SATHEE: Matrices and Determinants 2 Question 9

Matrices and Determinants 2 Question 9

11. Let $d \in R$ , and

$A =$ $| \begin{matrix} - 2 & 4 + d & (\sin θ) - 2 \\ 1 & (\sin θ) + 2 & d \\ 5 & (2 \sin θ) - d & (- \sin θ) + 2 + 2 d \end{matrix} |$ , $θ \in [θ, 2 π]$ .

If the minimum value of $\det (A)$ is 8 , then a value of $d$ is

(2019 Main, 10 Jan I)

(a) -5

(b) -7

(d) $2 (\sqrt{2} + 2)$

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Answer:

Correct Answer: 11. (a)

Solution:

Given,

$A = | \begin{matrix} - 2 & 4 + d & (\sin θ) - 2 \\ 1 & (\sin θ) + 2 & d \\ 5 & (2 \sin θ) - d & (- \sin θ) + 2 + 2 d \end{matrix} |$

$\begin{aligned} ∴ | A | & = | \begin{array}{c} - 2 & 4 + d & (\sin θ) - 2 \\ 1 & (\sin θ) + 2 & d \\ 5 & (2 \sin θ) - d & (- \sin θ) + 2 + 2 d \end{array} | \\ = | \begin{array}{c} - 2 & 4 + d & (\sin θ) - 2 \\ 1 & (\sin θ) + 2 & d \\ 1 & 0 & 0 \end{array} | \\ = 1 [(4 + d) d - (\sin θ + 2) (\sin θ - 2)] \\ = (d^{2} + 4 d - \sin^{2} θ + 4) \\ = (d^{2} + 4 d + 4) - \sin^{2} θ \\ = (d + 2)^{2} - \sin^{2} θ \end{aligned}$

Note that $| A |$ will be minimum if $\sin^{2} θ$ is maximum i.e. if $\sin^{2} θ$ takes value 1 .

$\begin{aligned} ∵ | A |_{min} = 8, \\ therefore (d + 2)^{2} - 1 = 8 \\ \Rightarrow (d + 2)^{2} = 9 \\ \Rightarrow d + 2 = \pm 3 \\ \Rightarrow d = 1, - 5 \end{aligned}$

Matrices and Determinants 2 Question 9

11. Let $d \in R$ , and

Answer:

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Matrices and Determinants 2 Question 9

11. Let d∈R, and

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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11. Let $d \in R$ , and