Question

If $A=\text{diag}(1-2-3)$ and $B=\text{diag}(-314)$, such that $6A+2B=\text{diag}\left(abc\right)$ then the value of $(a+b+c)$ is

• A. $20$
• B. $0$
• C. $-20$
• D. $-10$

Answer 1

The correct option is C $-20$

$6A+2B=6\text{diag}(1-2-3)+2\text{diag}(-314)=\text{diag}(6-12-18)+\text{diag}(-628)=\text{diag}(0-10-10)\Rightarrow a+b+c=-10-10=-20$