JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 13)
Let $$A + 2B = \left[ {\matrix{
1 & 2 & 0 \cr
6 & { - 3} & 3 \cr
{ - 5} & 3 & 1 \cr
} } \right]$$ and $$2A - B = \left[ {\matrix{
2 & { - 1} & 5 \cr
2 & { - 1} & 6 \cr
0 & 1 & 2 \cr
} } \right]$$. If Tr(A) denotes the sum of all diagonal elements of the matrix A, then Tr(A) $$-$$ Tr(B) has value equal to
1
2
0
3
Explanation
$$A = {1 \over 5}((A + 2B) + 2(2A - B))$$
$$ = {1 \over 5}\left( {\left[ {\matrix{ 1 & 2 & 0 \cr 6 & { - 3} & 3 \cr { - 5} & 3 & 1 \cr } } \right] + \left[ {\matrix{ 4 & { - 2} & {10} \cr 4 & { - 2} & {12} \cr 0 & 2 & 4 \cr } } \right]} \right)$$
$$ = {1 \over 5}\left[ {\matrix{ 5 & 0 & {10} \cr {10} & { - 5} & {15} \cr { - 5} & 5 & 5 \cr } } \right] \Rightarrow tr(A) = 1$$
Similarly,
$$B = {1 \over 5}(2(A + 2B) - (2A - B))$$
$$ = {1 \over 5}\left( {\left[ {\matrix{ 2 & 4 & 0 \cr {12} & { - 6} & 6 \cr { - 10} & 6 & 2 \cr } } \right] - \left[ {\matrix{ 2 & { - 1} & 5 \cr 2 & { - 1} & 6 \cr 0 & 1 & 2 \cr } } \right]} \right)$$
$$ = {1 \over 5}\left[ {\matrix{ 0 & 6 & { - 5} \cr {10} & { - 5} & 0 \cr { - 10} & 5 & 0 \cr } } \right] \Rightarrow tr(B) = - 1$$
$$Tr(A) - Tr(B) = 1 - ( - 1) = 2$$
$$ = {1 \over 5}\left( {\left[ {\matrix{ 1 & 2 & 0 \cr 6 & { - 3} & 3 \cr { - 5} & 3 & 1 \cr } } \right] + \left[ {\matrix{ 4 & { - 2} & {10} \cr 4 & { - 2} & {12} \cr 0 & 2 & 4 \cr } } \right]} \right)$$
$$ = {1 \over 5}\left[ {\matrix{ 5 & 0 & {10} \cr {10} & { - 5} & {15} \cr { - 5} & 5 & 5 \cr } } \right] \Rightarrow tr(A) = 1$$
Similarly,
$$B = {1 \over 5}(2(A + 2B) - (2A - B))$$
$$ = {1 \over 5}\left( {\left[ {\matrix{ 2 & 4 & 0 \cr {12} & { - 6} & 6 \cr { - 10} & 6 & 2 \cr } } \right] - \left[ {\matrix{ 2 & { - 1} & 5 \cr 2 & { - 1} & 6 \cr 0 & 1 & 2 \cr } } \right]} \right)$$
$$ = {1 \over 5}\left[ {\matrix{ 0 & 6 & { - 5} \cr {10} & { - 5} & 0 \cr { - 10} & 5 & 0 \cr } } \right] \Rightarrow tr(B) = - 1$$
$$Tr(A) - Tr(B) = 1 - ( - 1) = 2$$
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