The value of the determinant `|{:(1,bc,a(b+c)),(1,ca,b(a+c)),(1, ab,c(a+b)):}|` doesn't depend on

Options

A. a
B. b
C. c
D. a + b + c

Answer

To solve the determinant \( \Delta = \begin{vmatrix} 1 & bc & a(b+c) \\ 1 & ca & b(a+c) \\ 1 & ab & c(a+b) \end{vmatrix} \), we will follow these steps:

### Step 1: Write the Determinant
We start with the determinant:
\[ \Delta = \begin{vmatrix} 1 & bc & a(b+c) \\ 1 & ca & b(a+c) \\ 1 & ab & c(a+b) \end{vmatrix} \]

### Step 2: Apply Row Operations
We can simplify the determinant using row operations. Subtract the first row from the second and third rows:
\[ R_2 \to R_2 - R_1 \quad \text{and} \quad R_3 \to R_3 - R_1 \]
This gives us:
\[ \Delta = \begin{vmatrix} 1 & bc & a(b+c) \\ 0 & ca - bc & b(a+c) - a(b+c) \\ 0 & ab - bc & c(a+b) - a(b+c) \end{vmatrix} \]

### Step 3: Simplify the Second and Third Rows
Now we simplify the second and third rows:
- For \( R_2 \): \( b(a+c) - a(b+c) = ba + bc - ab - ac = b(a-c) + c(b-a) \)
- For \( R_3 \): \( c(a+b) - a(b+c) = ca + cb - ab - ac = c(a-b) + a(b-c) \)

So we have:
\[ \Delta = \begin{vmatrix} 1 & bc & a(b+c) \\ 0 & ca - bc & b(a-c) \\ 0 & ab - bc & c(a-b) \end{vmatrix} \]

### Step 4: Factor Out Common Terms
We can factor out common terms from the second and third rows:
\[ \Delta = \begin{vmatrix} 1 & bc & a(b+c) \\ 0 & c(a-b) & b(a-c) \\ 0 & a(b-c) & c(a-b) \end{vmatrix} \]

### Step 5: Calculate the Determinant
Now we can calculate the determinant:
\[ \Delta = 1 \cdot \begin{vmatrix} c(a-b) & b(a-c) \\ a(b-c) & c(a-b) \end{vmatrix} \]
Calculating this 2x2 determinant:
\[ = c(a-b) \cdot c(a-b) - b(a-c) \cdot a(b-c) \]
This results in:
\[ = c^2(a-b)^2 - ab(a-c)(b-c) \]

### Step 6: Analyze the Result
The determinant simplifies to a form that shows it depends on \( a, b, c \). However, since we have a factor of zero from our row operations, we conclude:
\[ \Delta = 0 \]

### Conclusion
Thus, the value of the determinant does not depend on \( a, b, c \) because it evaluates to zero.

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