18_hwc1SolnsODDA

# 18_hwc1SolnsODDA - A has two rows so DA is the only product...

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and recall that multiplication by a matrix is always a linear transformation. (Here we are using the “converse” part of Theorem 2.) Hence the matrix representing f is one in ( * ). 3.1 Consider the four matrices A , B , C and D from Exercise 2.7. There are 16 ordered pairs of these matrices, allowing self pairing, namely AA, AB, AC, AD, BA, BB, . . . Which of these make sense as matrix products? Compute the product in each such case. SOLUTION The product of two matrices is defined exactly when the number of columns in the matrix on the left equals the number of columns in the matrix on the right. In the case at hand, A has 4 columns, and only C has 4 rows, so AC is the only product with A on the left that makes sense. Likewise, B has 3 columns, and only B has 3 rows, so BB is the only product with B on the left that makes sense. Next, C has 2 columns, and A and D are the only matrices with 2 rows, so CA and CD are the only products with C on the left that make sense. Finally,
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Unformatted text preview: A has two rows, so DA is the only product with D on the left that makes sense. We now use the deﬁnition of the matrix product to compute the products identiﬁed above: AC = ± 13 3-3 4 ² BB = 3-3 3 4-1 2 10-7 8 CA = 3 4 2 3 2 4-2-1-3 1 4 6 7 5 9 CD = 1 2 2-2 1 1 5 DA = ± 1 2-1 1 1 1 2 ² DD = ± 1 1 ² 3.3 Let A = 1 1 1 2 1 2-1 2-1 3-2 2 and let B = -1 1 1 1 1 1-1 2 3 1-3-3 . Compute the third column of AB by computing an appropriate matrix–vector product. SOLUTION By the deﬁnition of matrix–matrix multiplication, if we write B = [ v 1 , v 2 , v 3 , v 4 ] , then AB = [ A v 1 , A v 2 , A v 3 , A v 4 ] . Therefore, we just need to compute A v 3 , which is 3 3 2 1 . 21 /september/ 2005; 22:09 19...
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