H21-1 - Math 330 Homework 2.1 (Pages 116-117) (2) Consider...

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Unformatted text preview: Math 330 Homework 2.1 (Pages 116-117) (2) Consider the following matrices: A= 2 0 -1 4 -5 2 1 2 -2 1 D= B= 3 5 -1 4 7 -5 1 1 -4 -3 E= -5 3 C= Form each of the following: (a) A + 2B SOLUTION: A + 2B = 2 0 -1 4 -5 2 +2 7 -5 1 1 -4 -3 = = 2 + 2 7 0 - 2 5 -1 + 2 1 4 + 2 1 -5 - 2 4 2 - 2 3 16 -10 1 6 -13 -4 (b) 3C - E SOLUTION: This is not defined because 3C is a 2 2 matrix and E is a 2 1 matrix. (c) CB SOLUTION: CB = 7 1 -2 + 2 1 -5 1 -2 1 2 -2 1 -4 2 1 7 -5 1 1 -4 -3 1 -2 -3 2 1 = 9 -13 -5 -13 6 -5 (d) EB SOLUTION: Since E is 2 1 and B is 2 3 the number of columns of the left operand does not match the number of rows of the right operand. Hence this is not defined. (10) Verify that AB = AC but yet B = C: A= 2 -3 -4 6 , B= 8 4 5 5 , C= 5 -2 3 1 SOLUTION: It is certainly clear that B = C because in fact none of their entries are equal. However, 1 AB = = 8 and AC = = 5 2 -4 +3 -3 6 2 -4 +5 -3 6 2 -3 -4 6 4 2 -4 +5 8 4 5 5 -3 6 = 1 -7 -2 14 2 -3 -4 6 -2 2 -4 5 -2 3 1 + -3 6 = 1 -7 -2 14 COMMENT: You should ask yourself - how can this happen? Notice that the columns of the resulting products are linear combinations of the columns of A. Since the columns of A are linearly dependent it means there is redundancy and hence many different linear combinations which produce the same results. This would not be possible if the columns of A were linearly independent. In this case all linear combinations would have unique weights. (12) Construct a 2 2 matrix B such that AB is the zero matrix. Use two different nonzero columns for B. A= 3 -6 -1 2 SOLUTION: Notice that the columns of A are linearly dependent. This means that there are many nontrivial linear combinations of the zero vector. Notice that column 1 (c1) and column 2 (c2) of A satisfy -2c1 = c2. This means that -2c1 - c2 = 0 and, perhaps also (there are many such choices), -4c1 - 2c2 = 0. Therefore, one of many solutions for B is B= (16) TRUE or FALSE? (a) If A and B are 3 3 and B = b1 b2 b3 , then AB = Ab1 + Ab2 + Ab3 . FALSE This could not possibly ever be TRUE since this would mean the result has only one column whereas it should have 3. The correct statement is that AB = Ab1 Ab2 Ab3 . (b) The second row of AB is the second row of A multiplied on the right by B TRUE In fact every entry in the second row of AB comes from the usual dot product of the second row of A with a different column of B. 2 -2 -4 -1 -2 (c) (AB)C = (AC)B FALSE Suppose that A, B, and C have respective dimensions of 1 2, 2 1, and 1 1. Then the left expression will be defined as a 1 1 matrix whereas the expression on the right (in particular AC) will be undefined. (d) (AB)T = AT B T FALSE Suppose that A, B have respective dimensions of 13 96 and 96 107. Then AT , B T will have respective dimensions of 96 13 and 107 96. Therefore, AT B T will be undefined. The correct result is that (AB)T = B T AT . (e) The transpose of a sum of matrices equals the sum of their transposes. TRUE This follows by the facts that the (i, j)th entry of the sum is the sum of the (i, j)th entries of each summand which produces the (j, i)th entry of the transpose of the sum which in turn comes from the sum of the (j, i)th entries of the transpose of each each summand. All summand matrices have the same dimensions. (20) Suppose that the second column of B is all zeroes. What can you say about the second column of AB? SOLUTION: If we let b2 denote the second column of B, then the second column of AB is simply Ab2. Therefore, we have that this is Ab2 = A0 = 0 although we should note that the zero vectors appearing on the two sides of this equation could indeed have different dimensions. 3 ...
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This note was uploaded on 09/19/2009 for the course MATH 330 taught by Professor Johnson,j during the Spring '08 term at Nevada.

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