IIT-JEE 2007 · Paper 1 · Mathematics Q64

MathematicsMatrix match+6Question 64 of 66

Consider the following linear equations
$ax+by+cz=0,\qquad bx+cy+az=0,\qquad cx+ay+bz=0.$
Match the conditions/expressions in Column I with statements in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column I

A
$a+b+c\ne0$ and $a^2+b^2+c^2=ab+bc+ca$
B
$a+b+c=0$ and $a^2+b^2+c^2\ne ab+bc+ca$
C
$a+b+c\ne0$ and $a^2+b^2+c^2\ne ab+bc+ca$
D
$a+b+c=0$ and $a^2+b^2+c^2=ab+bc+ca$

Column II

P
the equations represent planes meeting only at a single point.
Q
the equations represent the line $x=y=z$ .
R
the equations represent identical planes.
S
the equations represent the whole of the three dimensional space.

Answer and worked solution

Official answer: A → R; B → Q; C → P; D → S

Determinant. The coefficient matrix is circulant:
$\Delta=-(a^3+b^3+c^3-3abc)=-(a+b+c)\left(a^2+b^2+c^2-ab-bc-ca\right),$
and the second factor is $\frac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$ , zero only when $a=b=c$ .

A: $a=b=c\ne0$ : all three equations are $x+y+z=0$ , identical planes. → R.
B: $\Delta=0$ but $a,b,c$ not all equal, so the rank is 2 and the solutions form a line. Adding the equations gives $(a+b+c)(x+y+z)=0$ , which holds anyway; $x=y=z$ satisfies each equation since $a+b+c=0$ . So the solution set is the line $x=y=z$ . → Q.
C: $\Delta\ne0$ : only the trivial solution, planes meeting at a single point. → P.
D: $a=b=c$ and $a+b+c=0$ force $a=b=c=0$ : every point satisfies the equations, the whole space. → S.

Answer: A → R; B → Q; C → P; D → S

Topics:Systems of linear equations Determinants and properties

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IIT-JEE 2007 · Paper 1, Mathematics question 64