Properties of determinants Questions in English

(B) Let $\Delta = \left| \begin{array}{ccc} a - b & b - c & c - a \\ x - y & y - z & z - x \\ p - q & q - r & r - p \end{array} \right|$.
Applying the column operation $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left| \begin{array}{ccc} (a - b) + (b - c) + (c - a) & b - c & c - a \\ (x - y) + (y - z) + (z - x) & y - z & z - x \\ (p - q) + (q - r) + (r - p) & q - r & r - p \end{array} \right|$
$\Delta = \left| \begin{array}{ccc} 0 & b - c & c - a \\ 0 & y - z & z - x \\ 0 & q - r & r - p \end{array} \right|$
Since all elements of the first column are $0$,the value of the determinant is $0$.

(A) Let $\Delta = \left| \begin{matrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \end{matrix} \right|$.
Applying row operations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$:
$\Delta = \left| \begin{matrix} 0 & a-b & (a^2-bc) - (b^2-ac) \\ 0 & b-c & (b^2-ac) - (c^2-ab) \\ 1 & c & {{c}^{2}}-ab \end{matrix} \right|$
Simplify the third column elements:
$(a^2-bc) - (b^2-ac) = (a^2-b^2) + (ac-bc) = (a-b)(a+b) + c(a-b) = (a-b)(a+b+c)$
$(b^2-ac) - (c^2-ab) = (b^2-c^2) + (ab-ac) = (b-c)(b+c) + a(b-c) = (b-c)(a+b+c)$
Substituting these back:
$\Delta = \left| \begin{matrix} 0 & a-b & (a-b)(a+b+c) \\ 0 & b-c & (b-c)(a+b+c) \\ 1 & c & {{c}^{2}}-ab \end{matrix} \right|$
Taking $(a-b)$ common from $R_1$ and $(b-c)$ common from $R_2$:
$\Delta = (a-b)(b-c) \left| \begin{matrix} 0 & 1 & a+b+c \\ 0 & 1 & a+b+c \\ 1 & c & {{c}^{2}}-ab \end{matrix} \right|$
Since $R_1$ and $R_2$ are identical,the value of the determinant is $0$.

(D) Let the determinant be $\Delta = \left| {\begin{array}{*{20}{c}} 1 & 5 & \pi \\ {{\log }_e}e & 5 & {\sqrt 5 } \\ {{\log }_{10}}10 & 5 & e \end{array}} \right|$.
Since ${{\log }_e}e = 1$ and ${{\log }_{10}}10 = 1$,we can rewrite the determinant as:
$\Delta = \left| {\begin{array}{*{20}{c}} 1 & 5 & \pi \\ 1 & 5 & {\sqrt 5 } \\ 1 & 5 & e \end{array}} \right|$.
In this determinant,the first column $C_1$ and the second column $C_2$ are related by the operation $C_2 = 5 \times C_1$.
Since two columns are proportional,the value of the determinant is $0$.

(A) Let $D = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{array} \right|$.
Applying $C_1 \to C_1 + C_2$,we get $D = \left| \begin{array}{ccc} 2 & 1 & 1 \\ 3 & 2 & 3 \\ 4 & 3 & 6 \end{array} \right|$,which is option $B$.
Applying $C_2 \to C_2 + C_3$ to the original determinant,we get $D = \left| \begin{array}{ccc} 1 & 2 & 1 \\ 1 & 5 & 3 \\ 1 & 9 & 6 \end{array} \right|$,which is option $C$.
Applying $C_1 \to C_1 + C_2 + C_3$ to the original determinant,we get $D = \left| \begin{array}{ccc} 3 & 1 & 1 \\ 6 & 2 & 3 \\ 10 & 3 & 6 \end{array} \right|$,which is option $D$.
Option $A$ is $\left| \begin{array}{ccc} 2 & 1 & 1 \\ 2 & 2 & 3 \\ 2 & 3 & 6 \end{array} \right| = 2 \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 6 \end{array} \right| = 2D \neq D$.
Thus,the determinant is not equal to option $A$.

(B) Let $\Delta = \left| {\begin{array}{ccc} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{array}} \right|$.
Apply the operation $R_1 \to R_1 + R_2 + R_3$:
$\Delta = \left| {\begin{array}{ccc} a + b + c & a + b + c & a + b + c \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{array}} \right|$.
Take $(a + b + c)$ common from $R_1$:
$\Delta = (a + b + c) \left| {\begin{array}{ccc} 1 & 1 & 1 \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{array}} \right|$.
Apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = (a + b + c) \left| {\begin{array}{ccc} 1 & 0 & 0 \\ 2b & -(a + b + c) & 0 \\ 2c & 0 & -(a + b + c) \end{array}} \right|$.
Expanding along $R_1$:
$\Delta = (a + b + c) [1 \cdot (-(a + b + c)) \cdot (-(a + b + c)) - 0] = (a + b + c)^3$.

(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}} \right|$.
Applying the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$R_2 - R_1 = (a+2b)-(a+b) = b, (a+3b)-(a+2b) = b, (a+4b)-(a+3b) = b$.
$R_3 - R_2 = (a+4b)-(a+2b) = 2b, (a+5b)-(a+3b) = 2b, (a+6b)-(a+4b) = 2b$.
Thus,$\Delta = \left| {\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\b&b&b\\2b&2b&2b\end{array}} \right|$.
Since $R_2$ and $R_3$ are proportional (specifically,$R_3 = 2R_2$),the determinant is $0$.
Alternatively,by substituting $a = 1$ and $b = 1$,the determinant becomes $\left| {\begin{array}{*{20}{c}}2&3&4\\3&4&5\\5&6&7\end{array}} \right| = 2(28-30) - 3(21-25) + 4(18-20) = 2(-2) - 3(-4) + 4(-2) = -4 + 12 - 8 = 0$.

(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}} \right|$.
Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_2$:
$\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{1}&{2}\\{x + 3}&{2}&{3}\\{x + 7}&{3}&{4}\end{array}} \right|$.
Applying $C_3 \to C_3 - C_2$:
$\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{1}&{1}\\{x + 3}&{2}&{1}\\{x + 7}&{3}&{1}\end{array}} \right|$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\Delta = \left| {\begin{array}{*{20}{c}}{x + 1}&{1}&{1}\\{2}&{1}&{0}\\{4}&{1}&{0}\end{array}} \right|$.
Expanding along $C_3$:
$\Delta = 1 \times (2 - 4) = -2$.
Thus,the value of the determinant is $-2$.

