Question

If the traces of A and B are 20 and -8, then the traces of A + B are
a. 12
b. -12
c. 28
d. -28

Answer 1

Hint: In order to solve this question, we should know that traces of a matrix is nothing but the algebraic sum of the diagonal elements of any square matrix. Also, we need to know that the sum of matrices like, $A=\left[ {{a}_{ij}} \right]$ and $B=\left[ {{b}_{ij}} \right]$ is given by $A+B=\left[ {{a}_{ij}}+{{b}_{ij}} \right]$. By using these concepts we can solve the question.

Complete step-by-step answer:
In this question, we have been given that the traces of A and B are 20 and -8 and we are asked to find the traces of A + B. For that, we will first consider two matrices of the order $n\times n$ as,
\[A=\left[ \begin{matrix}
   {{x}_{11}} & {{x}_{12}} & .. & {{x}_{1n}} \\
   {{x}_{21}} & {{x}_{22}} & .. & {{x}_{2n}} \\
   : & : & ... & : \\
   {{x}_{n1}} & {{x}_{n2}} & .. & {{x}_{nn}} \\
\end{matrix} \right]\]
\[B=\left[ \begin{matrix}
   {{y}_{11}} & {{y}_{12}} & .. & {{y}_{1n}} \\
   {{y}_{21}} & {{y}_{22}} & .. & {{y}_{2n}} \\
   : & : & ... & : \\
   {{y}_{n1}} & {{y}_{n2}} & .. & {{y}_{nn}} \\
\end{matrix} \right]\]
Now, we know that trace of a matrix is the sum of the diagonal elements of that matrix. So, we can say that trace of $A={{x}_{11}}+{{x}_{22}}+{{x}_{33}}+......+{{x}_{nn}}=20$ and the trace of $B={{y}_{11}}+{{y}_{22}}+{{y}_{33}}+......+{{y}_{nn}}=-8$.
Now, we know that, when two matrices are added then the elements of a matrix are added with the respective elements of the other matrix. So, we can say,
\[A+B=\left[ \begin{matrix}
   {{x}_{11}} & {{x}_{12}} & .. & {{x}_{1n}} \\
   {{x}_{21}} & {{x}_{22}} & .. & {{x}_{2n}} \\
   : & : & ... & : \\
   {{x}_{n1}} & {{x}_{n2}} & .. & {{x}_{nn}} \\
\end{matrix} \right]+\left[ \begin{matrix}
   {{y}_{11}} & {{y}_{12}} & .. & {{y}_{1n}} \\
   {{y}_{21}} & {{y}_{22}} & .. & {{y}_{2n}} \\
   : & : & ... & : \\
   {{y}_{n1}} & {{y}_{n2}} & .. & {{y}_{nn}} \\
\end{matrix} \right]\]
\[\Rightarrow A+B=\left[ \begin{matrix}
   {{x}_{11}}+{{y}_{11}} & {{x}_{12}}+{{y}_{12}} & .. & {{x}_{1n}}+{{y}_{1n}} \\
   {{x}_{21}}+{{y}_{21}} & {{x}_{22}}+{{y}_{22}} & .. & {{x}_{2n}}+{{y}_{2n}} \\
   : & : & ... & : \\
   {{x}_{n1}}+{{y}_{n1}} & {{x}_{n2}}+{{y}_{n2}} & .. & {{x}_{nn}}+{{y}_{nn}} \\
\end{matrix} \right]\]
Now, the trace of A + B can be written as follows,
Trace of (A + B) = ${{x}_{11}}+{{y}_{11}}+{{x}_{22}}+{{y}_{22}}+{{x}_{33}}+{{y}_{33}}+......+{{x}_{nn}}+{{y}_{nn}}$
We can also write it as,
Trace of (A + B) = $\left( {{x}_{11}}+{{x}_{22}}+{{x}_{33}}+......+{{x}_{nn}} \right)+\left( {{y}_{11}}+{{y}_{22}}+{{y}_{33}}+......+{{y}_{nn}} \right)$
And we know that, $\left( {{x}_{11}}+{{x}_{22}}+{{x}_{33}}+......+{{x}_{nn}} \right)$ = trace of A and $\left( {{y}_{11}}+{{y}_{22}}+{{y}_{33}}+......+{{y}_{nn}} \right)$ = trace of B. So, we can say that,
Trace of (A + B) = trace of A + trace of B
We will substitute the values of trace of A and trace of B in the above equation, so we get,
Trace of (A + B) = 20 + (-8)
Trace of (A + B) = 20 – 8 = 12
Hence, we can say that if the traces of A and B are 20 and -8, then the trace of (A + B) is 12.
Therefore, option (a) is the correct answer.


Note:While solving this question, we can directly use the formula, trace of (A + B) = trace of A + trace of B. But we may forget this formula, so it is better to follow the conventional method to solve this question. Also, a possible mistake that the students can make while solving this question is that, they might write trace of A + trace of B as 20 + 8 = 28, because they tend to miss out the negative sign of (-8) so the students may end up writing option (c) as the answer, but it is wrong.