- #1

Mathman23

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Hi

When one is trying to multiply two matrices of different sizes e.g. a 2x3 and a 3x3. I know that one has to use the column-row-rule which states:

[tex]AB_{ij} = a_{i1} b_{1j} + a_{i2} b_{2j} + \cdots + a_{i n} b_{m j} [/tex]

Looking at the following example:

[tex] A= \left[ \begin{array}{ccc} 7 & 2 & 0 \\ -4 & 6 & 2 \\ -1 & \frac{2}{5} & \frac{7}{5} \end{array} \right] \ \ \ B = \left[ \begin{array}{ccc} 7 & 2 & 0 \\ -4 & 6 & 2 \\ \end{array} \right][/tex]

Using the column-row-rule I calculate the matrix-product AB:

[tex] AB= \left[ \begin{array}{ccc} 7 & 2 & 0 \\ \\ \\ \end{array} \right] \cdot \left[ \begin{array}{cc} -4 &7 \\ 6 &2 \\ 2 & 0 \end{array} \right] = \left[ \begin{array}{cc} -16 & 53 \\ \\ \\ \end{array} \right] [/tex]

But if I then write the B-matrix upside-down I get:

[tex] AB= \left[ \begin{array}{ccc} 7 & 2 & 0 \\ \\ \\ \end{array} \right] \cdot \left[ \begin{array}{cc} 0 &2 \\ 2 & 6 \\ 7 & -4 \end{array} \right] = \left[ \begin{array}{cc} 4 & 30 \\ \\ \\ \end{array} \right] [/tex]

Which of the two results is the correct approach to compute the matrix-product AB ?

Does there exist a rule in linear algebra which allows me to predetermain if the product of two matrices A and B both not of the same size ( A is n x n and B is m x n ) gives the resulting matrix C which has a different size than A and B ?

Many thanks in advance

Sincerley

Fred

When one is trying to multiply two matrices of different sizes e.g. a 2x3 and a 3x3. I know that one has to use the column-row-rule which states:

[tex]AB_{ij} = a_{i1} b_{1j} + a_{i2} b_{2j} + \cdots + a_{i n} b_{m j} [/tex]

Looking at the following example:

[tex] A= \left[ \begin{array}{ccc} 7 & 2 & 0 \\ -4 & 6 & 2 \\ -1 & \frac{2}{5} & \frac{7}{5} \end{array} \right] \ \ \ B = \left[ \begin{array}{ccc} 7 & 2 & 0 \\ -4 & 6 & 2 \\ \end{array} \right][/tex]

Using the column-row-rule I calculate the matrix-product AB:

[tex] AB= \left[ \begin{array}{ccc} 7 & 2 & 0 \\ \\ \\ \end{array} \right] \cdot \left[ \begin{array}{cc} -4 &7 \\ 6 &2 \\ 2 & 0 \end{array} \right] = \left[ \begin{array}{cc} -16 & 53 \\ \\ \\ \end{array} \right] [/tex]

But if I then write the B-matrix upside-down I get:

[tex] AB= \left[ \begin{array}{ccc} 7 & 2 & 0 \\ \\ \\ \end{array} \right] \cdot \left[ \begin{array}{cc} 0 &2 \\ 2 & 6 \\ 7 & -4 \end{array} \right] = \left[ \begin{array}{cc} 4 & 30 \\ \\ \\ \end{array} \right] [/tex]

Which of the two results is the correct approach to compute the matrix-product AB ?

Does there exist a rule in linear algebra which allows me to predetermain if the product of two matrices A and B both not of the same size ( A is n x n and B is m x n ) gives the resulting matrix C which has a different size than A and B ?

Many thanks in advance

Sincerley

Fred

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