If omega is one of the imaginary cube roots o | vedclass

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

Question

(C) Given the determinant $\Delta = \left| \begin{array}{ccc} 1 & \omega^3 & \omega^2 \ \omega^3 & 1 & \omega \ \omega^2 & \omega & 1 \end{array}

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given the determinant $\Delta = \left| \begin{array}{ccc} 1 & \omega^3 & \omega^2 \ \omega^3 & 1 & \omega \ \omega^2 & \omega & 1 \end{array} ight|$.Since $\omega$ is an imaginary cube root of unity,we know that $\omega^3 = 1$.Substituting $\omega^3 = 1$ into the determinant,we get:$\Delta = \left| \begin{array}{ccc} 1 & 1 & \omega^2 \ 1 & 1 & \omega \ \omega^2 & \omega & 1 \end{array} ight|$.Observe that the first two columns are identical $(C_1 = C_2)$.According to the properties of determinants,if any two columns or rows are identical,the value of the determinant is $0$.Therefore,$\Delta = 0$.

If $\omega$ is one of the imaginary cube roots of unity,then the value of the determinant $\left| \begin{array}{ccc} 1 & \omega^3 & \omega^2 \\ \omega^3 & 1 & \omega \\ \omega^2 & \omega & 1 \end{array} \right|$ is:

Similar Questions

Let $A = [a_{ij}]$ and $B = [b_{ij}]$ be two $3 \times 3$ real matrices such that $b_{ij} = (3)^{(i+j-2)} a_{ji}$,where $i, j = 1, 2, 3$. If the determinant of $B$ is $81$,then the determinant of $A$ is:

Using the property of determinants and without expanding,prove that $\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$.

If $A$ is a square matrix of order $3 \times 3$,then $|KA|$ is equal to

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b)\end{array}\right|=0$

If $D = \begin{vmatrix} a^2 + 1 & ab & ac \\ ba & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix}$,then $D =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module