Expansion of determinants, Solution of equation in the form of determinants and area of triangle and Equation of Line Questions in English

(D) Given the determinant: $\Delta = \left|\begin{array}{ccc}x & 3x+2 & 2x-1 \\ 2x-1 & 4x & 3x+1 \\ 7x-2 & 17x+6 & 12x-1\end{array}\right| = 0$.
Apply the row operation $R_{3} \rightarrow R_{3} - 3R_{1} - 2R_{2}$:
For the third row elements:
$R_{3,1} = (7x-2) - 3(x) - 2(2x-1) = 7x - 2 - 3x - 4x + 2 = 0$.
$R_{3,2} = (17x+6) - 3(3x+2) - 2(4x) = 17x + 6 - 9x - 6 - 8x = 0$.
$R_{3,3} = (12x-1) - 3(2x-1) - 2(3x+1) = 12x - 1 - 6x + 3 - 6x - 2 = 0$.
Since all elements of the third row are $0$,the value of the determinant is $0$ for all values of $x$.
Therefore,the equation is true for infinitely many values of $x$.

(D) To find the determinant of matrix $A$,we expand along the third row because it contains the most zeros:
$\det(A) = 0 \cdot \begin{vmatrix} 1 & 0 \\ 3-t & 1 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 3-t & 0 \\ -1 & 1 \end{vmatrix} + 0 \cdot \begin{vmatrix} 3-t & 1 \\ -1 & 3-t \end{vmatrix}$
$\det(A) = 1 \cdot ((3-t)(1) - (0)(-1))$
$\det(A) = 3-t$
Given that $\det(A) = 5$,we set up the equation:
$3-t = 5$
$-t = 5 - 3$
$-t = 2$
$t = -2$
Therefore,the correct option is $D$.

(A) We have,$|A| = \begin{vmatrix} 1 & \cos \theta & 0 \\ -\cos \theta & 1 & \cos \theta \\ -1 & -\cos \theta & 1 \end{vmatrix}$
Expanding along the first row:
$|A| = 1[1 - (-\cos \theta)(\cos \theta)] - \cos \theta[-\cos \theta - (-\cos \theta)] + 0[\cos^2 \theta + 1]$
$|A| = 1[1 + \cos^2 \theta] - \cos \theta[0] + 0$
$|A| = 1 + \cos^2 \theta$
Now,we know that $-1 \leq \cos \theta \leq 1$.
Therefore,$0 \leq \cos^2 \theta \leq 1$.
Adding $1$ to all parts,we get $1 \leq 1 + \cos^2 \theta \leq 2$.
Thus,$1 \leq |A| \leq 2$.
Therefore,the value of $|A|$ lies in the closed interval $[1, 2]$.

(A) Given $f(x) = \begin{vmatrix} 1 & x & x+1 \\ 2x & x(x-1) & x(x+1) \\ 3x(x-1) & x(x-1)(x-2) & x(x+1)(x-1) \end{vmatrix}$.
Taking $x$ common from $R_2$ and $x(x-1)$ common from $R_3$,we get:
$f(x) = x \cdot x(x-1) \begin{vmatrix} 1 & x & x+1 \\ 2 & x-1 & x+1 \\ 3 & x-2 & x+1 \end{vmatrix}$.
Taking $(x+1)$ common from $C_3$,we get:
$f(x) = x^2(x-1)(x+1) \begin{vmatrix} 1 & x & 1 \\ 2 & x-1 & 1 \\ 3 & x-2 & 1 \end{vmatrix}$.
Applying $R_1 \rightarrow R_1 - R_2$ and $R_2 \rightarrow R_2 - R_3$:
$f(x) = x^2(x-1)(x+1) \begin{vmatrix} -1 & 1 & 0 \\ -1 & 1 & 0 \\ 3 & x-2 & 1 \end{vmatrix}$.
Since two rows ($R_1$ and $R_2$) are identical,the value of the determinant is $0$.
Therefore,$f(x) = 0$ for all $x$,which implies $f(100) = 0$.

