SATHEE: Matrices and Determinants 2 Question 2

Matrices and Determinants 2 Question 2

2. The sum of the real roots of the equation $| \begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix} | = 0$ , is equal to

(a) 0

(b) -4

(d) 1

(2019 Main, 10 April II)

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Answer:

Correct Answer: 2. (a)

Solution:

Given equation

$| \begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix} | = 0$

On expansion of determinant along $R_{1}$ , we get

$x [(- 3 x) (x + 2) - 2 x (x - 3)] + 6 [2 (x + 2) + 3 (x - 3)]$

$- 1 [2 (2 x) - (- 3 x) (- 3)] = 0$

$\Rightarrow x [- 3 x^{2} - 6 x - 2 x^{2} + 6 x] + 6 [2 x + 4 + 3 x - 9]$

$- 1 [4 x - 9 x] = 0$

$\Rightarrow x (- 5 x^{2}) + 6 (5 x - 5) - 1 (- 5 x) = 0$

$\Rightarrow - 5 x^{3} + 30 x - 30 + 5 x = 0$

$\Rightarrow 5 x^{3} - 35 x + 30 = 0 \Rightarrow x^{3} - 7 x + 6 = 0$ .

Since all roots are real

$∴$ Sum of roots $= - \frac{coefficient of x^{2}}{coefficient of x^{3}} = 0$

Matrices and Determinants 2 Question 2

2. The sum of the real roots of the equation $| \begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix} | = 0$ , is equal to

Answer:

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Matrices and Determinants 2 Question 2

2. The sum of the real roots of the equation |x−6−12−3xx−3−32xx+2|=0, is equal to

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2. The sum of the real roots of the equation $| \begin{matrix} x & - 6 & - 1 \\ 2 & - 3 x & x - 3 \\ - 3 & 2 x & x + 2 \end{matrix} | = 0$ , is equal to