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Official PYQ · 2019Q44Eigenvalues and Eigenvectors, LU DecompositionNAT2 marks

GATE CS 2019 Q44 — Eigenvalues and Eigenvectors, LU Decomposition

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Consider the following matrix:

R=[1248139271416641525125]R = \begin{bmatrix} 1 & 2 & 4 & 8 \\ 1 & 3 & 9 & 27 \\ 1 & 4 & 16 & 64 \\ 1 & 5 & 25 & 125 \end{bmatrix}

The absolute value of the product of Eigen values of RR is _______
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Answer: 12

The product of the eigenvalues of a matrix is equal to its determinant. Thus, we need to find the absolute value of the determinant of RR, i.e., ∣det⁡(R)∣|\det(R)|.

The given matrix RR is a Vandermonde matrix of the form:

R=[1x1x12x131x2x22x231x3x32x331x4x42x43]R = \begin{bmatrix} 1 & x_1 & x_1^2 & x_1^3 \\ 1 & x_2 & x_2^2 & x_2^3 \\ 1 & x_3 & x_3^2 & x_3^3 \\ 1 & x_4 & x_4^2 & x_4^3 \end{bmatrix}

where x1=2x_1 = 2, x2=3x_2 = 3, x3=4x_3 = 4, and x4=5x_4 = 5.

The determinant of a Vandermonde matrix is given by the formula:

det⁡(R)=∏1≤i<j≤4(xj−xi)\det(R) = \prod_{1 \le i < j \le 4} (x_j - x_i)

Substituting the values:

det⁡(R)=(x2−x1)(x3−x1)(x4−x1)(x3−x2)(x4−x2)(x4−x3)=(3−2)(4−2)(5−2)(4−3)(5−3)(5−4)=1×2×3×1×2×1=12\begin{aligned} \det(R) &= (x_2 - x_1)(x_3 - x_1)(x_4 - x_1)(x_3 - x_2)(x_4 - x_2)(x_4 - x_3) \\ &= (3 - 2)(4 - 2)(5 - 2)(4 - 3)(5 - 3)(5 - 4) \\ &= 1 \times 2 \times 3 \times 1 \times 2 \times 1 \\ &= 12 \end{aligned}

Since the determinant is 1212, the absolute value of the product of the eigenvalues is ∣12∣=12|12| = 12.

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