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Show that if A has a row of zeros and B is any matrix for which AB is defined, then AB also has a row of zeros.

Here, I know that we can multiply A with B. I also know A has a row of zeros. In matrix multiplication, we multiply the rows by the columns and that's how we get the resulting matrix. Is that how I can prove it?

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  • $\begingroup$ Is this homework? $\endgroup$ – Aaron Maroja Mar 1 '15 at 17:19
  • $\begingroup$ ya its for my linear algebra class $\endgroup$ – user220177 Mar 1 '15 at 17:23
  • $\begingroup$ So basically you want us to do all the work for you? $\endgroup$ – Aaron Maroja Mar 1 '15 at 17:25
  • $\begingroup$ no im trying to understand. thing is we haven't practiced this sort of proof writing in class and the teacher believes we'll understand more by trying it on our own but if i tried and couldn't understand, it's normal to ask for help $\endgroup$ – user220177 Mar 1 '15 at 17:35
  • $\begingroup$ Then show what you have tried and thought. $\endgroup$ – Aaron Maroja Mar 1 '15 at 17:36
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Hint: Use the definition of matrix multiplication. If you have $A = (a_{ij})\in \Bbb R^{m \times n}$ and $B = (b_{ij})\in \Bbb R^{n \times p}$, then $AB \in \Bbb R^{m \times p}$ and its elements are given by $c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}$. The hypothesis of a row of zeros is translated as: exists $i_0$ between $1$ and $m$ such that $a_{i_0j} = 0$ for all $j$. Look at $c_{i_0j}$.

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  • $\begingroup$ I know that if a matrix has a row or a column of zeros and its multiplied with another matrix that has non-zero rows or columns then the resulting matrix will have either a column or a row of zeros since anything multiplied by zero gives zero $\endgroup$ – user220177 Mar 1 '15 at 17:49
  • $\begingroup$ This is not a proof, though. $\endgroup$ – Ivo Terek Mar 1 '15 at 17:57
  • $\begingroup$ For all columns j that i = 0, entries of the resulting matrix for that row will also be zero? $\endgroup$ – user220177 Mar 1 '15 at 18:49
  • $\begingroup$ Mainly: the same row or column that is made only of zeros in $A$, will be made of zeros in $AB$. $\endgroup$ – Ivo Terek Mar 1 '15 at 18:53

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