Question

Let the numbers 2, b, c be in an A.P and [1 1 1] A = [2 b c] .
[4 b² c²]
If det (A) ∈ [2, 16] then ce lies in the interval
(A) [3, 2 + 2²/⁴] (B) (2 + 2³/⁴, 4)
(C) {2,3) (D) [4, 6]

Answer 1

c ∈ [4,6]

Explanation:

Given :-

Matrix A = $\left[\begin{array}{ccc}1&1&1\\2&b&c\\4&b^{2} &c^{2} \end{array}\right]$

Det (A) ∈ [2 , 16]

Find :-

c lies in which interval

Matrix A = $\left[\begin{array}{ccc}1&1&1\\2&b&c\\4&b^{2} &c^{2} \end{array}\right]$

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

Apply column operations for matrix A

On applying :-

c2⇒ c2 - c1

c3⇒c3 - c1

we get,

A = $\left[\begin{array}{ccc}1&0&0\\2&b-2&c-2\\4&(b-2)(b+2)&(c-2)(c+2)\end{array}\right]$

Now take the first row and multiply the other two

A = 1 $\left[\begin{array}{ccc}b-2&c-2\\(b-2)(b+2)&(c-2)(c+2)\\\end{array}\right]$

Take (b-2) and (c-2) common

A = (b-2)(c-2)$\left[\begin{array}{ccc}1&1\\(b+2)&(c+2)\\\end{array}\right]$

By cross multiplying the values inside the matrix we get

det (A) = (b-2)(c-2) [(c+2 - b - 2)]

det ( A) = (b-2)(c-2)(c-b)------(1)

2,b,c are in AP

b-2=d

⇒ b = 2d------(2)

c - b = d

substitute equation 2 in above equation

we get,

c - b = d

⇒ c - 2 = 2d------(3)

⇒c = 2+2d

substitute the above values in equation 1

det ( A) = (b-2)(c-2)(c-b)------(1)

det (A) = d(2d)d

det (A) = $2d^{3}$ ∈ [2,16]

           = $d^{3}$ ∈ [1,8]

           = $d^{3}$ ∈ [1,2]

from the det ( A)

2d ∈ [2,4]

2+2d ∈ [4,6]

c ∈ [4,6]

Therefore, c lies in the interval [4,6]