Question

The value of the determinant \(\begin{vmatrix}bc & ca & ab\\ a^{3} & b^{3} & c^{3}\\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{vmatrix}\) is

• A. $a^{5}-1$
• B. $a^{2}bc+ab^{2}c+abc^{2}$
• C. $ab(a+b+c)$
• D. $a^{4}b^{4}c^{4}(a+b+c)$
• E. 0

Hint:

Determinants Tip: If you see cyclic symmetry like $bc, ca, ab$ accompanied by $1/a, 1/b, 1/c$, multiplying $C_1, C_2, C_3$ by $a, b, c$ respectively almost always forces two identical rows!

Answer 1

Answer: (E)

Concept:
The properties of determinants allow us to multiply a row or column by a scalar $k$, provided we divide the entire determinant by $k$ to maintain equality. Furthermore, if any two rows or columns of a determinant are identical or proportional, the value of the entire determinant is exactly $0$.

Step 1: Set up the determinant.
Let $\Delta$ equal the given determinant: $$\Delta = \begin{vmatrix}bc & ca & ab
a^{3} & b^{3} & c^{3}
\frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{vmatrix}$$

Step 2: Apply column operations.
To eliminate the fractions in the third row, multiply Column 1 ($C_1$) by $a$, Column 2 ($C_2$) by $b$, and Column 3 ($C_3$) by $c$. To balance this, we must divide the determinant by $abc$: $$\Delta = \frac{1}{abc} \begin{vmatrix}a(bc) & b(ca) & c(ab)
a(a^{3}) & b(b^{3}) & c(c^{3})
a(\frac{1}{a}) & b(\frac{1}{b}) & c(\frac{1}{c})\end{vmatrix}$$

Step 3: Simplify the elements inside the determinant.
Perform the multiplication for each term: $$\Delta = \frac{1}{abc} \begin{vmatrix}abc & abc & abc
a^{4} & b^{4} & c^{4}
1 & 1 & 1\end{vmatrix}$$

Step 4: Extract the common factor from Row 1.
Every element in the first row ($R_1$) is $abc$. We can factor this out to the front of the determinant: $$\Delta = \frac{abc}{abc} \begin{vmatrix}1 & 1 & 1
a^{4} & b^{4} & c^{4}
1 & 1 & 1\end{vmatrix}$$

Step 5: Analyze the rows and conclude.
The coefficient $\frac{abc}{abc}$ simplifies to $1$. Notice that Row 1 ($1, 1, 1$) and Row 3 ($1, 1, 1$) are identical. According to the properties of determinants, if two rows are identical, the determinant is zero. $$\Delta = 1 \times 0 = 0$$ Hence the correct answer is (E) 0.