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Multiplying matrices (pre-defined sizes in answers)
\nZero matrices AND AB = 0 does not imply that either A = 0 or B = 0.
", "licence": "Creative Commons Attribution-NonCommercial 4.0 International"}, "statement": "An $n \\times p$ matrix $A$ can be multiplied by a $p \\times n$ matrix $B$ to form an $n \\times m$ matrix $AB=C$.
\nThe number of columns of $A$ must match the number of rows of $B$.
\n\nThe 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": "\n
The matrix $\\var{Z2}$ is called the zerot matrix of order $2$, and is usually denoted by the symbol $\\underline{0}$.
\nAnd:
\n$ \\Large A \\times \\underline{0}=\\underline{0} \\times A=\\underline{0}$ for any matrix $A$
\n\n\nThis explains why, for the first four multiplications, we get fairly predictable results:
\n$A B=\\var{A1}\\var{Z2} $
\n$AB=\\var{Z2}$
\n\n$B A=\\var{Z2}\\var{A1}$
\n$B A=\\var{Z2}$
\n\n
$C D=\\var{B1}\\var{Z3}$
\n$C D=\\var{Z3}$
\n\n
$DC=\\var{Z3}\\var{B1}$
\n$DC=\\var{Z3}$
\n\n
But now we get some more interesting results:
\n$EF=\\var{X1}\\var{Y1}$
\n$EF=\\var{Z2}$
\n\n
$FE=\\var{Y1}\\var{X1}$
\n$FE=\\var{Z2}$
\n\n
In the multiplication of \"normal\" numbers
\n$\\large ab=0$
\nwould imply that either $a=0$, or $b=0$ or both are zero. This is not necessarily true for matrices.
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$A B=\\var{A1}\\var{Z2} $
\n$AB=$ [[0]]
\n\n$B A=\\var{Z2}\\var{A1}$
\n$B A=$ [[1]]
\n\n
$C D=\\var{B1}\\var{Z3}$
\n$C D=$ [[2]]
\n\n
$DC=\\var{Z3}\\var{B1}$
\n$DC=$ [[3]]
\n\n
$EF=\\var{X1}\\var{Y1}$
\n$EF=$ [[4]]
\n\n
$FE=\\var{Y1}\\var{X1}$
\n$FE=$ [[5]]
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