8. When we explored matrix multiplication in Section 2.2, we saw that some properties that are true for real numbers are not true for matrices. This exercise will investigate that in some more depth. a. Suppose that A and B are two matrices and that AB = 0. If B # 0, what can you say about the linear independence of the columns of A? b. Suppose that we have matrices A, B and C such that AB = AC. We have seen that we cannot generally conclude that B = C. If we assume additionally that A is a matrix whose columns are linearly independent, explain why B = C. You may wish to begin by rewriting the equation AB = AC as AB – AC = A(B – C) = 0.

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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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please prove using examples and language of proofs.
8. When we explored matrix multiplication in Section 2.2, we saw that some
properties that are true for real numbers are not true for matrices. This
exercise will investigate that in some more depth.
a. Suppose that A and B are two matrices and that AB = 0. If B+ 0, what
can you say about the linear independence of the columns of A?
b. Suppose that we have matrices A, B and C such that AB = AC. We
have seen that we cannot generally conclude that B = C. If we assume
additionally that A is a matrix whose columns are linearly independent,
explain why B = C. You may wish to begin by rewriting the equation
AB = AC as AB – AC = A(B – C) = 0.
Transcribed Image Text:please prove using examples and language of proofs. 8. When we explored matrix multiplication in Section 2.2, we saw that some properties that are true for real numbers are not true for matrices. This exercise will investigate that in some more depth. a. Suppose that A and B are two matrices and that AB = 0. If B+ 0, what can you say about the linear independence of the columns of A? b. Suppose that we have matrices A, B and C such that AB = AC. We have seen that we cannot generally conclude that B = C. If we assume additionally that A is a matrix whose columns are linearly independent, explain why B = C. You may wish to begin by rewriting the equation AB = AC as AB – AC = A(B – C) = 0.
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