SATHEE: Matrices and Determinants 2 Question 7

Matrices and Determinants 2 Question 7

8. If

$| \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$

$= (a + b + c) (x + a + b + c)^{2}, x \neq 0$ and $a + b + c \neq 0$ , then $x$ is equal to

(2019 Main, 11 Jan II)

(a) $- (a + b + c)$

(b) $- 2 (a + b + c)$

(d) $a b c$

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Answer:

Correct Answer: 8. (b)

Solution:

Let $Δ =$ $| \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} |$

Applying $R_{1} \to R_{1} + R_{2} + R_{3}$ , we get

$\begin{aligned} Δ & = | \begin{array}{c} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | \\ = (a + b + c) | \begin{array}{c} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{array} | \end{aligned}$

(taking common $(a + b + c)$ from $R_{1})$

Applying $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ , we get

$Δ = (a + b + c) | \begin{matrix} 1 & 0 & 0 \\ 2 b & - (a + b + c) & 0 \\ 2 c & 0 & - (a + b + c) \end{matrix} |$

Now, expanding along $R_{1}$ , we get

$Δ = (a + b + c) 1 . (a + b + c)^{2} - 0$

$= (a + b + c)^{3} = (a + b + c) (x + a + b + c)^{2} (given)$

$\Rightarrow (x + a + b + c)^{2} = (a + b + c)^{2}$

$\Rightarrow x + a + b + c = \pm (a + b + c)$

$\Rightarrow x = - 2 (a + b + c)$

Matrices and Determinants 2 Question 7

8. If

Answer:

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Matrices and Determinants 2 Question 7

8. If

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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