A) \[{{T}_{r}}(kA)=k{{T}_{r}}(A)\]\[(k\]is a scalar)
B) \[{{T}_{r}}(A+B)={{T}_{r}}(A)+{{T}_{r}}(B)\]
C) \[{{T}_{r}}({{I}_{3}})=A\]
D) \[{{T}_{r}}({{A}^{2}})={{T}_{r}}{{(A)}^{2}}\]
Correct Answer: D
Solution :
[a]\[{{T}_{r}}(kA)=k({{a}_{11}}+{{a}_{22}}+{{a}_{33}})=k{{T}_{r}}(1)\] [b]\[{{T}_{r}}(A+B)={{a}_{11}}+{{b}_{11}}+{{a}_{22}}+{{b}_{22}}+{{a}_{33}}+{{b}_{33}}\]\[={{T}_{r}}(1)+{{T}_{r}}(2)\] [c]\[{{T}_{r}}({{I}_{3}})=1+1+1=3\] [d]\[{{T}_{r}}({{A}^{2}})=\Sigma a_{11}^{2}+\Sigma a_{12}^{2}\ne {{({{a}_{11}}+{{a}_{22}}+{{a}_{33}})}^{2}}\]
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