Cho (A = left( {begin{array}{*{20}{c}}

0&2&2\

2&3&{ – 1}\

2&{ – 1}&3

end{array}} right)) Tìm ma trận trực giao P sao cho P$^{t}$ AP có dạng chéo:

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),mathop Pnolimits^{ – 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),mathop Pnolimits^{ – 1} AP = left( {begin{array}{*{20}{c}}

{ – 2}&0&0\

0&4&0\

0&0&4

end{array}} right))

C. (A = left( {begin{array}{*{20}{c}}

{frac{1}{{sqrt 3 }}}&{frac{1}{{sqrt 2 }}}&{frac{1}{{sqrt 6 }}}\

{frac{1}{{sqrt 3 }}}&{frac{{ – 1}}{{sqrt 2 }}}&{frac{1}{{sqrt 6 }}}\

{frac{1}{{sqrt 3 }}}&0&{frac{{ – 2}}{{sqrt 6 }}}

end{array}} right),mathop Pnolimits^{ – 1} AP = left( {begin{array}{*{20}{c}}

0&0&0\

0&3&0\

0&0&3

end{array}} right))

D. (A = left( {begin{array}{*{20}{c}}

{frac{1}{{sqrt 6 }}}&{frac{1}{{sqrt 2 }}}&{frac{1}{{sqrt 3 }}}\

{frac{1}{{sqrt 6 }}}&{frac{{ – 1}}{{sqrt 2 }}}&{frac{1}{{sqrt 3 }}}\

{frac{{ – 2}}{{sqrt 6 }}}&0&{frac{1}{{sqrt 3 }}}

end{array}} right),mathop Pnolimits^{ – 1} AP = left( {begin{array}{*{20}{c}}

0&0&0\

0&6&0\

0&0&6

end{array}} right))

Hướng dẫn

Chọn B là đáp án đúng