Question
If \[A = \left[ {\begin{array}{*{20}{c}}
0&c&{ - b} \\
{ - c}&0&a \\
b&{ - a}&0
\end{array}} \right]\] and \[B = \left[ {\begin{array}{*{20}{c}}
{{a^2}}&{ab}&{ac} \\
{ab}&{{b^2}}&{bc} \\
{ac}&{bc}&{{c^2}}
\end{array}} \right]\] then $$AB$$ is equal to
A.
$$0$$
B.
$$I$$
C.
$$2I$$
D.
None of these
Answer :
$$0$$
Solution :
\[\begin{array}{l}
A = \left[ {\begin{array}{*{20}{c}}
0&c&{ - b}\\
{ - c}&0&a\\
b&{ - a}&0
\end{array}} \right],\,B = \left[ {\begin{array}{*{20}{c}}
{{a^2}}&{ab}&{ac}\\
{ab}&{{b^2}}&{bc}\\
{ac}&{bc}&{{c^2}}
\end{array}} \right]\\
AB = \left[ {\begin{array}{*{20}{c}}
0&c&{ - b}\\
{ - c}&0&a\\
b&{ - a}&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{a^2}}&{ab}&{ac}\\
{ab}&{{b^2}}&{bc}\\
{ac}&{bc}&{{c^2}}
\end{array}} \right]\\
AB = \left[ \begin{array}{l}
0 + abc - abc\,\,\,\,\,\,\,\,\,\,0 + {b^2}c - {b^2}c\,\,\,\,\,\,\,\,\,\,\,\,0 + b{c^2} - b{c^2}\\
- {a^2}c + 0 + {a^2}c\,\,\,\,\,\,\,\,\, - abc + 0 + abc\,\,\,\,\,\,\,\, - a{c^2} + 0 + a{c^2}\\
{a^2}b - {a^2}b + 0\,\,\,\,\,\,\,\,\,\,a{b^2} - a{b^2} + 0\,\,\,\,\,\,\,\,\,\,\,abc - abc + 0
\end{array} \right]\\
AB = \left[ \begin{array}{l}
0\,\,\,\,\,0\,\,\,\,\,0\\
0\,\,\,\,\,0\,\,\,\,\,0\\
0\,\,\,\,\,0\,\,\,\,\,0
\end{array} \right]\\
AB = {0_{3 \times 3}}......\left( 1 \right)\\
BA = \left[ {\begin{array}{*{20}{c}}
{{a^2}}&{ab}&{ac}\\
{ab}&{{b^2}}&{bc}\\
{ac}&{bc}&{{c^2}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
0&c&{ - b}\\
{ - c}&0&a\\
b&{ - a}&0
\end{array}} \right]\\
BA = \left[ \begin{array}{l}
0 + abc - abc\,\,\,\,\,\,\,\,\,\,0 + {b^2}c - {b^2}c\,\,\,\,\,\,\,\,\,\,\,\,0 + b{c^2} - b{c^2}\\
- {a^2}c + 0 + {a^2}c\,\,\,\,\,\,\,\,\, - abc + 0 + abc\,\,\,\,\,\,\,\, - a{c^2} + 0 + a{c^2}\\
{a^2}b - {a^2}b + 0\,\,\,\,\,\,\,\,\,\,a{b^2} - a{b^2} + 0\,\,\,\,\,\,\,\,\,\,\,abc - abc + 0
\end{array} \right]\\
BA = \left[ \begin{array}{l}
0\,\,\,\,\,0\,\,\,\,\,0\\
0\,\,\,\,\,0\,\,\,\,\,0\\
0\,\,\,\,\,0\,\,\,\,\,0
\end{array} \right]\\
BA = {0_{3 \times 3}}......\left( 2 \right)\\
{\rm{From\, equation\, }}\left( 1 \right){\rm{ \,and\, }}\left( 2 \right)\\
AB = BA = {0_{3 \times 3}}
\end{array}\]