Proof: Let A & B be Matrices; Show BA has a Row of Zeros

AI Thread Summary
If matrix A has a row of zeros, and matrix B is an n x p matrix, then the product BA will also have a row of zeros. The proof involves recognizing that the zero row in A leads to zero contributions in the multiplication process. Specifically, any entry in the resulting matrix BA corresponding to the zero row in A will be zero, confirming that BA contains a row of zeros. The discussion highlights the importance of correctly interpreting the matrix multiplication rules. This conclusion is essential for understanding the implications of zero rows in matrix operations.
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Homework Statement


28. Let A be an m x n matrix with a row consisting entirely of zeros. Show that if B is an n x p matrix, then BA has a row of zeros.

Homework Equations


N/A

The Attempt at a Solution


A = (aij)_{mxn} and B = (bij)_{nxp}. Assuming that the entries for jth column of A are all zeros, I infer that the jth column of BA is also a row of zeros. Let cij be an entry in jth column. By the definition of multiplication:
cij = b_{i1}a_{j1} + b_{i2}a_{j2} + ...+ b_{in}a_{nj} = \sum_{k=1}^n b_{ik}a_{kj}

Since the jth column of A is zero, then there is:
a_{1j} = a_{2j} = ... = a_{nj} = 0. Hence, cij = 0 and, therefore, the jth column of BA is a column of zeros.

Did I do this proof correctly? Thanks in advance.
 
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Retribution said:

Homework Statement


28. Let A be an m x n matrix with a row consisting entirely of zeros. Show that if B is an n x p matrix, then BA has a row of zeros.

Homework Equations


N/A


The Attempt at a Solution


A = (aij)_{mxn} and B = (bij)_{nxp}. Assuming that the entries for jth column of A are all zeros,
You're given that one row of A consists of zeros. You can't also assume that one column (column j) is all zeros.
Retribution said:
I infer that the jth column of BA is also a row of zeros. Let cij be an entry in jth column. By the definition of multiplication:
cij = b_{i1}a_{j1} + b_{i2}a_{j2} + ...+ b_{in}a_{nj} = \sum_{k=1}^n b_{ik}a_{kj}

Since the jth column of A is zero, then there is:
a_{1j} = a_{2j} = ... = a_{nj} = 0. Hence, cij = 0 and, therefore, the jth column of BA is a column of zeros.

Did I do this proof correctly? Thanks in advance.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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