推 BreathWay : 推~ 01/18 02:37
課程名稱︰線性代數一
課程性質︰數學系必修
課程教師︰朱樺
開課學院：理學院
開課系所︰數學系
考試日期（年月日）︰2015/1/16
考試時限（分鐘）：160分鐘
試題 :
(1) Let A∈M (F) be of rank n and B∈M (F) be of rank p.
5% m*n n*p
Determine the rank of AB.
(2) Let V = M (F) and B∈V. If T is a linear operator on V defined by
5% n*n
T(A) = AB - BA for all A∈V, and if f is the trace function,
t
what is T f ?
(3) Describe explicitly al 3*3 row reduced echelon forms.
10%
(4) Let V = {(x_1,x_2,x_3,x_4,x_5)}∈R^5 : 2x_1 - 2x_2 + 3x_3 - x_4 + x_5 = 0}.
10% Extend {(0,1,1,1,0)} to a basis for V.
(5) (a) Let y(t) be a real-valued function. Solve the differential equation
10%
(7) (4)
y -2y + y' = 0
(b) Let y(t) be the polynomial of degree at most 5. Solve the differential
equation
(7) (4)
y -2y + y' = 5t^4 - 8t^3 + 9t^2 - 248t + 101
┌ ┐
(6) Let A = │ 1 -1 -2 3 │
15% │ 0 1 1 -1 │
│-1 3 4 -5 │
│ 2 -1 -1 1 │
│ 1 0 3 -6 │
└ ┘
(a) Find a rank 1 matrix B∈M (R) such that BA = O
4*5 4
(b) Find a rank 2 matrix B∈M (R) such that BA = O
4*5 4
(c) Find a rank 2 matrix C∈M (R) such that the equation AX=C is sovable,
4*5
where X = [x_ij] is a 4*4 matrix.
(d) Find a rank 3 matrix C∈M (R) such that the equation AX=C is sovable,
4*5
where X = [x_ij] is a 4*4 matrix.
┌ ┐
(7) (a) Let C = │-1 0 -2 2 │. Find an integer d such that the incerse of C
20% │ 3 0 d -1 │ belings to in M (Z).
│ 0 1 -1 1 │ 4*4
│ 0 0 2 -3 │
└ ┘
(b) For such d, find the inverse of C.
┌ ┐
(c) Let A = │ a_11 a_12 0 0 a_15 0 1 │
│ a_21 a_22 0 0 a_25 1 0 │
│ a_31 a_32 0 -1 a_35 0 1 │
│ a_41 a_42 0 1 a_45 0 0 │
│ a_51 a_52 1 0 a_55 0 0 │
└ ┘
┌ ┐
and B = │ 1 0 -1 0 -2 2 │ be the row reduced echolon form of A.
│ 0 1 3 0 d -1 │
│ 0 0 0 1 -1 1 │
│ 0 0 0 0 2 -3 │
│ 0 0 0 0 0 0 │
└ ┘
Find A.
(8) Let V be a vector space over C. Suppose that f and g are linear functionals
10% on V such that the function h(x) = f(x)g(x) is also a linear function on V.
Prove that either f=0 or g=0.
[Hint] There are x,y∈V such that f(x)≠0 and g(y)≠0.
Then use the linearity of h.
(9) Let A∈M be nonzero. Let k denote the largest integer such that some
10% n*n
k*k matrix has a nonzero determonant. Prove that rank(A) = k.
┌ ┐
(10) Let A = │-a_n 0 ... 0 a_1 │
15% │ 0 -a_n ... 0 a_2 │
│ . . 0 . │
│ : : : : │
│ 0 0 .. -a_n a_n-1 │
└ ┘(n-1)*n
┌ ┐
B = │b_11 b_12 ... .. b_1,n-1│
│b_21 b_22 ... .. b_2,n-2│
│ . . . │
│ : : : │
│b_n1 n_n2 ... .. b_n,n-1│
└ ┘n*(n-1)
┌ ┐
and C = │a_1 b_11 b_12 ... .. b_1,n-1│
│a_2 b_21 b_22 ... .. b_2,n-2│
│ . . . . │
│ : : : : │
│a_n b_n1 n_n2 ... .. b_n,n-1│
└ ┘n*n
n-2
Show that det(AB) = a_n det(C).
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※ 編輯: starsnight (119.14.43.11), 01/17/2015 22:36:05