u、v为点对应三角形的重心坐标，以下哪种情况为点在三角形内部 A: 0 <= u <= 1 且 0 <= v <= 1 B: 1 <= u 且 1 <= v C: 0 = u 且 0 = v D: 0 >= u 且 0 >= v

求下面矩阵的 Cholesky 分解 （다음 행렬의 Cholesky factorization을 구하시오）. \begin{bmatrix}<br/>1\ \,\, 3\ \,\, 7\\ <br/>3\ 10\ 26\\ <br/>7\ 26\ 75\\<br/>\end{bmatrix} A: \(U=\begin{bmatrix}<br/>1\ 3\ 7\\ <br/>0\ 1\ 5\\ <br/>0\ 0\ 1\\<br/>\end{bmatrix}\) B: \(U=\begin{bmatrix}<br/>1\ 2\ 7\\ <br/>0\ 3\ 5\\ <br/>0\ 0\ 1\\<br/>\end{bmatrix}\) C: \(U=\begin{bmatrix}<br/>1\ 3\ 7\\ <br/>0\ 2\ 5\\ <br/>0\ 0\ 1\\<br/>\end{bmatrix}\) D: \(U=\begin{bmatrix}<br/>1\ 3\ 1\\ <br/>0\ 1\ 5\\ <br/>0\ 0\ 7\\<br/>\end{bmatrix}\) E: \(U=\begin{bmatrix}<br/>1\ 2\ 7\\ <br/>0\ 3\ 1\\ <br/>0\ 0\ 1\\<br/>\end{bmatrix}\)

【单选题】Which of the following matrices does not have the same determinant of matrix B: [1, 3, 0, 2; -2, -5, 7, 4; 3, 5, 2, 1; -1, 0, -9,-5] A. [1, 3, 0, 2; -2, -5, 7, 4; 0, 0, 0, 0; -1, 0, -9, -5] B. [1, 3, 0, 2; -2, -5, 7, 4; 1, 0, 9, 5; -1, 0, -9, -5] C. [1, 3, 0, 2; -2, -5, 7, 4; 3, 5, 2, 1; -3, -5, -2, -1] D. [1, 3, 0, 2; -2, -5, 7, 4; 0, 0, 0, 1; -1, 0, -9, -5]

对信码1000100100001000011000011进行HDB3编码，结果可能是（ ）。 A: -1 0 0 0 +1 0 0 -1 0 0 0 -V +1 0 0 0 +V -1 +1 -B 0 0 -V +1 -1 B: +1 0 0 0 -1 0 0 +1 0 0 0 +V -1 0 0 0 -V +1 -1 +B 0 0 +V -1 +1 C: +1 0 0 0 -1 0 0 +1 0 0 0 +1 -1 0 0 0 -1 +1 -1 +1 0 0 +1 -1 +1 D: -1 0 0 0 +1 0 0 -1 0 0 0 +V +1 0 0 0 +V -1 +1 +B 0 0 -V +1 -1

设G=<;V,E>;为无向图，u,v∈V，若u,v连通，则( ) A: d(u,v)>;0 B: d(u,v)=0 C: d(u,v)<;0 D: d(u,v)≥0

矩阵[left[ {egin{array}{*{20}{c}} {m{0}}&{m{0}}&{m{5}}&{m{2}}\ {m{0}}&{m{0}}&{m{2}}&{m{1}}\ {m{4}}&{m{2}}&{m{0}}&{m{0}}\ {m{1}}&{m{1}}&{m{0}}&{m{0}} end{array}} ight]]的逆矩阵为 ()

矩阵[left[ {egin{array}{*{20}{c}} { m{0}}&{ m{0}}&{ m{5}}&{ m{2}}\ { m{0}}&{ m{0}}&{ m{2}}&{ m{1}}\ { m{4}}&{ m{2}}&{ m{0}}&{ m{0}}\ { m{1}}&{ m{1}}&{ m{0}}&{ m{0}} end{array}} ight]]的逆矩阵为 （ ） </p></p>

$u$, $v\in E^3$, $u=(0,-1,1),v=(1,-1,-1)$. What is $||u \times v||^2$:<br/>______

$u$, $v\in E^3$, $u=(0,-1,1),v=(1,-1,-1)$. What is $\langle{u,v}\rangle$:<br/>______

逻辑函数的最小项表达式为() A: F=Σm（0、2、5、7） B: C: F=Σm（1、3、6） D: F=Σm（0、1、2、6、7）