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

(B) Given the determinant equation: $\left| \begin{array}{ccc} a & a^2 & a^3 - 1 \\ b & b^2 & b^3 - 1 \\ c & c^2 & c^3 - 1 \end{array} \right| = 0$
We can split the determinant into two: $\left| \begin{array}{ccc} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{array} \right| - \left| \begin{array}{ccc} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array} \right| = 0$
Taking $abc$ common from the first determinant and rearranging columns in the second: $abc \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| - \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = 0$
Factoring out the common determinant: $(abc - 1) \left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = 0$
Since $a, b, c$ are distinct,the Vandermonde determinant $\left| \begin{array}{ccc} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array} \right| = (a-b)(b-c)(c-a) \neq 0$.
Therefore,$abc - 1 = 0$,which implies $abc = 1$.

(C) Given determinant is $\Delta = \left| {\begin{array}{*{20}{c}}{ - {a^2}}&{ab}&{ac}\\{ab}&{ - {b^2}}&{bc}\\{ac}&{bc}&{ - {c^2}}\end{array}} \right|$.
Taking $a, b, c$ common from $R_1, R_2, R_3$ respectively:
$\Delta = (abc) \left| {\begin{array}{*{20}{c}}{ - a}&b&c\\a&{ - b}&c\\a&b&{ - c}\end{array}} \right|$.
Taking $a, b, c$ common from $C_1, C_2, C_3$ respectively:
$\Delta = (abc)(abc) \left| {\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right| = {a^2}{b^2}{c^2} \left| {\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right|$.
Now,expanding the determinant:
$\left| {\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right| = -1(1 - 1) - 1(-1 - 1) + 1(1 - (-1)) = 0 + 2 + 2 = 4$.
Therefore,$\Delta = 4{a^2}{b^2}{c^2}$.
Comparing with $K{a^2}{b^2}{c^2}$,we get $K = 4$.

(D) Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1-x & 1 \\ 1 & 1 & 1+y \end{array} \right|$.
Applying column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:
$\Delta = \left| \begin{array}{ccc} 1 & 1-1 & 1-1 \\ 1 & (1-x)-1 & 1-1 \\ 1 & 1-1 & (1+y)-1 \end{array} \right|$
$\Delta = \left| \begin{array}{ccc} 1 & 0 & 0 \\ 1 & -x & 0 \\ 1 & 0 & y \end{array} \right|$
Expanding along the first row:
$\Delta = 1 \times ((-x)(y) - (0)(0)) - 0 + 0$
$\Delta = -xy$.

(B) Given the determinant equation: $\left| \begin{array}{ccc} a & b & a\alpha - b \\ b & c & b\alpha - c \\ 2 & 1 & 0 \end{array} \right| = 0$
Expanding along the third row:
$2(b(b\alpha - c) - c(a\alpha - b)) - 1(a(b\alpha - c) - b(a\alpha - b)) + 0 = 0$
$2(b^2\alpha - bc - ac\alpha + bc) - (ab\alpha - ac - ab\alpha + b^2) = 0$
$2(b^2\alpha - ac\alpha) - (b^2 - ac) = 0$
$2\alpha(b^2 - ac) - (b^2 - ac) = 0$
$(2\alpha - 1)(b^2 - ac) = 0$
Since it is given that $\alpha \neq \frac{1}{2}$,then $2\alpha - 1 \neq 0$.
Therefore,we must have $b^2 - ac = 0$,which implies $b^2 = ac$.
This condition indicates that $a, b, c$ are in $G.P.$

(C) Given the equation: ${\left| {\begin{array}{cc} 4 & 1 \\ 2 & 1 \end{array}} \right|^2} = \left| {\begin{array}{cc} 3 & 2 \\ 1 & x \end{array}} \right| - \left| {\begin{array}{cc} x & 3 \\ -2 & 1 \end{array}} \right|$
First,calculate the determinant on the left side:
$\left| {\begin{array}{cc} 4 & 1 \\ 2 & 1 \end{array}} \right| = (4 \times 1) - (1 \times 2) = 4 - 2 = 2$
So,the left side is $2^2 = 4$.
Next,calculate the determinants on the right side:
$\left| {\begin{array}{cc} 3 & 2 \\ 1 & x \end{array}} \right| = (3 \times x) - (2 \times 1) = 3x - 2$
$\left| {\begin{array}{cc} x & 3 \\ -2 & 1 \end{array}} \right| = (x \times 1) - (3 \times -2) = x + 6$
Substitute these values back into the equation:
$4 = (3x - 2) - (x + 6)$
$4 = 3x - 2 - x - 6$
$4 = 2x - 8$
$2x = 4 + 8$
$2x = 12$
$x = 6$

(B) Given the determinant equation: $\left| \begin{array}{ccc} 3x - 8 & 3 & 3 \\ 3 & 3x - 8 & 3 \\ 3 & 3 & 3x - 8 \end{array} \right| = 0$
Applying the column operation $C_1 \to C_1 + C_2 + C_3$,we get:
$(3x - 2) \left| \begin{array}{ccc} 1 & 3 & 3 \\ 1 & 3x - 8 & 3 \\ 1 & 3 & 3x - 8 \end{array} \right| = 0$
Applying row operations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$,we get:
$(3x - 2) \left| \begin{array}{ccc} 0 & -3x + 11 & 0 \\ 0 & 3x - 11 & -3x + 11 \\ 1 & 3 & 3x - 8 \end{array} \right| = 0$
Expanding along the first column:
$(3x - 2) \cdot 1 \cdot [(-3x + 11)(-3x + 11) - 0] = 0$
$(3x - 2)(3x - 11)^2 = 0$
Setting each factor to zero:
$3x - 2 = 0 \implies x = 2/3$
$(3x - 11)^2 = 0 \implies x = 11/3$
Thus,the values of $x$ are $2/3$ and $11/3$.

