Solucionario-de-algebra-lineal-Kolman-octava-edicion.pdf

T16 show that if a is n n with a skew symmetric a t a

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T16. Show that if A is n × n , with A skew symmetric ( A T = - A , see Section 1.4, Exercise T.24) and n is odd, then det( A ) = 0. Solution. By Theorem 3.1, det( A T ) = det( A ). By the Exercise T3 above, det( - A ) = det(( - 1) A ) = ( - 1) n det( A ) = - det( A ) because n is odd. Hence det( A ) = - det( A ) and so 2 det( A ) = 0 and det( A ) = 0. Page 210. T7. Show that if A is singular, then adj A is singular. Solution. If A is singular, then det( A ) = 0. Since A (adj A ) = det( A ) I n (this is Theorem 3.11), A (adj A ) = O . First of all, if A = O then adj A = O by the definition of the adjoint matrix: all cofactors are zero. In the remaining case, if A 6 = O then adj A cannot be nonsingular because, otherwise, multiplying to the right with (adj A ) - 1 the equality A (adj A ) = O we obtain A = O . So adjA is singular. T8. Show that if A is an n × n matrix, then det(adj A ) = det( A ) n - 1 . Solution. Use the equality given in Theorem 3.11: A (adj A ) = det( A ) I n . Taking the determinants of both members of the equality one obtains: det( A ) det(adj A ) = (det( A )) n (notice that det( A ) I n is a scalar matrix hav- ing n copies of the real number det( A ) on the diagonal; Theorem 3.7, p. 188, is used).
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31 If det( A ) = 0, according to the previous Exercise, together with A , adj A is also singular, and so det(adj A ) = 0 and the formula holds. If det( A ) 6 = 0 we can divide both members in the last equality by det( A ) and thus det(adj A ) = (det( A )) n ÷ det( A ) = [det( A )] n - 1 . T10. Let AB = AC . Show that if det( A ) 6 = 0, then B = C . Solution. Using Theorem 3.12 (p. 203), A is nonsingular and so, A - 1 exists. Multiplying AB = AC to the left with A - 1 we obtain A - 1 ( AB ) = A - 1 ( AC ) and finally B = C . T12. Show that if A is nonsingular, then adj A is nonsingular and (adj A ) - 1 = 1 det( A ) A = adj( A - 1 ) . Solution. First use Exercise T8 previously solved: if A is nonsingular then det( A ) 6 = 0 and so det(adj A ) = [det( A )] n - 1 6 = 0. Thus adj A is also nonsingular. Further, the formulas in Theorem 3.11, A (adj A ) = (adj A ) A = det( A ) I n , give (adj A )( 1 det( A ) A ) = ( 1 det( A ) A )(adj A ) = I n and so (adj A ) - 1 = 1 det( A ) A , by definition. Finally, one can write the equalities given in Theorem 3.11 for A - 1 : A - 1 (adj( A - 1 )) = det( A - 1 ) I n = 1 det( A ) I n (by Corollary 3.2, p. 191). Hence, by left multiplication with A , one finds adj( A - 1 ) = 1 det( A ) A . Supplementary Exercises (Bonus): 1) the Proof of Theorem 1.11, Section 1.7 Suppose that A and B are n × n matrices. (a) If AB = I n then BA = I n (b) If BA = I n then AB = I n . Proof. (a) If AB = I n , taking determinants and using suitable prop- erties, det( A ) det( B ) = det( AB ) = det( I n ) = 1 shows that det( A ) 6 = 0 and det( B ) 6 = 0 and so A and B are nonsingular. By left multiplication with
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32 CHAPTER 3. DETERMINANTS A - 1 (which exists, A being nonsingular), we obtain A - 1 ( AB ) = A - 1 I n and B = A - 1 . Hence BA = A - 1 A = I n . (b) Similarly. 2) Exercise T3 . Show that if A is symmetric, then adj A is also symmetric. Solution. Take an arbitrary cofactor A ij = ( - 1) i + j det( M ij ) where M ij is the submatrix of A obtained by deleting the i -th row and j -th column.
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