Matrices And Determinants Ques 43
The sum of the real roots of the equation $\left|\begin{matrix}x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2\end{matrix}\right|=0$, is equal to
(a) $0$
(b) $-4$
(c) $6$
(d) $1$
(2019 Main, 10 April II)
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Answer:
Correct Answer: 43.(a)
Solution:
Formula:
Evaluation of the Determinant:
- Given equation
$ \left|\begin{matrix} x & -6 & -1 \\ 2 & -3 x & x-3 \\ -3 & 2 x & x+2 \end{matrix}\right|=0 $
On expansion of determinant along $R_{1}$, we get
$x[(-3 x)(x+2)-2 x(x-3)]+6[2(x+2)+3(x-3)]$
$ -1[2(2 x)-(-3 x)(-3)]=0 $
$\Rightarrow x\left[-3 x^{2}-6 x-2 x^{2}+6 x\right]+6[2 x+4+3 x-9]$
$ -1[4 x-9 x]=0 $
$\Rightarrow x\left(-5 x^{2}\right)+6(5 x-5)-1(-5 x)=0$
$\Rightarrow \quad-5 x^{3}+30 x-30+5 x=0$
$\Rightarrow \quad 5 x^{3}-35 x+30=0 \quad \Rightarrow \quad x^{3}-7 x+6=0$.
Since all roots are real
$\therefore$ Sum of roots $=-\frac{\text { coefficient of } x^{2}}{\text { coefficient of } x^{3}}=0$
BETA
