Question

If

a_{1}, a_{2}, a_{3}

are distinct real roots of the equation

p x^{3} + p x^{2} + q x + r = 0

such that

|\begin{matrix} 1 + a_{1} & 1 & 1 \\ 1 & 1 + a_{2} & 1 \\ 1 & 1 & 1 + a_{3} \end{matrix}| = 0,

then Prove that

\frac{r}{p} < 0

Accepted Answer

It is given that $a_{1}, a_{2}, a_{3}$ are distinct roots of the equation $p x^{3} + p x^{2} + q x + r = 0$

So, $a_{1} + a_{2} + a_{3} = - \frac{p}{p}, a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{1} = \frac{q}{p} & a_{1} a_{2} a_{3} = - \frac{r}{p}$

Now given that $|\begin{matrix} 1 + a_{1} & 1 & 1 \\ 1 & 1 + a_{2} & 1 \\ 1 & 1 & 1 + a_{3} \end{matrix}| = 0$

Now using the property $C_{2} \to C_{2} - C_{3}$

$\Rightarrow |\begin{matrix} 1 + a_{1} & 0 & 1 \\ 1 & a_{2} & 1 \\ 1 & - a_{3} & 1 + a_{3} \end{matrix}| = 0$

Now expanding determinant then

$\Rightarrow (1 + a_{1}) \{a_{2} (1 + a_{3}) + a_{3}\} - 0 (1 + a_{3} - 1) + 1 \{- a_{3} - a_{2}\} = 0$

$\Rightarrow (1 + a_{1}) \{a_{2} + a_{2} a_{3} + a_{3}\} - a_{3} - a_{2} = 0$

$\Rightarrow 1 \{a_{2} + a_{2} a_{3} + a_{3}\} + a_{1} a_{2} + a_{1} a_{2} a_{3} + a_{1} a_{3} - a_{3} - a_{2} = 0$

$\Rightarrow \{a_{1} a_{2} + a_{2} a_{3} + a_{1} a_{2} a_{3} + a_{1} a_{3}\} = 0$

$\Rightarrow \frac{q}{p} - \frac{r}{p} = 0$

$\Rightarrow q = r$

So, above equation can written as $p x^{3} + p x^{2} + r x + r = 0$

$\Rightarrow p x^{2} (x + 1) + r (x + 1) = 0$

$\Rightarrow x = - 1, x^{2} = - \frac{r}{p}$

We know that $x^{2} \geq 0$ for real $x$

And for $x = 0$ given equation becomes $r = 0$ which does not gives the distinct real roots so

$x^{2} > 0$ implies $- \frac{r}{p} > 0 \Rightarrow \frac{r}{p} < 0$

So it is possible for real solution $\frac{r}{p} < 0$ .

Proved

If $a_{1}, a_{2}, a_{3}$ are distinct real roots of the equation $p x^{3} + p x^{2} + q x + r = 0$ such that $|\begin{matrix} 1 + a_{1} & 1 & 1 \\ 1 & 1 + a_{2} & 1 \\ 1 & 1 & 1 + a_{3} \end{matrix}| = 0,$ then Prove that $\frac{r}{p} < 0$

Important Questions on Matrices and Determinants