Adding a Multiple of One Column to Another

Given   $$\begin{vmatrix} 1 & \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3} & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$

First operation will be $$C_2$$ $$\Rightarrow$$ $$C_2-C_3$$

$$\therefore$$ $$\begin{vmatrix} 1 & \frac{3}{2}- \alpha - \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3}-\alpha - \frac{1}{3} & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$

$$\therefore$$ $$\begin{vmatrix} 1 & - \alpha  & \alpha + \frac{3}{2} \\ 1 & -\alpha & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1-0 & 0 \end{vmatrix} = 0$$

Now we will operate $$R_1$$ $$\Rightarrow$$ $$R_1-R_2$$

$$\therefore$$  $$\begin{vmatrix} 1-1 & \alpha- \alpha & \alpha + \frac{3}{2}-\alpha - \frac{1}{3} \\ 1 & -\alpha & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$

$$\therefore$$ $$\begin{vmatrix} 0 & 0 & \frac{7}{6} \\ 1 & -\alpha & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$

$$\therefore$$ $$\dfrac{7}{6}\left(\left(1\right)\left(3\alpha\ +1\right)-\left(-\alpha\ \right)\left(2\alpha\ +3\right)\right)=0$$

$$\therefore$$ $$2\alpha\ ^2+6\alpha\ +1=0$$

$$\therefore$$ $$\alpha\ =\dfrac{\left(-6\pm\ \sqrt{\ 28}\right)}{4}$$

$$\therefore$$ $$\alpha_1\ =-1.5+\dfrac{\sqrt{\ 7}}{2}$$ and $$\alpha_2\ =-1.5-\dfrac{\sqrt{\ 7}}{2}$$

$$\sqrt{\ 7}\simeq\ 2.645$$

$$\therefore$$ $$\alpha\ _1=-0.177$$ and $$\alpha\ _2=-2.8228$$

$$\therefore$$ $$\alpha\ \in\ \left(0,-3\right)$$

Hence, correct option is B.

We begin with the determinant-valued function

$$f(x)=\begin{vmatrix} \sin^{2}x & -2+\cos^{2}x & \cos 2x\\[4pt] 2+\sin^{2}x & \cos^{2}x & \cos 2x\\[4pt] \sin^{2}x & \cos^{2}x & 1+\cos 2x \end{vmatrix},\qquad x\in[0,\pi].$$

For compactness let us put

$$s=\sin^{2}x,\qquad c=\cos^{2}x.$$

Using the well-known double-angle identity $$\cos 2x=\cos^{2}x-\sin^{2}x=c-s,$$ we can rewrite every entry of the matrix in terms of these symbols.

The third-row, third-column entry becomes

$$1+\cos 2x = 1 + (c-s)=1+c-s.$$

Consequently the matrix takes the form

$$ \begin{vmatrix} s & -2+c & c-s\\[4pt] 2+s & c & c-s\\[4pt] s & c & 1+c-s \end{vmatrix}. $$

To simplify the determinant, we apply elementary row operations, remembering that adding or subtracting a multiple of one row from another does not change the value of the determinant.

First subtract the first row from the second:

$$R_2 \longleftarrow R_2-R_1,$$

which produces

$$R_2:\;(2+s)-s=2$$, $$\quad c-(-2+c)=2$$, $$\quad (c-s)-(c-s)=0.$$

$$ \begin{vmatrix} s & -2+c & c-s\\[4pt] 2 & 2 & 0\\[4pt] s & c & 1+c-s \end{vmatrix}. $$

Next, subtract the first row from the third:

$$R_3 \longleftarrow R_3-R_1,$$

to obtain

$$R_3:\;s-s=0$$, $$\quad c-(-2+c)=2$$, $$\quad (1+c-s)-(c-s)=1.$$

After these two operations the matrix becomes

$$ \begin{vmatrix} s & -2+c & c-s\\[4pt] 2 & 2 & 0\\[4pt] 0 & 2 & 1 \end{vmatrix}. $$

Now we will operate $$C_2 \longleftarrow C_2-C_1,$$

$$ \begin{vmatrix} s & -2+c-S & c-s\\[4pt] 2 & 2-2 & 0\\[4pt] 0 & 2-0 & 1 \end{vmatrix}. $$

$$\therefore$$ we get  $$\begin{vmatrix} s & -2+c-S & c-s\\[4pt] 2 & 0 & 0\\[4pt] 0 & 2 & 1 \end{vmatrix}. $$

Now we expand the determinant along the second row, using the cofactor expansion rule “$$+\;-\;+$$”:

$$ \begin{aligned} f(x) &= -2 \;\begin{vmatrix}2+c-s & c-s\\ 2 & 1\end{vmatrix} \; \end{aligned} $$

Now we expand it we get $$f(x)=-2\left(\left(-2+c-s\right)-2\left(c-s\right)\right)$$

$$\Rightarrow$$ $$f(x)=4-2s+2c$$

Recall that $$s-c=1-2c$$ 

Therefore, $$ f(x)=2+4c $$

The cosine function satisfies $$0\le\cos ^2x\le 1$$ for every real $$x$$, and this remains true on the interval $$[0,\pi]$$. Consequently

$$ 2+4(0)\le f(x)\le 2+4(1)\;\Longrightarrow\;2\le f(x)\le 6. $$

Hence the maximum value attained by $$f(x)$$ is

$$ f_{\max}=4+2\cdot1=6. $$

So, the answer is $$6$$.

Column operation $$C_1 \to C_1+C_2+C_3$$

All rows of new $$C_1$$ equal $$x+6$$, so:

$$f(x) = (x+6)\begin{vmatrix} 1 & 2 & 3 \\ 1 & x+2 & 3 \\ 1 & 2 & x+3 \end{vmatrix}$$

Row operations $$R_2 \to R_2-R_1$$, $$R_3 \to R_3-R_1$$

$$f(x) = (x+6)\begin{vmatrix} 1 & 2 & 3 \\ 0 & x & 0 \\ 0 & 0 & x \end{vmatrix} = x^2(x+6)$$

$$\alpha = 0,\quad \beta = 0,\quad \gamma = -6$$

$$\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)\left[(\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha)\right]$$

$$\alpha+\beta+\gamma = -6$$, $$;\alpha\beta+\beta\gamma+\gamma\alpha = 0$$, $$;\alpha\beta\gamma = 0$$:

$$= (-6)\left[(-6)^2 - 3(0)\right] = (-6)(36) = \boxed{-216}$$