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Aij is the minor obtained by removing row i and

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Unformatted text preview: +j Aij]T. 5 A −A A . 15. The adjoint formula for matrix inverse. ∙ adj( ) = det( ) = adj( ) ∙ . The (1,1)-entry in A adj(A) is equal to − + which is the same as det(A). Similarly, we can see that the diagonal entries of A adj(A) are all equal to det(A). On the other hand, the (2,1)-entry in the product A adj(A) is − + − . By the Laplace expansion on the second row, this is equal to the following 3x3 determinant . The first and second rows are identical, and hence by (13C), this determinant is equal to zero. By similarly calculations, we can show that for i not equal to j, the (i,j)-entry of A adj(A) is equal to a 3x3 determinant, in which two rows are identical, and hence equal to zero. This proves that A adj(A) = det(A) I3. From the property that det(M) = det(MT), we can prove that adj(A) A is equal to det(A) I3. If det(A) is nonzero, we can divide both sides of ∙ adj( ) = det( ) by det(A) and obtain the following adjoint formula for matrix inverse = 1 adj( ) det( ) This proves that reverse direction of (*) for 3x3 matrix. (The Kreyszig ’s book, the above formula for matrix inverse is written in terms of the cofactors, which is defined as (-1)i+j Aij.) 16. For the forward direction of (*), we need to establish the multiplicative property of the determinant function for 3x3 matrices. Now, we consider the determinant as a function of the three columns. To emphasize this view point, we let a1, a2, and a3 be the three columns of A, and write the determinant as 6 . det( ) = det[ is a linear function in each From (13B), we know that det[ input column vector. For any real number p and q, we have det[ = det[ det[ + + det[ , det[ + det[ , = + det[ = + det[ + det[ . Let A and B be 3x3 matrices. The three columns in AB are respectively = + + , = + + , = + + , where aj and bj represent respectively the jth row of A and B, for j=1,2,3. = = det ( det [ det [ ) + + + + + + Using the linear property (13B) of determinant, we expand det(AB) to det[ The above calculation is essentially the same as the expansion ( )( )( + + + + + =∑ ∑ ∑ . + ) Among the twenty seven terms, lots of them are zero. Because if any two are identical, the determinant is zero. There columns of det[ are six remaining terms, one for each permutation of 1, 2 and 3,+ det[ det[+ + det[ det[+. + det[ det[ The determinant in each term is very “close” to the determinant of A, except that the order of the columns is not right. For the second term, we exchange column 2 and column 3, and by the exchange property in (13D), 7 = −. det[ det[ After exchanging column 1 and column 2 in the third term, we obtain= −. det[ det[ Repeat the same procedure for the last three terms, we finally get− det( ) = det[ det[+ − det[ det[−. + det[ det[ is a common factor, we can factor it out, and Because det[ what remains inside the parenthesis is precisely the determinant of B, det ( ) = ( − = − + ( ) det ( ) det − + ) det ( ) This completes the proof of the multiplicative property of determinant for 3x3 matrix. Once we...
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