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Let’s break this down step by step. --- **Step 1: Understanding the problem** We have a 3x3 matrix \( A \) with entries \( a_{ij} \) given by: \[ a_{ij} = \frac{1}{i+j-1} \] So: \[ A = \begin{bmatrix} \frac{1}{1+1-1} & \frac{1}{1+2-1} & \frac{1}{1+3-1} \\ \frac{1}{2+1-1} & \frac{1}{2+2-1} & \frac{1}{2+3-1} \\ \frac{1}{3+1-1} & \frac{1}{3+2-1} & \frac{1}{3+3-1} \end{bmatrix} = \begin{bmatrix} \frac{1}{1} & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \end{bmatrix} = \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} \end{bmatrix}. \] This is a famous matrix known as the **Hilbert matrix** of order 3. It is known to be symmetric and positive definite but ill-conditioned. --- **Step 2: Determinant of the Hilbert matrix** For a Hilbert matrix \( H_n \) with entries \( H_{ij} = \frac{1}{i+j-1} \), there is a known formula for the determinant: \[ \det(H_n) = \frac{\left[1!\, 2!\, \dots\, (n-1)!\right]^4}{1!\, 2!\, \dots\, (2n-1)!}. \] For \( n = 3 \): \[ \det(H_3) = \frac{(1! \cdot 2!)^4}{1! \cdot 2! \cdot 3! \cdot 4! \cdot 5!}. \] Let’s compute step-by-step: - \( 1! = 1 \), \( 2! = 2 \), so \( (1! \cdot 2!)^4 = (1 \cdot 2)^4 = 2^4 = 16 \). Denominator: \( 1! \cdot 2! \cdot 3! \cdot 4! \cdot 5! \). - \( 1! = 1 \) - \( 2! = 2 \) - \( 3! = 6 \) - \( 4! = 24 \) - \( 5! = 120 \) Multiply: \( 1 \times 2 = 2 \), \( 2 \times 6 = 12 \), \( 12 \times 24 = 288 \), \( 288 \times 120 = 34560 \). So: \[ \det(H_3) = \frac{16}{34560} = \frac{1}{2160}. \] --- **Step 3: Check with direct computation** Let’s compute determinant directly for \( A \): \[ A = \begin{bmatrix} 1 & \frac12 & \frac13 \\ \frac12 & \frac13 & \frac14 \\ \frac13 & \frac14 & \frac15 \end{bmatrix}. \] Determinant formula for 3x3: \( a(ei − fh) − b(di − fg) + c(dh − eg) \). Let: \[ a = 1,\ b = \frac12,\ c = \frac13, \quad d = \frac12,\ e = \frac13,\ f = \frac14, \quad g = \frac13,\ h = \frac14,\ i = \frac15. \] First term: \( a(ei - fh) \): \( ei = \frac13 \cdot \frac15 = \frac{1}{15} \), \( fh = \frac14 \cdot \frac14 = \frac{1}{16} \). So \( ei - fh = \frac{1}{15} - \frac{1}{16} = \frac{16 - 15}{240} = \frac{1}{240} \). Multiply by \( a = 1 \) → \( \frac{1}{240} \). Second term: \( -b(di - fg) \): \( di = \frac12 \cdot \frac15 = \frac{1}{10} \), \( fg = \frac14 \cdot \frac13 = \frac{1}{12} \). \( di - fg = \frac{1}{10} - \frac{1}{12} = \frac{6 - 5}{60} = \frac{1}{60} \). Multiply by \( -b = -\frac12 \) → \( -\frac12 \cdot \frac{1}{60} = -\frac{1}{120} \). Third term: \( c(dh - eg) \): \( dh = \frac12 \cdot \frac14 = \frac{1}{8} \), \( eg = \frac13 \cdot \frac13 = \frac{1}{9} \). \( dh - eg = \frac{1}{8} - \frac{1}{9} = \frac{9 - 8}{72} = \frac{1}{72} \). Multiply by \( c = \frac13 \) → \( \frac{1}{3} \cdot \frac{1}{72} = \frac{1}{216} \). Sum: \( \frac{1}{240} - \frac{1}{120} + \frac{1}{216} \). Common denominator? Let’s compute numerically: \( \frac{1}{240} = 0.0041667 \), \( -\frac{1}{120} = -0.0083333 \), \( \frac{1}{216} \approx 0.0046296 \). Sum: \( 0.0041667 - 0.0083333 + 0.0046296 = 0.0004630 \). \( 0.0004630 = \frac{1}{2160} \) since \( 1/2160 \approx 0.00046296 \). Matches the formula. --- **Step 4: Conclusion** The determinant is \( \frac{1}{2160} \). \[ \boxed{\frac{1}{2160}} \]