Matrices and Determinants 2 Question 6
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6. Let the numbers $2, b, c$ be in an $AP$ and $A=2 \quad b \quad c$. $\begin{array}{lll}4 & b^{2} & c^{2}\end{array}$
======= ####6. Let the numbers $2, b, c$ be in an $AP$ and $A=$ $\begin{bmatrix} 1 & 1 & 1\\ 2 & b & c\\ 4 & b^{2} & c^{2} \end{bmatrix}$
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If $\operatorname{det}(A) \in[2,16]$, then $c$ lies in the interval
(2019 Main, 8 April II)
(a) $\left[3,2+2^{3 / 4}\right]$
(b) $\left(2+2^{3 / 4}, 4\right)$
(c) $[4,6]$
(d) $[2,3]$
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Answer:
Correct Answer: 6. (c)
Solution:
- Given, matrix $A=$ $\begin{bmatrix} 1 & 1 & 1\\ 2 & b & c\\ 4 & b^{2} & c^{2} \end{bmatrix}$ , so
$ \operatorname{det}(A)=\begin{bmatrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^{2} & c^{2} \end{bmatrix} $
On applying, $C _2 \rightarrow C _2-C _1$ and $C _3 \rightarrow C _3-C _1$,
we $\operatorname{get} \operatorname{det}(A)=$ $ \begin{bmatrix} 1 & 0 & 0\\ 2 & b-2 & c-2\\ 4 & b^{2} -4 & c^{2}-4 \end{bmatrix} $
$ =\begin{bmatrix} b-2 & c-2\\ b^{2} -4 & c^{2}-4 \end{bmatrix} $
$ \begin{aligned} & =\begin{bmatrix} b-2 & c-2 \\ (b-2)(b+2) & (c-2)(c+2) \end{bmatrix}\\ & =(b-2)(c-2) \begin{bmatrix} 1 & 1 \\ b+2 & c+2 \end{bmatrix} \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 \text { and } c=2+2 d $
So, $\quad|A|=d(2 d) d=2 d^{3} \in[2,16$] $\quad$ [given]
$\Rightarrow \quad d^{3} \in[1,8] \Rightarrow d \in[1,2]$
$\therefore \quad 2+2 d \in[2+2,2+4]$
$ =[4,6] \Rightarrow c \in[4,6] $
