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Matrix Algebra. Networks
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In the above answer, note, for example, that we cannot factor a 2 out of the rst column; such factoring out can be done only if a number factors out of all the elements of a matrix, as discussed at the end of section 11.1. Important Note: Suppose, for example, that A is a 3 by 2 matrix and B is a 2 by 4 matrix. Then the product of the two can be taken in the order AB because the rst factor A has 2 columns and the second factor B has 2 rows; that is, in this case A3;2 B2;4 C3;4 But notice that multiplication in the order BA cannot be done, because if we attempt the product BA we have B2;4 A3;2 where now the rst factor B has 4 columns and the second factor A has 3 rows, and so they are not conformable in the order BA. This illustrates the fact that, in general, Matrix multiplication is not commutative; that is, in general, AB does not equal BA. This is true even if A and B are conformable in both AB and BA form, as the following example illustrates. Let ! ! 2 5 3 2 A and B 1 4 3 1 Note that the products AB and BA both exist, because the number of columns of the rst factor equals the number of rows of the second factor regardless of whether we write AB or BA. But note that AB whereas BA 2 3 5 1 ! 3 1 2 4 ! 6 5 9 1 4 20 6 4 ! 11 24 10 10 ! 3 1 2 4 ! 2 3 5 1 ! 6 6 2 12 15 2 5 4 ! 12 17 14 9 !
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which, since the answers are not equal, shows that in general AB and BA do not represent equal matrix products, even if A and B are conformable in either order of multiplication.
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CHAPTER 11 Matrix Algebra. Networks
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Problem 224 Find matrix C, given that 2 6 Problem 225 Given A 2 3 1 2 0 4 ! and 4 4
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Matrix Algebra. Networks
The Inverse of a Square Matrix
Let us now consider the important case where two sets of unknowns, which we ll denote by xs and ys, are related by means of n simultaneous linear equations. Let us take, as an example, the case for n 3, as shown below, where the as are constant coe cients. Note that the subscript of each a coe cient gives the location, row, and column of that particular a. 9 a11 x1 a12 x2 a13 x3 y1 = > 475 a21 x1 a22 x2 a23 x3 y2 > ; a31 x1 a32 x2 a33 x3 y3 Now notice, as a result of the de nition of multiplication of matrices laid down in section 11.2, that the above set of equations can be written in the form of a matrix product; thus 32 3 2 3 2 a11 a12 a13 x1 y1 76 7 6 7 6 476 4 a21 a22 a23 54 x2 5 4 y2 5 a31 or, in abbreviated form A X Y or, more simply, we may merely write AX Y 477 a32 a33 x3 y3
where, in eq. (477), A represents the square matrix formed from the constant a coe cients, and X and Y represent the column matrices formed from the values of the unknowns in eq. (476). Now suppose we wish to have the X matrix alone, by itself, on the left-hand side of eq. (477). To indicate this we change matrix (477) into the form X A 1 Y
478
where the matrix A is called the INVERSE of the square matrix A, or simply as the inverse of matrix A, since only a square matrix can have an inverse. Let us now nd the actual form that A 1 must have in order to legitimately transform (477) into (478). To do this, let us go back and solve the given set of simultaneous equations, eq. (475), for the x values, using the standard procedure of determinants from Chap. 3. Doing this, we have        y1 a12 a13   a11 y1 a13   a11 a12 y1               y2 a22 a23   a21 y2 a23   a21 a22 y2         y3 a32 a33   a31 y3 a33   a31 a32 y3  ; x2 ; x3 x1 D D D where, as usual, D (delta) is the value of the determinant formed from the constant a coe cients. Now recall, from Chap. 3, that the value of any determinant of order 3 or more can be found by expanding the determinant in terms of the minors of any row or any column of the determinant. Since we have to deal with determinants when nding the inverse matrix, let us review, just brie y, some details from Chap. 3.