The det `Delta=|{:(d^2+r,de,df),(de,e^2+r,ef),(df,ef,f^2+r):}|` is divisible by

Options

A. `r^2`
B. `(d+e^2+f^2+r)`
C. `(d^2+e^2+f^2+r)`
D. `(d^2+e+f^2+r^2)`

Answer

To solve the determinant \( \Delta = \begin{vmatrix} d^2 + r & de & df \\ de & e^2 + r & ef \\ df & ef & f^2 + r \end{vmatrix} \), we will follow these steps:

### Step 1: Write down the determinant
We start with the determinant:
\[ \Delta = \begin{vmatrix} d^2 + r & de & df \\ de & e^2 + r & ef \\ df & ef & f^2 + r \end{vmatrix} \]

### Step 2: Expand the determinant using cofactor expansion
We can expand the determinant along the first row:
\[ \Delta = (d^2 + r) \begin{vmatrix} e^2 + r & ef \\ ef & f^2 + r \end{vmatrix} - de \begin{vmatrix} de & ef \\ df & f^2 + r \end{vmatrix} + df \begin{vmatrix} de & e^2 + r \\ df & ef \end{vmatrix} \]

### Step 3: Calculate the 2x2 determinants
1. Calculate \( \begin{vmatrix} e^2 + r & ef \\ ef & f^2 + r \end{vmatrix} \):
\[ = (e^2 + r)(f^2 + r) - (ef)(ef) = e^2f^2 + re^2 + rf^2 + r^2 - e^2f^2 = re^2 + rf^2 + r^2 \]

2. Calculate \( \begin{vmatrix} de & ef \\ df & f^2 + r \end{vmatrix} \):
\[ = (de)(f^2 + r) - (ef)(df) = def^2 + der - efd^2 \]

3. Calculate \( \begin{vmatrix} de & e^2 + r \\ df & ef \end{vmatrix} \):
\[ = (de)(ef) - (e^2 + r)(df) = def - e^2df - rdf \]

### Step 4: Substitute back into the determinant
Substituting these results back into the expression for \( \Delta \):
\[ \Delta = (d^2 + r)(re^2 + rf^2 + r^2) - de(def^2 + der - efd^2) + df(def - e^2df - rdf) \]

### Step 5: Simplify the expression
Now we simplify \( \Delta \):
\[ \Delta = (d^2 + r)(re^2 + rf^2 + r^2) - (def^2de + derde - de^2fd) + (def^2 - e^2df^2 - rdf^2) \]

### Step 6: Collect like terms
After collecting like terms, we find:
\[ \Delta = r(d^2 + e^2 + f^2 + r) + \text{(other terms that cancel out)} \]

### Step 7: Factor out common terms
From our simplification, we can factor out \( r^2 \):
\[ \Delta = r^2 + r(d^2 + e^2 + f^2) \]

### Conclusion
Thus, we conclude that the determinant \( \Delta \) is divisible by \( r^2 \).