Question

The value of the determinant $\begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ac \\ 1 & c & c^2-ab \end{vmatrix}$

• A. a+b+c
• B. $a^2+b^2+c^2$
• C. abc
• D. 0

Answer 1

Answer: (D)

0

Answer 2

To find the value of the determinant $D = \begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ac \\ 1 & c & c^2-ab \end{vmatrix}$, we use properties of determinants.

Step 1: Apply row operations to simplify the determinant. Perform the following row operations:

1. $R_2 \to R_2 - R_1$
2. $R_3 \to R_3 - R_1$

Let's calculate the new elements for $R_2$:

• First element: $1 - 1 = 0$
• Second element: $b - a$
• Third element: $(b^2 - ac) - (a^2 - bc) = b^2 - a^2 - ac + bc$ $= (b-a)(b+a) + c(b-a)$ $= (b-a)(b+a+c)$

Now, let's calculate the new elements for $R_3$:

• First element: $1 - 1 = 0$
• Second element: $c - a$
• Third element: $(c^2 - ab) - (a^2 - bc) = c^2 - a^2 - ab + bc$ $= (c-a)(c+a) + b(c-a)$ $= (c-a)(c+a+b)$

After these operations, the determinant becomes: $D = \begin{vmatrix} 1 & a & a^2-bc \\ 0 & b-a & (b-a)(a+b+c) \\ 0 & c-a & (c-a)(a+b+c) \end{vmatrix}$

Step 2: Factor out common terms from rows. We can take out $(b-a)$ as a common factor from the second row ($R_2$) and $(c-a)$ as a common factor from the third row ($R_3$).

$D = (b-a)(c-a) \begin{vmatrix} 1 & a & a^2-bc \\ 0 & 1 & a+b+c \\ 0 & 1 & a+b+c \end{vmatrix}$

Step 3: Evaluate the simplified determinant. Observe the determinant obtained in Step 2. The second row ($R_2$) and the third row ($R_3$) are identical: $R_2 = [0 \quad 1 \quad a+b+c]$ $R_3 = [0 \quad 1 \quad a+b+c]$

A property of determinants states that if any two rows (or columns) of a determinant are identical, the value of the determinant is zero.

Therefore, the determinant $\begin{vmatrix} 1 & a & a^2-bc \\ 0 & 1 & a+b+c \\ 0 & 1 & a+b+c \end{vmatrix}$ is equal to 0.

So, $D = (b-a)(c-a) \times 0 = 0$.

The final answer is $\boxed{0}$.