Let 1, omega and omega^2 be the cube roots of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The matrix is $M = \begin{bmatrix} 1 & a & b \ \omega & 1 & c \ \omega^2 & \omega & 1 \end{bmatrix}$.For the matrix to be non-singular,its deter

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The matrix is $M = \begin{bmatrix} 1 & a & b \ \omega & 1 & c \ \omega^2 & \omega & 1 \end{bmatrix}$.For the matrix to be non-singular,its determinant must be non-zero: $\det(M) 
eq 0$.Calculating the determinant: $\det(M) = 1(1 - \omega c) - a(\omega - \omega^2 c) + b(\omega^2 - \omega^2) = 1 - \omega c - a\omega + a\omega^2 c = 1 - \omega(a + c) + a\omega^2 c$.Given $a, c \in \{\omega, \omega^2\}$,we test the combinations:$1$. If $a = \omega, c = \omega$: $\det(M) = 1 - \omega(2\omega) + \omega(\omega^2)(\omega) = 1 - 2\omega^2 + \omega^4 = 1 - 2\omega^2 + \omega = 1 - 2\omega^2 + (-1 - \omega^2) = -3\omega^2 
eq 0$.$2$. If $a = \omega^2, c = \omega^2$: $\det(M) = 1 - \omega(2\omega^2) + \omega^2(\omega^2)(\omega^2) = 1 - 2\omega^3 + \omega^6 = 1 - 2(1) + 1 = 0$.$3$. If $a = \omega, c = \omega^2$: $\det(M) = 1 - \omega(\omega + \omega^2) + \omega(\omega^2)(\omega^2) = 1 - \omega(-1) + \omega^5 = 1 + \omega + \omega^2 = 0$.$4$. If $a = \omega^2, c = \omega$: $\det(M) = 1 - \omega(\omega^2 + \omega) + \omega^2(\omega^2)(\omega) = 1 - \omega(-1) + \omega^5 = 1 + \omega + \omega^2 = 0$.Since $b$ can be either $\omega$ or $\omega^2$ in the case where $\det(M) 
eq 0$,and only the case $(a, c) = (\omega, \omega)$ yields a non-zero determinant,there are $2$ possible values for $b$ (namely $\omega$ and $\omega^2$).Thus,the number of elements in $S$ is $2$.

List-$I$	List-$II$
$(A)$ If $A$ is a non-singular matrix of order $3$ and $\|A\|=a$,then $\|\text{adj}(A)\|=$	$(I)$ null matrix
$(B)$ $A$ is a non-singular matrix of order $3$ and $B$ is any matrix of order $3$ such that $AB=O$,then $B$ is	$(II)$ $a^2$
$(C)$ $\begin{vmatrix} 1 & x & x^2 \\ \cos(a-b)y & \cos ay & \cos(a+b)y \\ \sin(a-b)y & \sin ay & \sin(a+b)y \end{vmatrix}$ does not depend on	$(III)$ $b$
$(D)$ $A$ is a square matrix of order $3$ and $B=A-A^T$,then $B$ is	$(IV)$ $a$
	$(V)$ $0$

Let $1, \omega$ and $\omega^2$ be the cube roots of unity. If $S$ is the set of all non-singular matrices of the form $M = \begin{bmatrix} 1 & a & b \\ \omega & 1 & c \\ \omega^2 & \omega & 1 \end{bmatrix}$ where $a, b, c \in \{\omega, \omega^2\}$,then the number of elements in $S$ is

Similar Questions

If $A = \begin{bmatrix} 2 & 3 & 4 \\ 1 & k & 2 \\ 4 & 1 & 5 \end{bmatrix}$ is a singular matrix,then the quadratic equation having the roots $k$ and $\frac{1}{k}$ is

If $A$ is a square matrix of order $3$ and $A^2+A+2I=0$,then

The total number of $3 \times 3$ matrices $A$ having entries from the set $\{0, 1, 2, 3\}$ such that the sum of all the diagonal entries of $AA^{T}$ is $9$,is equal to........

Let $X = \left\{\begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \mathbb{R} \right\}$. Define $f: X \rightarrow \mathbb{R}$ by $f(A) = \operatorname{det}(A), \forall A \in X$. Then, $f$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module