(C) Let the determinant be $\Delta$.
Factor out common terms from the columns:
$\Delta = \left| {\begin{array}{*{20}{c}}b(b - a)&(b - c)&c(b - a)\\a(b - a)&(a - b)&b(a - b)\\c(b - a)&(c - a)&a(b - a)\end{array}} \right|$
Taking $(b-a)$ common from $C_1$ and $(a-b)$ common from $C_3$:
$\Delta = (b-a)(a-b) \left| {\begin{array}{*{20}{c}}b&(b - c)&-c\\a&(a - b)&-b\\c&(c - a)&-a\end{array}} \right|$
Since $(b-a)(a-b) = -(a-b)^2$:
$\Delta = -(a-b)^2 \left| {\begin{array}{*{20}{c}}b&(b - c)&-c\\a&(a - b)&-b\\c&(c - a)&-a\end{array}} \right|$
Applying $C_2 \to C_2 + C_3$:
$\Delta = -(a-b)^2 \left| {\begin{array}{*{20}{c}}b&b&-c\\a&a&-b\\c&c&-a\end{array}} \right|$
Since $C_1$ and $C_2$ are identical,the value of the determinant is $0$.

(D) To evaluate the determinant $\Delta = \left| {\begin{array}{ccc} 1/a & a^2 & bc \\ 1/b & b^2 & ca \\ 1/c & c^2 & ab \end{array}} \right|$,we multiply and divide by $abc$:
$\Delta = \frac{1}{abc} \left| {\begin{array}{ccc} 1 & a^3 & abc \\ 1 & b^3 & abc \\ 1 & c^3 & abc \end{array}} \right|$
Now,take $abc$ common from the third column:
$\Delta = \frac{abc}{abc} \left| {\begin{array}{ccc} 1 & a^3 & 1 \\ 1 & b^3 & 1 \\ 1 & c^3 & 1 \end{array}} \right|$
Since the first column and the third column are identical,the value of the determinant is $0$.

(C) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}} \right|$.
Apply the operation ${R_1} \to {R_1} - ({R_2} + {R_3})$:
$\Delta = \left| {\begin{array}{*{20}{c}}{-2{c^2}}&{-2{c^2}}&{0}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}} \right|$.
Alternatively,taking common factors from columns or rows:
$\Delta = 2{a^2}{b^2}{c^2} \left| {\begin{array}{*{20}{c}}{\frac{{{b^2} + {c^2}}}{{{a^2}}}}&1&1\\1&{\frac{{{c^2} + {a^2}}}{{{b^2}}}}&1\\1&1&{\frac{{{a^2} + {b^2}}}{{{c^2}}}}\end{array}} \right|$.
By simplifying the determinant,we get:
$\Delta = 4{a^2}{b^2}{c^2}$.
Trick: Put $a=1, b=1, c=1$ is not ideal as options might overlap. Let $a=1, b=2, c=3$:
$\Delta = \left| {\begin{array}{*{20}{c}}{13}&1&1\\4&10&4\\9&9&5\end{array}} \right| = 13(50-36) - 1(20-36) + 1(36-90) = 13(14) + 16 - 54 = 182 + 16 - 54 = 144$.
Checking options: $4{a^2}{b^2}{c^2} = 4(1)^2(2)^2(3)^2 = 4 \times 1 \times 4 \times 9 = 144$. Hence,option $C$ is correct.

(B) Let $\Delta = \left| \begin{array}{ccc} y + z & x & y \\ z + x & z & x \\ x + y & y & z \end{array} \right|$.
Applying $R_1 \to R_1 + R_2 + R_3$,we get:
$\Delta = \left| \begin{array}{ccc} 2(x + y + z) & x + y + z & x + y + z \\ z + x & z & x \\ x + y & y & z \end{array} \right|$
$= (x + y + z) \left| \begin{array}{ccc} 2 & 1 & 1 \\ z + x & z & x \\ x + y & y & z \end{array} \right|$.
Applying $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$:
$= (x + y + z) \left| \begin{array}{ccc} 1 & 0 & 1 \\ x - z & z - x & x \\ x - y & y - z & z \end{array} \right|$
$= (x + y + z) [1((z - x)z - x(y - z)) + 1((x - z)(y - z) - (x - y)(z - x))]$.
Simplifying the expression,we get $\Delta = (x + y + z)(x - z)^2$.
Comparing this with $k(x + y + z)(x - z)^2$,we find $k = 1$.

(A) We are given the determinant $\Delta = \left| \begin{array}{ccc} a & a^2 & a^3 + 1 \\ b & b^2 & b^3 + 1 \\ c & c^2 & c^3 + 1 \end{array} \right|$.
By the property of determinants,we can split the third column into two parts:
$\Delta = \left| \begin{array}{ccc} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array} \right| + \left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right|$.
Taking $abc$ common from the first determinant,we get:
$\Delta = abc \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| + \left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right|$.
In the second determinant,perform column swaps to match the first: $C_3 \leftrightarrow C_2$ then $C_2 \leftrightarrow C_1$. This results in two sign changes,so the sign remains positive:
$\Delta = abc \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| + \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = (1 + abc) \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right|$.
The value of the determinant $\left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right|$ is $(a - b)(b - c)(c - a)$.
Thus,$\Delta = (1 + abc)(a - b)(b - c)(c - a) = 0$.
Since $a, b, c$ are unequal,$(a - b)(b - c)(c - a) \neq 0$.
Therefore,the condition is $1 + abc = 0$.

(C) Let $\Delta = \left| \begin{array}{ccc} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{array} \right|$.
Apply the column operation $C_3 \to C_3 + C_2$:
$\Delta = \left| \begin{array}{ccc} 1 & a & a + b + c \\ 1 & b & a + b + c \\ 1 & c & a + b + c \end{array} \right|$.
Taking $(a + b + c)$ as a common factor from $C_3$:
$\Delta = (a + b + c) \left| \begin{array}{ccc} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{array} \right|$.
Since column $C_1$ and column $C_3$ are identical $(C_1 = C_3)$,the value of the determinant is $0$.
Therefore,$\Delta = (a + b + c) \times 0 = 0$.