(B) Let $\Delta = \left|\begin{array}{ccc}-1 & \cos R & \cos Q \\ \cos R & -1 & \cos P \\ \cos Q & \cos P & -1\end{array}\right|$.
Expanding the determinant along the first row:
$\Delta = -1(1 - \cos^2 P) - \cos R(-\cos R - \cos Q \cos P) + \cos Q(\cos R \cos P + \cos Q)$
$\Delta = -(1 - \cos^2 P) + \cos^2 R + \cos R \cos Q \cos P + \cos Q \cos R \cos P + \cos^2 Q$
$\Delta = -1 + \cos^2 P + \cos^2 R + \cos^2 Q + 2 \cos P \cos Q \cos R$.
Since $P, Q, R$ are angles of a triangle,$P+Q+R = \pi$,so $R = \pi - (P+Q)$.
Using the identity for the sum of squares of cosines in a triangle: $\cos^2 P + \cos^2 Q + \cos^2 R = 1 - 2 \cos P \cos Q \cos R$.
Substituting this into the expression for $\Delta$:
$\Delta = -1 + (1 - 2 \cos P \cos Q \cos R) + 2 \cos P \cos Q \cos R = 0$.

(A) Let $\Delta = \left| \begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array} \right|$.
Applying the column operation $C_1 \rightarrow C_1 + C_2 + C_3$:
$\Delta = \left| \begin{array}{lll} (a-b)+(b-c)+(c-a) & b-c & c-a \\ (b-c)+(c-a)+(a-b) & c-a & a-b \\ (c-a)+(a-b)+(b-c) & a-b & b-c \end{array} \right|$.
Simplifying the elements in the first column:
$\Delta = \left| \begin{array}{lll} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \end{array} \right|$.
Since all elements in the first column are $0$,the value of the determinant is $0$.

(B) Given the determinant equation: $\left|\begin{array}{ccc}x+\omega^2 & \omega & 1 \\ \omega & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2\end{array}\right|=0$.
Applying the column operation $C_1 \rightarrow C_1 + C_2 + C_3$:
Since $1 + \omega + \omega^2 = 0$,the first column becomes:
$C_1 = \begin{bmatrix} x + \omega^2 + \omega + 1 \\ \omega + \omega^2 + 1 + x \\ 1 + x + \omega + \omega^2 \end{bmatrix} = \begin{bmatrix} x \\ x \\ x \end{bmatrix}$.
Thus,the determinant is $x \left|\begin{array}{ccc} 1 & \omega & 1 \\ 1 & \omega^2 & 1+x \\ 1 & x+\omega & \omega^2 \end{array}\right| = 0$.
This implies $x = 0$ is one of the solutions.

(B) We know that $\det(AB) = \det(A) \cdot \det(B)$.
Given $\det(AB) = 0$,we have $\det(A) \cdot \det(B) = 0$.
Calculating $\det(A) = (x+2)^2 - 9x = x^2 + 4x + 4 - 9x = x^2 - 5x + 4 = (x-1)(x-4)$.
Calculating $\det(B) = x(x+2) - 0 = x(x+2)$.
Thus,the equation becomes $(x-1)(x-4) \cdot x(x+2) = 0$.
Setting each factor to zero,we get $x-1=0, x-4=0, x=0, x+2=0$.
Therefore,the solutions are $x = 1, 4, 0, -2$.

(C) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.
The determinant is given by:
$\left|\begin{array}{ccc}1 & 4a & a \\ 1 & 3b & b \\ 1 & 2c & c\end{array}\right|=0$
Expanding along the first column:
$1(3bc - 2bc) - 1(4ac - 2ac) + 1(4ab - 3ab) = 0$
Simplifying the expression:
$(bc) - (2ac) + (ab) = 0$
$bc + ab = 2ac$
Dividing both sides by $abc$ (assuming $a, b, c \neq 0$):
$\frac{bc}{abc} + \frac{ab}{abc} = \frac{2ac}{abc}$
$\frac{1}{a} + \frac{1}{c} = \frac{2}{b}$
This condition implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $AP$,which means $a, b, c$ are in $HP$.

