predictable results:

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$A B=\\var{A1}\\var{Z2} $

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$AB=\\var{Z2}$ 

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$B A=\\var{Z2}\\var{A1}$

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$B A=\\var{Z2}$ 

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$C D=\\var{B1}\\var{Z3}$

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$C D=\\var{Z3}$ 

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$DC=\\var{Z3}\\var{B1}$

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$DC=\\var{Z3}$ 

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But now we get some more interesting results:

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$EF=\\var{X1}\\var{Y1}$

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$EF=\\var{Z2}$ 

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$FE=\\var{Y1}\\var{X1}$

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$FE=\\var{Z2}$ 

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In the multiplication of \"normal\" numbers

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$\\large ab=0$

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would imply that either $a=0$, or $b=0$ or both are zero. This is not necessarily true for matrices. 

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Carry out the following multiplications:

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$A B=\\var{A1}\\var{Z2} $

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$AB=$ [[0]]

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$B A=\\var{Z2}\\var{A1}$

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$B A=$ [[1]]

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$C D=\\var{B1}\\var{Z3}$

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$C D=$ [[2]]

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$DC=\\var{Z3}\\var{B1}$

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$DC=$ [[3]]

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$EF=\\var{X1}\\var{Y1}$

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$EF=$ [[4]]

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$FE=\\var{Y1}\\var{X1}$

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$FE=$ [[5]]

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