SATHEE: Matrices and Determinants 2 Question 10

Matrices and Determinants 2 Question 10

12. If

$\begin{array}{r} | \begin{array}{c} x - 4 & 2 x & 2 x \\ 2 x & x - 4 & 2 x \\ 2 x & 2 x & x - 4 \end{array} | = (A + B x) (x - A)^{2} \end{array}$ , then the ordered pair $(A, B)$ is equal to

(2018 Main)

(a) $(- 4, - 5)$

(b) $(- 4, 3)$

(d) $(4, 5)$

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Answer:

Correct Answer: 12. (c)

Solution:

Given,

$\begin{array}{r} | \begin{array}{c} x - 4 & 2 x & 2 x \\ 2 x & x - 4 & 2 x \\ 2 x & 2 x & x - 4 \end{array} | = (A + B x) (x - A)^{2} \end{array}$

$\Rightarrow Apply C_{1} \to C_{1} + C_{2} + C_{3}$

$\begin{array}{r} | \begin{array}{c} 5 x - 4 & 2 x & 2 x \\ 5 x - 4 & x - 4 & 2 x \\ 5 x - 4 & 2 x & x - 4 \end{array} | \end{array}$

$= (A + B x) (x - A)^{2}$

Taking common $(5 x - 4)$ from $C_{1}$ , we get

(5x-4) $| \begin{matrix} 1 & 2 x & 2 x \\ x - 4 & 2 x \\ 1 & 2 x & x - 4 \end{matrix} | = (A + B x) (x - A)^{2}$

Apply $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$

$∴ (5 x - 4) | \begin{matrix} 1 & 2 x & 0 \\ 0 & - x - 4 & 0 \\ 0 & 0 & - x - 4 \end{matrix} | = (A + B x) (x - A)^{2}$

Expanding along $C_{1}$ , we get

$(5 x - 4) (x + 4)^{2} = (A + B x) (x - A)^{2}$

Equating, we get, $A = - 4$ and $B = 5$

Matrices and Determinants 2 Question 10

12. If

Answer:

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Matrices and Determinants 2 Question 10

12. 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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