(D) To evaluate $\Delta = \left| \begin{array}{ccc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \end{array} \right|$,we multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$ and divide the determinant by $abc$:
$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} ab^2c^2 & abc & a(b+c) \\ a^2bc^2 & abc & b(c+a) \\ a^2b^2c & abc & c(a+b) \end{array} \right|$
Taking $abc$ common from $C_1$ and $abc$ common from $C_2$:
$\Delta = \frac{abc \cdot abc}{abc} \left| \begin{array}{ccc} bc & 1 & ab+ac \\ ac & 1 & bc+ab \\ ab & 1 & ac+bc \end{array} \right| = abc \left| \begin{array}{ccc} bc & 1 & ab+ac \\ ac & 1 & bc+ab \\ ab & 1 & ac+bc \end{array} \right|$
Applying $C_3 \to C_3 + C_1$ is not directly helpful here,instead observe that if we add $C_1$ to $C_3$,we get $ab+bc+ca$ in each element of $C_3$:
$\Delta = abc \left| \begin{array}{ccc} bc & 1 & ab+bc+ca \\ ac & 1 & ab+bc+ca \\ ab & 1 & ab+bc+ca \end{array} \right|$
Taking $(ab+bc+ca)$ common from $C_3$:
$\Delta = abc(ab+bc+ca) \left| \begin{array}{ccc} bc & 1 & 1 \\ ac & 1 & 1 \\ ab & 1 & 1 \end{array} \right|$
Since $C_2$ and $C_3$ are identical,the value of the determinant is $0$.

(D) Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b+c-a & c+a-b & a+b-c \end{array} \right|$.
Applying the column operations $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$:
$\Delta = \left| \begin{array}{ccc} 1-1 & 1-1 & 1 \\ (b+c)-(c+a) & (c+a)-(a+b) & a+b \\ (b+c-a)-(c+a-b) & (c+a-b)-(a+b-c) & a+b-c \end{array} \right|$
$\Delta = \left| \begin{array}{ccc} 0 & 0 & 1 \\ b-a & c-b & a+b \\ 2b-2a & 2c-2b & a+b-c \end{array} \right|$
$\Delta = \left| \begin{array}{ccc} 0 & 0 & 1 \\ b-a & c-b & a+b \\ 2(b-a) & 2(c-b) & a+b-c \end{array} \right|$
Since the first two columns are proportional (the third row is $2$ times the second row in the first two columns),the value of the determinant is $0$.

(D) By the property of determinants,if each element of a row (or column) of a determinant is multiplied by a constant $k$,then its value gets multiplied by $k$.
Since there are $3$ rows in the given $3 \times 3$ determinant,we can factor out $k$ from each of the three rows.
$\begin{vmatrix} ka & kb & kc \\ kx & ky & kz \\ kp & kq & kr \end{vmatrix} = k \cdot k \cdot k \begin{vmatrix} a & b & c \\ x & y & z \\ p & q & r \end{vmatrix} = k^3 \Delta$.

(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}{a - 1}&a&{bc}\\{b - 1}&b&{ca}\\{c - 1}&c&{ab}\end{array}} \right|$.
Using the property of determinants,we can split the determinant as:
$\Delta = \left| {\begin{array}{*{20}{c}}a&a&{bc}\\b&b&{ca}\\c&c&{ab}\end{array}} \right| - \left| {\begin{array}{*{20}{c}}1&a&{bc}\\1&b&{ca}\\1&c&{ab}\end{array}} \right|$.
The first determinant is $0$ because the first two columns are identical.
So,$\Delta = 0 - \left| {\begin{array}{*{20}{c}}1&a&{bc}\\1&b&{ca}\\1&c&{ab}\end{array}} \right| = - \left| {\begin{array}{*{20}{c}}1&a&{bc}\\1&b&{ca}\\1&c&{ab}\end{array}} \right|$.
Multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$ and divide by $abc$:
$\Delta = - \frac{1}{abc} \left| {\begin{array}{*{20}{c}}a&{a^2}&{abc}\\b&{b^2}&{abc}\\c&{c^2}&{abc}\end{array}} \right| = - \frac{abc}{abc} \left| {\begin{array}{*{20}{c}}a&{a^2}&1\\b&{b^2}&1\\c&{c^2}&1\end{array}} \right| = - \left| {\begin{array}{*{20}{c}}a&{a^2}&1\\b&{b^2}&1\\c&{c^2}&1\end{array}} \right|$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = - \left| {\begin{array}{*{20}{c}}a&{a^2}&1\\b-a&{b^2-a^2}&0\\c-a&{c^2-a^2}&0\end{array}} \right| = - (b-a)(c-a) \left| {\begin{array}{*{20}{c}}a&{a^2}&1\\1&{b+a}&0\\1&{c+a}&0\end{array}} \right|$.
Expanding along the third column:
$\Delta = - (b-a)(c-a) [1 \cdot ((c+a) - (b+a))] = - (b-a)(c-a)(c-b) = (a-b)(b-c)(c-a)$.
Since this result is not among the options $A, B, C$,the correct answer is $D$.

(A) Given the determinant $\Delta = \left| \begin{array}{ccc} a_1 & m a_1 & b_1 \\ a_2 & m a_2 & b_2 \\ a_3 & m a_3 & b_3 \end{array} \right|$.
By taking the common factor $m$ from the second column $(C_2)$,we get:
$\Delta = m \left| \begin{array}{ccc} a_1 & a_1 & b_1 \\ a_2 & a_2 & b_2 \\ a_3 & a_3 & b_3 \end{array} \right|$.
Since the first column $(C_1)$ and the second column $(C_2)$ are identical $(C_1 = C_2)$,the value of the determinant is $0$.
Therefore,$\Delta = m \times 0 = 0$.

(B) Let $\Delta = \left| \begin{array}{ccc} 2 & 8 & 4 \\ -5 & 6 & -10 \\ 1 & 7 & 2 \end{array} \right|$.
We can observe that the third column $C_3$ is related to the first column $C_1$ by the factor $2$,i.e.,$C_3 = 2 \times C_1$.
Taking $2$ as a common factor from the third column $C_3$,we get:
$\Delta = 2 \times \left| \begin{array}{ccc} 2 & 8 & 2 \\ -5 & 6 & -5 \\ 1 & 7 & 1 \end{array} \right|$.
Since two columns ($C_1$ and $C_3$) are identical,the value of the determinant is $0$.
Therefore,$\Delta = 2 \times 0 = 0$.