(D) Let $\Delta = \left| \begin{array}{ccc} x+2 & x+3 & x+a \\ x+4 & x+5 & x+b \\ x+6 & x+7 & x+c \end{array} \right|$.
Applying the column operation $C_2 \to C_2 - C_1$,we get:
$\Delta = \left| \begin{array}{ccc} x+2 & 1 & x+a \\ x+4 & 1 & x+b \\ x+6 & 1 & x+c \end{array} \right|$.
Now,applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$,we get:
$\Delta = \left| \begin{array}{ccc} x+2 & 1 & x+a \\ 2 & 0 & b-a \\ 4 & 0 & c-a \end{array} \right|$.
Expanding along the second column:
$\Delta = -1 \times [2(c-a) - 4(b-a)]$
$\Delta = -1 \times [2c - 2a - 4b + 4a]$
$\Delta = -1 \times [2c + 2a - 4b]$
$\Delta = 2(2b - a - c)$.
Since $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $2b - a - c = 0$.
Therefore,$\Delta = 2(0) = 0$.

(C) Given the determinant equation: $\left| \begin{array}{ccc} a & 2b & 2c \\ 3 & b & c \\ 4 & a & b \end{array} \right| = 0$.
Applying the row operation $R_1 \to R_1 - 2R_2$:
$\left| \begin{array}{ccc} a - 6 & 2b - 2b & 2c - 2c \\ 3 & b & c \\ 4 & a & b \end{array} \right| = 0$
$\left| \begin{array}{ccc} a - 6 & 0 & 0 \\ 3 & b & c \\ 4 & a & b \end{array} \right| = 0$.
Expanding along the first row:
$(a - 6)(b^2 - ac) = 0$.
Since $a \ne 6$,we must have $b^2 - ac = 0$,which implies $ac = b^2$.
Multiplying both sides by $b$,we get $abc = b^3$.

(B) Given the determinant $\Delta = \left| {\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = \lambda $.
Applying column operations ${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_1}$,we get:
$\Delta = \left| {\begin{array}{*{20}{c}}{1 + a}&{-a}&{-a}\\1&b&0\\1&0&c\end{array}} \right| = \lambda $.
Expanding along the third row $({R_3})$:
$1(0 - (-ab)) + 0 + c((1+a)b - (-a)) = \lambda $
$ab + c(b + ab + a) = \lambda $
$ab + bc + abc + ac = \lambda $
$ab + bc + ca + abc = \lambda \quad \dots (i)$
Given that ${a^{ - 1}} + {b^{ - 1}} + {c^{ - 1}} = 0$,we have:
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \implies \frac{bc + ac + ab}{abc} = 0$.
This implies $ab + bc + ca = 0$.
Substituting this into equation $(i)$,we get $\lambda = 0 + abc = abc$.
Thus,the value of $\lambda$ is $abc$.

(C) To find the value of the determinant $\Delta = \left| \begin{matrix} 1 & 2 & 3 \\ 3 & 5 & 7 \\ 8 & 14 & 20 \end{matrix} \right|$,we expand along the first row:
$\Delta = 1(5 \times 20 - 7 \times 14) - 2(3 \times 20 - 7 \times 8) + 3(3 \times 14 - 5 \times 8)$
$\Delta = 1(100 - 98) - 2(60 - 56) + 3(42 - 40)$
$\Delta = 1(2) - 2(4) + 3(2)$
$\Delta = 2 - 8 + 6$
$\Delta = 0$
Thus,the correct option is $C$.

(B) Given the determinant: $\left| {\begin{array}{*{20}{c}}a&b&{a + b}\\b&c&{b + c}\\{a + b}&{b + c}&0\end{array}} \right| = 0$.
Applying the column operation $C_3 \to C_3 - (C_1 + C_2)$:
$\left| {\begin{array}{*{20}{c}}a&b&0\\b&c&0\\{a + b}&{b + c}&{-(a + 2b + c)}\end{array}} \right| = 0$.
Expanding along the third column:
$-(a + 2b + c) \times (ac - b^2) = 0$.
For the determinant to be zero,either $(a + 2b + c) = 0$ or $(ac - b^2) = 0$.
If $ac - b^2 = 0$,then $b^2 = ac$,which implies that $a, b, c$ are in $G.P.$

(B) Let the determinant be $\Delta = \left| \begin{array}{ccc} 5^2 & 5^3 & 5^4 \\ 5^3 & 5^4 & 5^5 \\ 5^4 & 5^5 & 5^7 \end{array} \right|$.
Taking out $5$ as a common factor from row $R_2$:
$\Delta = 5 \times \left| \begin{array}{ccc} 5^2 & 5^3 & 5^4 \\ 5^2 & 5^3 & 5^4 \\ 5^4 & 5^5 & 5^7 \end{array} \right|$.
Since row $R_1$ and row $R_2$ are identical,the value of the determinant is $0$.

(A) Given the determinant: $\Delta = \left| \begin{array}{ccc} x + \omega^2 & \omega & 1 \\ \omega & \omega^2 & 1 + x \\ 1 & x + \omega & \omega^2 \end{array} \right| = 0$.
Applying the column operation $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left| \begin{array}{ccc} x + \omega^2 + \omega + 1 & \omega & 1 \\ \omega + \omega^2 + 1 + x & \omega^2 & 1 + x \\ 1 + x + \omega + \omega^2 & x + \omega & \omega^2 \end{array} \right|$.
Since $1 + \omega + \omega^2 = 0$,the first column becomes:
$\Delta = \left| \begin{array}{ccc} x & \omega & 1 \\ x & \omega^2 & 1 + x \\ x & x + \omega & \omega^2 \end{array} \right| = x \left| \begin{array}{ccc} 1 & \omega & 1 \\ 1 & \omega^2 & 1 + x \\ 1 & x + \omega & \omega^2 \end{array} \right|$.
For the determinant to be $0$,either $x = 0$ or the remaining determinant must be $0$.
Checking $x = 0$ in the original determinant:
$\left| \begin{array}{ccc} \omega^2 & \omega & 1 \\ \omega & \omega^2 & 1 \\ 1 & \omega & \omega^2 \end{array} \right| = \omega^2(\omega^4 - \omega) - \omega(\omega^3 - 1) + 1(\omega^2 - \omega^2) = \omega^2(\omega - \omega) - \omega(1 - 1) + 0 = 0$.
Thus,$x = 0$ is the correct value.

