A matrix $A$ is given such that $A = \left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  1&3&4 \\
  1&{ - 1}&3
\end{array}} \right]$ . If it is also given that $B = adjA$ and $C = 3A$ , then what is the value of $\dfrac{{\left| {adjB} \right|}}{{\left| C \right|}}$ ?
A. 16
B. 2
C. 8
D. 72

Hint: When a matrix $A$ is multiplied by a constant value $k$, the value of its determinant also gets multiplied by a factor of ${k^n}$, where $n$ is the order of the matrix. This gives $\left| {kA} \right| = {k^n}\left| A \right|$ . The adjoint of the matrix is the transpose of a matrix of its cofactors. It has the following property: $adj(adjA) = {\left| A \right|^{n - 2}}.A$

Formula used:
1. $adj(adjA) = {\left| A \right|^{n - 2}}.A$

Complete step by step Solution:
Matrix $A$ is given such that
$A = \left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  1&3&4 \\
  1&{ - 1}&3
\end{array}} \right]$
It is also given that:
$B = adjA$ … (1)
And $C = 3A$ … (2)
Now, using equation (1),
$adjB = adj(adjA)$
Calculating its determinant,
$\left| {adjB} \right| = \left| {adj(adjA)} \right|$
From the properties of adjoints and determinants of matrices we know that $adj(adjA) = {\left| A \right|^{n - 2}}.A$ , where $n$ is the order of the matrix.
The given matrix is of order 3, therefore, $n = 3$ … (3)
Now, we’ll calculate the value of its determinant.
$\left| A \right| = 1(9 + 4) - 1(3 - 4) + 2( - 1 - 3)$
On simplifying further,
$\left| A \right| = 13 + 1 - 8 = 6$ … (4)
Using this, we’ll now calculate the value of $adj(adjA)$ .
$adj(adjA) = {\left| A \right|^{n - 2}}.A$
Substituting all the values, we get
$adj(adjA) = {(6)^{3 - 2}}A = 6A$
Therefore, $adjB = 6A$
Calculating its determinant,
$\left| {adjB} \right| = \left| {6A} \right|$
We also know that $\left| {kA} \right| = {k^n}\left| A \right|$ … (5)
Therefore, using this property we get
$\left| {adjB} \right| = {6^n}\left| A \right|$
Substituting the required values from (3) and (4) in the above equation we get
$\left| {adjB} \right| = {6^3}(6) = {6^4}$ … (6)
Using (3), and calculating its determinant,
$\left| C \right| = \left| {3A} \right|$
Using (5) and substituting all the required values from (3) and (4),
$\left| C \right| = {3^3}\left| A \right| = {3^3}(6)$ … (7)
Dividing equation (6) by equation (7),
$\dfrac{{\left| {adjB} \right|}}{{\left| C \right|}} = \dfrac{{{6^4}}}{{{3^3}(6)}} = {2^3}$
On simplifying further,
$\dfrac{{\left| {adjB} \right|}}{{\left| C \right|}} = 8$

Therefore, the correct option is C.

Note: In the above question, we have calculated the value of $adj(adjA)$ first, followed by the value of its determinant. However, you can also calculate the value of $\left| {adj(adjA)} \right|$ in the first step itself using the formula $\left| {adj(adjA)} \right| = {\left| A \right|^{{n^2} - 2n + 1}}$ , where $n$ is the order of the square matrix, respectively.