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CHAPTER 11 Matrix Algebra. Networks
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Problem 221 ! 0 4 2 3 2 1 ! 3 2
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Product of Two Matrices
Let us now de ne the matrix product AB, that is, matrix A times matrix B. First of all we ll nd that, in matrix multiplication, the ORDER in which the factors are written is important. Thus, in general, the matrix products AB and BA will NOT be equal. The reason for the seemingly peculiar way that matrix multiplication is de ned will become clear to us later on in our work. The rst requirement, in the de nition of matrix multiplication, is that
A matrix product, in the order AB, exists only if the number of columns of the rst matrix A is equal to the number of rows of the second matrix B. Thus if A is an m n matrix and B is an n p matrix, then the product of the two in the order AB does exist, because A has n columns and B has the same number of n rows. If this requirement is not satis ed, then the product in the order AB cannot be taken. Any two such matrices, in which the number of columns of A is equal to the number of rows of B, are said to be conformable in the order AB. Thus the matrix product AB exists only of A and B are conformable matrices in the order AB. If A and B are two matrices conformable in the order AB, then the product AB is itself a matrix C, whose elements are found according to the following rule:
Matrix Algebra. Networks
If AB C, the element cij , at the intersection of the i th row and the j th column of C, is equal to the sum of the products of corresponding pairs of the elements of the i th row of A and the j th column of B. The above de nition can be expressed as a general formula as follows. Let A be an m n and B an n p matrix. Note that A has n columns and B has n rows, so that the product in the order AB C does exist. The procedure for nding the value of any element cij in the product C as stated in the above rule, is illustrated in Fig. 275.
Fig. 275
It follows from the de nition and from inspection of Fig. 275 that the SUM OF THE PRODUCTS OF CORRESPONDING PAIRS of the i th row of A and the j th column of B gives cij ai1 b1j ai2 b2j ain bnj 474
Since A is an m by n matrix we have that i 1; 2; 3; . . . ; m, and since B is an n by p matrix we have that j 1; 2; 3; . . . ; p, and therefore the product matrix C will consist of m rows and p columns of elements; that is If A is an m n matrix and B is an n p matrix, the product AB is an m p matrix C, as illustrated in Fig. 276.
Fig. 276
CHAPTER 11 Matrix Algebra. Networks
The rst step in nding the above product, AB C, is to note the values of m, n, and p, for the given problem. The second step is then to calculate the value of each of the elements in the matrix C, which is done by making use of eq. (474). Consider now the following. Example
Find the matrix product AB C if 2 7 6 A 4 4 25 1 5 2 3 3 and B 2 3 4 1 !
Solution Since the rst factor A has 2 columns and the second factor B has 2 rows, a product in the order AB does exist. Note that since A is a 3 2 matrix and B is a 2 2 matrix, the product C will be a 3 2 matrix; that is, C will have 3 rows and 2 columns of elements. Hence the solution, AB C, is the form 3 3 2 2 c11 c12 2 3 ! 7 2 3 7 6 6 4 c21 c22 5 4 4 25 4 1 1 5 c31 c32 Let us begin by nding the value of element c11 . By de nition, c11 is equal to the sum of the products of corresponding pairs or row 1 of A and column 1 of B, and hence c11 2 2 3 4 16 Let us next nd the value of element c21 . By de nition, this is equal to the sum of the products of corresponding pairs of row 2 of A and column 1 of B, and hence c21 4 2 2 4 16 In the same way, the value of c31 is the sum of the products of corresponding pairs of row 3 of A and column 1 of B; thus c31 1 2 5 4 18 We ve now found the values of the elements of the rst column of C; the next step is to nd the values of the elements of the second column of C, beginning with element c12 . Again, by de nition, the value of c12 is equal to the sum of the products of corresponding pairs of row 1 of A and column 2 of B, and hence c12 2 3 3 1 3 Next, the value of c22 is the sum of the products of corresponding pairs of row 2 of A and column 2 of B; thus c22 4 3 2 1 10 Finally, in the same way, the value of element c32 is the sum of the products by pairs of row 3 of A and column 2 of B, and hence c32 1 3 5 1 8
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