(B) Let the determinant be $\Delta = \left| \begin{array}{ccc} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{array} \right|$.
Expanding along the first row:
$\Delta = -1(1 - \cos^2 A) - \cos C(-\cos C - \cos A \cos B) + \cos B(\cos C \cos A + \cos B)$
$\Delta = -\sin^2 A + \cos^2 C + \cos A \cos B \cos C + \cos A \cos B \cos C + \cos^2 B$
$\Delta = -\sin^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C$.
Since $A+B+C = \pi$,we have $A = \pi - (B+C)$,so $\cos A = -\cos(B+C) = -(\cos B \cos C - \sin B \sin C) = \sin B \sin C - \cos B \cos C$.
Substituting $\cos A$ into the expression:
$\Delta = -\sin^2 A + \cos^2 B + \cos^2 C + 2 \cos B \cos C (\sin B \sin C - \cos B \cos C)$
$\Delta = -\sin^2 A + \cos^2 B + \cos^2 C + 2 \cos B \cos C \sin B \sin C - 2 \cos^2 B \cos^2 C$
Using $\sin^2 A = \sin^2(B+C) = (\sin B \cos C + \cos B \sin C)^2 = \sin^2 B \cos^2 C + \cos^2 B \sin^2 C + 2 \sin B \cos C \cos B \sin C$.
Substituting this back,all terms cancel out to give $\Delta = 0$.

(A) Let $D = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & \omega^2 & \omega \\ 1 & \omega & \omega^2 \end{array} \right|$.
Applying $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$:
$D = \left| \begin{array}{ccc} 0 & 0 & 1 \\ 1-\omega^2 & \omega^2-\omega & \omega \\ 1-\omega & \omega-\omega^2 & \omega^2 \end{array} \right|$.
Expanding along the first row:
$D = 1 \cdot [(1-\omega^2)(\omega-\omega^2) - (1-\omega)(\omega^2-\omega)]$
$D = (\omega - \omega^2 - \omega^2 + \omega^4) - (\omega^2 - \omega - \omega^3 + \omega^2)$
Since $\omega^3 = 1$ and $\omega^4 = \omega$:
$D = (\omega - 2\omega^2 + \omega) - (2\omega^2 - \omega - 1)$
$D = 2\omega - 2\omega^2 - 2\omega^2 + \omega + 1 = 3\omega - 4\omega^2 + 1$.
Using $1 + \omega + \omega^2 = 0$,we have $1 = -\omega - \omega^2$:
$D = 3\omega - 4\omega^2 - \omega - \omega^2 = 2\omega - 5\omega^2$.
Alternatively,calculating directly:
$D = 1(\omega^4 - \omega^2) - 1(\omega^2 - \omega) + 1(\omega - \omega^2) = \omega - \omega^2 - \omega^2 + \omega + \omega - \omega^2 = 3\omega - 3\omega^2 = 3(\omega - \omega^2)$.
Substituting $\omega = \frac{-1 + i\sqrt{3}}{2}$ and $\omega^2 = \frac{-1 - i\sqrt{3}}{2}$:
$D = 3 \left( \frac{-1 + i\sqrt{3}}{2} - \frac{-1 - i\sqrt{3}}{2} \right) = 3 \left( \frac{2i\sqrt{3}}{2} \right) = 3i\sqrt{3}$.

(A) Given the determinant $\Delta = \left| \begin{array}{ccc} a & b & a - b \\ b & c & b - c \\ 2 & 1 & 0 \end{array} \right| = 0$.
Expanding along the third row:
$2 \cdot \left| \begin{array}{cc} b & a - b \\ c & b - c \end{array} \right| - 1 \cdot \left| \begin{array}{cc} a & a - b \\ b & b - c \end{array} \right| + 0 = 0$.
$2(b(b - c) - c(a - b)) - (a(b - c) - b(a - b)) = 0$.
$2(b^2 - bc - ac + bc) - (ab - ac - ab + b^2) = 0$.
$2(b^2 - ac) - (b^2 - ac) = 0$.
$b^2 - ac = 0$.
$b^2 = ac$.
This condition implies that $a, b, c$ are in $G. P.$

(A) Given determinant is $\left| \begin{array}{ccc} x+1 & 1 & 1 \\ 2 & x+2 & 2 \\ 3 & 3 & x+3 \end{array} \right| = 0$.
Applying $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$:
$\left| \begin{array}{ccc} x & 0 & 1 \\ -x & x & 2 \\ 0 & -x & x+3 \end{array} \right| = 0$.
Taking $x$ common from $C_1$ and $C_2$:
$x^2 \left| \begin{array}{ccc} 1 & 0 & 1 \\ -1 & 1 & 2 \\ 0 & -1 & x+3 \end{array} \right| = 0$.
Expanding along $R_1$:
$x^2 [1(x+3+2) - 0 + 1(1-0)] = 0$.
$x^2 (x+5+1) = 0$.
$x^2 (x+6) = 0$.
Thus,$x = 0$ or $x = -6$.

