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Consider two n x nmatrices A and B whose entries are positive or zero. Suppose that all entries of A are less than or equal to 鈥s鈥, and all column sums of B are less than or equal to 鈥r鈥 (the column sum of a matrix is the sum of all the entries in its column). Show that all entries of the matrix AB are less than or equal to 鈥sr鈥.

Short Answer

Expert verified

If we have two nnmatrices A and B whose entries are positive or zero and all entries of A are less than or equal to 鈥s鈥, and all column sums of B are less than or equal to 鈥谤鈥 (the column sum of a matrix is the sum of all the entries in its column) then all entries of the matrix AB are less than or equal to 鈥sr鈥.

Step by step solution

01

  Given matrices A and B

We have given twonnmatrices A and B whose entries are positive or zero.

Since all entries of A are less than or equal to 鈥s鈥, then we have

aips (1)

Since all column sums of B are less than or equal to 鈥谤鈥, then we have

p=1nbpjr (2)

02

To show all entries of the matrix AB are less than or equal to ‘sr’

Now we have to show that all entries of the matrix AB are less than or equal to 鈥s谤鈥.

i.e. p=1naipbpjsr

Since we have from equation (1) and (2),

localid="1659346125962" aipsandp=1nbpjr

Now,

p=1naipbpjp=1nS.bpj(aips)p=1naipbpjsp=1nbpjp=1naipbpjsrp=1nbpjr

This shows that all entries of the matrix AB are less than or equal to 鈥s谤鈥.

03

Final Answer

If we have two nnmatrices A and B whose entries are positive or zero and all entries of A are less than or equal to 鈥s鈥, and all column sums of B are less than or equal to 鈥谤鈥 (the column sum of a matrix is the sum of all the entries in its column) then all entries of the matrix AB are less than or equal to 鈥sr鈥.

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