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
If \[\begin{vmatrix} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{vmatrix} = 0\] and vectors \((1, a, a^2)\), \((1, b, b^2)\) and \((1, c, c^2)\) are non-coplanar, then the product \(abc\) equals:
17. Let \(M_n = (a_{ij})\) where \(i, j = 1, 2, 3, \ldots, n\). We first find out \(a_{11}\) for the \(n^{\text{th}}\) matrix, which is the \(n^{\text{th}}\) term in the series: \(1, 2, 6, 15, \ldots\). The diagonal elements of the \(n^{\text{th}}\) matrix form an arithmetic progression with first term \(1 + \dfrac{n(n-1)(2n-1)}{6}\) and common difference \(n+1\). Find the required sum \(M_n\).