(A) Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & x \\ p+1 & p+1 & p+x \\ 3 & x+1 & x+2 \end{array} \right| = 0$.
Applying column operations $C_2 \to C_2 - C_1$:
$\Delta = \left| \begin{array}{ccc} 1 & 0 & x \\ p+1 & 0 & p+x \\ 3 & x-2 & x+2 \end{array} \right| = 0$.
Expanding along the second column:
$-(x-2) \left| \begin{array}{cc} 1 & x \\ p+1 & p+x \end{array} \right| = 0$.
$-(x-2) [ (p+x) - (p+1)x ] = 0$.
$-(x-2) [ p + x - px - x ] = 0$.
$-(x-2) [ p(1-x) ] = 0$.
$p(x-2)(x-1) = 0$.
Thus,the solutions are $x = 1$ and $x = 2$.

(D) Let $D = \left| \begin{array}{ccc} 1 & \cos(\alpha - \beta) & \cos \alpha \\ \cos(\alpha - \beta) & 1 & \cos \beta \\ \cos \alpha & \cos \beta & 1 \end{array} \right|$.
Expanding along the first row:
$D = 1(1 - \cos^2 \beta) - \cos(\alpha - \beta)(\cos(\alpha - \beta) - \cos \alpha \cos \beta) + \cos \alpha(\cos \beta \cos(\alpha - \beta) - \cos \alpha)$.
Using the identity $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:
$D = \sin^2 \beta - \cos(\alpha - \beta)(\sin \alpha \sin \beta) + \cos \alpha(\cos \beta(\cos \alpha \cos \beta + \sin \alpha \sin \beta) - \cos \alpha)$.
$D = \sin^2 \beta - \cos(\alpha - \beta)\sin \alpha \sin \beta + \cos \alpha(\cos \alpha \cos^2 \beta + \sin \alpha \sin \beta \cos \beta - \cos \alpha)$.
$D = \sin^2 \beta - \cos(\alpha - \beta)\sin \alpha \sin \beta + \cos^2 \alpha \cos^2 \beta + \sin \alpha \sin \beta \cos \alpha \cos \beta - \cos^2 \alpha$.
$D = \sin^2 \beta - \cos(\alpha - \beta)\sin \alpha \sin \beta + \cos^2 \alpha(\cos^2 \beta - 1) + \sin \alpha \sin \beta \cos \alpha \cos \beta$.
$D = \sin^2 \beta - \sin \alpha \sin \beta(\cos(\alpha - \beta) - \cos \alpha \cos \beta) - \cos^2 \alpha \sin^2 \beta$.
$D = \sin^2 \beta - \sin \alpha \sin \beta(\sin \alpha \sin \beta) - \cos^2 \alpha \sin^2 \beta$.
$D = \sin^2 \beta - \sin^2 \alpha \sin^2 \beta - \cos^2 \alpha \sin^2 \beta$.
$D = \sin^2 \beta - \sin^2 \beta(\sin^2 \alpha + \cos^2 \alpha) = \sin^2 \beta - \sin^2 \beta(1) = 0$.

(B) Let $\Delta = \left| {\begin{array}{ccc} {1^2} & {2^2} & {3^2} \\ {2^2} & {3^2} & {4^2} \\ {3^2} & {4^2} & {5^2} \end{array}} \right| = \left| {\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}} \right|$.
Apply row operations: ${R_2} \to {R_2} - {R_1}$ and ${R_3} \to {R_3} - {R_2}$.
$\Delta = \left| {\begin{array}{ccc} 1 & 4 & 9 \\ 4-1 & 9-4 & 16-9 \\ 9-4 & 16-9 & 25-16 \end{array}} \right| = \left| {\begin{array}{ccc} 1 & 4 & 9 \\ 3 & 5 & 7 \\ 5 & 7 & 9 \end{array}} \right|$.
Expanding along the first row:
$\Delta = 1(5 \times 9 - 7 \times 7) - 4(3 \times 9 - 7 \times 5) + 9(3 \times 7 - 5 \times 5)$
$\Delta = 1(45 - 49) - 4(27 - 35) + 9(21 - 25)$
$\Delta = 1(-4) - 4(-8) + 9(-4)$
$\Delta = -4 + 32 - 36 = -8$.

(D) Let $\Delta = \left| \begin{array}{ccc} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{array} \right| = 0$.
Applying $C_1 \to C_1 + C_2 + C_3$:
$\Delta = \left| \begin{array}{ccc} 3a-x & a-x & a-x \\ 3a-x & a+x & a-x \\ 3a-x & a-x & a+x \end{array} \right| = 0$.
Taking $(3a-x)$ common from $C_1$:
$(3a-x) \left| \begin{array}{ccc} 1 & a-x & a-x \\ 1 & a+x & a-x \\ 1 & a-x & a+x \end{array} \right| = 0$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$(3a-x) \left| \begin{array}{ccc} 1 & a-x & a-x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \end{array} \right| = 0$.
Expanding along $C_1$:
$(3a-x) [1(4x^2 - 0)] = 0$.
$(3a-x)(4x^2) = 0$.
Therefore,$x = 3a$ or $x^2 = 0$,which implies $x = 0$.
Thus,the values are $x = 0, x = 3a$.

(D) Expanding the determinant along the first row:
$(x - 1) \cdot \left| \begin{array}{cc} x - 3 & 4 \\ 5 & 6 \end{array} \right| - 3 \cdot \left| \begin{array}{cc} 2 & 4 \\ 3 & 6 \end{array} \right| + 0 \cdot \left| \begin{array}{cc} 2 & x - 3 \\ 3 & 5 \end{array} \right| = 0$
$(x - 1) \cdot [6(x - 3) - 20] - 3 \cdot [12 - 12] + 0 = 0$
$(x - 1) \cdot [6x - 18 - 20] - 3 \cdot [0] = 0$
$(x - 1)(6x - 38) = 0$
$2(x - 1)(3x - 19) = 0$
Thus,$x - 1 = 0$ or $3x - 19 = 0$.
Therefore,$x = 1$ or $x = \frac{19}{3}$.
Comparing with the given options,$x = 1$ is the correct choice.

