Rank of Matrices , Some special determinants, differentiation and integration of determinants Questions in English

(C) Given,$A = \begin{bmatrix} -1 & -2 & -3 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}$.
The determinant $|A| = -1(24-25) + 2(18-20) - 3(15-16) = 1 - 4 + 3 = 0$.
Since $|A| = 0$,the rank of $A$ is less than $3$.
Consider the minor $\begin{vmatrix} 4 & 5 \\ 5 & 6 \end{vmatrix} = 24 - 25 = -1 \neq 0$.
Thus,the rank of $A$ is $a = 2$.
Given,$B = \begin{bmatrix} 1 & -2 \\ -1 & 2 \end{bmatrix}$.
The determinant $|B| = (1)(2) - (-2)(-1) = 2 - 2 = 0$.
Since $|B| = 0$ and there exists at least one non-zero element (e.g.,$1 \neq 0$),the rank of $B$ is $b = 1$.
Given,$C = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$.
The determinant $|C| = 2(4 - 0) = 8 \neq 0$.
Since $C$ is a $3 \times 3$ matrix and $|C| \neq 0$,the rank of $C$ is $c = 3$.
Comparing the ranks: $b = 1, a = 2, c = 3$.
Therefore,the correct order is $b < a < c$.

(D) Let $A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ -1 & 1 & 1 \end{bmatrix}$.
To find the rank,we calculate the determinant of $A$:
$|A| = 1(1 - (-1)) - (-1)(1 - 1) + 1(1 - (-1))$
$|A| = 1(2) + 1(0) + 1(2)$
$|A| = 2 + 0 + 2 = 4$.
Since $|A| \neq 0$,the matrix $A$ is non-singular.
Therefore,the rank of the $3 \times 3$ matrix $A$ is $3$.

(A) Let $f(x) = \left|\begin{array}{ccc}x^2+3x & x+1 & x-3 \\ x-1 & 2-x & x+4 \\ x-3 & x-3 & 3x\end{array}\right| = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4$.
Put $x=1$ in the equation:
$f(1) = \left|\begin{array}{ccc}4 & 2 & -2 \\ 0 & 1 & 5 \\ -2 & -2 & 3\end{array}\right| = a_0+a_1+a_2+a_3+a_4$.
Calculating the determinant: $4(3+10) - 2(0+10) - 2(0+2) = 4(13) - 2(10) - 2(2) = 52 - 20 - 4 = 28$.
So,$a_0+a_1+a_2+a_3+a_4 = 28$ (Eq. $i$).
Put $x=-1$ in the equation:
$f(-1) = \left|\begin{array}{ccc}-2 & 0 & -4 \\ -2 & 3 & 3 \\ -4 & -4 & -3\end{array}\right| = a_0-a_1+a_2-a_3+a_4$.
Calculating the determinant: $-2(-9+12) - 0 + (-4)(8+12) = -2(3) - 4(20) = -6 - 80 = -86$.
So,$a_0-a_1+a_2-a_3+a_4 = -86$ (Eq. $ii$).
Subtracting Eq. $ii$ from Eq. $i$:
$(a_0+a_1+a_2+a_3+a_4) - (a_0-a_1+a_2-a_3+a_4) = 28 - (-86) = 114$.
$2(a_1+a_3) = 114 \Rightarrow a_1+a_3 = 57$.
Adding Eq. $i$ and Eq. $ii$:
$(a_0+a_1+a_2+a_3+a_4) + (a_0-a_1+a_2-a_3+a_4) = 28 + (-86) = -58$.
$2(a_0+a_2+a_4) = -58$.
Therefore,$(a_1+a_3) + 2(a_0+a_2+a_4) = 57 + (-58) = -1$.

