SECTION 11.4 Matrix Algebra 801 Notice that the product AB in Example 8 is a 2 by 4 matrix, as we expected. Also notice that, for the matrices given in Example 8, the product BA is not defined because B is 3 by 4 and A is 2 by 3. Now Work PROBLEM 27 Figure 14 Matrix multiplication Algebraic Solution Suppose that we want the entry in row 2, column 3 of AB. To find it, find the product of the row vector from row 2 of A and the column vector from column 3 of B. Column 3 of B [ ] ( ) − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = ⋅ + ⋅ + − = 5 8 0 1 0 2 5 1 8 0 0 2 5 So far, we have Graphing Solution Enter the matrices A and B into a graphing utility. Figure 14 shows the product AB using a TI-84 Plus CE. Row 2 of A Column 3 Row 2 = ↓ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ← AB 5 Now, to find the entry in row 1, column 4 of AB, find the product of row 1 of A and column 4 of B. [ ] ( )( ) − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = ⋅ + ⋅ + − − = 2 4 1 4 6 1 2 4 4 6 1 1 33 Continuing in this fashion, we find AB. Column 4 of B Row 1 of A ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( ) = ⎡ − ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ − − − ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ = ⋅+⋅+−− ⋅+⋅+− ⋅ + ⋅ + − − ⋅ + ⋅ + − ⋅ + ⋅ + ⋅ ⋅ + ⋅ + − ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ AB A A A A B B B B A A A A B B B B 2 4 1 5 8 0 2 5 1 4 4 8 0 6 3 1 2 1 Row 1 of Row 1 of Row 1 of Row 1 of times times times times column 1 of column 2 of column 3 of column 4 of Row 2 of Row 2 of Row 2 of Row 2 of times times times times column 1 of column 2 of column 3 of column 4 of 2 2 4 4 1 3 2 5 4 8 1 1 2 1 4 0 1 2 33 fromearlier 5 2 8 4 0 3 5 5 8 8 0 1 5 fromearlier 5 4 8 6 0 1 23 41 4 33 42 89 5 68 Multiplying Two Matrices If A B 2 1 3 1 1 0 and 1 0 2 1 3 2 = − ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ find: (a) AB (b) BA EXAMPLE 9 (continued)
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