(A) Given the determinant equation: $\left| \begin{matrix} x & 0 & 8 \\ 4 & 1 & 3 \\ 2 & 0 & x \end{matrix} \right| = 0$
Expanding the determinant along the second column (since it has two zeros):
$1 \times \left| \begin{matrix} x & 8 \\ 2 & x \end{matrix} \right| = 0$
Calculating the $2 \times 2$ determinant:
$x(x) - 8(2) = 0$
$x^2 - 16 = 0$
$x^2 = 16$
$x = \pm 4$
Thus,the roots are $4$ and $-4$.

(B) Given the determinant equation: $\left| \begin{matrix} -x & 1 & 0 \\ 1 & -x & 1 \\ 0 & 1 & -x \end{matrix} \right| = 0$
Expanding along the first row:
$-x((-x)(-x) - (1)(1)) - 1((1)(-x) - (1)(0)) + 0 = 0$
$-x(x^2 - 1) - 1(-x) = 0$
$-x^3 + x + x = 0$
$-x^3 + 2x = 0$
$-x(x^2 - 2) = 0$
Thus,$x = 0$ or $x^2 = 2$,which gives $x = \pm \sqrt{2}$.
Therefore,the values of $x$ are $0, \pm \sqrt{2}$.

(D) Given the determinant equation: $\left| \begin{array}{ccc} 5 & 3 & -1 \\ -7 & x & -3 \\ 9 & 6 & -2 \end{array} \right| = 0$
Expanding along the first row:
$5(x(-2) - (-3)(6)) - 3((-7)(-2) - (-3)(9)) + (-1)((-7)(6) - x(9)) = 0$
$5(-2x + 18) - 3(14 + 27) - 1(-42 - 9x) = 0$
$-10x + 90 - 3(41) + 42 + 9x = 0$
$-10x + 90 - 123 + 42 + 9x = 0$
$-x + 9 = 0$
$x = 9$
Therefore,the correct option is $D$.

(C) Let $\Delta = \left| \begin{array}{ccc} a & b\omega^2 & a\omega \\ b\omega & c & b\omega^2 \\ c\omega^2 & a\omega & c \end{array} \right|$.
Applying the column transformation $C_1 \to C_1 + C_2 + C_3$ is not directly helpful,so let us evaluate the determinant directly.
Expanding along the first row:
$\Delta = a(c^2 - ab\omega^3) - b\omega^2(bc\omega - bc\omega^4) + a\omega(ab\omega^2 - c^2\omega^2)$
Since $\omega^3 = 1$ and $\omega^4 = \omega$:
$\Delta = a(c^2 - ab) - b\omega^2(bc\omega - bc\omega) + a\omega(ab\omega^2 - c^2\omega^2)$
$\Delta = a(c^2 - ab) - b\omega^2(0) + a\omega(ab\omega^2 - c^2\omega^2)$
$\Delta = ac^2 - a^2b + a^2b\omega^3 - ac^2\omega^3$
Since $\omega^3 = 1$:
$\Delta = ac^2 - a^2b + a^2b - ac^2 = 0$.

(C) Let $\Delta = \left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$.
Applying the column operations $C_1 \to C_1 - C_2$ and $C_2 \to C_2 - C_3$:
$\Delta = \left| \begin{array}{ccc} 0 & 0 & 1 \\ bc-ca & ca-ab & ab \\ (b+c)-(c+a) & (c+a)-(a+b) & a+b \end{array} \right|$
$\Delta = \left| \begin{array}{ccc} 0 & 0 & 1 \\ -c(a-b) & -a(b-c) & ab \\ b-a & c-b & a+b \end{array} \right|$
Taking $(b-a)$ common from $C_1$ and $(c-b)$ common from $C_2$:
$\Delta = (b-a)(c-b) \left| \begin{array}{ccc} 0 & 0 & 1 \\ -c & -a & ab \\ 1 & 1 & a+b \end{array} \right|$
Expanding along the first row:
$\Delta = (b-a)(c-b) [1 \cdot (-c - (-a))] = (b-a)(c-b)(a-c)$
$= (-(a-b))(-(b-c))(-(c-a)) = (a-b)(b-c)(c-a)$.

(C) Let $\Delta = \left| \begin{array}{ccc} 0 & b^3 - a^3 & c^3 - a^3 \\ a^3 - b^3 & 0 & c^3 - b^3 \\ a^3 - c^3 & b^3 - c^3 & 0 \end{array} \right|$.
This is a skew-symmetric determinant of odd order $(3 \times 3)$.
For any skew-symmetric matrix $A$ of odd order $n$,the determinant is given by $\det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A)$.
Since $n = 3$ (odd),$\det(A) = -\det(A)$,which implies $2 \det(A) = 0$,so $\det(A) = 0$.
Alternatively,by observing the rows: $R_1 + R_2 + R_3 = [0, 0, 0]$,which means the rows are linearly dependent.
Therefore,the value of the determinant is $0$.

(A) Given the determinant equation: $\left| \begin{array}{ccc} x & 2 & -1 \\ 2 & 5 & x \\ -1 & 2 & x \end{array} \right| = 0$
Expanding along the first row:
$x(5x - 2x) - 2(2x - (-1)(x)) - 1(4 - (-5)) = 0$
$x(3x) - 2(2x + x) - 1(4 + 5) = 0$
$3x^2 - 2(3x) - 9 = 0$
$3x^2 - 6x - 9 = 0$
Dividing by $3$:
$x^2 - 2x - 3 = 0$
Factoring the quadratic equation:
$(x - 3)(x + 1) = 0$
Thus,the solutions are $x = 3$ and $x = -1$.