(B) To evaluate the determinant $\Delta = \begin{vmatrix} a^2 + x & ab & ac \\ ab & b^2 + x & bc \\ ac & bc & c^2 + x \end{vmatrix}$,we multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$ and divide the determinant by $abc$:
$\Delta = \frac{1}{abc} \begin{vmatrix} a(a^2 + x) & a^2b & a^2c \\ ab^2 & b(b^2 + x) & b^2c \\ ac^2 & bc^2 & c(c^2 + x) \end{vmatrix}$
Taking $a, b, c$ common from $C_1, C_2, C_3$ respectively:
$\Delta = \frac{abc}{abc} \begin{vmatrix} a^2 + x & a^2 & a^2 \\ b^2 & b^2 + x & b^2 \\ c^2 & c^2 & c^2 + x \end{vmatrix} = \begin{vmatrix} a^2 + x & a^2 & a^2 \\ b^2 & b^2 + x & b^2 \\ c^2 & c^2 & c^2 + x \end{vmatrix}$
Applying $R_1 \to R_1 + R_2 + R_3$:
$\Delta = \begin{vmatrix} a^2 + b^2 + c^2 + x & a^2 + b^2 + c^2 + x & a^2 + b^2 + c^2 + x \\ b^2 & b^2 + x & b^2 \\ c^2 & c^2 & c^2 + x \end{vmatrix}$
Taking $(a^2 + b^2 + c^2 + x)$ common from $R_1$:
$\Delta = (a^2 + b^2 + c^2 + x) \begin{vmatrix} 1 & 1 & 1 \\ b^2 & b^2 + x & b^2 \\ c^2 & c^2 & c^2 + x \end{vmatrix}$
Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = (a^2 + b^2 + c^2 + x) \begin{vmatrix} 1 & 0 & 0 \\ b^2 & x & 0 \\ c^2 & 0 & x \end{vmatrix} = (a^2 + b^2 + c^2 + x)(x^2)$
Thus,$\Delta$ is divisible by $x^2$.

(D) Using the property of determinants,we can split the determinant $D'$ into the sum of determinants.
$D' = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} + \begin{vmatrix} pb_1 & qc_1 & ra_1 \\ pb_2 & qc_2 & ra_2 \\ pb_3 & qc_3 & ra_3 \end{vmatrix}$
$D' = D + pqr \begin{vmatrix} b_1 & c_1 & a_1 \\ b_2 & c_2 & a_2 \\ b_3 & c_3 & a_3 \end{vmatrix}$
By applying column swaps,$\begin{vmatrix} b_1 & c_1 & a_1 \\ b_2 & c_2 & a_2 \\ b_3 & c_3 & a_3 \end{vmatrix} = - \begin{vmatrix} a_1 & c_1 & b_1 \\ a_2 & c_2 & b_2 \\ a_3 & c_3 & b_3 \end{vmatrix} = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = D$.
Thus,$D' = D + pqrD = D(1 + pqr)$.

(B) Let $\Delta = \left| {\begin{array}{ccc} a + b & b + c & c + a \\ b + c & c + a & a + b \\ c + a & a + b & b + c \end{array}} \right|$.
Applying $R_1 \to R_1 + R_2 + R_3$,we get:
$\Delta = \left| {\begin{array}{ccc} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ b + c & c + a & a + b \\ c + a & a + b & b + c \end{array}} \right| = 2(a+b+c) \left| {\begin{array}{ccc} 1 & 1 & 1 \\ b + c & c + a & a + b \\ c + a & a + b & b + c \end{array}} \right|$.
Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = 2(a+b+c) \left| {\begin{array}{ccc} 1 & 0 & 0 \\ b + c & a - b & a - c \\ c + a & b - c & b - a \end{array}} \right| = 2(a+b+c) [-(a-b)^2 - (a-c)(b-c)] = -2(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.
This simplifies to $-2(a^3+b^3+c^3-3abc)$.
Now,the determinant $D = \left| {\begin{array}{ccc} a & b & c \\ b & c & a \\ c & a & b \end{array}} \right| = a(bc-a^2) - b(b^2-ac) + c(ab-c^2) = -(a^3+b^3+c^3-3abc)$.
Comparing $\Delta = K \cdot D$,we get $-2(a^3+b^3+c^3-3abc) = K \cdot -(a^3+b^3+c^3-3abc)$,so $K = 2$.

(A) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\{{{(a + 1)}^2}}&{{{(b + 1)}^2}}&{{{(c + 1)}^2}}\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} \right|$.
Applying the operation $R_2 \to R_2 - R_3$:
Since $(x+1)^2 - (x-1)^2 = 4x$,the determinant becomes:
$\Delta = \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\4a&4b&4c\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} \right| = 4 \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\{{{(a - 1)}^2}}&{{{(b - 1)}^2}}&{{{(c - 1)}^2}}\end{array}} \right|$.
Now,apply $R_3 \to R_3 - R_1 + 2R_2$:
Note that $(x-1)^2 - x^2 + 2x = x^2 - 2x + 1 - x^2 + 2x = 1$.
Thus,the third row becomes $1, 1, 1$.
Therefore,$\Delta = 4 \left| {\begin{array}{*{20}{c}}{{a^2}}&{{b^2}}&{{c^2}}\\a&b&c\\1&1&1\end{array}} \right|$.
Hence,the correct option is $A$.

(B) Let $\Delta = \left| \begin{array}{ccc} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{array} \right|$.
Applying the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_2$:
$\Delta = \left| \begin{array}{ccc} 11 & 12-11 & 13-12 \\ 12 & 13-12 & 14-13 \\ 13 & 14-13 & 15-14 \end{array} \right|$
$\Delta = \left| \begin{array}{ccc} 11 & 1 & 1 \\ 12 & 1 & 1 \\ 13 & 1 & 1 \end{array} \right|$.
Since two columns ($C_2$ and $C_3$) are identical,the value of the determinant is $0$.

(D) Let $\Delta = \left| \begin{array}{ccc} x & 4 & y + z \\ y & 4 & z + x \\ z & 4 & x + y \end{array} \right|$.
Apply the column operation $C_1 \to C_1 + C_3$:
$\Delta = \left| \begin{array}{ccc} x + y + z & 4 & y + z \\ y + z + x & 4 & z + x \\ z + x + y & 4 & x + y \end{array} \right|$.
Take $(x + y + z)$ common from $C_1$:
$\Delta = (x + y + z) \left| \begin{array}{ccc} 1 & 4 & y + z \\ 1 & 4 & z + x \\ 1 & 4 & x + y \end{array} \right|$.
Since $C_1$ and $C_2$ are proportional (or specifically,$C_1$ is a multiple of the vector $(1, 1, 1)^T$ and $C_2$ is $4$ times the same vector),the two columns are linearly dependent.
Alternatively,since $C_1$ and $C_2$ are proportional,the value of the determinant is $0$.

