Solveeit Logo
Login

Question

Question: The value of \( {T_r}(A + {B^T} + 3C) \) equals – (where \( {T_r}(A) \) denotes the trace of matr...

The value of Tr(A+BT+3C){T_r}(A + {B^T} + 3C) equals –
(where Tr(A){T_r}(A) denotes the trace of matrix AA )
A. 1717
B. 1818
C. 2020
D. 2121

Explanation

Solution

Hint : In this question we have to find the trace of a matrix, where the trace of a matrix is defined as the sum of the diagonal elements of the matrix. We consider the 3 matrices and we have to find the trace of each matrix and then we have to find the total trace of the 3 matrices.

Complete step-by-step answer :
let us consider the 3 matrices namely A, B and C
A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 2&3&2 \\\ 3&5&1 \end{array}} \right] , B = \left[ {\begin{array}{*{20}{c}} { - 1}&4&2 \\\ 2&1&5 \\\ 3&2&3 \end{array}} \right] and C = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 3&{ - 1}&2 \\\ 1&4&3 \end{array}} \right]
In this we have to find Tr(A+BT+3C){T_r}(A + {B^T} + 3C) , we have the property Tr=(A+B)=Tr(A)+Tr(B){T_r} = (A + B) = {T_r}(A) + {T_r}(B) so we have Tr(A+BT+3C)=Tr(A)+Tr(BT)+Tr(3C){T_r}(A + {B^T} + 3C) = {T_r}(A) + {T_r}({B^T}) + {T_r}(3C) . Now we will find the trace of each matrix.
The trace of matrix A, it is given by Tr(A){T_r}(A) the main diagonal elements of the matrix A = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 2&3&2 \\\ 3&5&1 \end{array}} \right] is 1, 3 and 1, so we have
{T_r}(A) = {T_r}\left( {\left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 2&3&2 \\\ 3&5&1 \end{array}} \right]} \right)
⇒Tr(A)=(1+3+1)\Rightarrow {T_r}(A) = (1 + 3 + 1)
⇒Tr(A)=5\Rightarrow {T_r}(A) = 5
Now we find the trace of the matrix B transpose.
Given B = \left[ {\begin{array}{*{20}{c}} { - 1}&4&2 \\\ 2&1&5 \\\ 3&2&3 \end{array}} \right] ,
We have to find the transpose of B. transpose of matrix B is interchanging rows into columns.
Therefore, {B^T} = \left[ {\begin{array}{*{20}{c}} { - 1}&2&3 \\\ 4&1&2 \\\ 2&5&3 \end{array}} \right] ,
the main diagonal elements of the matrix B transpose are -1, 1 and 3
{T_r}({B^T}) = {T_r}\left( {\left[ {\begin{array}{*{20}{c}} { - 1}&2&3 \\\ 4&1&2 \\\ 2&5&3 \end{array}} \right]} \right)
⇒Tr(BT)=(−1+1+3)\Rightarrow {T_r}({B^T}) = ( - 1 + 1 + 3)
⇒Tr(BT)=3\Rightarrow {T_r}({B^T}) = 3
Now we find the trace of the matrix 3C.
Given C = \left[ {\begin{array}{*{20}{c}} 1&2&3 \\\ 3&{ - 1}&2 \\\ 1&4&3 \end{array}} \right] ,
we have to find 3C. we have to multiply the matrix C by 3
Therefore, 3C = \left[ {\begin{array}{*{20}{c}} 3&6&9 \\\ 9&{ - 3}&6 \\\ 3&{12}&9 \end{array}} \right]
the main diagonal elements of the matrix B transpose are 3, -3 and 9
{T_r}(3C) = {T_r}\left( {\left[ {\begin{array}{*{20}{c}} 3&6&9 \\\ 9&{ - 3}&6 \\\ 3&{12}&9 \end{array}} \right]} \right)
⇒Tr(3C)=(3+(−3)+9)\Rightarrow {T_r}(3C) = (3 + ( - 3) + 9)
⇒Tr(3C)=(3−3+9)\Rightarrow {T_r}(3C) = (3 - 3 + 9)
⇒Tr(3C)=9\Rightarrow {T_r}(3C) = 9
We have found traces of each matrix and now have to find the total trace of 3 matrices.
Therefore, Tr(A+BT+3C)=Tr(A)+Tr(BT)+Tr(3C){T_r}(A + {B^T} + 3C) = {T_r}(A) + {T_r}({B^T}) + {T_r}(3C)
⇒Tr(A+BT+3C)=5+3+9\Rightarrow {T_r}(A + {B^T} + 3C) = 5 + 3 + 9
⇒Tr(A+BT+3C)=17\Rightarrow {T_r}(A + {B^T} + 3C) = 17
Therefore Tr(A+BT+3C){T_r}(A + {B^T} + 3C) equals 17.
So, the correct answer is “Option A”.

Note : We use the property on the trace of matrix Tr=(A+B)=Tr(A)+Tr(B){T_r} = (A + B) = {T_r}(A) + {T_r}(B) and it is standard property and we have to find trace of each matrix and we will apply the property to get the answer. The trace of the matrix is defined as the sum of the main diagonal elements of the matrix and thus we obtain the answer.