Let $A=\begin{pmatrix}-\dfrac{1}{\sqrt{2}} & 1\\ 0 & 1\end{pmatrix}$ and $P=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix},\,\theta>0.$ If $B=PAP^{T},\,C=P^{T}B^{10}P$ and the sum of the diagonal elements of $C$ is $\dfrac{m}{n}$, where $\gcd(m,n)=1$, then $m+n$ is:
Let for any three distinct consecutive terms $a,b,c$ of an A.P., the lines $ax+by+c=0$ be concurrent at the point $P$ and $Q(\alpha,\beta)$ be a point such that the system of equations $x+y+z=6$, $2x+5y+\alpha z=\beta$ and $x+2y+3z=4$, has infinitely many solutions. Then $(PQ)^2$ is equal to
Let $A = \begin{bmatrix}\frac{1}{\sqrt{2}} & -2 \\ 0 & 1\end{bmatrix}$ and $P = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$, $\theta > 0$. If $B = PAP^T$, $C = P^TB^{10}P$ and the sum of the diagonal elements of $C$ is $\frac{m}{n}$, where $\gcd(m, n) = 1$, then $m + n$ is:
Let $A=\begin{bmatrix}2&0&1\\1&1&0\\1&0&1\end{bmatrix}$, $B=[B_1,B_2,B_3]$, where $B_1,B_2,B_3$ are column matrices, and $AB_1=\begin{bmatrix}1\\0\\0\end{bmatrix}$, $AB_2=\begin{bmatrix}2\\3\\0\end{bmatrix}$, $AB_3=\begin{bmatrix}3\\2\\1\end{bmatrix}$. If $\alpha=|B|$ and $\beta$ is the sum of all the diagonal elements of $B$, then $\alpha^3+\beta^3$ is equal to
Let \(\{D_1, D_2, D_3, \ldots, D_n\}\) be the set of third-order determinants that can be made with the distinct non-zero real numbers \(a_1, a_2, \ldots, a_9\). Then