(A) Let $\Delta = \left| \begin{array}{ccc} \sin^2 x & \cos^2 x & 1 \\ \cos^2 x & \sin^2 x & 1 \\ -10 & 12 & 2 \end{array} \right|$.
Applying the column operation $C_1 \to C_1 + C_2$:
$\Delta = \left| \begin{array}{ccc} \sin^2 x + \cos^2 x & \cos^2 x & 1 \\ \cos^2 x + \sin^2 x & \sin^2 x & 1 \\ -10 + 12 & 12 & 2 \end{array} \right|$
Since $\sin^2 x + \cos^2 x = 1$,we have:
$\Delta = \left| \begin{array}{ccc} 1 & \cos^2 x & 1 \\ 1 & \sin^2 x & 1 \\ 2 & 12 & 2 \end{array} \right|$
In this determinant,column $C_1$ and column $C_3$ are identical (both are $\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$).
According to the properties of determinants,if any two columns are identical,the value of the determinant is $0$.
Therefore,$\Delta = 0$.

(A) Let the given determinant be $\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ {a^2 - bc} & {b^2 - ac} & {c^2 - ab} \end{array}} \right|$.
We can split the determinant into two parts:
$\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}} \right| - \left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ bc & ac & ab \end{array}} \right|$.
For the second determinant,multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$ and divide by $abc$:
$\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}} \right| - \frac{1}{abc} \left| {\begin{array}{ccc} a & b & c \\ a^2 & b^2 & c^2 \\ abc & abc & abc \end{array}} \right|$.
Taking $abc$ common from $R_3$ in the second determinant:
$\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}} \right| - \frac{abc}{abc} \left| {\begin{array}{ccc} a & b & c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1 \end{array}} \right|$.
By swapping rows to match the first determinant,we see the two determinants are identical.
Thus,$\Delta = \left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}} \right| - \left| {\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}} \right| = 0$.
Therefore,$2 \times \Delta = 2 \times 0 = 0$.

(B) Given determinant is $\Delta = \left| {\begin{array}{ccc} 1 & 1+ac & 1+bc \\ 1 & 1+ad & 1+bd \\ 1 & 1+ae & 1+be \end{array}} \right|$.
Applying the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = \left| {\begin{array}{ccc} 1 & ac & bc \\ 1 & ad & bd \\ 1 & ae & be \end{array}} \right|$.
Now,taking $a$ common from $C_2$ and $b$ common from $C_3$:
$\Delta = ab \left| {\begin{array}{ccc} 1 & c & c \\ 1 & d & d \\ 1 & e & e \end{array}} \right|$.
Since column $C_2$ and column $C_3$ are identical,the value of the determinant is $0$.

(A) Let $\Delta = \left| \begin{array}{ccc} 13 & 16 & 19 \\ 14 & 17 & 20 \\ 15 & 18 & 21 \end{array} \right|$.
Applying the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_2$:
$C_2 \to C_2 - C_1 = \begin{bmatrix} 16-13 \\ 17-14 \\ 18-15 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 3 \end{bmatrix}$.
$C_3 \to C_3 - C_2 = \begin{bmatrix} 19-16 \\ 20-17 \\ 21-18 \end{bmatrix} = \begin{bmatrix} 3 \\ 3 \\ 3 \end{bmatrix}$.
Now the determinant becomes $\Delta = \left| \begin{array}{ccc} 13 & 3 & 3 \\ 14 & 3 & 3 \\ 15 & 3 & 3 \end{array} \right|$.
Since column $C_2$ and column $C_3$ are identical $(C_2 = C_3)$,the value of the determinant is $0$.

(B) Let $\Delta_1 = \begin{vmatrix} x & 2y & z \\ 2p & 4q & 2r \\ a & 2b & c \end{vmatrix}$.
We can factor out $2$ from the second column $(C_2)$:
$\Delta_1 = 2 \begin{vmatrix} x & y & z \\ 2p & 2q & 2r \\ a & b & c \end{vmatrix}$.
Next,we can factor out $2$ from the second row $(R_2)$:
$\Delta_1 = 2 \times 2 \begin{vmatrix} x & y & z \\ p & q & r \\ a & b & c \end{vmatrix}$.
Thus,$\Delta_1 = 4\Delta$.

(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{ab}&{ca}\\{ab}&{{b^2} + {x^2}}&{bc}\\{ca}&{bc}&{{c^2} + {x^2}}\end{array}} \right|$.
Multiply $C_1, C_2, C_3$ by $a, b, c$ respectively and divide the determinant by $abc$:
$\Delta = \frac{1}{{abc}}\left| {\begin{array}{*{20}{c}}{a({a^2} + {x^2})}&{ab^2}&{ac^2}\\{a^2b}&{b({b^2} + {x^2})}&{bc^2}\\{a^2c}&{b^2c}&{c({c^2} + {x^2})}\end{array}} \right|$.
Taking $a, b, c$ common from $R_1, R_2, R_3$ respectively:
$\Delta = \frac{abc}{abc}\left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{b^2}&{c^2}\\{a^2}&{{b^2} + {x^2}}&{c^2}\\{a^2}&{b^2}&{{c^2} + {x^2}}\end{array}} \right| = \left| {\begin{array}{*{20}{c}}{{a^2} + {x^2}}&{b^2}&{c^2}\\{a^2}&{{b^2} + {x^2}}&{c^2}\\{a^2}&{b^2}&{{c^2} + {x^2}}\end{array}} \right|$.
Applying $C_1 \to C_1 + C_2 + C_3$:
$\Delta = ({a^2} + {b^2} + {c^2} + {x^2})\left| {\begin{array}{*{20}{c}}1&{b^2}&{c^2}\\1&{{b^2} + {x^2}}&{c^2}\\1&{b^2}&{{c^2} + {x^2}}\end{array}} \right|$.
Subtracting $R_1$ from $R_2$ and $R_3$:
$\Delta = ({a^2} + {b^2} + {c^2} + {x^2})\left| {\begin{array}{*{20}{c}}1&{b^2}&{c^2}\\0&{x^2}&{0}\\0&{0}&{x^2}\end{array}} \right| = ({a^2} + {b^2} + {c^2} + {x^2}) \cdot x^4$.
Thus,the determinant is divisible by ${x^4}$.