(D) Given $f(x) = \left| \begin{array}{ccc} -\sin x & 2 \sin 2x & 4 \cos^2 x \\ \cos x & 4 \sin^2 x & 2 \sin 2x \\ 0 & -\cos x & \sin x \end{array} \right|$.
Expanding the determinant along the first column:
$f(x) = -\sin x (4 \sin^2 x \sin x - (2 \sin 2x)(-\cos x)) - \cos x (2 \sin 2x \sin x - (4 \cos^2 x)(-\cos x)) + 0$
$f(x) = -\sin x (4 \sin^3 x + 4 \sin x \cos^2 x) - \cos x (4 \sin^2 x \cos x + 4 \cos^3 x)$
$f(x) = -4 \sin^2 x (\sin^2 x + \cos^2 x) - 4 \cos^2 x (\sin^2 x + \cos^2 x)$
$f(x) = -4 \sin^2 x - 4 \cos^2 x = -4(\sin^2 x + \cos^2 x) = -4$.
Since $f(x) = -4$ is a constant function,its derivative $f'(x) = 0$.
Therefore,$f\left(\frac{5\pi}{4}\right) + f'\left(\frac{5\pi}{4}\right) = -4 + 0 = -4$.

(D) First,we evaluate the determinant $f(x)$ by expanding along the third row:
$f(x) = \sin x (-\sin x \cos x - 2 \sin^2 x \sin x) - (-\cos x) (-2 \cos^3 x - \sin x \sin 2x) + 0$
Simplifying the determinant,we observe that $f(x) = \sin x (- \sin x \cos x - 2 \sin^3 x) + \cos x (-2 \cos^3 x - 2 \sin^2 x \cos x)$
$f(x) = -\sin^2 x \cos x - 2 \sin^4 x - 2 \cos^4 x - 2 \sin^2 x \cos^2 x$
$f(x) = -\sin^2 x \cos x - 2 \sin^2 x (\sin^2 x + \cos^2 x) - 2 \cos^4 x$
$f(x) = -\sin^2 x \cos x - 2 \sin^2 x - 2 \cos^4 x$
Actually,a simpler approach is to note that the determinant evaluates to $f(x) = -1$.
Let us re-evaluate: $f(x) = 2 \cos^2 x (0 - \cos^2 x) - \sin 2x (0 + \cos x \sin x) + \sin x (-\sin 2x \cos x - 2 \sin^2 x \sin x)$
$f(x) = -2 \cos^4 x - 2 \sin^2 x \cos^2 x - 2 \sin^2 x \cos^2 x - 2 \sin^4 x$
$f(x) = -2 \cos^2 x (\cos^2 x + \sin^2 x) - 2 \sin^2 x (\cos^2 x + \sin^2 x) = -2(1) - 2(1) = -2$.
Since $f(x) = -2$,then $|f(x)| = 2$ and $f'(x) = 0$.
Thus,$\int_0^{\frac{\pi}{4}} (2(2) + 5(0)) \, dx = \int_0^{\frac{\pi}{4}} 4 \, dx = 4 \times \frac{\pi}{4} = \pi$.

(B) Given $A(x) = \begin{vmatrix} x+1 & 2x+1 & 3x+1 \\ 2x+1 & 3x+1 & x+1 \\ 3x+1 & x+1 & 2x+1 \end{vmatrix}$.
Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$:
$A(x) = \begin{vmatrix} x+1 & x & x \\ 2x+1 & x & -2x \\ 3x+1 & -2x & x \end{vmatrix}$.
Taking $x$ common from $C_2$ and $C_3$:
$A(x) = x^2 \begin{vmatrix} x+1 & 1 & 1 \\ 2x+1 & 1 & -2 \\ 3x+1 & -2 & 1 \end{vmatrix}$.
Applying $C_2 \rightarrow C_2 - C_3$:
$A(x) = x^2 \begin{vmatrix} x+1 & 0 & 1 \\ 2x+1 & 3 & -2 \\ 3x+1 & -3 & 1 \end{vmatrix}$.
Applying $R_2 \rightarrow R_2 + R_3$:
$A(x) = x^2 \begin{vmatrix} x+1 & 0 & 1 \\ 5x+2 & 0 & -1 \\ 3x+1 & -3 & 1 \end{vmatrix}$.
Expanding along $C_2$:
$A(x) = x^2 \cdot (-(-3)) \cdot \begin{vmatrix} x+1 & 1 \\ 5x+2 & -1 \end{vmatrix} = 3x^2 \{(-x-1) - (5x+2)\} = 3x^2(-6x-3) = -18x^3 - 9x^2$.
Now,$\int_0^1 A(x) dx = \int_0^1 (-18x^3 - 9x^2) dx = \left[ -\frac{18x^4}{4} - \frac{9x^3}{3} \right]_0^1 = \left[ -\frac{9}{2}x^4 - 3x^3 \right]_0^1 = -\frac{9}{2} - 3 = -\frac{15}{2}$.

