a) 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.
Pij = term in product matrix
Pij = sum over j of Aij Bji
If Akj = 0 for some i = k and all j
Then the product row Pkj = sum over j of AkjBkj
But Akj = 0 for all j
To show that if matrix A has a row of zeros and matrix B is such that the product AB is defined, we need to prove that the resulting matrix AB also has a row of zeros.
Let's denote A as an m x n matrix and B as an n x p matrix. Since A has a row of zeros, let's assume that this row is the k-th row, where 1 ≤ k ≤ m. This means that the k-th row of A is a vector of zeros: A(k,:) = [0, 0, ..., 0].
Now let's consider the product AB. The i-th row of AB, denoted as (AB)(i,:), is obtained by multiplying the i-th row of A with each column of B and summing the results element-wise.
For the k-th row of AB, the i-th element can be calculated as follows:
(AB)(k,i) = A(k,:) * B(:,i)
= [0, 0, ..., 0] * B(:,i)
= 0 * B(1,i) + 0 * B(2,i) + ... + 0 * B(n,i)
= 0
From the above calculation, we can observe that each element of the k-th row of AB will always be zero. Therefore, AB has a row of zeros, specifically the k-th row.
Hence, we have shown that if matrix A has a row of zeros and matrix B is such that the product AB is defined, then the resulting matrix AB also has a row of zeros.