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Systems of Linear Equations
Mathematics — Methods · hilbert-matrix-invertibility
Problem The result of Example 16 suggests that perhaps the matrix
A = [ 1 1 2 ⋯ 1 n 1 2 1 3 ⋯ 1 n + 1 ⋮ ⋮ ⋮ 1 n 1 n + 1 ⋯ 1 2 n − 1 ] A = \begin{bmatrix} 1 & \dfrac{1}{2} & \cdots & \dfrac{1}{n} \\[6pt] \dfrac{1}{2} & \dfrac{1}{3} & \cdots & \dfrac{1}{n+1} \\[6pt] \vdots & \vdots & & \vdots \\[6pt] \dfrac{1}{n} & \dfrac{1}{n+1} & \cdots & \dfrac{1}{2n-1} \end{bmatrix} A = 1 2 1 ⋮ n 1 2 1 3 1 ⋮ n + 1 1 ⋯ ⋯ ⋯ n 1 n + 1 1 ⋮ 2 n − 1 1 A − 1 A^{-1} A − 1 Browse the problem library