NIMCET 2010 — Mathematics PYQ

Solution

Properties used:

1. Sum of cube roots of unity: $1+\omega +{\omega }^2=0⟹{\omega }^2+1=-\omega$ and $1+{\omega }^2=-\omega$.

2. Product of cube roots: ${\omega }^3=1$.

3. Determinant row operations.

Step 1: Simplify the terms in the determinant.

Given ${\omega }^3=1$, the element in the third row, third column becomes $-{\omega }^3=-1$.

The determinant is:

Step 2: Apply Row Operation $R_1\to R_1+R_3$.

We add the third row to the first row to simplify the terms:

Simplifying the new $R_1$:

• $1+(-i)=1-i$

• $1+i+{\omega }^2-i+\omega -1={\omega }^2+\omega$

• ${\omega }^2-1$

Now the determinant looks like this:

Step 3: Apply Row Operation $R_1\to R_1-R_2$.

Notice that the first and third elements of $R_1$ and $R_2$ are now identical.

This simplifies to:

Step 4: Use the property $1+\omega +{\omega }^2=0$.

Substitute $0$ for the term in the first row, second column:

Since all elements of the first row are zero, the value of the determinant is $0$.

Correct Option:

(a) 0