# Basis maths-linear algebra-Q1

+1 vote

If $a> b> c$ and the system of equations

has a non trivial solution then the roots of the quadratic equation 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

reshown May 16

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

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   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  are real.

We know that if X,Y are roots of  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 (112,760 points)