Basis maths-linear algebra-Q1

+1 vote

If a> b> c and the system of equations

ax + by + cz = 0,

bx+cy + az = 0,

cx + ay + bz = 0

has a non trivial solution then the roots of the quadratic equation at^2 + bt + c = 0 are

(A). Both Roots are Real and Positive

(B). Both Roots are Real and Negative

(C). Complex But Not Real

(D). One Root is Positive and One is Negative

asked May 15 in Basic Maths by gbeditor (44,560 points)
reshown May 16 by gbeditor

2 Answers

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Best answer

A  n\times n homogeneous system of linear equations has a unique solution (the trivial solutionif and only if its determinant is non-zeroIf this determinant is zero, then the system has an infinite number of solutions.

So,  the system of equations

ax + by + cz = 0,

bx+cy + az = 0,

cx + ay + bz = 0

has a non trivial solution, it means that :

\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0

So, a(cb-a^2) - b(b^2 -ac) + c(ba-c^2) = 0

So   a^3+b^3+c^3 = 3abc

We know that  a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) 

Since a^3+b^3+c^3 = 3abc, So, 

(a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)=0

this means that either (a^2 + b^2 + c^2 -ab - ac -bc) = 0 \ or \ (a + b + c ) = 0

(a^2 + b^2 + c^2 -ab - ac -bc) cannot be zero because:
2a^2 + 2b^2 + 2c^2 -2ab - 2ac - 2bc = 0 \\ a^2 + b^2 -2ab + a^2 +b^2 +2c^2 - 2ac -2bc = 0 \\ (a-b)^2 + a^2 + c^2 -2ac + b^2 + c^2 -2bc = 0 \\ (a-b)^2 + (a-c)^2 + (b-c)^2 = 0 \\ (a-b)^2, (a-c)^2 ,(b-c)^2 >=0

As a \neq b \neq c , The given value cannot be zero.

This means (a + b + c ) has to be zero.

So, a+b+c =0, and   a>b>c  , so, 

So, \text{a has to be positive and c has to be negative }.

The quadratic equation at^2 + bt + c = 0  has real solutions iff b^2 -4ac \geq 0

Since, a,c are of different signs, so, b^2 -4ac \geq 0. Hence, the roots of the equation at^2 + bt + c = 0 are real. 

We know that if X,Y are roots of at^2 + bt + c = 0 then XY = \frac{c}{a}

Since a,c are of different signs, so, XY is negative, so, Both roots are of different signs.

Hence, answer is option D.

answered Jun 12 by deepak-gatebook (112,760 points)
0 votes
@deepak-gatebook, can we have the solution of this question...
answered Jun 12 by tsabhineet21 (7,690 points)
Answered. Check now.
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