(D) Let $\Delta = \left| {\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{y - z}&{z + x}&{y - x}\\{z - y}&{z - x}&{x + y}\end{array}} \right|$.
Apply row operations ${R_2} \to {R_2} + {R_1}$ and ${R_3} \to {R_3} + {R_1}$:
$\Delta = \left| {\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\{2y}&{2x}&0\\{2z}&0&{2x}\end{array}} \right|$.
Taking $2$ common from $R_2$ and $R_3$:
$\Delta = 4 \left| {\begin{array}{*{20}{c}}{y + z}&{x - z}&{x - y}\\y&x&0\\z&0&x\end{array}} \right|$.
Expanding along the first row:
$\Delta = 4[(y + z)(x^2 - 0) - (x - z)(xy - 0) + (x - y)(0 - zx)]$
$\Delta = 4[x^2y + zx^2 - (x^2y - xyz) - (xzx - xyz)]$
$\Delta = 4[x^2y + zx^2 - x^2y + xyz - x^2z + xyz]$
$\Delta = 4[2xyz] = 8xyz$.
Comparing with $kxyz$,we get $k = 8$.

(B) Given the determinant $\Delta = \begin{vmatrix} 1 & 2\omega \\ \omega & \omega^2 \end{vmatrix}$.
Expanding the determinant,we get $\Delta = (1)(\omega^2) - (2\omega)(\omega) = \omega^2 - 2\omega^2 = -\omega^2$.
Now,we need to find $\Delta^2$.
$\Delta^2 = (-\omega^2)^2 = \omega^4$.
Since $\omega$ is a cube root of unity,we know that $\omega^3 = 1$.
Therefore,$\Delta^2 = \omega^3 \times \omega = 1 \times \omega = \omega$.

(B) Given,${\Delta _1} = \left| {\begin{array}{*{20}{c}}1&0\\a&b\end{array}} \right| = (1 \times b) - (0 \times a) = b$.
Similarly,${\Delta _2} = \left| {\begin{array}{*{20}{c}}1&0\\c&d\end{array}} \right| = (1 \times d) - (0 \times c) = d$.
Therefore,the product ${\Delta _2}{\Delta _1} = d \times b = bd$.

(B) Let $D_1 = \left| {\begin{array}{*{20}{c}}{{{\log }_3}512}&{{{\log }_4}3}\\{{{\log }_3}8}&{{{\log }_4}9}\end{array}} \right|$ and $D_2 = \left| {\begin{array}{*{20}{c}}{{{\log }_2}3}&{{{\log }_8}3}\\{{{\log }_3}4}&{{{\log }_3}4}\end{array}} \right|$.
For $D_1$:
$D_1 = (\log_3 512)(\log_4 9) - (\log_4 3)(\log_3 8)$
$= (\frac{\log 512}{\log 3} \times \frac{\log 9}{\log 4}) - (\frac{\log 3}{\log 4} \times \frac{\log 8}{\log 3})$
$= (\frac{9 \log 2}{\log 3} \times \frac{2 \log 3}{2 \log 2}) - (\frac{\log 8}{\log 4}) = 9 - \frac{3 \log 2}{2 \log 2} = 9 - 1.5 = 7.5$.
For $D_2$:
$D_2 = (\log_2 3)(\log_3 4) - (\log_8 3)(\log_3 4)$
$= (\frac{\log 3}{\log 2} \times \frac{\log 4}{\log 3}) - (\frac{\log 3}{\log 8} \times \frac{\log 4}{\log 3})$
$= (\frac{\log 4}{\log 2}) - (\frac{\log 4}{\log 8}) = 2 - \frac{2 \log 2}{3 \log 2} = 2 - \frac{2}{3} = \frac{4}{3}$.
Therefore,$D_1 \times D_2 = 7.5 \times \frac{4}{3} = \frac{15}{2} \times \frac{4}{3} = 10$.

(C) According to Cramer's Rule,for a system of linear equations $a_1x + b_1y + c_1z = d_1$,$a_2x + b_2y + c_2z = d_2$,and $a_3x + b_3y + c_3z = d_3$,the value of $x$ is given by $x = \frac{D_x}{D}$.
Here,$D = \left| \begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{array} \right| = \left| \begin{array}{ccc} 2 & 3 & -5 \\ 1 & 1 & 1 \\ 3 & -4 & 2 \end{array} \right|$.
And $D_x$ is obtained by replacing the first column of $D$ with the constants $d_1, d_2, d_3$,so $D_x = \left| \begin{array}{ccc} 7 & 3 & -5 \\ 6 & 1 & 1 \\ 1 & -4 & 2 \end{array} \right|$.
Thus,$x = \frac{D_x}{D} = \left| \begin{array}{ccc} 7 & 3 & -5 \\ 6 & 1 & 1 \\ 1 & -4 & 2 \end{array} \right| \div \left| \begin{array}{ccc} 2 & 3 & -5 \\ 1 & 1 & 1 \\ 3 & -4 & 2 \end{array} \right|$.

(D) system of linear homogeneous equations has a non-trivial solution if and only if the determinant of the coefficient matrix is equal to $0$.
The coefficient matrix is given by:
$D = \begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(k(-4) - 3(-2)) - k(3(-4) - 2(-2)) + 3(3(3) - 2(k)) = 0$
$1(-4k + 6) - k(-12 + 4) + 3(9 - 2k) = 0$
$-4k + 6 + 8k + 27 - 6k = 0$
$-2k + 33 = 0$
$2k = 33$
$k = \frac{33}{2}$

(D) For a system of linear equations to have a non-zero (non-trivial) solution,the determinant of the coefficient matrix must be equal to zero,i.e.,$\Delta = 0$.
The system is given by:
$x - ky - z = 0$
$kx - y - z = 0$
$x + y - z = 0$
The determinant $\Delta$ is:
$\Delta = \begin{vmatrix} 1 & -k & -1 \\ k & -1 & -1 \\ 1 & 1 & -1 \end{vmatrix} = 0$
Expanding along the first row:
$1((-1)(-1) - (-1)(1)) - (-k)((k)(-1) - (-1)(1)) + (-1)((k)(1) - (-1)(1)) = 0$
$1(1 + 1) + k(-k + 1) - 1(k + 1) = 0$
$2 - k^2 + k - k - 1 = 0$
$1 - k^2 = 0$
$k^2 = 1$
$k = \pm 1$
Thus,the possible values of $k$ are $1$ and $-1$.

