Answer
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Hint: In the above problem, we have to find the incorrect option from the options given. ${{t}_{r}}$ represents trace in the above options and trace of any matrix is the sum of the elements present in the main diagonal. Now, checking options one by one and seeing which option is not correct. In checking the first option, first of all, assume two square matrices A and B then add the two matrices and take the trace. Then take separately the trace of two matrices A and B and then add them and then see whether they are coming equal or not. For the remaining two options, multiplying a constant to a matrix is done by multiplying every element of the matrix with that constant. And transpose of a matrix is done by interchanging rows with columns.
Complete step-by-step solution:
Let us assume two matrices A and B as follows:
$\begin{align}
& A=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right) \\
& B=\left( \begin{matrix}
{{b}_{11}} & {{b}_{12}} \\
{{b}_{21}} & {{b}_{22}} \\
\end{matrix} \right) \\
\end{align}$
Now, checking the options given in the above problem whether their L.H.S is equal to or not equal to R.H.S we get,
Checking option (a),
${{t}_{r}}\left( A+B \right)={{t}_{r}}\left( A \right)+{{t}_{r}}\left( B \right)$
In the above equation, ${{t}_{r}}$ represents the trace of the matrix.
Now, solving the left hand side of the above equation we get,
${{t}_{r}}\left( A+B \right)$
Adding matrices A and B we get,
$A+B=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)+\left( \begin{matrix}
{{b}_{11}} & {{b}_{12}} \\
{{b}_{21}} & {{b}_{22}} \\
\end{matrix} \right)$
Two matrices are added by adding their respective elements like first row and first column matrix is added with the first row and first column of another matrix. Adding the above matrices we get,
$A+B=\left( \begin{matrix}
{{a}_{11}}+{{b}_{11}} & {{a}_{12}}+{{b}_{12}} \\
{{a}_{21}}+{{b}_{21}} & {{a}_{22}}+{{b}_{22}} \\
\end{matrix} \right)$
Now, trace of the above matrix is the addition of the elements present in the main diagonal.
${{t}_{r}}\left( A+B \right)={{a}_{11}}+{{b}_{11}}+{{a}_{22}}+{{b}_{22}}$
Now, solving the R.H.S of this option we get,
${{t}_{r}}\left( A \right)+{{t}_{r}}\left( B \right)$
Trace of A is equal to:
${{t}_{r}}\left( A \right)={{a}_{11}}+{{a}_{22}}$
Trace of B is equal to:
${{t}_{r}}\left( B \right)={{b}_{11}}+{{b}_{22}}$
Adding trace of A and trace of B we get,
${{t}_{r}}\left( A \right)+{{t}_{r}}\left( B \right)={{a}_{11}}+{{b}_{11}}+{{a}_{22}}+{{b}_{22}}$
As you can see that L.H.S is equal to R.H.S of this option so this option is not an incorrect option.
Checking option (b),
${{t}_{r}}\left( \alpha A \right)=\alpha {{t}_{r}}\left( A \right),\alpha \in R$
The following is the square matrix that we have assumed above,
$A=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)$
Now, multiplying $\alpha $ by matrix A we get,
When we multiply a constant to a matrix then every element of the matrix gets multiplied by that constant.
$\alpha A=\left( \begin{matrix}
\alpha {{a}_{11}} & \alpha {{a}_{12}} \\
\alpha {{a}_{21}} & \alpha {{a}_{22}} \\
\end{matrix} \right)$
Taking trace of the above matrix we get,
${{t}_{r}}\left( \alpha A \right)=\alpha {{a}_{11}}+\alpha {{a}_{22}}$
Taking $\alpha $ as common from the left hand side of the above equation we get,
${{t}_{r}}\left( \alpha A \right)=\alpha \left( {{a}_{11}}+{{a}_{22}} \right)$
Hence, we have reduced the L.H.S of this option to ${{t}_{r}}\left( \alpha A \right)=\alpha \left( {{a}_{11}}+{{a}_{22}} \right)$.
