qr code vb.net source a (3 2) matrix; thus 3 2 2 c11 0 2 " # 7 7 4 6 6 60 47 6 c21 5 4 4 6 11 6 1 c31 3 2 0 12 in .NET

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a (3 2) matrix; thus 3 2 2 c11 0 2 " # 7 7 4 6 6 60 47 6 c21 5 4 4 6 11 6 1 c31 3 2 0 12
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227. This is a matrix product of the form ABC D. The procedure is to rst nd the product AB, then take that result times C. The work is as follows. First, ! ! ! 8 14 1 2 0 2 ; AB 16 18 3 4 4 6 a 2 2 matrix, which we now multiply the 2 3 matrix C by. The result is a 2 3 matrix D; thus # " " #" # c11 c12 c13 7 6 0 8 14 16 18 2 1 4 c21 c22 c23 " # 56 28 48 14 0 56 112 36 96 18 0 72 " # 84 62 56 ; answer: 148 114 72 228. The rst factor is a 4 4 matrix and the second factor is a 4 1 matrix, so the product of the two, in the order shown, does exist and will be a 4 1 matrix; thus 3 2 3 32 3 2 2 3 2 c11 6 29 1 2 2 3 6 4 0 27 60 0 6 1 76 2 7 6 c21 7 6 0 0 0 9 7 6 9 7 7 6 7 7 6 76 6 7 6 7 6 7; answer: 76 7 6 6 7 6 40 0 7 2 54 0 5 4 c31 5 4 0 0 0 18 5 4 18 5 9 110 24 4 0 90 4 2 5 10 c41 229. The square of any (n n) square matrix is also an (n n) square matrix; thus, the square of the given 3 3 matrix is a 3 3 square matrix, A2 AA C; thus 32 2 3 2 3 1 1 4 1 1 4 c11 c12 c13 76 6 7 6 7 6 0 3 2 7 6 c21 c22 c23 7 2 76 0 3 4 54 5 4 5 9 6 5 9 6 5 c31 c32 c33 2 3 1 0 36 1 3 24 4 2 20 6 7 6 0 0 18 0 9 12 0 6 10 7 4 5 9 0 45 9 18 30 36 12 25 2 3 37 22 14 6 7 6 18 21 16 7; answer: 4 5 36 39 73
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230. Since A is a (3 4) matrix and B is a (4 2) matrix, their product AB does exist and is a (3 2) matrix, C AB, whose value is found as follows. 3 2 1 2 3 3 2 2 c c 2 0 1 3 6 7 7 6 11 12 7 6 3 07 76 6 61 2 0 4 76 7 6 c21 c22 7 56 5 4 4 57 5 4 2 c31 c32 3 2 6 1 6 1 2 3 2 0 2 18 4 0 5 3 6 7 6 1 6 0 24 2 0 0 4 7 4 5 3 6 12 6 6 0 30 1 3 2 22 2 7 6 6 17 2 7; answer: 5 4 9 35 231. The rst step is to nd D, the value of the third-order determinant formed from the elements of the given matrix. Using, for example, the elements of the second row, we nd that        0 4  6 2 4  76 D 5  3 2   1 2  Next, in the manner of eq. (483), we replace each element in the given matrix with its cofactor, to get 3     2   5    6 0 6     5 0   3 1 7 2  3 2 6  1 2 3 6  12 10 13 7    7 6  0 4  2  2 4  0 7 6 7 6      16 25 A 0 6   3 2  3 1 7 4 4  1 2 7 6 6  7    24 20 12 4  0 4  2    0 5     2 4   5  5 0  6 0 6 Now, in the last expression, interchange the rows and columns; that is, let the rst row become the rst column, the second row become the second column, and so on. Doing this, and remembering to multiply by 1/D, the nal answer can be written in either of the forms 3 2 3 2 12=76 4=76 24=76 12 4 24 1 6 7 6 20=76 7 A 1 20 5 4 10=76 16=76 5 4 10 16 76 13=76 2=76 12=76 13 2 12 the second form being in accordance with the rule for multiplication of a matrix by a constant, as laid down at the end of section 11.1. 232. First, D 20 21 1. Now replace each element in the given matrix with its cofactor to get A0 ; thus ! 5 3 A0 7 4
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