If `A=[[1,1,2],[1,3,4],[1,-1,3]], B=adj(A)` and `C=3A` then `(|adj(B)|)/(|C|)` is

Options

A. `2`
B. `16`
C. `8`
D. `72`

Answer

To solve the problem, we need to find the value of \(\frac{|\text{adj}(B)|}{|C|}\) where \(B = \text{adj}(A)\) and \(C = 3A\).

### Step 1: Understand the properties of determinants
We know that for any square matrix \(A\) of order \(n\):
1. \(|\text{adj}(A)| = |A|^{n-1}\)
2. If \(C = kA\) where \(k\) is a constant, then \(|C| = k^n |A|\)

### Step 2: Determine the order of matrix \(A\)
Matrix \(A\) is given as:
\[ A = \begin{bmatrix} 1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & -1 & 3 \end{bmatrix} \]
This is a \(3 \times 3\) matrix, so \(n = 3\).

### Step 3: Calculate \(|B|\)
Since \(B = \text{adj}(A)\), we can use the property of determinants:
\[ |B| = |\text{adj}(A)| = |A|^{n-1} = |A|^{3-1} = |A|^2 \]

### Step 4: Calculate \(|C|\)
Given \(C = 3A\), we apply the second property:
\[ |C| = |3A| = 3^3 |A| = 27 |A| \]

### Step 5: Calculate \(|\text{adj}(B)|\)
Using the property again:
\[ |\text{adj}(B)| = |B|^{n-1} = |B|^{3-1} = |B|^2 \]
Substituting \(|B| = |A|^2\):
\[ |\text{adj}(B)| = (|A|^2)^{2} = |A|^4 \]

### Step 6: Substitute into the expression
Now we can substitute \(|\text{adj}(B)|\) and \(|C|\) into our expression:
\[ \frac{|\text{adj}(B)|}{|C|} = \frac{|A|^4}{27 |A|} \]
This simplifies to:
\[ \frac{|A|^4}{27 |A|} = \frac{|A|^3}{27} \]

### Step 7: Calculate \(|A|\)
Now we need to calculate \(|A|\):
\[ |A| = 1 \cdot (3 \cdot 3 - 4 \cdot (-1)) - 1 \cdot (1 \cdot 3 - 4 \cdot 1) + 2 \cdot (1 \cdot (-1) - 3 \cdot 1) \]
Calculating each term:
1. First term: \(1 \cdot (9 + 4) = 13\)
2. Second term: \(-1 \cdot (3 - 4) = -1 \cdot (-1) = 1\)
3. Third term: \(2 \cdot (-1 - 3) = 2 \cdot (-4) = -8\)

Adding these together:
\[ |A| = 13 + 1 - 8 = 6 \]

### Step 8: Substitute \(|A|\) back into the expression
Now substituting \(|A| = 6\) into our expression:
\[ \frac{|A|^3}{27} = \frac{6^3}{27} = \frac{216}{27} = 8 \]

### Final Answer
Thus, the final answer is:
\[ \frac{|\text{adj}(B)|}{|C|} = 8 \]