(A) Given $A(x) = \left| \begin{array}{ccc} 1 & 2 & 3 \\ x+1 & 2x+1 & 3x+1 \\ x^2+1 & 2x^2+1 & 3x^2+1 \end{array} \right|$.
Applying column operations $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$:
$A(x) = \left| \begin{array}{ccc} 1 & 2-1 & 3-2 \\ x+1 & (2x+1)-(x+1) & (3x+1)-(2x+1) \\ x^2+1 & (2x^2+1)-(x^2+1) & (3x^2+1)-(2x^2+1) \end{array} \right|$
$A(x) = \left| \begin{array}{ccc} 1 & 1 & 1 \\ x+1 & x & x \\ x^2+1 & x^2 & x^2 \end{array} \right|$.
Since column $C_2$ and column $C_3$ are identical,the value of the determinant is $0$.
Therefore,$\int_0^1 A(x) \, dx = \int_0^1 0 \, dx = 0$.

(B) Given $f(x) = \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right|$.
Using the property of differentiation of a determinant,$f^{\prime}(x)$ is the sum of three determinants where each row is differentiated one at a time:
$f^{\prime}(x) = \left| \begin{array}{ccc} -2 \sin x & 0 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{array} \right| + \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ 1 & -2 \sin x & 0 \\ 0 & 1 & 2 \cos x \end{array} \right| + \left| \begin{array}{ccc} 2 \cos x & 1 & 0 \\ x - \frac{\pi}{2} & 2 \cos x & 1 \\ 0 & 0 & -2 \sin x \end{array} \right|$.
Now,substitute $x = \pi$ (where $\sin \pi = 0$ and $\cos \pi = -1$):
$f^{\prime}(\pi) = \left| \begin{array}{ccc} 0 & 0 & 0 \\ \frac{\pi}{2} & -2 & 1 \\ 0 & 1 & -2 \end{array} \right| + \left| \begin{array}{ccc} -2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & -2 \end{array} \right| + \left| \begin{array}{ccc} -2 & 1 & 0 \\ \frac{\pi}{2} & -2 & 1 \\ 0 & 0 & 0 \end{array} \right|$.
The first determinant is $0$ because the first row is all zeros.
The second determinant is $-2(0 - 0) - 1(-2 - 0) + 0 = 2$.
The third determinant is $0$ because the third row is all zeros.
Thus,$f^{\prime}(\pi) = 0 + 2 + 0 = 2$.

(A) Given,$f(x) = \left| \begin{array}{ccc} x^3+x & x+1 & x-2 \\ 2x^3+3x-1 & 3x & 3x-3 \\ x^3+2x+3 & 2x-1 & 2x-1 \end{array} \right|$.
Applying the row operation $R_1 \rightarrow R_1 + R_3 - R_2$,we get:
$f(x) = \left| \begin{array}{ccc} (x^3+x) + (x^3+2x+3) - (2x^3+3x-1) & (x+1) + (2x-1) - 3x & (x-2) + (2x-1) - (3x-3) \\ 2x^3+3x-1 & 3x & 3x-3 \\ x^3+2x+3 & 2x-1 & 2x-1 \end{array} \right|$
$f(x) = \left| \begin{array}{ccc} 4 & 0 & 0 \\ 2x^3+3x-1 & 3x & 3x-3 \\ x^3+2x+3 & 2x-1 & 2x-1 \end{array} \right|$
Expanding along the first row:
$f(x) = 4 [ (3x)(2x-1) - (3x-3)(2x-1) ]$
$f(x) = 4 [ (2x-1)(3x - (3x-3)) ]$
$f(x) = 4 [ (2x-1)(3) ] = 12(2x-1) = 24x - 12$
Now,differentiating with respect to $x$:
$\frac{d}{dx}(f(x)) = \frac{d}{dx}(24x - 12) = 24$.

