# Solve this following

Question:

Let $A+2 B=\left[\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right]$

and $2 \mathrm{~A}-\mathrm{B}=\left[\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right]$. If $\operatorname{Tr}(\mathrm{A})$ denotes the

sum of all diagonal elements of the matrix A, then $\operatorname{Tr}(\mathrm{A})-\operatorname{Tr}(\mathrm{B})$ has value equal to

1. 1

2. 2

3. 0

4. 3

Correct Option: 2,

Solution:

$A+2 B=\left(\begin{array}{ccc}1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1\end{array}\right)$  ......(1)

$2 \mathrm{~A}-\mathrm{B}=\left(\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right)$

$\Rightarrow 4 \mathrm{~A}-2 \mathrm{~B}=\left(\begin{array}{ccc}4 & -2 & 10 \\ 4 & -2 & 12 \\ 0 & 2 & 4\end{array}\right) \quad \ldots(2)$

$(1)+(2) \Rightarrow 5 \mathrm{~A}=\left(\begin{array}{ccc}5 & 0 & 10 \\ 10 & -5 & 15 \\ -5 & 5 & 5\end{array}\right)$

$A=\left(\begin{array}{ccc}1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1\end{array}\right)$ and $2 A=\left(\begin{array}{ccc}2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2\end{array}\right)$

$\therefore B=\left(\begin{array}{ccc}2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2\end{array}\right)-\left(\begin{array}{ccc}2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2\end{array}\right)$

$B=\left(\begin{array}{ccc}0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0\end{array}\right)$

$\operatorname{tr}(\mathrm{A})=1-1+1=1$

$\operatorname{tr}(\mathrm{B})=-1$

$\operatorname{tr}(\mathrm{A})=1$ and $\operatorname{tr}(\mathrm{B})=-1$

$\therefore \operatorname{tr}(\mathrm{A})-\operatorname{tr}(\mathrm{B})=2$