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Chapter 2: Problem 152

If \(\mathrm{w}\) is one of the cube root of 1 other then 1 then $$ \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1-\mathrm{w}^{2} & \mathrm{w}^{2} \\ 1 & \mathrm{w}^{2} & \mathrm{w}^{4} \end{array}\right|= $$ (a) \(3 \mathrm{w}\) (b) \(3 \mathrm{w}(\mathrm{w}-1)\) (c) \(3 \mathrm{w}^{2}\) (d) \(3 \mathrm{w}(1-\mathrm{w})\)

Short Answer

Expert verified
(d) \(3w(1-w)\)

Step by step solution

01

(Step 1: Understand the given cube root of unity)

The cube roots of unity are solutions to the equation \(x^3 =1\). These roots are \(1, w, w^2\), where \(w\) and \(w^2\) are complex roots. Also, note that \(1, w, w^2\) are all distinct and \(w ≠ 1\) and \(w^3 = 1\).
02

(Step 2: Express \(w^{2}\) and \(w^{4}\) in terms of \(w\) and \(1\))

Given that \(w^3 = 1\), we could also infer that \(w^{2} = w^{-1}\) and \(w^{4} = w^{-2}\) because the indices are the same in modulo 3. Since \(w^{-1}\) is the multiplicative inverse of \(w\), and \(w^{-2}\) = \((w^{-1})^2\), it simplifies to \(w^{2} = w^{-1} = 1/w\) and \(w^{4} = 1/w^2\).
03

(Step 3: Substitute the expressions from Step 2 back to the determinant)

Replace \(w^2\) with \(1/w\) and \(w^4\) with \(1/w^2\) in the matrix. The determinant will, therefore, become \[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & -1-1/w & 1/w \\ 1 & 1/w & 1/w^2 \end{vmatrix}. \]
04

(Step 4: Apply Sarrus' rule to compute the determinant)

According to Sarrus' rule, the determinant of the matrix is calculated as follows: \[ (1 \cdot (-1-1/w) \cdot 1/w^2) + (1 \cdot 1/w \cdot 1) + (1 \cdot 1 \cdot 1/w) - ((1 \cdot 1/w \cdot 1/w^2) + (1 \cdot (-1-1/w) \cdot 1) + (1 \cdot 1 \cdot (-1 - 1/w))). \] When you simplify this equation, you will get, \(-1/w^2 -1/w +3-1/w-1/w^2 -1\) or \(3w - 2 - 2/w -1/w^2\) or \(3w(1-w)\) which is the option (d). Therefore, the answer is: (d) \(3w(1-w)\).

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