Solving R.H.S of this option which is given as:
$\alpha {{t}_{r}}\left( A \right)$
Now, trace of matrix A we have already solved in the previous option i.e.
${{t}_{r}}\left( A \right)={{a}_{11}}+{{a}_{22}}$
Multiplying $\alpha $ to the above trace we get,
$\alpha {{t}_{r}}\left( A \right)=\alpha \left( {{a}_{11}}+{{a}_{22}} \right)$
As you can see that again we are getting L.H.S as equal to R.H.S. Hence, this option is also not an incorrect option.
Checking option (c) we get,
${{t}_{r}}\left( {{A}^{T}} \right)={{t}_{r}}\left( A \right)$
To solve this option, we need to take the transpose of a square matrix A.
The following is the square matrix A that we have assumed above,
$A=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)$
The transpose of a matrix is the exchange of rows with columns or vice versa. Taking transpose of a square matrix A we get,
${{A}^{T}}=\left( \begin{matrix}
{{a}_{11}} & {{a}_{21}} \\
{{a}_{12}} & {{a}_{22}} \\
\end{matrix} \right)$
Taking trace of the above matrix we get,
${{t}_{r}}\left( {{A}^{T}} \right)={{a}_{11}}+{{a}_{22}}$
R.H.S of this option is given as ${{t}_{r}}\left( A \right)$ which we have already solved above and its value is:
${{t}_{r}}\left( A \right)={{a}_{11}}+{{a}_{22}}$
As you can see that, L.H.S is equal to R.H.S of this option. Hence, this option is also not an incorrect option.
From the above options, we have found that none of the three options are incorrect options. Hence, the correct option is (d).
Note: The mistake that could happen in this problem is that you might think ${{t}_{r}}$ as the transpose of the matrix. The possibility of making this mistake is high because its initials are matching with the initials of “transpose” i.e. “tr” so make sure you won’t commit such mistakes in exams.
The other problem that you could face is that in multiplying a constant $\alpha $ to the matrix because when we multiply a constant to a determinant then either we multiply a row or a column by that constant but multiplying a constant to a matrix we multiply every element of the matrix by that constant so make sure you won’t mix both these multiplications of a constant by matrix and the determinant.
Complete step-by-step solution:
Let us assume two matrices A and B as follows:
$\begin{align}
& A=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right) \\
& B=\left( \begin{matrix}
{{b}_{11}} & {{b}_{12}} \\
{{b}_{21}} & {{b}_{22}} \\
\end{matrix} \right) \\
\end{align}$
Now, checking the options given in the above problem whether their L.H.S is equal to or not equal to R.H.S we get,
Checking option (a),
${{t}_{r}}\left( A+B \right)={{t}_{r}}\left( A \right)+{{t}_{r}}\left( B \right)$
In the above equation, ${{t}_{r}}$ represents the trace of the matrix.
Now, solving the left hand side of the above equation we get,
${{t}_{r}}\left( A+B \right)$
Adding matrices A and B we get,
$A+B=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)+\left( \begin{matrix}
{{b}_{11}} & {{b}_{12}} \\
{{b}_{21}} & {{b}_{22}} \\
\end{matrix} \right)$
Two matrices are added by adding their respective elements like first row and first column matrix is added with the first row and first column of another matrix. Adding the above matrices we get,
$A+B=\left( \begin{matrix}
{{a}_{11}}+{{b}_{11}} & {{a}_{12}}+{{b}_{12}} \\
{{a}_{21}}+{{b}_{21}} & {{a}_{22}}+{{b}_{22}} \\
\end{matrix} \right)$
Now, trace of the above matrix is the addition of the elements present in the main diagonal.
