与矩阵 \[ \left( {\begin{array}{*20{c}} 1&0&0\\ 0&1&0\\ 0&0&2 \end{array}} \right) 相似的矩阵是(\ \ ) \]
A. \[ \left( {\begin{array}{*20{c}}1&1&0\\0&1&0\\0&0&2\end{array}} \right);\]
B. \[ \left( {\begin{array}{*20{c}}1&0&0\\0&1&1\\0&0&2\end{array}} \right);\]
C. \[\left( {\begin{array}{*20{c}}1&1&1\\0&1&0\\0&0&2\end{array}} \right);\]
D. \[\left( {\begin{array}{*20{c}}1&0&0\\1&1&0\\1&-1&2\end{array}} \right).\]
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设\[A = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}}&{{a_{14}}}\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}}&{{a_{24}}}\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}}&{{a_{34}}}\\ {{a_{41}}}&{{a_{42}}}&{{a_{43}}}&{{a_{44}}} \end{array}} \right),B = \left( {\begin{array}{*{20}{c}} {{a_{14}}}&{{a_{13}}}&{{a_{12}}}&{{a_{11}}}\\ {{a_{24}}}&{{a_{23}}}&{{a_{22}}}&{{a_{21}}}\\ {{a_{34}}}&{{a_{33}}}&{{a_{32}}}&{{a_{31}}}\\ {{a_{44}}}&{{a_{43}}}&{{a_{42}}}&{{a_{41}}} \end{array}} \right)\],\[{P_1} = \left( {\begin{array}{*{20}{c}} 0&0&0&1\\ 0&1&0&0\\ 0&0&1&0\\ 1&0&0&0 \end{array}} \right),{P_1} = \left( {\begin{array}{*{20}{c}} 1&0&0&0\\ 0&0&1&0\\ 0&1&0&0\\ 0&0&0&1 \end{array}} \right)\],若`\A`可逆,则`\B^{-1}=` ( )
A. \[{A^{ - 1}}{P_1}{P_2}\]
B. \[{P_2}{A^{ - 1}}{P_1}\]
C. \[{P_1}{P_2}{A^{ - 1}}\]
D. \[{P_1}{A^{ - 1}}{P_2}\]
设`\n`阶方阵`\A`满足`\|A|=2`,`\A^**`是`\A`的伴随阵, 则`\|A^{-1}(A^**+A^{-1})A|=` ( )
A. \[\frac{{\rm{3}}}{{\rm{2}}}\]
B. \[\frac{{\rm{3}}}{{{{\rm{2}}^n}}}\]
C. \[\frac{{{{\rm{3}}^{\rm{n}}}}}{{{{\rm{2}}^n}}}\]
D. \[\frac{{{{\rm{3}}^{\rm{n}}}}}{{\rm{2}}}\]
设`\A`为`\n`阶方阵,且`\A`的行列式`\| A | = a \ne 0`,而`\A^**`是`\A`的伴随矩阵,则`\|2A^**|`等于 ( )
A. `\2a`
B. `\2(2a)^(n-1)`
C. `\(2a)^{n-1}`
D. `\2^na`
已知`\A,B`均是三阶矩阵,将`\A`的第三行的-2倍加到第二行得矩阵`\A_1`,将`\B`中第一列和第二列对换得到`\B_1`,又\[{A_1}{B_1} = \left( {\begin{array}{*{20}{c}} 1&1&1\\ 1&0&2\\ 2&1&3 \end{array}} \right)\],则`\AB=` ( )
A. \[\left( {\begin{array}{*{20}{c}}2&1&1\\2&5&8\\2&2&3\end{array}} \right)\]
B. \[\left( {\begin{array}{*{20}{c}}1&1&1\\2&5&8\\3&2&1\end{array}} \right)\]
C. \[\left( {\begin{array}{*{20}{c}}1&1&1\\2&5&8\\1&2&3\end{array}} \right)\]
D. \[\left( {\begin{array}{*{20}{c}}1&1&1\\2&5&1\\3&2&1\end{array}} \right)\]