Exams Joint Entrance Examination Mathematics Matrices and Determinants Question 383 Question 383 of 472 Medium MCQ
Let \[A + 2B = \left[ { ( 1 2 0 6 − 3 3 − 5 3 1 ) \begin{pmatrix} 1 & 2 & 0 \\
6 & { - 3} & 3 \\
{ - 5} & 3 & 1 \\ \end{pmatrix} 1 6 − 5 2 − 3 3 0 3 1 } \right]\] and \[2A - B = \left[ { ( 2 − 1 5 2 − 1 6 0 1 2 ) \begin{pmatrix} 2 & { - 1} & 5 \\
2 & { - 1} & 6 \\
0 & 1 & 2 \\ \end{pmatrix} 2 2 0 − 1 − 1 1 5 6 2 } \right]\]. 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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\[A = {1 \over 5}((A + 2B) + 2(2A - B))\]
\[ = {1 \over 5}\left( {\left[ { ( 1 2 0 6 − 3 3 − 5 3 1 ) \begin{pmatrix} 1 & 2 & 0 \\
6 & { - 3} & 3 \\
{ - 5} & 3 & 1 \\
\end{pmatrix} 1 6 − 5 2 − 3 3 0 3 1 } \right] + \left[ { ( 4 − 2 10 4 − 2 12 0 2 4 ) \begin{pmatrix} 4 & { - 2} & {10} \\
4 & { - 2} & {12} \\
0 & 2 & 4 \\
\end{pmatrix} 4 4 0 − 2 − 2 2 10 12 4 } \right]} \right)\]
\[ = {1 \over 5}\left[ { ( 5 0 10 10 − 5 15 − 5 5 5 ) \begin{pmatrix} 5 & 0 & {10} \\
{10} & { - 5} & {15} \\
{ - 5} & 5 & 5 \\
\end{pmatrix} 5 10 − 5 0 − 5 5 10 15 5 } \right] \Rightarrow tr(A) = 1\]
Similarly,
\[B = {1 \over 5}(2(A + 2B) - (2A - B))\]
\[ = {1 \over 5}\left( {\left[ { ( 2 4 0 12 − 6 6 − 10 6 2 ) \begin{pmatrix} 2 & 4 & 0 \\
{12} & { - 6} & 6 \\
{ - 10} & 6 & 2 \\
\end{pmatrix} 2 12 − 10 4 − 6 6 0 6 2 } \right] - \left[ { ( 2 − 1 5 2 − 1 6 0 1 2 ) \begin{pmatrix} 2 & { - 1} & 5 \\
2 & { - 1} & 6 \\
0 & 1 & 2 \\
\end{pmatrix} 2 2 0 − 1 − 1 1 5 6 2 } \right]} \right)\]
\[ = {1 \over 5}\left[ { ( 0 6 − 5 10 − 5 0 − 10 5 0 ) \begin{pmatrix} 0 & 6 & { - 5} \\
{10} & { - 5} & 0 \\
{ - 10} & 5 & 0 \\
\end{pmatrix} 0 10 − 10 6 − 5 5 − 5 0 0 } \right] \Rightarrow tr(B) = - 1\]
T r ( A ) − T r ( B ) = 1 − ( − 1 ) = 2 Tr(A) - Tr(B) = 1 - ( - 1) = 2 T r ( A ) − T r ( B ) = 1 − ( − 1 ) = 2 Answer the question to unlock