${{t}_{r}}\left( A+B \right)={{a}_{11}}+{{b}_{11}}+{{a}_{22}}+{{b}_{22}}$
Now, solving the R.H.S of this option we get,
${{t}_{r}}\left( A \right)+{{t}_{r}}\left( B \right)$
Trace of A is equal to:
${{t}_{r}}\left( A \right)={{a}_{11}}+{{a}_{22}}$
Trace of B is equal to:
${{t}_{r}}\left( B \right)={{b}_{11}}+{{b}_{22}}$
Adding trace of A and trace of B we get,
${{t}_{r}}\left( A \right)+{{t}_{r}}\left( B \right)={{a}_{11}}+{{b}_{11}}+{{a}_{22}}+{{b}_{22}}$
As you can see that L.H.S is equal to R.H.S of this option so this option is not an incorrect option.
Checking option (b),
${{t}_{r}}\left( \alpha A \right)=\alpha {{t}_{r}}\left( A \right),\alpha \in R$
The following is the square matrix that we have assumed above,
$A=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)$
Now, multiplying $\alpha $ by matrix A we get,
When we multiply a constant to a matrix then every element of the matrix gets multiplied by that constant.
$\alpha A=\left( \begin{matrix}
\alpha {{a}_{11}} & \alpha {{a}_{12}} \\
\alpha {{a}_{21}} & \alpha {{a}_{22}} \\
\end{matrix} \right)$
Taking trace of the above matrix we get,
${{t}_{r}}\left( \alpha A \right)=\alpha {{a}_{11}}+\alpha {{a}_{22}}$
Taking $\alpha $ as common from the left hand side of the above equation we get,
${{t}_{r}}\left( \alpha A \right)=\alpha \left( {{a}_{11}}+{{a}_{22}} \right)$
Hence, we have reduced the L.H.S of this option to ${{t}_{r}}\left( \alpha A \right)=\alpha \left( {{a}_{11}}+{{a}_{22}} \right)$.
Solving R.H.S of this option which is given as:
$\alpha {{t}_{r}}\left( A \right)$
Now, trace of matrix A we have already solved in the previous option i.e.
${{t}_{r}}\left( A \right)={{a}_{11}}+{{a}_{22}}$
Multiplying $\alpha $ to the above trace we get,
$\alpha {{t}_{r}}\left( A \right)=\alpha \left( {{a}_{11}}+{{a}_{22}} \right)$
As you can see that again we are getting L.H.S as equal to R.H.S. Hence, this option is also not an incorrect option.
Checking option (c) we get,
${{t}_{r}}\left( {{A}^{T}} \right)={{t}_{r}}\left( A \right)$
To solve this option, we need to take the transpose of a square matrix A.
The following is the square matrix A that we have assumed above,
$A=\left( \begin{matrix}
{{a}_{11}} & {{a}_{12}} \\
{{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)$
The transpose of a matrix is the exchange of rows with columns or vice versa. Taking transpose of a square matrix A we get,
${{A}^{T}}=\left( \begin{matrix}
{{a}_{11}} & {{a}_{21}} \\
{{a}_{12}} & {{a}_{22}} \\
\end{matrix} \right)$
Taking trace of the above matrix we get,
${{t}_{r}}\left( {{A}^{T}} \right)={{a}_{11}}+{{a}_{22}}$
R.H.S of this option is given as ${{t}_{r}}\left( A \right)$ which we have already solved above and its value is:
${{t}_{r}}\left( A \right)={{a}_{11}}+{{a}_{22}}$
As you can see that, L.H.S is equal to R.H.S of this option. Hence, this option is also not an incorrect option.
From the above options, we have found that none of the three options are incorrect options. Hence, the correct option is (d).
Note: The mistake that could happen in this problem is that you might think ${{t}_{r}}$ as the transpose of the matrix. The possibility of making this mistake is high because its initials are matching with the initials of “transpose” i.e. “tr” so make sure you won’t commit such mistakes in exams.
The other problem that you could face is that in multiplying a constant $\alpha $ to the matrix because when we multiply a constant to a determinant then either we multiply a row or a column by that constant but multiplying a constant to a matrix we multiply every element of the matrix by that constant so make sure you won’t mix both these multiplications of a constant by matrix and the determinant.
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