SATHEE: Matrices and Determinants 2 Question 6

Matrices and Determinants 2 Question 6

6. Let the numbers $2, b, c$ be in an $AP$ and $A =$

$[\begin{matrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^{2} & c^{2} \end{matrix}]$

If $\det (A) \in [2, 16]$ , then $c$ lies in the interval

(2019 Main, 8 April II)

(a) $[3, 2 + 2^{3 / 4}]$

(b) $(2 + 2^{3 / 4}, 4)$

(d) $[2, 3]$

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Answer:

Correct Answer: 6. (c)

Solution:

Given, matrix $A =$ $[\begin{matrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^{2} & c^{2} \end{matrix}]$ , so

$\det (A) = [\begin{matrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^{2} & c^{2} \end{matrix}]$

On applying, $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$ ,

we $get \det (A) =$ $[\begin{matrix} 1 & 0 & 0 \\ 2 & b - 2 & c - 2 \\ 4 & b^{2} - 4 & c^{2} - 4 \end{matrix}]$

$= [\begin{matrix} b - 2 & c - 2 \\ b^{2} - 4 & c^{2} - 4 \end{matrix}]$

$\begin{aligned} = [\begin{array}{c} b - 2 & c - 2 \\ (b - 2) (b + 2) & (c - 2) (c + 2) \end{array}] \\ = (b - 2) (c - 2) [\begin{array}{c} 1 & 1 \\ b + 2 & c + 2 \end{array}] \end{aligned}$

[taking common $(b - 2)$ from $C_{1}$ and $(c - 2)$ from $C_{2}$ ]

$= (b - 2) (c - 2) (c - b)$

Since, $2, b$ and $c$ are in AP, if assume common difference of $AP$ is $d$ , then

$b = 2 + d and c = 2 + 2 d$

So, $| A | = d (2 d) d = 2 d^{3} \in [2, 16$ ] [given]

$\Rightarrow d^{3} \in [1, 8] \Rightarrow d \in [1, 2]$

$∴ 2 + 2 d \in [2 + 2, 2 + 4]$

$= [4, 6] \Rightarrow c \in [4, 6]$

Matrices and Determinants 2 Question 6

6. Let the numbers $2, b, c$ be in an $AP$ and $A =$

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Matrices and Determinants 2 Question 6

6. Let the numbers 2,b,c be in an AP and A=

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

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NCERT Exercise Video Solution

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6. Let the numbers $2, b, c$ be in an $AP$ and $A =$