Question #102857

Let A be an m x r matrix, B be an m x s matrix, C be an r x n matrix, and D be an s x n matrix. Prove that

(A B) (C

D) -> C on top of D

= AC + BD

Expert's answer

I think there is a mistake in the problem, for example, multiplying the m x r matrix AA by the m x s matrix BB.

Let AA be 2x3 matrix and BB be 2x1 matrix, then the product of AA by BB won't be defined:


(a11a12a13a21a22a23)\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{pmatrix} *(b11b21)\begin{pmatrix} b_{11} \\ b_{21} \end{pmatrix} =?=?


To multiply matrices, the number of columns of one of them must be equal to the number of rows of another one!

For example,


(a11a12a21a22a31a32)\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{pmatrix} * (b11b12b21b22)\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} =(a11b11+a12b21a11b12+a12b22a21b11+a22b21a21b12+a22b22a31b11+a32b21a31b12+a32b22)\begin{pmatrix} a_{11}*b_{11}+ a_{12}*b_{21}& a_{11}*b_{12}+a_{12}*b_{22} \\ a_{21}*b_{11}+ a_{22}*b_{21}& a_{21}*b_{12}+a_{22}*b_{22} \\ a_{31}*b_{11}+ a_{32}*b_{21}& a_{31}*b_{12}+a_{32}*b_{22} \end{pmatrix}


In matrices AA and BB we can't do like this!


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