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I can remove a row from matrix

$M=\left( \begin{array}{cccc} m_{11} & m_{12} & m_{13} & m_{14} \\ m_{21} & m_{22} & m_{23} & m_{24} \\ m_{31} & m_{32} & m_{33} & m_{34} \\ m_{41} & m_{42} & m_{43} & m_{44} \end{array} \right)$

with factor like

$X=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right)$

by multiplying by it from left. The $X$ removes 3rd row:

$X M = \left( \begin{array}{cccc} m_{11} & m_{12} & m_{13} & m_{14} \\ m_{21} & m_{22} & m_{23} & m_{24} \\ m_{41} & m_{42} & m_{43} & m_{44} \end{array} \right)$

I can also remove column by

$Y=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{array} \right)$

by multiplying from right

$M Y = \left( \begin{array}{ccc} m_{11} & m_{12} & m_{14} \\ m_{21} & m_{22} & m_{24} \\ m_{31} & m_{32} & m_{34} \\ m_{41} & m_{42} & m_{44} \end{array} \right)$

Can I remove column by multiplying from left and remove row by multiplying from right?

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The product of two matrices always has as many rows as the left-hand factor, and as many columns as the right-hand factor. (See my diagram here.) So, the answer is "no". –  Blue Jun 23 '13 at 19:20

1 Answer 1

up vote 2 down vote accepted
+50

For $A \in M_{p\times q}\left(\Bbb K\right)$ and $B \in M_{q\times r}\left(\Bbb K\right)$, $AB \in M_{p\times r}\left(\Bbb K\right)$$.

Your matrix is either $A$ or $B$ and the other one of those is the matrix you use to "remove" a line / column. So either $p$ and $q$ are fixed (meaning you modify your matrix by multiplying it by something on the right) and then you can only change $r$, the number of columns, or it's the other twi that are fixed so you can only change $p$, the number of columns.

So no, you can't.

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