设矩阵\(A = \left( {\matrix{ \matrix{ x \cr 0 \cr y \cr} & \matrix{ 0 \cr 2 \cr 0 \cr} & \matrix{ y \cr 0 \cr - 2 \cr} \cr } } \right)\)的一个特征值为\(-3\)，且\(A\)的三个特征值之积为\(-12\)，则\(x =\)______

1 ． K 2 Cr 2 O 7 中 Cr 的氧化数是 ______ ， Cr 3 (SO 4 ) 2 中 Cr 的氧化数是 ______ 。

曲线$\left\{ \matrix{ {x^2} + {y^2} + {z^2} = 9 \cr y = x \cr} \right.$的参数方程为( ). A: $$\left\{ \matrix{ x = \sqrt 3 \cos t \cr y = \sqrt 3 \cos t \cr z = \sqrt 3 \sin t \cr} \right.(0 \le t \le 2\pi )$$ B: $$\left\{ \matrix{ x = {3 \over {\sqrt 2 }}\cos t\cr y = {3 \over {\sqrt 2 }}\cos t \cr z = 3\sin t \cr} \right.(0 \le t \le 2\pi )$$ C: $$\left\{ \matrix{ x = \cos t\cr y = \cos t\cr z = \sin t \cr} \right.(0 \le t \le 2\pi )$$ D: $$\left\{ \matrix{ x = {{\sqrt 3 } \over 3}\cos t\cr y = {{\sqrt 3 } \over 3}\cos t \cr z = {{\sqrt 3 } \over 3}\sin t\cr} \right.(0 \le t \le 2\pi )$$

【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]

向量组\(\left( {\matrix{ { - 1} \cr 3 \cr 1 \cr } } \right),\left( {\matrix{ 2 \cr 1 \cr 0 \cr } } \right),\left( {\matrix{ 1 \cr 4 \cr 1 \cr } } \right) \)线性相关.

下列矩阵中，不是初等矩阵的是（ ） A: \( \left( {\matrix{ 1 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right) \) B: \( \left( {\matrix{ 1 & 0 & 0 \cr 0 & { - 3} & 0 \cr 0 & 0 & 1 \cr } } \right) \) C: \( \left( {\matrix{ 1 & 3 & 0 \cr 0 & 0 & 1 \cr 0 & 1 & 0 \cr } } \right) \) D: \( \left( {\matrix{ 1 & 0 & 3 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right) \)

下图所示机构自由度计算,( )是正确的。 A: mg src="http://p.ananas.chaoxing.com/star3/origin/cb07ca0fb12be985c301490389c1e187.jpg" B: F=3×7 –(2×9 + 2 – 2)– 2 = 1 C: F=3×7 –(2×9+ 2– 0)– 0 = 1 D: F=3×7 –(2×8+ 2 – 0)– 2 = 1 E: F=3×5 –(2×6+ 2– 0)– 0 = 1

线性方程组\(\left\{ \matrix{ {x_1} + {x_2} + 2{x_3} - {x_4} = 0, \cr 2{x_1} + {x_2} + {x_3} - {x_4} = 0, \cr 2{x_1} + 2{x_2} + {x_3} + 2{x_4} = 0; \cr} \right.\)有无穷多解.

曲线\( \left\{ {\matrix{ { { x^2} + {y^2} = {z^2}} \cr { { z^2} = y} \cr } } \right. \)在坐标面\( yoz \) 上的投影曲线方程为（ ） A: \( \left\{ {\matrix{ { { x^2} + { { \left( {y - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \) B: \( \left\{ {\matrix{ { { z^2} = y} \cr {x = 0} \cr } } \right. \) C: \( \left\{ {\matrix{ {z = {y^2}} \cr {x = 0} \cr } } \right. \) D: \( \left\{ {\matrix{ { { y^2} + { { \left( {x - {1 \over 2}} \right)}^2} = {1 \over 4}} \cr {z = 0} \cr } } \right. \)

已知a=[1 2 3;5 6 7];b=[0 2 1;0 7 7];c=a==b，则c等于