Let $b$ and $c$ be two real numbers such that the numbers $3, b, c$ are in a geometric progression. For the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & b & c \\ 9 & b^2 & c^2 \end{bmatrix}$, if $\det[A] = 162$, then the value of $c$ is

Since $3, b, c$ are in a geometric progression, there's a common ratio $r$ such that:

Term	Value
First term	$3$
Second term	$b = 3r$
Third term	$c = 3r^2$

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Substituting these values into the matrix:

$A = \begin{bmatrix} 1 & 1 & 1 \\ 3 & 3r & 3r^2 \\ 9 & 9r^2 & 9r^4 \end{bmatrix}$

To find the determinant, we can factor out common factors from each row. From row 2, factor out $3$, and from row 3, factor out $9$:

$\det[A] = 3 \times 9 \times \begin{vmatrix} 1 & 1 & 1 \\ 1 & r & r^2 \\ 1 & r^2 & r^4 \end{vmatrix} = 27 \times \begin{vmatrix} 1 & 1 & 1 \\ 1 & r & r^2 \\ 1 & r^2 & r^4 \end{vmatrix}$

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To simplify the remaining determinant, apply row operations: $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:

$\begin{vmatrix} 1 & 1 & 1 \\ 0 & r-1 & r^2-1 \\ 0 & r^2-1 & r^4-1 \end{vmatrix}$

Expanding along the first column:

$= (r-1)(r^4-1) - (r^2-1)(r^2-1)$

Using the factorizations $r^2-1 = (r-1)(r+1)$ and $r^4-1 = (r-1)(r+1)(r^2+1)$:

$= (r-1)(r-1)(r+1)(r^2+1) - (r-1)^2(r+1)^2$

$= (r-1)^2(r+1)[(r^2+1) - (r+1)]$

$= (r-1)^2(r+1)(r^2-r)$

$= r(r-1)^3(r+1)$

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Now we have:

$\det[A] = 27 \times r(r-1)^3(r+1) = 162$

$r(r-1)^3(r+1) = 6$

Testing $r = 2$:

$2(2-1)^3(2+1) = 2 \times 1 \times 3 = 6$ ✓

Therefore, $c = 3r^2 = 3 \times 2^2 = 3 \times 4 = 12$

Answer: Option 4 (12)