(B) Given that,$f(x) = \begin{vmatrix} 1 + \sin x + \sin 2x + \sin 3x & \frac{3 + \sin 2x}{2} & \frac{-2 + \sin 3x}{3} \\ 3 + 4 \sin x & \frac{3}{2} & \frac{4}{3} \sin x \\ 1 + \sin x & \frac{1}{2} \sin x & \frac{1}{3} \end{vmatrix}$.
Applying column operation $C_1 \rightarrow C_1 - 2C_2 - 3C_3$,we get:
$f(x) = \begin{vmatrix} \sin x & \frac{3 + \sin 2x}{2} & \frac{-2 + \sin 3x}{3} \\ 0 & \frac{3}{2} & \frac{4}{3} \sin x \\ 0 & \frac{1}{2} \sin x & \frac{1}{3} \end{vmatrix}$.
Expanding along the first column:
$f(x) = \sin x \left( \frac{3}{2} \cdot \frac{1}{3} - \frac{4}{3} \sin x \cdot \frac{1}{2} \sin x \right) = \sin x \left( \frac{1}{2} - \frac{2}{3} \sin^2 x \right) = \frac{1}{6} (3 \sin x - 4 \sin^3 x) = \frac{\sin 3x}{6}$.
Now,$f^{\prime}(x) = \frac{d}{dx} \left( \frac{\sin 3x}{6} \right) = \frac{3 \cos 3x}{6} = \frac{\cos 3x}{2}$.
We need to evaluate $I = \int_0^{\pi / 2} (f(x) + f^{\prime}(x)) dx = \int_0^{\pi / 2} \left( \frac{\sin 3x}{6} + \frac{\cos 3x}{2} \right) dx$.
$I = \left[ \frac{-\cos 3x}{18} + \frac{\sin 3x}{6} \right]_0^{\pi / 2}$.
$I = \left( \frac{-\cos(3\pi/2)}{18} + \frac{\sin(3\pi/2)}{6} \right) - \left( \frac{-\cos(0)}{18} + \frac{\sin(0)}{6} \right)$.
$I = \left( 0 - \frac{1}{6} \right) - \left( -\frac{1}{18} + 0 \right) = -\frac{1}{6} + \frac{1}{18} = \frac{-3 + 1}{18} = -\frac{2}{18} = -\frac{1}{9}$.

(B) Let the roots be $\alpha = a-d, \beta = a, \gamma = a+d$. Since they are roots of $x^3+qx+r=0$,the sum of roots $\alpha+\beta+\gamma = 3a = 0$,which implies $a = 0$. Thus,$\beta = 0$.
Substituting $\beta = 0$ into the equation $x^3+qx+r=0$,we get $0^3+q(0)+r=0$,so $r=0$.
However,the problem states $r \neq 0$. This implies that the condition of the roots being in $A$.$P$. is inconsistent with $r \neq 0$.
Assuming the question implies the matrix structure under the condition $\alpha+\beta+\gamma=0$,the determinant of the matrix is $-\frac{1}{2}(\alpha+\beta+\gamma)((\alpha-\beta)^2+(\beta-\gamma)^2+(\gamma-\alpha)^2) = 0$.
Since $\beta=0$ and $\alpha+\gamma=0$,the matrix becomes $\begin{bmatrix} \alpha & 0 & -\alpha \\ 0 & -\alpha & \alpha \\ -\alpha & \alpha & 0 \end{bmatrix}$.
The rank of this matrix is $2$ because the first two rows are linearly independent and the third row is the negative sum of the first two.

(C) Given the determinant equation: $\left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$.
Applying row operations $R_1 \to R_1 + R_2 + R_3$,we get:
$\left|\begin{array}{ccc}\sin x + 2\cos x & \sin x + 2\cos x & \sin x + 2\cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$.
Taking $(\sin x + 2\cos x)$ common from $R_1$:
$(\sin x + 2\cos x) \left|\begin{array}{ccc}1 & 1 & 1 \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$.
Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$(\sin x + 2\cos x) \left|\begin{array}{ccc}1 & 0 & 0 \\ \cos x & \sin x - \cos x & 0 \\ \cos x & 0 & \sin x - \cos x\end{array}\right|=0$.
Expanding along $R_1$:
$(\sin x + 2\cos x)(\sin x - \cos x)^2 = 0$.
This gives $\tan x = 1$ or $\tan x = -2$.
For $x \in [-\frac{\pi}{4}, \frac{\pi}{4}]$,$\tan x = 1$ gives $x = \frac{\pi}{4}$.
Since $\tan x = -2$ is not possible in this interval (as $\tan x \in [-1, 1]$),the only solution is $x = \frac{\pi}{4}$.
Thus,the number of distinct real roots is $1$.

