Cho dạng toàn phương Q: R³ -> R  xác định bởi \[(x,y) = 2{x^2} - 6xy + {y^2}\].Tìm ma trận của Q trong cơ sở \[\left\{ {v1 = \left( {1,0} \right),v2 = \left( {1,1} \right)} \right\}\]

A. p = 1, q = 2

B. p = 2, q = 1

C. p = 1, q = 1

D. p = 0, q = 2

A. \[A = \left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 3 }}}&0&{\frac{2}{{\sqrt 6 }}}\\{\frac{1}{{\sqrt 3 }}}&{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 6 }}}\\2&{ - 1}&3\end{array}} \right),{P^{ - 1}}AP = \left( {\begin{array}{*{20}{c}}3&0&0\\0&5&0\\0&0&9\end{array}} \right)\]

B. \[A = \left( {\begin{array}{*{20}{c}}{\frac{{ - 2}}{{\sqrt 6 }}}&0&{\frac{1}{{\sqrt 3 }}}\\{\frac{1}{{\sqrt 6 }}}&{\frac{1}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 3 }}}\\{\frac{{ - 1}}{{\sqrt 6 }}}&{\frac{{ - 1}}{{\sqrt 2 }}}&{\frac{1}{{\sqrt 3 }}}\end{array}} \right),{P^{ - 1}}AP = \left( {\begin{array}{*{20}{c}}{ - 2}&0&0\\0&4&0\\0&0&4\end{array}} \right)\]

A. \[I = \frac{2}{3}x\sqrt x + 2\sqrt x + C\]

B. \[I = \frac{1}{3}x\sqrt x + 2\sqrt x + C\]

C. \[I = \frac{2}{3}x\sqrt x + 3\sqrt x + C\]

D. \[I = \frac{2}{3}{x^2}\sqrt x + 3\sqrt x + C\]

A. p=1,q=3p=1,q=3

B. p=3,q=1p=3,q=1

C. p=2,q=2p=2,q=2

D. p=1,q=2p=1,q=2

A. \[A = \left( {\begin{array}{*{20}{c}}3&2&{ - 4}\\2&2&2\\{ - 4}&2&{ - 1}\end{array}} \right)\]

B. \[A = \left( {\begin{array}{*{20}{c}}3&2&{ - 4}\\0&2&2\\0&0&{ - 1}\end{array}} \right)\]

C. \[A = \left( {\begin{array}{*{20}{c}}3&2&{ - 2}\\1&2&1\\{ - 2}&1&{ - 1}\end{array}} \right)\]

D. \[A = \left( {\begin{array}{*{20}{c}}{ - 3}&2&{ - 4}\\2&{ - 2}&2\\{ - 4}&2&1\end{array}} \right)\]

A. \[I = 60\frac{2}{7}\]

B. \[I = 66\frac{2}{7}\]

C. \[I = - 60\frac{2}{7}\]

D. \[I = - 66\frac{2}{7}\]