Solveeit Logo
Login

Question

Question: Let A = $[a_{ij}]$ be a 3 × 3 matrix such that $a_{ij}$ = cos(iθ + jθ) for 1≤i, j ≤ 3 where θ = $\fr...

Let A = [aij][a_{ij}] be a 3 × 3 matrix such that aija_{ij} = cos(iθ + jθ) for 1≤i, j ≤ 3 where θ = 2π3\frac{2π}{3}, then the determinant of the matrix C (where C = A + I3I_3) is

A

34\frac{3}{4}

B

−54\frac{-5}{4}

C

94\frac{9}{4}

D

−74\frac{-7}{4}

Answer

−54\frac{-5}{4}

Explanation

Solution

Here's how to solve this problem:

  1. Construct the matrix A:

    • Calculate the elements of matrix A using the formula aij=cos⁡((i+j)θ)a_{ij} = \cos((i+j)\theta), where θ=2π3\theta = \frac{2\pi}{3}.

    A=[−121−121−12−12−12−121]A = \begin{bmatrix} -\frac{1}{2} & 1 & -\frac{1}{2} \\ 1 & -\frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} & 1 \end{bmatrix}

  2. Construct the matrix C:

    • Add the identity matrix I3I_3 to matrix A: C=A+I3C = A + I_3

    C=[121−12112−12−12−122]C = \begin{bmatrix} \frac{1}{2} & 1 & -\frac{1}{2} \\ 1 & \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} & 2 \end{bmatrix}

  3. Calculate the determinant of C:

    • Use cofactor expansion along the first row to find the determinant of C.

    det⁡(C)=12∣12−12−122∣−1∣1−12−122∣+(−12)∣112−12−12∣\det(C) = \frac{1}{2} \begin{vmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & 2 \end{vmatrix} - 1 \begin{vmatrix} 1 & -\frac{1}{2} \\ -\frac{1}{2} & 2 \end{vmatrix} + (-\frac{1}{2}) \begin{vmatrix} 1 & \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} \end{vmatrix}

    • Calculate the 2x2 determinants:

      • ∣12−12−122∣=(12)(2)−(−12)(−12)=1−14=34\begin{vmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & 2 \end{vmatrix} = (\frac{1}{2})(2) - (-\frac{1}{2})(-\frac{1}{2}) = 1 - \frac{1}{4} = \frac{3}{4}

      • ∣1−12−122∣=(1)(2)−(−12)(−12)=2−14=74\begin{vmatrix} 1 & -\frac{1}{2} \\ -\frac{1}{2} & 2 \end{vmatrix} = (1)(2) - (-\frac{1}{2})(-\frac{1}{2}) = 2 - \frac{1}{4} = \frac{7}{4}

      • ∣112−12−12∣=(1)(−12)−(12)(−12)=−12+14=−14\begin{vmatrix} 1 & \frac{1}{2} \\ -\frac{1}{2} & -\frac{1}{2} \end{vmatrix} = (1)(-\frac{1}{2}) - (\frac{1}{2})(-\frac{1}{2}) = -\frac{1}{2} + \frac{1}{4} = -\frac{1}{4}

    • Substitute these values back into the determinant formula:

      det⁡(C)=12(34)−1(74)−12(−14)=38−74+18=38−148+18=−108=−54\det(C) = \frac{1}{2} (\frac{3}{4}) - 1 (\frac{7}{4}) - \frac{1}{2} (-\frac{1}{4}) = \frac{3}{8} - \frac{7}{4} + \frac{1}{8} = \frac{3}{8} - \frac{14}{8} + \frac{1}{8} = \frac{-10}{8} = -\frac{5}{4}

Therefore, the determinant of matrix C is −54-\frac{5}{4}.