© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand that 2, b, c form an Arithmetic Progression (A.P.)
When three numbers 2, b, c are in an A.P., it means there is a common difference d such that
b = 2 + d and c = 2 + 2d.
Step 2: Write the given matrix A and its determinant
The matrix A is:
$$
A =
\begin{bmatrix}
1 & 1 & 1 \\
2 & b & c \\
4 & b^2 & c^2
\end{bmatrix}.
$$
We need to find $ \det(A) $.
Step 3: Perform column transformations to simplify the determinant
Apply the operations:
1) $C_2 \leftarrow C_2 - C_1$
2) $C_3 \leftarrow C_3 - C_1$
This transforms $A$ into:
$$
\begin{vmatrix}
1 & 0 & 0 \\
2 & b-2 & c-2 \\
4 & b^2 - 4 & c^2 - 4
\end{vmatrix}.
$$
Step 4: Factor out common factors to further simplify
Notice that $b^2 - 4 = (b-2)(b+2)$ and $c^2 - 4 = (c-2)(c+2)$. Factoring out $(b-2)(c-2)$ from the second and third columns leads us to a simpler determinant form. Eventually, we obtain:
$$
\det(A) = (b - 2)(c - 2)(c - b).
$$
Step 5: Substitute b and c in terms of d
Since $b = 2 + d$ and $c = 2 + 2d$, we have:
$b - 2 = d$,
$c - 2 = 2d$,
$c - b = (2 + 2d) - (2 + d) = d.
Therefore,
$$
\det(A) = (b - 2)(c - 2)(c - b) = d \times 2d \times d = 2d^3.
$$
Step 6: Apply the given range for $ \det(A) $
It is given that
$$
\det(A) \in [2, 16].
$$
Substituting $ \det(A) = 2d^3 $, we get
$$
2d^3 \in [2, 16].
$$
Dividing through by 2:
$$
d^3 \in [1, 8].
$$
Taking the cube root on both sides:
$$
d \in [1, 2].
$$
Step 7: Find the interval for c
Recall $c = 2 + 2d$. If $d \in [1, 2]$, then
$$
c = 2 + 2d \in [2 + 2(1),\; 2 + 2(2)] = [4, 6].
$$
Thus, $c$ lies in the interval $[4, 6]$.
