The determinant `Delta =|(lamda a , lamda^(2)+a^(2),1),(lamda b,lamda^(2)+b^(2),1),(lamdac,lamda^(2)+c^(2),1)|` equals

Options

A. `lamda(a-b)(b-c)(c-a)`
B. `lamda(a^(2)+b^(2)+c^(2))`
C. `lamda(a+b+c)`
D. `lamda^(2)(a-b)(b-c)(c-a)`

Answer

To solve the determinant

\[ \Delta = \begin{vmatrix} \lambda a & \lambda^2 + a^2 & 1 \\ \lambda b & \lambda^2 + b^2 & 1 \\ \lambda c & \lambda^2 + c^2 & 1 \end{vmatrix} \]

we can follow these steps:

### Step 1: Apply Row Operations
We will perform the operation \( C_2 = C_2 - \lambda^2 C_3 \). This means we will subtract \(\lambda^2\) times the third column from the second column.

\[ \Delta = \begin{vmatrix} \lambda a & (\lambda^2 + a^2 - \lambda^2) & 1 \\ \lambda b & (\lambda^2 + b^2 - \lambda^2) & 1 \\ \lambda c & (\lambda^2 + c^2 - \lambda^2) & 1 \end{vmatrix} \]

This simplifies to:

\[ \Delta = \begin{vmatrix} \lambda a & a^2 & 1 \\ \lambda b & b^2 & 1 \\ \lambda c & c^2 & 1 \end{vmatrix} \]

### Step 2: Factor Out \(\lambda\)
Now, we can factor out \(\lambda\) from the first column:

\[ \Delta = \lambda \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} \]

### Step 3: Recognize the Standard Determinant
The determinant

\[ \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} \]

is a standard determinant that can be evaluated using the formula for the determinant of a matrix of this form, which is given by:

\[ (a-b)(b-c)(c-a) \]

### Step 4: Combine the Results
Thus, we can write:

\[ \Delta = \lambda \cdot (a-b)(b-c)(c-a) \]

### Final Answer
The final value of the determinant \(\Delta\) is:

\[ \Delta = \lambda (a-b)(b-c)(c-a) \]

---