(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be equal to zero.
The coefficient matrix is given by:
$A = \begin{bmatrix} \lambda & 1 & 1 \\ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{bmatrix}$
Setting the determinant to zero:
$\begin{vmatrix} \lambda & 1 & 1 \\ -1 & \lambda & 1 \\ -1 & -1 & \lambda \end{vmatrix} = 0$
Expanding along the first row:
$\lambda(\lambda^2 - (-1)) - 1(-\lambda - (-1)) + 1(1 - (-\lambda)) = 0$
$\lambda(\lambda^2 + 1) - 1(-\lambda + 1) + 1(1 + \lambda) = 0$
$\lambda^3 + \lambda + \lambda - 1 + 1 + \lambda = 0$
$\lambda^3 + 3\lambda = 0$
$\lambda(\lambda^2 + 3) = 0$
This gives $\lambda = 0$ or $\lambda^2 = -3$. Since $\lambda$ must be a real value,$\lambda^2 = -3$ has no real solutions.
Therefore,the only real value for $\lambda$ is $0$.

(C) Let $A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$.
Given conditions imply that the rows (or columns) of the matrix $A$ are orthonormal vectors.
Specifically,the condition $a_i^2 + b_i^2 + c_i^2 = 1$ means the sum of squares of elements in each column is $1$.
The condition $a_i a_j + b_i b_j + c_i c_j = 0$ for $i \ne j$ means the dot product of any two distinct columns is $0$.
Thus,$A^T A = I$,where $I$ is the identity matrix of order $3$.
Taking the determinant on both sides,$|A^T A| = |I| = 1$.
Since $|A^T| = |A|$,we have $|A|^2 = 1$.
Therefore,the value of the square of the determinant is $1$.

(B) Given $\Delta = \left| \begin{array}{ccc} x! & (x+1)! & (x+2)! \\ (x+1)! & (x+2)! & (x+3)! \\ (x+2)! & (x+3)! & (x+4)! \end{array} \right|$.
Taking $x!$ common from $C_1$,$(x+1)!$ common from $C_2$,and $(x+2)!$ common from $C_3$:
$\Delta = x!(x+1)!(x+2)! \left| \begin{array}{ccc} 1 & (x+1) & (x+2)(x+1) \\ 1 & (x+2) & (x+3)(x+2) \\ 1 & (x+3) & (x+4)(x+3) \end{array} \right|$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\Delta = x!(x+1)!(x+2)! \left| \begin{array}{ccc} 1 & x+1 & (x+2)(x+1) \\ 0 & 1 & (x+2)(x+3-x-1) \\ 0 & 1 & (x+3)(x+4-x-2) \end{array} \right|$
$= x!(x+1)!(x+2)! \left| \begin{array}{ccc} 1 & x+1 & (x+2)(x+1) \\ 0 & 1 & 2(x+2) \\ 0 & 1 & 2(x+3) \end{array} \right|$.
Expanding along $C_1$:
$\Delta = x!(x+1)!(x+2)! [1 \cdot (2(x+3) - 2(x+2))]$
$= x!(x+1)!(x+2)! [2x + 6 - 2x - 4]$
$= x!(x+1)!(x+2)! [2]$
$= 2(x!)(x+1)!(x+2)!$.

(C) Let the first term be $A$ and the common difference be $D$.
Then,$a = A + (p - 1)D$,$b = A + (q - 1)D$,and $c = A + (r - 1)D$.
The given determinant is $\Delta = \left| \begin{array}{ccc} A + (p - 1)D & p & 1 \\ A + (q - 1)D & q & 1 \\ A + (r - 1)D & r & 1 \end{array} \right|$.
Applying the column operation $C_1 \to C_1 - (A - D)C_3$,we get:
$\Delta = \left| \begin{array}{ccc} pD & p & 1 \\ qD & q & 1 \\ rD & r & 1 \end{array} \right|$.
Taking $D$ common from $C_1$,we get $\Delta = D \left| \begin{array}{ccc} p & p & 1 \\ q & q & 1 \\ r & r & 1 \end{array} \right|$.
Since column $C_1$ and $C_2$ are identical,the value of the determinant is $0$.

(D) For the system of equations to be consistent,the determinant of the coefficient matrix must be zero.
The system is:
$(\alpha + 1)^3 x + (\alpha + 2)^3 y = (\alpha + 3)^3$
$(\alpha + 1) x + (\alpha + 2) y = (\alpha + 3)$
$x + y = 1$
The determinant of the augmented matrix must be zero:
$\begin{vmatrix} (\alpha + 1)^3 & (\alpha + 2)^3 & -(\alpha + 3)^3 \\ \alpha + 1 & \alpha + 2 & -(\alpha + 3) \\ 1 & 1 & -1 \end{vmatrix} = 0$
Applying column operations $C_1 \to C_1 + C_2$ and $C_2 \to C_2 + C_3$ (or evaluating directly),we find:
$6\alpha + 12 = 0$
$6\alpha = -12$
$\alpha = -2$
Thus,the correct option is $D$.

(B) matrix is singular if its determinant is equal to $0$.
Given the matrix $A = \begin{bmatrix} 1 & 3 & \lambda + 2 \\ 2 & 4 & 8 \\ 3 & 5 & 10 \end{bmatrix}$.
Setting the determinant $|A| = 0$:
$|A| = 1(4 \times 10 - 8 \times 5) - 3(2 \times 10 - 8 \times 3) + (\lambda + 2)(2 \times 5 - 4 \times 3) = 0$
$|A| = 1(40 - 40) - 3(20 - 24) + (\lambda + 2)(10 - 12) = 0$
$|A| = 1(0) - 3(-4) + (\lambda + 2)(-2) = 0$
$0 + 12 - 2(\lambda + 2) = 0$
$12 - 2\lambda - 4 = 0$
$8 - 2\lambda = 0$
$2\lambda = 8$
$\lambda = 4$.

