Hi(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Question regarding matrix multiplication

**Physics Forums | Science Articles, Homework Help, Discussion**