设\( A \)是3阶矩阵，若\( \left| {3A} \right| = 3 \)，则\( \left| {2A} \right| = \)（ ）
A: 1
B: 2
C: \( {2 \over 3} \)
D: \( {8 \over 9} \)
A: 1
B: 2
C: \( {2 \over 3} \)
D: \( {8 \over 9} \)
举一反三
- 设\( {\alpha _1} = {\left( {1,2, - a, - 3} \right)^T},{\alpha _2} = {\left( { - 3,2,4,1} \right)^T} \)且\( \left( { { \alpha _1},{\alpha _2}} \right) = - 1 \)，则\( a = \)（ ） A: \( - {2 \over 3} \) B: \( - {3 \over 4} \) C: \( - {1 \over 4} \) D: \( {1 \over 2} \)
- 设\( A \) 是\( 3 \) 阶矩阵，且\( \left| A \right| = 3 \) ，则 \( \left| { - 3A} \right| = \)（ ） A: -9 B: 9 C: -81 D: 81
- 设 \( A \)是 \( 3 \times 3 \)矩阵， \( B \)是 \( 4 \times 4 \)矩阵，且\( \left| A \right| = 1,\,\left| B \right| = - 2, \) 则\( \left| {\left| B \right|A} \right| = \) ______
- 求函数$y = \root 3 \of {x + \sqrt x } $的导数$y' = $( ) A: ${{1 + 2\sqrt x } \over {\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ B: $ {{1 + 2\sqrt x } \over {6\root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ C: $ {{1 + 2\sqrt x } \over {6\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$ D: $ {{1 + 2\sqrt x } \over {\sqrt x \cdot \root 3 \of {{{\left( {x + \sqrt x } \right)}^2}} }}$
- 设\(z = z\left( {x,y} \right)\)是由方程\({z^3}{\rm{ + }}3xyz - 3\sin xy = 1\)确定的隐函数，则\( { { \partial z} \over {\partial y}}=\)( ) A: \( { { y\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) B: \( { { y\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\) C: \( { { x\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) D: \( { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\)