Question

Let $\alpha$ and $\beta$ be the roots of the equation $x^2+x+1=0$. Then

$\begin{vmatrix} 3 & \alpha & \beta \\ \alpha & 2+\beta & 1 \\ \beta & 1 & 2+\alpha \end{vmatrix}$ is equal to

• A. 0
• B. 1
• C. 4
• D. 8

Answer 1

Answer: (D)

8

Answer 2

The given equation is $x^2+x+1=0$. Let its roots be $\alpha$ and $\beta$. From Vieta's formulas, we have:

1. Sum of roots: $\alpha + \beta = -1$
2. Product of roots: $\alpha \beta = 1$

Also, since $\alpha$ and $\beta$ are roots of $x^2+x+1=0$, they satisfy the equation: $\alpha^2+\alpha+1=0 \implies \alpha^2 = -\alpha-1$ $\beta^2+\beta+1=0 \implies \beta^2 = -\beta-1$ These roots are the complex cube roots of unity, usually denoted as $\omega$ and $\omega^2$. So, we can consider $\alpha = \omega$ and $\beta = \omega^2$. This also means $\alpha^3 = 1$ and $\beta^3 = 1$.

Let the given determinant be $D$. $D = \begin{vmatrix} 3 & \alpha & \beta \\ \alpha & 2+\beta & 1 \\ \beta & 1 & 2+\alpha \end{vmatrix}$

Apply the column operation $C_1 \to C_1 + C_2 + C_3$. The new elements in the first column will be: $3 + \alpha + \beta = 3 + (-1) = 2$ $\alpha + (2+\beta) + 1 = \alpha + \beta + 3 = (-1) + 3 = 2$ $\beta + 1 + (2+\alpha) = \alpha + \beta + 3 = (-1) + 3 = 2$

So the determinant becomes: $D = \begin{vmatrix} 2 & \alpha & \beta \\ 2 & 2+\beta & 1 \\ 2 & 1 & 2+\alpha \end{vmatrix}$

Take 2 common from the first column: $D = 2 \begin{vmatrix} 1 & \alpha & \beta \\ 1 & 2+\beta & 1 \\ 1 & 1 & 2+\alpha \end{vmatrix}$

Now, apply row operations to create zeros in the first column: $R_2 \to R_2 - R_1$ $R_3 \to R_3 - R_1$

The new elements for $R_2$: $1-1 = 0$ $(2+\beta) - \alpha = 2+\beta-\alpha$ $1 - \beta$

The new elements for $R_3$: $1-1 = 0$ $1 - \alpha$ $(2+\alpha) - \beta = 2+\alpha-\beta$

The determinant simplifies to: $D = 2 \begin{vmatrix} 1 & \alpha & \beta \\ 0 & 2+\beta-\alpha & 1-\beta \\ 0 & 1-\alpha & 2+\alpha-\beta \end{vmatrix}$

Expand the determinant along the first column: $D = 2 \times 1 \times \left[ (2+\beta-\alpha)(2+\alpha-\beta) - (1-\beta)(1-\alpha) \right]$

Let's evaluate the two terms inside the bracket:

Term 1: $(2+\beta-\alpha)(2+\alpha-\beta)$ We know $\beta = -1-\alpha$. Substitute this into $2+\beta-\alpha$: $2+(-1-\alpha)-\alpha = 1-2\alpha$. We know $\alpha = -1-\beta$. Substitute this into $2+\alpha-\beta$: $2+(-1-\beta)-\beta = 1-2\beta$. So, the product is $(1-2\alpha)(1-2\beta)$. Expand this product: $1 - 2\alpha - 2\beta + 4\alpha\beta = 1 - 2(\alpha+\beta) + 4\alpha\beta$. Substitute $\alpha+\beta=-1$ and $\alpha\beta=1$: $1 - 2(-1) + 4(1) = 1 + 2 + 4 = 7$.

Term 2: $(1-\beta)(1-\alpha)$ Expand this product: $1 - \alpha - \beta + \alpha\beta = 1 - (\alpha+\beta) + \alpha\beta$. Substitute $\alpha+\beta=-1$ and $\alpha\beta=1$: $1 - (-1) + 1 = 1 + 1 + 1 = 3$.

Now substitute these values back into the expression for $D$: $D = 2 [7 - 3]$ $D = 2 [4]$ $D = 8$

The final answer is $\boxed{8}$.