## 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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