Question ID: #485
Let $A=\begin{bmatrix}\frac{1}{\sqrt{2}}&-2\\ 0&1\end{bmatrix}$ and $P=\begin{bmatrix}\cos \theta&-\sin \theta\\ \sin \theta&\cos \theta\end{bmatrix}$.
If $B=PAP^{T}$, $C=P^{T}B^{10}P$ and the sum of the diagonal elements of $C$ is $\frac{m}{n}$ where $\gcd(m,n)=1$, then $m+n$ is equal to:
- (1) 65
- (2) 127
- (3) 258
- (4) 2049
Solution:
We are given $B = PAP^T$. Note that $P$ is an orthogonal matrix, so $P^T = P^{-1}$ and $P P^T = I$.
Calculate $B^{10}$:
$$B^{10} = (PAP^T)^{10} = (PAP^T)(PAP^T)…(PAP^T)$$
$$B^{10} = P A (P^T P) A (P^T P) … A P^T$$
$$B^{10} = P A^{10} P^T$$
Now, we are given $C = P^T B^{10} P$. Substitute $B^{10}$:
$$C = P^T (P A^{10} P^T) P$$
$$C = (P^T P) A^{10} (P^T P)$$
$$C = I A^{10} I = A^{10}$$
So, we just need to find the trace (sum of diagonal elements) of $A^{10}$.
$$A = \begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix}$$
Since $A$ is an upper triangular matrix, its eigenvalues are the diagonal entries: $\lambda_1 = \frac{1}{\sqrt{2}}$ and $\lambda_2 = 1$.
The eigenvalues of $A^{10}$ will be $\lambda_1^{10}$ and $\lambda_2^{10}$. The diagonal elements of an upper triangular matrix raised to a power $n$ are just the diagonal elements raised to power $n$.
Diagonal elements of $C = A^{10}$ are $(\frac{1}{\sqrt{2}})^{10}$ and $(1)^{10}$.
$$d_1 = \left( 2^{-1/2} \right)^{10} = 2^{-5} = \frac{1}{32}$$
$$d_2 = 1^{10} = 1$$
Sum of diagonal elements $= \frac{1}{32} + 1 = \frac{33}{32}$.
Given sum $= \frac{m}{n}$, so $\frac{m}{n} = \frac{33}{32}$.
$\gcd(33, 32) = 1$, which satisfies the condition.
We need $m+n$:
$$m+n = 33 + 32 = 65$$
Ans. (1)
Was this solution helpful?
YesNo