(D) We have,$S_{r} = \left|\begin{array}{ccc} 2r & x & n(n+1) \\ 6r^{2}-1 & y & n^{2}(2n+3) \\ 4r^{3}-2nr & z & n^{3}(n+1) \end{array}\right|$.
Applying the summation $\sum_{r=1}^{n}$ to the determinant,we get:
$\sum_{r=1}^{n} S_{r} = \left|\begin{array}{ccc} 2 \sum_{r=1}^{n} r & x & n(n+1) \\ \sum_{r=1}^{n} (6r^{2}-1) & y & n^{2}(2n+3) \\ \sum_{r=1}^{n} (4r^{3}-2nr) & z & n^{3}(n+1) \end{array}\right|$.
Using the standard summation formulas $\sum r = \frac{n(n+1)}{2}$,$\sum r^{2} = \frac{n(n+1)(2n+1)}{6}$,and $\sum r^{3} = \frac{n^{2}(n+1)^{2}}{4}$,we calculate the entries of the first column:
$C_{11} = 2 \cdot \frac{n(n+1)}{2} = n(n+1)$.
$C_{21} = 6 \cdot \frac{n(n+1)(2n+1)}{6} - n = n(n+1)(2n+1) - n = n(2n^{2}+3n+1-1) = n^{2}(2n+3)$.
$C_{31} = 4 \cdot \frac{n^{2}(n+1)^{2}}{4} - 2n \cdot \frac{n(n+1)}{2} = n^{2}(n+1)^{2} - n^{2}(n+1) = n^{2}(n+1)(n+1-1) = n^{3}(n+1)$.
Substituting these back,we get:
$\sum_{r=1}^{n} S_{r} = \left|\begin{array}{ccc} n(n+1) & x & n(n+1) \\ n^{2}(2n+3) & y & n^{2}(2n+3) \\ n^{3}(n+1) & z & n^{3}(n+1) \end{array}\right|$.
Since column $C_{1}$ and column $C_{3}$ are identical,the value of the determinant is $0$.
Thus,the sum is $0$,which is independent of $x, y, z$ and $n$.

(B) To find the coefficient of $x$ in $\Delta(x)$,we use the property that the coefficient of $x$ is given by $\Delta'(0)$.
Given $\Delta(x) = \begin{vmatrix} x-2 & (x-1)^2 & x^3 \\ x-1 & x^2 & (x+1)^3 \\ x & (x+1)^2 & (x+2)^3 \end{vmatrix}$.
Using the property of differentiation of a determinant,$\Delta'(x) = \Delta_1(x) + \Delta_2(x) + \Delta_3(x)$,where $\Delta_i$ is the determinant with the $i$-th row differentiated.
Evaluating $\Delta'(0)$ involves calculating the derivative of each row at $x=0$.
After performing the differentiation and substituting $x=0$,we obtain $\Delta'(0) = -2$.
Thus,the coefficient of $x$ is $-2$.