(B) Let $\Delta = \left| {\begin{array}{*{20}{c}}1&1&1\\{\cos nx}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin nx}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}} \right|$.
Applying the column transformation ${C_3} \to {C_3} - (2\cos x){C_2} + {C_1}$ is not ideal,so let us use the property $\cos(A+B) + \cos(A-B) = 2\cos A \cos B$ and $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$.
Applying ${C_1} \to {C_1} + {C_3}$:
$\Delta = \left| {\begin{array}{*{20}{c}}{1 + \cos (n + 2)x}&1&{\cos (n + 2)x}\\{\cos nx + \cos (n + 2)x}&{\cos (n + 1)x}&{\cos (n + 2)x}\\{\sin nx + \sin (n + 2)x}&{\sin (n + 1)x}&{\sin (n + 2)x}\end{array}} \right|$.
Using $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$ and $\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$:
$\cos nx + \cos (n + 2)x = 2\cos(n+1)x \cos x$ and $\sin nx + \sin (n + 2)x = 2\sin(n+1)x \cos x$.
Subtracting $2\cos x$ times $C_2$ from $C_1$ and $C_3$ makes the determinant zero or simplifies it.
Alternatively,calculating the determinant directly:
$\Delta = 1[\cos(n+1)x \sin(n+2)x - \sin(n+1)x \cos(n+2)x] - 1[\cos nx \sin(n+2)x - \sin nx \cos(n+2)x] + 1[\cos nx \sin(n+1)x - \sin nx \cos(n+1)x]$.
Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$\Delta = \sin((n+2)x - (n+1)x) - \sin((n+2)x - nx) + \sin((n+1)x - nx)$.
$\Delta = \sin(x) - \sin(2x) + \sin(x) = 2\sin x - \sin 2x$.
Since the result $2\sin x - \sin 2x$ does not contain $n$,the determinant is independent of $n$.

(D) Given that $\left| \begin{array}{ccc} a & b & c \\ m & n & p \\ x & y & z \end{array} \right| = k$.
We need to evaluate $\Delta = \left| \begin{array}{ccc} 6a & 2b & 2c \\ 3m & n & p \\ 3x & y & z \end{array} \right|$.
Taking out common factors from the rows and columns:
From the first row,we can factor out $2$: $\Delta = 2 \left| \begin{array}{ccc} 3a & b & c \\ 3m & n & p \\ 3x & y & z \end{array} \right|$.
From the first column,we can factor out $3$: $\Delta = 2 \times 3 \left| \begin{array}{ccc} a & b & c \\ m & n & p \\ x & y & z \end{array} \right|$.
Substituting the given value $k$: $\Delta = 6 \times k = 6k$.

(B) To find the factors of the determinant $\Delta = \begin{vmatrix} a + x & b & c \\ b & x + c & a \\ c & a & x + b \end{vmatrix}$,we apply the column operation $C_1 \to C_1 + C_2 + C_3$.
Performing this operation,the first column becomes:
$C_1 \to \begin{bmatrix} (a + x) + b + c \\ b + (x + c) + a \\ c + a + (x + b) \end{bmatrix} = \begin{bmatrix} x + a + b + c \\ x + a + b + c \\ x + a + b + c \end{bmatrix}$.
Now,we can factor out $(x + a + b + c)$ from the first column:
$\Delta = (x + a + b + c) \begin{vmatrix} 1 & b & c \\ 1 & x + c & a \\ 1 & a & x + b \end{vmatrix}$.
Since $(x + a + b + c)$ is a common factor of the determinant,the correct option is $B$.

(B) Given determinant $\Delta = \left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{{(c + a)}^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{{(a + b)}^2}}\end{array}} \right|$.
Applying $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = \left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{{a^2} - {{(b + c)}^2}}&{{a^2} - {{(b + c)}^2}}\\{{b^2}}&{{{(c + a)}^2} - {b^2}}&0\\{{c^2}}&0&{{{(a + b)}^2} - {c^2}}\end{array}} \right|$
$= \left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{(a - b - c)(a + b + c)}&{(a - b - c)(a + b + c)}\\{{b^2}}&{(c + a - b)(c + a + b)}&0\\{{c^2}}&0&{(a + b - c)(a + b + c)}\end{array}} \right|$
Taking out $(a + b + c)$ from $C_2$ and $C_3$:
$\Delta = {(a + b + c)^2} \left| {\begin{array}{*{20}{c}}{{{(b + c)}^2}}&{a - b - c}&{a - b - c}\\{{b^2}}&{c + a - b}&0\\{{c^2}}&0&{a + b - c}\end{array}} \right|$
Applying $R_1 \to R_1 - R_2 - R_3$:
$\Delta = {(a + b + c)^2} \left| {\begin{array}{*{20}{c}}{2bc}&{ - 2c}&{ - 2b}\\{{b^2}}&{c + a - b}&0\\{{c^2}}&0&{a + b - c}\end{array}} \right|$
Expanding along $R_1$:
$\Delta = {(a + b + c)^2} [2bc((c + a - b)(a + b - c)) + 2c(b^2(a + b - c)) - 2b(b^2(c + a - b)) ]$
After simplification,we get $\Delta = 2abc{(a + b + c)^3}$.
Comparing with $k\,abc{(a + b + c)^3}$,we get $k = 2$.

(C) Given,$\Delta = \left| \begin{array}{ccc} 41 & 42 & 43 \\ 44 & 45 & 46 \\ 47 & 48 & 49 \end{array} \right|$.
Applying the column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_2$:
$C_2 - C_1 = \begin{bmatrix} 42-41 \\ 45-44 \\ 48-47 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$
$C_3 - C_2 = \begin{bmatrix} 43-42 \\ 46-45 \\ 49-48 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$
Thus,$\Delta = \left| \begin{array}{ccc} 41 & 1 & 1 \\ 44 & 1 & 1 \\ 47 & 1 & 1 \end{array} \right|$.
Since two columns ($C_2$ and $C_3$) are identical,the value of the determinant is $0$.

(A) Let $\Delta = \left| \begin{array}{ccc} 1/a & 1 & bc \\ 1/b & 1 & ca \\ 1/c & 1 & ab \end{array} \right|$.
Multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$,and divide the determinant by $abc$:
$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} 1 & a & abc \\ 1 & b & abc \\ 1 & c & abc \end{array} \right|$.
Take $abc$ common from the third column:
$\Delta = \frac{abc}{abc} \left| \begin{array}{ccc} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{array} \right|$.
Since the first and third columns are identical,the value of the determinant is $0$.

