## Linear Algebra and Its Applications, exercise 1.4.20

Exercise 1.4.20. If A and B are two nxn matrices with all entries equal to 1, what are the entries $(AB)_{ij}$ of their product AB? Use the following summation formula to find the answer:

$(AB)_{ij} = \sum_{k=1}^{k=n} a_{ik}b_{kj}$

Also, if C is a third nxn matrix with all entries equal to 2, compute the entries of (AB)C = A(BC), using the corresponding summation formulas for the two sides of that equation

$\sum_{j=1}^{j=n} (\sum_{k=1}^{k=n} a_{ik}b_{kj}) c_{jl} = \sum_{k=1}^{k=n} a_{ik} (\sum_{j=1}^{j=n} b_{kj}c_{jl})$

Answer: If $a_{ij} = b_{ij} = 1$ for all i and j, then we have

$(AB)_{ij} = \sum_{k=1}^{k=n} a_{ik}b_{kj} = \sum_{k=1}^{k=n} 1 \cdot 1 = \sum_{k=1}^{k=n} 1 = n$

for all i and j.

If $c_{ij} = 2$ for all i and j, then the i-jth entry of (AB)C is

$\sum_{j=1}^{j=n} (\sum_{k=1}^{k=n} a_{ik}b_{kj}) c_{jl} = \sum_{j=1}^{j=n} (\sum_{k=1}^{k=n} 1 \cdot 1) \cdot 2 = \sum_{j=1}^{j=n} n \cdot 2 = 2n^2$

and the i-jth entry of A(BC) is

$\sum_{k=1}^{k=n} a_{ik} (\sum_{j=1}^{j=n} b_{kj}c_{jl}) = \sum_{k=1}^{k=n} 1 \cdot (\sum_{j=1}^{j=n} 1 \cdot 2) = \sum_{k=1}^{k=n} 2n = 2n^2$

with (AB)C = A(BC) as expected.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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