Tr(A) ndash 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:

Answer: (d)

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

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

On solving (1) and (2) we get

t_r (A) = 1, t_r(B) = –1

t_r (A) – t_r(B) = 1 + 1 = 2

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Tr(A) – Tr(B) has value equal to:

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