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

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Detailed Solution

$A + 2 B = [\begin{array}{ccc} 1 & 2 & 0 \\ 6 & - 3 & 3 \\ 5 & 3 & 1 \end{array}]; 2 A - B = [\begin{array}{ccc} 2 & - 1 & 5 \\ 2 & - 1 & 6 \\ 0 & 1 & 2 \end{array}] 5 A = A + 2 B + 2 (2 A - B) = [\begin{array}{ccc} 1 & 2 & 0 \\ 6 & - 3 & 3 \\ 5 & 3 & 1 \end{array}] + [\begin{array}{ccc} 4 & - 2 & 10 \\ 4 & - 2 & 12 \\ 0 & 2 & 4 \end{array}] 5 A = [\begin{array}{ccc} 5 & 0 & 10 \\ 10 & - 5 & 15 \\ 5 & 5 & 5 \end{array}] \Rightarrow A = [\begin{array}{ccc} 1 & 0 & 2 \\ 2 & - 1 & 3 \\ 1 & 1 & 1 \end{array}] t r (A) = 1 5 B = 2 (A + 2 B) - (2 A - B) = [\begin{array}{ccc} 2 & 4 & 0 \\ 12 & - 6 & 6 \\ 10 & 6 & 2 \end{array}] - [\begin{array}{ccc} 2 & - 1 & 5 \\ 2 & - 1 & 6 \\ 0 & 1 & 2 \end{array}] 5 B = [\begin{array}{ccc} 0 & 5 & - 5 \\ 10 & - 5 & 0 \\ 10 & 5 & 0 \end{array}] \Rightarrow B = [\begin{array}{ccc} 0 & 1 & - 1 \\ 2 & - 1 & 0 \\ 2 & 1 & 0 \end{array}] t r (B) = - 1 t r (A) - t r (B) = 1 + 1 = 2$