If $\mathrm{A}=\left[{\mathrm{a}}_{\mathrm{ij}}\right]$ is a scalar matrix of order n×nsuch that ${\mathrm{a}}_{\mathrm{ii}}=\mathrm{k}$ for all i, then trace of A =

for a scaler matrix all the principle diagonal elements are same.

 Given that ${\mathrm{a}}_{\mathrm{ii}}=\mathrm{K}$

$$\begin{gathered} \therefore \mathrm{tr} \left(\mathrm{A}\right)={\mathrm{a}}_{11}+{\mathrm{a}}_{22}+....+{\mathrm{a}}_{\mathrm{nn}} \\ = \mathrm{K}+\mathrm{K}+....\mathrm{n} \mathrm{times} \\ =\mathrm{nK} \end{gathered}$$