# Tr(A) – Tr(B) has value equal to:

Let

$$\begin{array}{l}A+2B=\begin{bmatrix} 1&2 &0 \\ 6&-3 &3 \\ -5&3 &1 \end{bmatrix}\end{array}$$
and
$$\begin{array}{l}2A-B=\begin{bmatrix} 2&-1 &5 \\ 2&-1 &6 \\ 0&1 &2 \end{bmatrix}\end{array}$$
. If Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A) – Tr(B) has value equal to:

a. 0

b. 1

c. 3

d. 2

Solution:

tr (A + 2B) ≡ tr (A) + 2 tr (B) = –1 …..(1)

and tr (2A – B) ≡ 2tr (A) – tr (B) = 3 …..(2)

On solving (1) and (2) we get

tr (A) = 1, tr(B) = –1

tr (A) – tr(B) = 1 + 1 = 2