Let omega be a complex cube root of unity wit | vedclass

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(B,C,D) The matrix $P$ is given by $p_{ij} = \omega^{i+j}$.We can write $P$ as the product of two column vectors: $P = uv^T$,where $u = [\omega^1, \om

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(B,C,D) The matrix $P$ is given by $p_{ij} = \omega^{i+j}$.We can write $P$ as the product of two column vectors: $P = uv^T$,where $u = [\omega^1, \omega^2, \dots, \omega^n]^T$ and $v = [\omega^1, \omega^2, \dots, \omega^n]^T$.Then $P^2 = (uv^T)(uv^T) = u(v^Tu)v^T$.Since $v^Tu = \sum_{k=1}^n \omega^{k+k} = \sum_{k=1}^n \omega^{2k}$,we have $P^2 = 0$ if and only if $v^Tu = 0$.The sum $S = \sum_{k=1}^n \omega^{2k}$ is a geometric series with first term $\omega^2$ and common ratio $\omega^2$.$S = \omega^2 \frac{1-(\omega^2)^n}{1-\omega^2} = \omega^2 \frac{1-\omega^{2n}}{1-\omega^2}$.$S = 0$ if and only if $1 - \omega^{2n} = 0$,which means $\omega^{2n} = 1$.Since $\omega^3 = 1$,this occurs when $2n$ is a multiple of $3$,i.e.,$n$ is a multiple of $3$.Thus,$P^2 
eq 0$ when $n$ is not a multiple of $3$.Checking the options:$(A) 57 = 3 	imes 19$ (Multiple of $3$)$(B) 55$ (Not a multiple of $3$)$(C) 58$ (Not a multiple of $3$)$(D) 56$ (Not a multiple of $3$)Therefore,$P^2 
eq 0$ for $n = 55, 58, 56$.

$List-I$	$List-II$
$(P)$ The number of matrices $M=(a_{ij})_{3 \times 3}$ with all entries in $T$ such that $R_i=C_j=0$ for all $i, j$ is	$(1)$ $1$
$(Q)$ The number of symmetric matrices $M=(a_{ij})_{3 \times 3}$ with all entries in $T$ such that $C_j=0$ for all $j$ is	$(2)$ $2$
$(R)$ Let $M=(a_{ij})_{3 \times 3}$ be a skew-symmetric matrix such that $a_{ij} \in T$ for $i>j$. Then the number of elements in the set $\{\begin{bmatrix} x \\ y \\ z \end{bmatrix}: x, y, z \in \mathbb{R}, M\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} a_{12} \\ 0 \\ -a_{23} \end{bmatrix}\}$ is	$(3)$ $\text{Infinite}$
$(S)$ Let $M=(a_{ij})_{3 \times 3}$ be a matrix with all entries in $T$ such that $R_i=0$ for all $i$. Then the absolute value of the determinant of $M$ is	$(4)$ $6$
	$(5)$ $0$

Let $\omega$ be a complex cube root of unity with $\omega \neq 1$ and $P = [p_{ij}]$ be an $n \times n$ matrix with $p_{ij} = \omega^{i+j}$. Then $P^2 \neq 0$ when $n =$

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