+4 marks−1 if incorrectSingle correctPrevious-year question
Singular Matrix and Cyclic Symmetric Rational Expression
Let α be a root of the equation (a - c)x^2 + (b - a)x + (c - b) = 0 where a, b, c are distinct real numbers such that the matrix [ α^2 α 1; 1 1 1; a b c ] is singular. Then, the value of ((a-c)^2)/((b-a)(c-b)) + ((b-a)^2)/((a-c)(c-b)) + ((c-b)^2)/((a-c)(b-a)) is
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Hint 1 of 2
Let X=a−c, Y=b−a, and Z=c−b. What is the sum X+Y+Z?
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Step-by-step solutionView
Correct answer
Setting X=a−c, Y=b−a, and Z=c−b gives X+Y+Z=0, so YZX2+XZY2+XYZ2=XYZX3+Y3+Z3=XYZ3XYZ=3.
Option analysis
Why each option works or fails
A · 12
Believing that the condition X+Y+Z=0 implies an extra factor of 4 in the symmetric sum identity. Use the identity X3+Y3+Z3−3XYZ=(X+Y+Z)(X2+Y2+Z2−XY−YZ−ZX); when X+Y+Z=0, X3+Y3+Z3=3XYZ directly.
B · 9
Squaring the coefficient 3 when expanding the identity or misapplying (X+Y+Z)2 substitutions. Recognize that the numerator simplifies to X3+Y3+Z3=3XYZ, yielding XYZ3XYZ=3, not 32.
C · 3
The algebraic simplification correctly uses the condition that the sum of the variables is zero. This is the correct option.
D · 6
Assuming that the two roots of the quadratic equation contribute an extra factor of 2 to the cyclic sum. Notice that the given target expression is independent of α and depends only on X+Y+Z=(a−c)+(b−a)+(c−b)=0.
Reviewed route
Solution
StepWorking
01Given
Quadratic equation (a−c)x2+(b−a)x+(c−b)=0 with distinct real numbers a,b,c, root α, and detα21aα1b11c=0.
02Goal
Find the value of the cyclic expression S=(b−a)(c−b)(a−c)2+(a−c)(c−b)(b−a)2+(a−c)(b−a)(c−b)2.
03Approach
Notice that the target expression is purely an algebraic identity in variables X=a−c, Y=b−a, Z=c−b. Since X+Y+Z=(a−c)+(b−a)+(c−b)=0, the identity X3+Y3+Z3=3XYZ will directly simplify the sum independent of α, provided the common denominator is formed.
04Execute
Let X=a−c, Y=b−a, and Z=c−b. Notice that X+Y+Z=(a−c)+(b−a)+(c−b)=0. The given expression can be written with a common denominator XYZ by multiplying the numerator and denominator of each term by its missing variable: S=YZX2+XZY2+XYZ2=XYZX3+Y3+Z3. Using the standard conditional identity, if X+Y+Z=0, then X3+Y3+Z3=3XYZ. Substituting this yields S=XYZ3XYZ=3.
✓Verify
Pick specific values satisfying all conditions: let a=1,b=2,c=3. Then a−c=−2, b−a=1, c−b=1. The sum X+Y+Z=−2+1+1=0. Then S=(1)(1)(−2)2+(−2)(1)12+(−2)(1)12=4−21−21=3. This confirms the result holds consistently.
Hints that build this answer step by step
Let X=a−c, Y=b−a, and Z=c−b. What is the sum X+Y+Z?
X+Y+Z=0
When X+Y+Z=0, what does the algebraic sum YZX2+XZY2+XYZ2 simplify to?
Did we even need the condition on α and the singular matrix?
No, the algebraic target expression identically equals 3 for any distinct real numbers a,b,c such that (a−c)+(b−a)+(c−b)=0. The quadratic and matrix conditions are consistent (having root α=1), but the value of the algebraic expression is independent of α.
