A) \[{{\text{T}}_{\text{r}}}\text{(kA)=k}{{\text{T}}_{\text{r}}}\text{(A)}\](k is a scalar)
B) \[{{\text{T}}_{\text{r}}}\text{(A+B)}\,\text{=}\,{{\text{T}}_{\text{r}}}\text{(A)+}{{\text{T}}_{\text{r}}}(\text{B)}\]
C) \[{{\text{T}}_{\text{r}}}\text{(}{{\text{I}}_{\text{3}}}\text{)}\,\text{=3}\]
D) \[{{\text{T}}_{\text{r}}}\text{(}{{\text{A}}^{\text{2}}}\text{)}\,\text{=}{{\text{T}}_{\text{r}}}{{\text{(A)}}^{\text{2}}}\]
Correct Answer: D
Solution :
[a] \[{{T}_{r}}(kA)=k({{a}_{11}}+{{a}_{22}}+{{a}_{33}})=k{{T}_{r}}(A)\] [b] \[{{T}_{r}}(A+B)={{a}_{11}}+{{b}_{11}}+{{a}_{22}}+{{b}_{22}}+{{a}_{33}}+{{b}_{33}}\]\[={{T}_{r}}(A)+{{T}_{r}}(B)\] [c] \[{{T}_{r}}({{I}_{3}})=1+1+1=3\] [d] \[{{T}_{r}}({{A}^{2}})=\sum a_{11}^{2}+\sum a_{12}^{2}\ne {{({{a}_{11}}+{{a}_{22}}+{{a}_{33}})}^{2}}\]You need to login to perform this action.
You will be redirected in
3 sec