Let \(a-2b+c=1\). If \(f(x)=\begin{vmatrix}x+a & x+2 & x+1\\x+b & x+3 & x+2\\x+c & x+4 & x+3\end{vmatrix}\) then:
f(50) = 1
Apply the column operation \(C_2\to C_2-C_3\): new \(C_2=[(x{+}2)-(x{+}1),\,(x{+}3)-(x{+}2),\,(x{+}4)-(x{+}3)]^T=[1,1,1]^T\).
Apply \(C_1\to C_1-C_3\): new \(C_1=[(x{+}a)-(x{+}1),\,(x{+}b)-(x{+}2),\,(x{+}c)-(x{+}3)]^T=[a{-}1,\,b{-}2,\,c{-}3]^T\).
Now apply \(C_3\to C_3-(x+1)C_2\) (subtracting \((x+1)\) times the new all-ones column): new \(C_3=[0,\,(x{+}2)-(x{+}1),\,(x{+}3)-(x{+}1)]^T=[0,1,2]^T\).
The determinant is now completely independent of x: \(f(x)=\begin{vmatrix}a-1 & 1 & 0\\b-2 & 1 & 1\\c-3 & 1 & 2\end{vmatrix}\).
Expanding along the first row: \(f(x)=(a-1)(1\cdot2-1\cdot1)-1\big((b-2)\cdot2-1\cdot(c-3)\big)+0\)
\(=(a-1)-\big(2b-4-c+3\big)=(a-1)-(2b-c-1)=a-1-2b+c+1=a-2b+c\)
Given \(a-2b+c=1\), so \(f(x)=1\) for every value of x. In particular \(f(50)=1\) and \(f(-50)=1\).
The value of the determinant \(\left| {\begin{array}{*{20}{c}} {1 - {\rm{\alpha }}}&{{\rm{\alpha }} - {{\rm{\alpha }}^2}}&{{{\rm{\alpha }}^2}}\\ {1 - {\rm{\beta }}}&{{\rm{\beta }} - {{\rm{\beta }}^2}}&{{{\rm{\beta }}^2}}\\ {1 - {\rm{\gamma }}}&{{\rm{\gamma }} - {{\rm{\gamma }}^2}}&{{{\rm{\gamma }}^2}} \end{array}} \right|\) is equal to
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Let p, q and r be three distinct positive real numbers. If \(\rm D = \left| {\begin{array}{*{20}{c}} \rm p&\rm q&\rm r\\ \rm q&\rm r&\rm p\\ \rm r&\rm p&\rm q \end{array}} \right|,\) then which one of the following is correct?