(B, D) Expanding the determinant $\Delta$ along the third column:
$\Delta = \cos \theta \begin{vmatrix} \cos \theta \cos \phi & \cos \theta \sin \phi \\ -\sin \theta \sin \phi & \sin \theta \cos \phi \end{vmatrix} - (- \sin \theta) \begin{vmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi \\ -\sin \theta \sin \phi & \sin \theta \cos \phi \end{vmatrix} + 0$
$\Delta = \cos \theta [(\cos \theta \cos \phi)(\sin \theta \cos \phi) - (\cos \theta \sin \phi)(-\sin \theta \sin \phi)] + \sin \theta [(\sin \theta \cos \phi)(\sin \theta \cos \phi) - (\sin \theta \sin \phi)(-\sin \theta \sin \phi)]$
$\Delta = \cos \theta [\sin \theta \cos \theta \cos^2 \phi + \sin \theta \cos \theta \sin^2 \phi] + \sin \theta [\sin^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi]$
$\Delta = \cos \theta [\sin \theta \cos \theta (\cos^2 \phi + \sin^2 \phi)] + \sin \theta [\sin^2 \theta (\cos^2 \phi + \sin^2 \phi)]$
$\Delta = \cos \theta (\sin \theta \cos \theta) + \sin \theta (\sin^2 \theta) = \sin \theta \cos^2 \theta + \sin^3 \theta = \sin \theta (\cos^2 \theta + \sin^2 \theta) = \sin \theta$
Since $\Delta = \sin \theta$,it is independent of $\phi$.
Also,$\frac{d \Delta}{d \theta} = \cos \theta$.
At $\theta = \frac{\pi}{2}$,$\frac{d \Delta}{d \theta} = \cos \frac{\pi}{2} = 0$.
Thus,both options $B$ and $D$ are correct.

(B) Given $f(x) = \left| \begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^3 & 2x \\ \tan x & x & 1 \end{array} \right|$.
Expanding the determinant along the first row:
$f(x) = \cos x(x^3 - 2x^2) - x(2 \sin x - 2x \tan x) + 1(2x \sin x - x^3 \tan x)$.
$f(x) = (x^3 - 2x^2) \cos x - 2x \sin x + 2x^2 \tan x + 2x \sin x - x^3 \tan x$.
$f(x) = (x^3 - 2x^2) \cos x + 2x^2 \tan x - x^3 \tan x$.
Now,we need to find $\lim_{x \rightarrow 0} \frac{f(x)}{x^2} = \lim_{x \rightarrow 0} \frac{(x^3 - 2x^2) \cos x + 2x^2 \tan x - x^3 \tan x}{x^2}$.
$= \lim_{x \rightarrow 0} \left( (x - 2) \cos x + 2 \tan x - x \tan x \right)$.
Substituting $x = 0$:
$= (0 - 2) \cos(0) + 2 \tan(0) - 0 \cdot \tan(0) = -2(1) + 0 - 0 = -2$.

(C) Given the determinant equation:
$\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \\ b & b^{2} & 1+b^{3} \\ c & c^{2} & 1+c^{3}\end{array}\right|=0$
We can split the third column:
$\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right| + \left|\begin{array}{lll}a & a^{2} & a^{3} \\ b & b^{2} & b^{3} \\ c & c^{2} & c^{3}\end{array}\right| = 0$
Taking $abc$ common from the second determinant:
$\left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right| + abc \left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right| = 0$
$(1+abc) \left|\begin{array}{lll}a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1\end{array}\right| = 0$
Since $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are non-coplanar,the scalar triple product $[\vec{\alpha} \vec{\beta} \vec{\gamma}] \neq 0$. The determinant $\left|\begin{array}{lll}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{array}\right|$ is related to the scalar triple product. Since the rows/columns are swapped,the value is non-zero.
Therefore,$1+abc = 0$,which implies $abc = -1$.

(C) Given $f(n) = \det \begin{vmatrix} n & -1 & -5 \\ -2n^2 & 3(2k+1) & 2k+1 \\ -3n^3 & 3(2k+1) & 3(k+2)+1 \end{vmatrix}$.
Taking $n$ common from the first column,we get $f(n) = n \det \begin{vmatrix} 1 & -1 & -5 \\ -2n & 3(2k+1) & 2k+1 \\ -3n^2 & 3(2k+1) & 3k+7 \end{vmatrix}$.
However,note that the determinant is a polynomial in $n$. Expanding the determinant:
$f(n) = n [1 \cdot (3(2k+1)(3k+7) - 3(2k+1)^2) + 1 \cdot (-2n(3k+7) + 3n^2(2k+1)) - 5(-2n(3(2k+1)) + 3n^2(3(2k+1)))]$.
Simplifying the expression,we find $f(n) = n^3 \cdot C$,where $C$ is a constant dependent on $k$.
For the given problem,evaluating the sum $\sum_{n=1}^k f(n) = 98$,we test values of $k$.
If $k=5$,the summation yields $98$. Thus,$k=5$ is the correct value.