(A) Let the given determinant be $\Delta$.
Using the identity $(p+q)^2 - (p-q)^2 = 4pq$,we apply the column operation $C_1 \to C_1 - C_2$:
$(a^x + a^{-x})^2 - (a^x - a^{-x})^2 = 4(a^x)(a^{-x}) = 4(a^0) = 4$.
Similarly,for the second and third rows,we get $4$ and $4$.
Thus,the determinant becomes:
$\Delta = \left| \begin{array}{ccc} 4 & (a^x - a^{-x})^2 & 1 \\ 4 & (b^x - b^{-x})^2 & 1 \\ 4 & (c^x - c^{-x})^2 & 1 \end{array} \right|$.
Since column $C_1$ and column $C_3$ are proportional (specifically,$C_1 = 4C_3$),the value of the determinant is $0$.

(B) Let $\Delta = \left| \begin{array}{ccc} 441 & 442 & 443 \\ 445 & 446 & 447 \\ 449 & 450 & 451 \end{array} \right|$.
Applying the column operations $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$:
$C_1 - C_2 = \begin{bmatrix} 441-442 \\ 445-446 \\ 449-450 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}$.
$C_2 - C_3 = \begin{bmatrix} 442-443 \\ 446-447 \\ 450-451 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \\ -1 \end{bmatrix}$.
Thus,the determinant becomes $\Delta = \left| \begin{array}{ccc} -1 & -1 & 443 \\ -1 & -1 & 447 \\ -1 & -1 & 451 \end{array} \right|$.
Since column $C_1$ and column $C_2$ are identical,the value of the determinant is $0$.

(A) Given the determinant equation: $\left| \begin{array}{ccc} a & a^3 & a^4 - 1 \\ b & b^3 & b^4 - 1 \\ c & c^3 & c^4 - 1 \end{array} \right| = 0$.
Splitting the determinant into two: $\left| \begin{array}{ccc} a & a^3 & a^4 \\ b & b^3 & b^4 \\ c & c^3 & c^4 \end{array} \right| + \left| \begin{array}{ccc} a & a^3 & -1 \\ b & b^3 & -1 \\ c & c^3 & -1 \end{array} \right| = 0$.
Taking $abc$ common from the first determinant: $abc \left| \begin{array}{ccc} 1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3 \end{array} \right| - \left| \begin{array}{ccc} a & a^3 & 1 \\ b & b^3 & 1 \\ c & c^3 & 1 \end{array} \right| = 0$.
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$,we factor out $(a-b)(b-c)(c-a)$.
After simplification,the expression becomes $(abc)(ab + bc + ca) - (a + b + c) = 0$ is not correct,rather the expansion leads to $(abc)(ab + bc + ca) = a + b + c$.

(C) Given the determinant equation: $\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$
Apply column operations: $C_1 \to C_1 - C_3$ and $C_2 \to C_2 - C_3$:
$\left| {\begin{array}{*{20}{c}}1&0&{{{\sin }^2}\theta }\\{ - 1}&1&{{{\cos }^2}\theta }\\{ - 1}&{ - 1}&{1 + 4\sin 4\theta }\end{array}} \right| = 0$
Expanding along the first row:
$1(1(1 + 4\sin 4\theta) - (-1)(\cos^2 \theta)) - 0 + \sin^2 \theta((-1)(-1) - 1(-1)) = 0$
$1(1 + 4\sin 4\theta + \cos^2 \theta) + \sin^2 \theta(1 + 1) = 0$
$1 + 4\sin 4\theta + \cos^2 \theta + 2\sin^2 \theta = 0$
Since $\cos^2 \theta + \sin^2 \theta = 1$,we have:
$1 + 4\sin 4\theta + 1 + \sin^2 \theta = 0$ (Wait,let us re-evaluate the determinant expansion).
Correct expansion:
$1(1 + 4\sin 4\theta + \cos^2 \theta) - 0 + \sin^2 \theta(1 + 1) = 0$
$1 + 4\sin 4\theta + \cos^2 \theta + 2\sin^2 \theta = 0$
$2 + 4\sin 4\theta = 0$
$4\sin 4\theta = -2$
$\sin 4\theta = -1/2$.

(C) Let $r$ be the common ratio of the $G$.$P$. Then ${a_n} = {a_1}{r^{n-1}}$.
Taking logarithm on both sides,we get $\log {a_n} = \log {a_1} + (n-1)\log r$.
Let $A = \log {a_1}$ and $R = \log r$. Then $\log {a_n} = A + (n-1)R$.
Now,the terms in the determinant are of the form $\log {a_{n+k}} = A + (n+k-1)R$.
Applying column operations ${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_2}$:
${C_2} - {C_1}$ results in entries: $(n+2-1)R - (n-1)R = 2R$,$(n+8-1)R - (n+6-1)R = 2R$,and $(n+14-1)R - (n+12-1)R = 2R$.
${C_3} - {C_2}$ results in entries: $(n+4-1)R - (n+2-1)R = 2R$,$(n+10-1)R - (n+8-1)R = 2R$,and $(n+16-1)R - (n+14-1)R = 2R$.
Since columns ${C_2}$ and ${C_3}$ become identical (both consist of $2R$),the value of the determinant $\Delta = 0$.

(C) Given ${D_r} = \left| \begin{array}{ccc} {2^{r - 1}} & {2 \cdot 3^{r - 1}} & {4 \cdot 5^{r - 1}} \\ x & y & z \\ {2^n} - 1 & {3^n} - 1 & {5^n} - 1 \end{array} \right|$.
Applying the summation $\sum\limits_{r = 1}^n$ to the determinant,we get:
$\sum\limits_{r = 1}^n {D_r} = \left| \begin{array}{ccc} \sum\limits_{r = 1}^n {2^{r - 1}} & \sum\limits_{r = 1}^n {2 \cdot 3^{r - 1}} & \sum\limits_{r = 1}^n {4 \cdot 5^{r - 1}} \\ x & y & z \\ {2^n} - 1 & {3^n} - 1 & {5^n} - 1 \end{array} \right|$.
Using the geometric series sum formula $\sum\limits_{k=0}^{n-1} ar^k = a\frac{r^n - 1}{r - 1}$:
$\sum\limits_{r = 1}^n {2^{r - 1}} = \frac{2^n - 1}{2 - 1} = 2^n - 1$.
$2\sum\limits_{r = 1}^n {3^{r - 1}} = 2 \cdot \frac{3^n - 1}{3 - 1} = 3^n - 1$.
$4\sum\limits_{r = 1}^n {5^{r - 1}} = 4 \cdot \frac{5^n - 1}{5 - 1} = 5^n - 1$.
Substituting these values into the determinant:
$\sum\limits_{r = 1}^n {D_r} = \left| \begin{array}{ccc} 2^n - 1 & 3^n - 1 & 5^n - 1 \\ x & y & z \\ 2^n - 1 & 3^n - 1 & 5^n - 1 \end{array} \right|$.
Since row $1$ $(R_1)$ and row $3$ $(R_3)$ are identical,the value of the determinant is $0$.

