Capital Institute of Physical Education 623 Sports Artificial Intelligence Major Basic Postgraduate Examination Materials
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Let a, b, and c all be nth order matrices, and if ab c and b are reversible, then ( ) is numbered.
1. Number. Second, the number three 2013 research].
The row vector group of matrix A c is equivalent to the row vector group of matrix A.
b The column vector group of matrix c is equivalent to the column vector group of matrix a.
The row vector group of the C matrix C is equivalent to the row vector group of the matrix B.
The column vector group of D matrix c is equivalent to the column vector group of matrix b.
Answer] B Analysis] Divide the matrix A and C into columns as follows: A ( 1, 2,...，n），c＝（γ1，γ2，…,n) Since ab c, then i b1i 1 b2i 2 ....＋bniαn（i＝1，2，…,n), the column vector group of matrix c can be linearly represented by the column vector group of matrix a, and since b is reversible, i.e., a cb-1, it can be seen that the column vector group of matrix a can be represented linearly by the column vector group of matrix c, so the column vector group of matrix c is equivalent to the column vector group of matrix a.
Let the nth-order matrix a [ 1, 2,..n]，b＝[αn，α1，..n 1], if the determinant |a|1, then |a－b|＝（
a．0b．2
c．1＋（－1）n＋1
d．1＋（－1）n
Answer] A Analysis] From the known conditions a b [ 1 n, 2 1 ,..n n 1], which will be |a－b|add to the first column to get |a－b|＝[0，α2－α1，..n－αn－1]＝0。
Let a2 e,e be the identity matrix, then the following conclusion is correct ( ).
a a e reversible.
b When a ≠ e, a e is reversible.
c a e reversible.
d When a ≠ e, a e is irreversible.
Answer] D Analysis] From a2 e, (a e) (a e) 0 can be obtained. Take the determinant on both sides of the equation to get |a＋e||a－e|0, if a≠e, i.e. |a－e|≠0, there is |a＋e|0, i.e., the matrix a e is irreversible.
Let a be a fourth-order phalanx and satisfy a2 a, then rank r(a) rank r(a e) (
a．4b．3
c．2d．1
Answer] A Analysis] Since a(a e) a2 a 0, then r(a) r(a e) 4, and e (e a) a, so 4 r(e) r(e a a) r(e a) r(a) r(a e) r(a e) r(a) r(a e) 4.
Let a be the m n matrix, then m n is a system of homogeneous linear equations atax 0 with a non-zero solution ( ).
a Required.
b Sufficient conditions.
c Sufficient conditions.
d None of the above.
Answer] B Analysis] Because r(ata) r(a) m n, where n is the order of ata, that is, the number of unknowns of the system of equations atax 0, the system of equations atax 0 has a non-zero solution, but it is not necessary, because when m n, r(ata) n m, then the system of equations may have only zero solutions, or there may be non-zero solutions.
It is known that a is a matrix of order 4 and a* is the adjoint matrix of a, and if the eigenvalues of a* are 1, 1, 2, 4, then the irreversible matrix is ( ).
a．a－eb．2a－e
c．a＋2e
d．a－4e
Answer] C Analysis] The eigenvalue of a* is L, 1,2,4a*|8, and because of |a*|＝a|n 1, and know |a|3 8, and so|a|＝－2。Then, the eigenvalues of matrix a are: 2,2, l, 1 2. Therefore, the eigenvalues of a e are 3,1, 2, 3 2. Because the eigenvalues are not 0, the matrix a e is reversible. Similarly, it can be seen that the eigenvalues of matrix A2e contain 0, so matrix A2e is irreversible.
Knowing that a is an invertible matrix of order n, then a matrix with the same eigenvalue as a is ( ).
a．atb．a2
c．a－1d．a－e
Answer: A Analysis] Due to |λe－at|＝|e－at）|＝e－a|, a has the same eigenpolynomial as at, so a has the same eigenvalue as at.
From a 0, we get: a2 2 , a 1 (1) a e) 1), indicating that a2, a 1, a e and a e are not the same as a (but a eigenvector is also their eigenvector).
It is known that the third-order matrix a is related to the three-dimensional non-zero column vector, if the vector group , a , a2 is linearly independent, and a3 3a 2a2 , then the eigenvector of matrix a belongs to eigenvalue 3 is ( ).
a．αb．aα＋2α
c．a2α－aα
d．a2α＋2aα－3α
Answer] C analysis] It is known that a3 2a2 3a 0, that is, there is (a 3e) (a2 a ) 0(a2 a ) 0, because , a , a2 is linearly unrelated, then there must be a2 a ≠0, so a2 a is the eigenvector of the matrix a 3e belongs to the eigenvalue 0, that is, the matrix a belongs to the eigenvalue 3 eigenvector.
For the n-ary quadratic xtax, the correct one of the following propositions is ( ).
a The coordinate transformation that turns xtax into a standard shape is unique.
b The coordinate transformation that turns xtax into a canonical shape is unique.
The standard form of c xtax is unique.
The canonical form of d xtax is unique.
Answer] D Analysis] AC two items, the quadratic form can be transformed into a standard shape by either the orthogonal transformation method or the matching method, and the formation of the standard form and the coordinate transformation used are not unique.
In the bd two, the canonical shape is determined by the positive and negative inertia indices of the quadratic type, and the positive and negative inertia indices are invariant under the coordinate transformation.
Let a be a real symmetry matrix of order n, and swap column i and column j of a to get b, and then swap row i and row j of b to get c, then a and c ( ).
a Equivalent but not similar.
b Contracts but not similar.
c Similar but not contractual.
d Equivalent, contractual and similar.
Answer] D Analysis] Transform the elementary rows and columns, and use the left and right multiplication of the elementary matrix to form the table, and set aeij b and eijb c by the question. Therefore c eijb eijaeij, because eij eijt eij 1, so c eijaeij eij 1aeij eijtaeij, so a b, c a, and c a.
Let a and b be nth-order matrices, consider the following propositions: a is equivalent to b; A is similar to B; A and B contracts; A and b are positive definite matrices. Use "p q" to indicate that the proposition p can deduce the proposition q, then ( ).
a．①⇒b．②⇒
c．④⇒d．③⇒
Answer] C Analysis] If A and B are positive definite matrices, then A and B are both contracted in the identity matrix, so that A and B are contract matrices, and the rank of the matrix of the contract is the same, then A and B are equivalent, so the C term is true, and the other three options can construct counterexamples to show that they are not valid.
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