【题解】同济线代习题一.6.3
题目
证明：
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\begin{vmatrix} a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2 \\ \end{vmatrix} = 0
∣
∣a2b2c2d2(a+1)2(b+1)2(c+1)2(d+1)2(a+2)2(b+2)2(c+2)2(d+2)2(a+3)2(b+3)2(c+3)2(d+3)2∣
∣=0
解答
∣ a 2 ( a + 1 ) 2 ( a + 2 ) 2 ( a + 3 ) 2 b 2 ( b + 1 ) 2 ( b + 2 ) 2 ( b + 3 ) 2 c 2 ( c + 1 ) 2 ( c + 2 ) 2 ( c + 3 ) 2 d 2 ( d + 1 ) 2 ( d + 2 ) 2 ( d + 3 ) 2 ∣ = c 2 − c 1 c 3 − c 1 c 4 − c 1 ∣ a 2 2 a + 1 4 a + 4 6 a + 9 b 2 2 b + 1 4 b + 4 6 b + 9 c 2 2 c + 1 4 c + 4 6 c + 9 d 2 2 d + 1 4 d + 4 6 d + 9 ∣ = c 3 − 2 c 2 c 4 − 3 c 2 ∣ a 2 2 a + 1 2 6 b 2 2 b + 1 2 6 c 2 2 c + 1 2 6 d 2 2 d + 1 2 6 ∣ = 0 \begin{vmatrix} a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2 \\ \end{vmatrix} \xlongequal{\begin{align*} c_2 - c_1 \\ c_3 - c_1 \\ c_4 - c_1 \end{align*}} \begin{vmatrix} a^2 & 2a+1 & 4a+4 & 6a+9 \\ b^2 & 2b+1 & 4b+4 & 6b+9 \\ c^2 & 2c+1 & 4c+4 & 6c+9 \\ d^2 & 2d+1 & 4d+4 & 6d+9 \\ \end{vmatrix} \xlongequal{\begin{align*} c_3 - 2c_2 \\ c_4 - 3c_2 \end{align*}} \begin{vmatrix} a^2 & 2a+1 & 2 & 6 \\ b^2 & 2b+1 & 2 & 6 \\ c^2 & 2c+1 & 2 & 6 \\ d^2 & 2d+1 & 2 & 6 \\ \end{vmatrix} = 0 ∣ ∣a2b2c2d2(a+1)2(b+1)2(c+1)2(d+1)2(a+2)2(b+2)2(c+2)2(d+2)2(a+3)2(b+3)2(c+3)2(d+3)2∣ ∣c2−c1c3−c1c4−c1∣ ∣a2b2c2d22a+12b+12c+12d+14a+44b+44c+44d+46a+96b+96c+96d+9∣ ∣c3−2c2c4−3c2∣ ∣a2b2c2d22a+12b+12c+12d+122226666∣ ∣=0
得证。