WBJEE 2022 - Question 3 - Find the value of a+d.

Question

If $\mathrm{A}=\left(\begin{array}{ll}1 & 1 \\ 0 & i\end{array}\right)$ and $\mathrm{A}^{2018}=\left(\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{d}\end{array}\right)$, then $(\mathrm{a}+\mathrm{d})$ equals

$$\begin{array}{llll}\text { (A) } 1+\mathrm{i} & \text { (B) } 0 & \text { (C) } 2 & \text { (D) } 2018\end{array}$$

Answer 1

In order to solve this question quickly we need two recall two concepts. 

1. If $\lambda$ is an eigenvalue of $A$ then $\lambda^k$ is an eigenvalue of $A^k$. 

2. The sum of all eigenvalue of $A$ is equals to the trace of $A$. 

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Let $A$ be the given $2\times 2$ matrix. Let $\lambda_1$ and $\lambda_2$ be two eigenvalues of $A$. Note that the given matrix is a upper triangular matrix. So its eigenvalues are nothing but its diagonal entries. Therefore, 

$$\lambda_1 =1 \qquad\text{and}\qquad \lambda_2=i$$

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Therefore, the eigenvalues of $A^k$ are 

$$\lambda_1^k =1 \qquad\text{and}\qquad \lambda_2^k=i^k$$

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Note that $trace(A^{2018})=a+d$, that is sum of the eigenvalue of $A^{2018}$. It follows 

$$a+d = trace(A^{2018}) = \lambda_1^{2018}+\lambda_2^{2018} =1+i^{2018}=1-1=0.$$

Therefore, the correct option is (B).