Cho ánh xạ tuyến tính \[f = {R^3} \to {R^2}\] có ma trận chính tắc \[\left( {\begin{array}{*{20}{c}}\begin{array}{l}4\\6\end{array}&\begin{array}{l}1\\2\end{array}&\begin{array}{l}2\\3\end{array}\end{array}} \right)\] Vectơ nào sau đây thuộc Ker f:

A. A=B=0

B. A tùy ý, B=0

C. B tùy ý, A=0

D. A B, tùy ý

A. \[\left\{ {\left( {2,1, - 1} \right),\left( {3,2, - 5} \right),\left( {1, - 1,1} \right)} \right\}\]

B. \[\left\{ {\left( {1,0, - 1} \right),\left( {1,1,1} \right),\left( { - 1,2,2} \right),\left( {1,0,3} \right)} \right\}\]

C. \[\left\{ {\left( {1,0, - 1} \right),\left( {1,1,1} \right)} \right\}\]

D. \[\left\{ {\left( {2,1, - 1} \right),\left( {3,2, - 5} \right),\left( {1, - 1,10} \right)} \right\}\]

A. \[f\left( {x,y,z} \right) = \left( {x + z,y} \right)\]

B. \[f\left( {x,y,z} \right) = \left( {2x + 3y + 4z,0} \right)\]

C. \[f\left( {x,y,z} \right) = \left( {x + 2y + z} \right)\]

D. \[f\left( {x,y,z} \right) = \left( {xy,yz} \right)\]

A. \[A = \left( {\begin{array}{*{20}{c}}\begin{array}{l}1\\3\\4\end{array}&\begin{array}{l}1\\0\\ - 7\end{array}\end{array}} \right)\]

B. \[A = \left( {\begin{array}{*{20}{c}}\begin{array}{l}1\\1\end{array}&\begin{array}{l}3\\0\end{array}&\begin{array}{l}4\\ - 7\end{array}\end{array}} \right)\]

C. \[A = \left( {\begin{array}{*{20}{c}}\begin{array}{l}2\\ - 8\end{array}&\begin{array}{l} - 1\\4\end{array}\end{array}} \right)\]

D. \[A = \left( {\begin{array}{*{20}{c}}\begin{array}{l}2\\ - 8\end{array}&\begin{array}{l} - 1\\4\end{array}&\begin{array}{l}4\\ - 7\end{array}\end{array}} \right)\]

A. (2,1, 3)

B. (0,1, 0)

C. (1,1, 0)

D. (0,1,-1)

A. m≠−1

B. m≠1

C. m≠2

D. m≠−2