(C) Given that $A$ is a square matrix of order $n$ and $A = kB$,where $k$ is a scalar.
By the property of determinants,if we multiply a matrix of order $n$ by a scalar $k$,the determinant of the resulting matrix is $k^n$ times the determinant of the original matrix.
Therefore,$|A| = |kB| = k^n|B|$.

(A) For a square matrix $A$ of order $n$,the property of the determinant is $det(kA) = k^n det(A)$.
Given that $A$ is a square matrix of order $n = 3$,we have $det(-A) = det((-1)A)$.
Using the property $det(kA) = k^n det(A)$ with $k = -1$ and $n = 3$,we get:
$det(-A) = (-1)^3 det(A) = -1 \times det(A) = -det(A)$.
Therefore,the statement $det(-A) = -det(A)$ is true.

(A) We are given two square matrices $A = \begin{bmatrix} 2 & 5 \\ 3 & 7 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 3 \\ 4 & 1 \end{bmatrix}$.
According to the fundamental properties of determinants for square matrices of the same order,the determinant of the product of two matrices is equal to the product of their individual determinants.
Mathematically,this is expressed as $|AB| = |A| \times |B|$.
Therefore,the correct property is $|AB| = |A||B|$.

(A) Given that $A$ is a $3 \times 3$ matrix and $|A| = 6$.
We are given $B = 5A^2$.
Using the property of determinants,for a $n \times n$ matrix $A$ and a scalar $k$,$|kA| = k^n |A|$.
Here,$n = 3$,so $|B| = |5A^2| = 5^3 |A^2|$.
Since $|A^2| = |A|^2$,we have $|B| = 125 \times |A|^2$.
Substituting $|A| = 6$,we get $|B| = 125 \times (6)^2 = 125 \times 36$.
Calculating the product,$125 \times 36 = 4500$.
Thus,the determinant of $B$ is $4500$.

(D) Given $\Delta = \begin{vmatrix} 1 + a & 1 & 1 \\ 1 & 1 + b & 1 \\ 1 & 1 & 1 + c \end{vmatrix} = 0$.
Taking $a, b, c$ common from $C_1, C_2, C_3$ respectively:
$\Delta = abc \begin{vmatrix} \frac{1}{a} + 1 & \frac{1}{b} & \frac{1}{c} \\ \frac{1}{a} & \frac{1}{b} + 1 & \frac{1}{c} \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c} + 1 \end{vmatrix}$.
Applying $C_1 \to C_1 + C_2 + C_3$:
$\Delta = abc \begin{vmatrix} 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & \frac{1}{b} & \frac{1}{c} \\ 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & \frac{1}{b} + 1 & \frac{1}{c} \\ 1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} & \frac{1}{b} & \frac{1}{c} + 1 \end{vmatrix}$.
Taking $(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ common from $C_1$:
$\Delta = abc (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \begin{vmatrix} 1 & \frac{1}{b} & \frac{1}{c} \\ 1 & \frac{1}{b} + 1 & \frac{1}{c} \\ 1 & \frac{1}{b} & \frac{1}{c} + 1 \end{vmatrix}$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta = abc (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \begin{vmatrix} 1 & \frac{1}{b} & \frac{1}{c} \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$.
Expanding along $C_1$,we get $\Delta = abc (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \times 1 = 0$.
Since $a, b, c \neq 0$,$abc \neq 0$. Therefore,$1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$,which implies $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = -1$.

(B) Given the determinant $\Delta = \begin{vmatrix} p & b & c \\ p + a & q + b & 2c \\ a & b & r \end{vmatrix} = 0$.
Applying $R_2 \to R_2 - R_1$,we get:
$\begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \end{vmatrix} = 0$.
Now,applying $R_2 \to R_2 - R_3$ and $R_1 \to R_1 - R_3$,we get:
$\begin{vmatrix} p - a & 0 & c - r \\ 0 & q - b & c - r \\ a & b & r \end{vmatrix} = 0$.
Expanding the determinant:
$(p - a)[(q - b)r - b(c - r)] + (c - r)[0 - a(q - b)] = 0$.
$(p - a)(q - b)r - (p - a)b(c - r) - a(q - b)(c - r) = 0$.
Dividing by $(p - a)(q - b)(r - c)$ (since $p \ne a, q \ne b, r \ne c$):
$\frac{r}{r - c} + \frac{b}{q - b} + \frac{a}{p - a} = 0$.
Using $\frac{x}{x - y} = 1 + \frac{y}{x - y}$,we rewrite the terms:
$(1 + \frac{p}{p - a} - 1) + (1 + \frac{q}{q - b} - 1) + (1 + \frac{r}{r - c} - 1) = 0$ is not the direct path.
Actually,$\frac{p}{p - a} = 1 + \frac{a}{p - a}$.
Thus,$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = (1 + \frac{a}{p - a}) + (1 + \frac{b}{q - b}) + (1 + \frac{c}{r - c}) = 3 + (\frac{a}{p - a} + \frac{b}{q - b} + \frac{c}{r - c})$.
From the expansion,we found $\frac{a}{p - a} + \frac{b}{q - b} + \frac{c}{r - c} = -1$.
Therefore,$3 - 1 = 2$.

(A) For any square matrix $A$ of order $n$,the property of determinants states that $|kA| = k^n |A|$.
Given that the order of the matrix $A$ is $n = 3$ and the scalar constant is $k = -2$.
Substituting these values into the formula:
$|-2A| = (-2)^3 |A|$
$|-2A| = -8 |A|$
Therefore,the correct option is $A$.