GATE Overflow Test Series | Mixed Subjects | Test 4 | Question: 21

Question

Consider the $2 \times 2$ matrix, $A = \begin{bmatrix}
3 & 1\\
-1 & 5
\end{bmatrix}.$ What is the trace of $A^{20} + $ trace of $A^{21}?$ (Mark all the appropriate choices)

• A. $5 \times 2^{41}$
• B. $2^{20} + 2^{21}$
• C. $2^{40} + 2^{42}$
• D. $2^{41} + 2^{43}$

Answer 1

Cayley-hamilton theorem:  Every square matrix satisfies its own characteristic equation.

Characteristic equation $\mid A -\lambda I \mid = 0$

$\implies \begin{bmatrix}
3 -\lambda & 1\\
-1 & 5 -\lambda
\end{bmatrix} = 0$

$\implies (3 - \lambda)(5 - \lambda) + 1 = 0$

$\implies \lambda = 4,4$

The eigenvalues of $A$ are $4, 4,$ so that the eigenvalues of $A^{20}$ are $4^{20},4^{20}$ and the trace of $A^{20}$ is $2\times 4^{20} = 2\times (2^{2})^{20} = 2^{1} \times 2^{40} = 2^{41}.$

The eigenvalues of $A$ are $4, 4,$ so that the eigenvalues of $A^{21}$ are $4^{21},4^{21}$ and the trace of $A^{21}$ is $2\times 4^{21} = 2\times (2^{2})^{21} = 2^{1} \times 2^{42} = 2^{43}.$

The trace of $A^{20} + $ trace of $A^{21} = 2^{41} + 2^{43} = 2^{41}(1 + 2^{2}) = 2^{41}(1 + 4) = 5 \times 2^{41}.$

$\textbf{PS:}$ A little few calculations give us $\text{tr}(A)=2^3, \text{tr}(A^2) = 2^5, \text{tr}(A^3)=2^7, \dots $ So, if we define the trace sequence $:a_{n} = 2^{2n + 1},$ then $a_{20} = 2^{41}$ and $a_{21} = 2^{43}.$

So, the correct answer is $A;D.$