(D) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$.
Given $A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ \lambda & -3 & 0 \end{bmatrix}$.
Calculating the determinant along the first row:
$|A| = 0(0 - (-9)) - 1(0 - 3\lambda) + (-2)(3 - 0) = 0$
$|A| = 0 - 1(-3\lambda) - 2(3) = 0$
$|A| = 3\lambda - 6 = 0$
$3\lambda = 6$
$\lambda = 2$.

(D) matrix $A$ is non-singular if and only if its determinant $|A| \neq 0$.
Given the matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & \lambda & 5 \end{bmatrix}$,we calculate its determinant:
$|A| = 1(5 \times 5 - 6 \times \lambda) - 2(4 \times 5 - 6 \times 3) + 3(4 \times \lambda - 5 \times 3)$
$|A| = 1(25 - 6\lambda) - 2(20 - 18) + 3(4\lambda - 15)$
$|A| = 25 - 6\lambda - 2(2) + 12\lambda - 45$
$|A| = 25 - 6\lambda - 4 + 12\lambda - 45$
$|A| = 6\lambda - 24$
For the matrix to be non-singular,we require $|A| \neq 0$:
$6\lambda - 24 \neq 0$
$6\lambda \neq 24$
$\lambda \neq 4$
Therefore,$\lambda$ should not be equal to $4$.

(B) matrix is singular if its determinant is equal to $0$.
Given,$\begin{vmatrix} 2 + x & 3 & 4 \\ 1 & -1 & 2 \\ x & 1 & -5 \end{vmatrix} = 0$.
Expanding along the first row:
$(2 + x)((-1)(-5) - (2)(1)) - 3((1)(-5) - (2)(x)) + 4((1)(1) - (-1)(x)) = 0$
$(2 + x)(5 - 2) - 3(-5 - 2x) + 4(1 + x) = 0$
$(2 + x)(3) + 15 + 6x + 4 + 4x = 0$
$6 + 3x + 15 + 6x + 4 + 4x = 0$
$13x + 25 = 0$
$13x = -25$
$x = -\frac{25}{13}$.

(A) matrix $A$ is non-singular if its determinant $|A| \ne 0$.
Given $A = \left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}\end{array}} \right]$.
Calculating the determinant $|A|$:
$|A| = 2[3(-3) - 4(-2)] - \lambda[(-1)(-3) - 4(1)] + (-4)[(-1)(-2) - 3(1)]$
$|A| = 2[-9 + 8] - \lambda[3 - 4] - 4[2 - 3]$
$|A| = 2(-1) - \lambda(-1) - 4(-1)$
$|A| = -2 + \lambda + 4$
$|A| = \lambda + 2$
For the matrix to be non-singular,$|A| \ne 0$,so $\lambda + 2 \ne 0$,which implies $\lambda \ne -2$.

(B) matrix $A$ is singular if its determinant is zero,i.e.,$|A| = 0$.
Given $A = \begin{bmatrix} a & 2 \\ 2 & 4 \end{bmatrix}$.
The determinant $|A| = (a \times 4) - (2 \times 2) = 4a - 4$.
For the matrix to be singular,set $|A| = 0$:
$4a - 4 = 0$
$4a = 4$
$a = 1$.
Therefore,the value of $a$ is $1$.

(B) matrix $A$ is not invertible if its determinant is zero,i.e.,$|A| = 0$.
Given the matrix $A = \begin{bmatrix} 1 & a & 2 \\ 1 & 2 & 5 \\ 2 & 1 & 1 \end{bmatrix}$,we calculate the determinant:
$|A| = 1(2 \times 1 - 5 \times 1) - a(1 \times 1 - 5 \times 2) + 2(1 \times 1 - 2 \times 2) = 0$
$|A| = 1(2 - 5) - a(1 - 10) + 2(1 - 4) = 0$
$|A| = 1(-3) - a(-9) + 2(-3) = 0$
$-3 + 9a - 6 = 0$
$9a - 9 = 0$
$9a = 9$
$a = 1$
Thus,the matrix is not invertible when $a = 1$.

(B) square matrix $A$ is invertible if and only if its determinant is non-zero,i.e.,$|A| \neq 0$.
Given the matrix $A = \begin{bmatrix} \lambda & -1 & 4 \\ -3 & 0 & 1 \\ -1 & 1 & 2 \end{bmatrix}$.
Calculating the determinant $|A|$:
$|A| = \lambda(0 \times 2 - 1 \times 1) - (-1)(-3 \times 2 - 1 \times (-1)) + 4(-3 \times 1 - 0 \times (-1))$
$|A| = \lambda(0 - 1) + 1(-6 + 1) + 4(-3 - 0)$
$|A| = -\lambda - 5 - 12$
$|A| = -\lambda - 17$
For the matrix to be invertible,$|A| \neq 0$:
$-\lambda - 17 \neq 0$
$-\lambda \neq 17$
$\lambda \neq -17$.

(C) Given $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} = 2I$,where $I$ is the identity matrix of order $3 \times 3$.
We know that the determinant of the product of two matrices is the product of their determinants,i.e.,$|AB| = |A| \cdot |B|$.
First,calculate $|A|$:
$|A| = \begin{vmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{vmatrix} = 2(2 \times 2 - 0 \times 0) = 2(4) = 8$.
Next,calculate $|B|$:
Since $B$ is an upper triangular matrix,its determinant is the product of its diagonal elements:
$|B| = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{vmatrix} = 1 \times 1 \times 2 = 2$.
Therefore,$|AB| = |A| \times |B| = 8 \times 2 = 16$.