(B) The determinant of the matrix is calculated as:
$\begin{vmatrix} \cos^2\theta & -\sin^2\theta \\ \sin^2\theta & \cos^2\theta \end{vmatrix} = (\cos^2\theta)(\cos^2\theta) - (-\sin^2\theta)(\sin^2\theta) = \cos^4\theta + \sin^4\theta$.
Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$,we get:
$\cos^4\theta + \sin^4\theta = (\cos^2\theta + \sin^2\theta)^2 - 2\sin^2\theta\cos^2\theta = 1 - 2\sin^2\theta\cos^2\theta$.
Since $\sin 2\theta = 2\sin\theta\cos\theta$,we have $\sin^2 2\theta = 4\sin^2\theta\cos^2\theta$,so $2\sin^2\theta\cos^2\theta = \frac{1}{2}\sin^2 2\theta$.
Thus,the expression becomes $1 - \frac{1}{2}\sin^2 2\theta$.
Using the identity $\sin^2 2\theta = \frac{1 - \cos 4\theta}{2}$,we substitute:
$1 - \frac{1}{2}(\frac{1 - \cos 4\theta}{2}) = 1 - \frac{1}{4} + \frac{1}{4}\cos 4\theta = \frac{3}{4} + \frac{1}{4}\cos 4\theta = \frac{1}{4}(3 + \cos 4\theta)$.
Therefore,option $(B)$ is correct.

(D) The area of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ is given by the formula: $\text{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)| = 35$.
Substituting the given vertices $(2, -6)$,$(5, 4)$,and $(k, 4)$ into the formula:
$\frac{1}{2} |2(4-4) + 5(4 - (-6)) + k(-6-4)| = 35$
$\frac{1}{2} |2(0) + 5(10) + k(-10)| = 35$
$|50 - 10k| = 70$
This gives two possible cases:
Case $1$: $50 - 10k = 70 \implies -10k = 20 \implies k = -2$.
Case $2$: $50 - 10k = -70 \implies -10k = -120 \implies k = 12$.
Thus,the possible values for $k$ are $12$ and $-2$.

(D) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero.
$\begin{vmatrix} \cos 3\theta & -8 & -12 \\ \cos 2\theta & 3 & 3 \\ 1 & 1 & 3 \end{vmatrix} = 0$
Expanding along the first row:
$\cos 3\theta(9 - 3) + 8(3\cos 2\theta - 3) - 12(\cos 2\theta - 3) = 0$
$6\cos 3\theta + 24\cos 2\theta - 24 - 12\cos 2\theta + 36 = 0$
$6\cos 3\theta + 12\cos 2\theta + 12 = 0$
Dividing by $6$: $\cos 3\theta + 2\cos 2\theta + 2 = 0$
Using trigonometric identities $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ and $\cos 2\theta = 2\cos^2\theta - 1$:
$(4\cos^3\theta - 3\cos\theta) + 2(2\cos^2\theta - 1) + 2 = 0$
$4\cos^3\theta + 4\cos^2\theta - 3\cos\theta = 0$
$\cos\theta(4\cos^2\theta + 4\cos\theta - 3) = 0$
$\cos\theta(2\cos\theta - 1)(2\cos\theta + 3) = 0$
Since $\cos\theta$ cannot be $-3/2$,we have $\cos\theta = 0$ or $\cos\theta = 1/2$.
For $\theta \in [0, 2\pi]$,$\cos\theta = 0 \Rightarrow \theta = \pi/2, 3\pi/2$.
For $\cos\theta = 1/2 \Rightarrow \theta = \pi/3, 5\pi/3$.
The sum of all values is $\pi/2 + 3\pi/2 + \pi/3 + 5\pi/3 = 2\pi + 2\pi = 4\pi$.