Solution of the Linear equations using Matrices Questions in English

(D) For a homogeneous system of linear equations $AX = 0$ to have a non-trivial solution (a solution other than $x = y = z = 0$),the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.
The coefficient matrix is:
$A = \begin{bmatrix} 3 & -2 & 1 \\ \lambda & -14 & 15 \\ 1 & 2 & -3 \end{bmatrix}$
Setting the determinant to zero:
$\begin{vmatrix} 3 & -2 & 1 \\ \lambda & -14 & 15 \\ 1 & 2 & -3 \end{vmatrix} = 0$
Expanding along the first row:
$3((-14)(-3) - (15)(2)) - (-2)((\lambda)(-3) - (15)(1)) + 1((\lambda)(2) - (-14)(1)) = 0$
$3(42 - 30) + 2(-3\lambda - 15) + 1(2\lambda + 14) = 0$
$3(12) - 6\lambda - 30 + 2\lambda + 14 = 0$
$36 - 6\lambda - 30 + 2\lambda + 14 = 0$
$-4\lambda + 20 = 0$
$4\lambda = 20$
$\lambda = 5$

(C) 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:
$A = \begin{bmatrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \end{bmatrix}$
Setting the determinant to zero:
$\begin{vmatrix} 1 & k & -1 \\ 3 & -k & -1 \\ 1 & -3 & 1 \end{vmatrix} = 0$
Expanding along the first row:
$1(-k - 3) - k(3 - (-1)) - 1(-9 - (-k)) = 0$
$1(-k - 3) - k(4) - 1(-9 + k) = 0$
$-k - 3 - 4k + 9 - k = 0$
$-6k + 6 = 0$
$6k = 6$
$k = 1$

(D) The given system of homogeneous linear equations is:
$x + y - z = 0$
$3x - y - z = 0$
$x - 3y + z = 0$
To find the number of solutions,we calculate the determinant $\Delta$ of the coefficient matrix:
$\Delta = \begin{vmatrix} 1 & 1 & -1 \\ 3 & -1 & -1 \\ 1 & -3 & 1 \end{vmatrix}$
Expanding along the first row:
$\Delta = 1((-1)(1) - (-1)(-3)) - 1((3)(1) - (-1)(1)) + (-1)((3)(-3) - (-1)(1))$
$\Delta = 1(-1 - 3) - 1(3 + 1) - 1(-9 + 1)$
$\Delta = 1(-4) - 1(4) - 1(-8)$
$\Delta = -4 - 4 + 8 = 0$
Since the determinant $\Delta = 0$ for a homogeneous system of equations,the system has non-trivial solutions,which implies there are infinitely many solutions.

(D) The given system of homogeneous linear equations is:
$x + y - z = 0$
$3x - \alpha y - 3z = 0$
$x - 3y + z = 0$
For a homogeneous system of equations to have a non-zero solution,the determinant of the coefficient matrix must be equal to zero,i.e.,$\Delta = 0$.
$\Delta = \begin{vmatrix} 1 & 1 & -1 \\ 3 & -\alpha & -3 \\ 1 & -3 & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(-\alpha + 9) - 1(3 + 3) - 1(-9 + \alpha) = 0$
$-\alpha + 9 - 6 + 9 - \alpha = 0$
$-2\alpha + 12 = 0$
$2\alpha = 12$
$\alpha = 6$
Wait,re-evaluating the determinant calculation:
$1(-\alpha - 9) - 1(3 - (-3)) - 1(-9 - (-\alpha)) = 0$
$1(-\alpha - 9) - 1(6) - 1(-9 + \alpha) = 0$
$-\alpha - 9 - 6 + 9 - \alpha = 0$
$-2\alpha - 6 = 0$
$-2\alpha = 6$
$\alpha = -3$
Thus,the correct option is $D$.

(B) The given system of homogeneous linear equations is:
$x + 4y - z = 0$
$3x - 4y - z = 0$
$x - 3y + z = 0$
To find the number of solutions,we calculate the determinant $\Delta$ of the coefficient matrix:
$\Delta = \begin{vmatrix} 1 & 4 & -1 \\ 3 & -4 & -1 \\ 1 & -3 & 1 \end{vmatrix}$
Expanding along the first row:
$\Delta = 1((-4)(1) - (-1)(-3)) - 4((3)(1) - (-1)(1)) + (-1)((3)(-3) - (-4)(1))$
$\Delta = 1(-4 - 3) - 4(3 + 1) - 1(-9 + 4)$
$\Delta = 1(-7) - 4(4) - 1(-5)$
$\Delta = -7 - 16 + 5 = -18$
Since $\Delta \neq 0$,the system of homogeneous equations has only the trivial solution $(x, y, z) = (0, 0, 0)$.
Therefore,there is exactly $1$ solution.

(D) The given system of equations is a homogeneous system of linear equations of the form $AX = 0$,where $A$ is the coefficient matrix and $X = [x, y, z]^T$.
For a homogeneous system $AX = 0$,the nature of the solution depends on the determinant of the coefficient matrix,denoted by $|A|$.
If $|A| \neq 0$,the system has only the trivial solution $(x = 0, y = 0, z = 0)$.
If $|A| = 0$,the system has infinitely many non-trivial solutions in addition to the trivial solution.
Since the problem states that $\left| \begin{matrix} {a_1} & {b_1} & {c_1} \\ {a_2} & {b_2} & {c_2} \\ {a_3} & {b_3} & {c_3} \end{matrix} \right| = 0$,the determinant of the coefficient matrix is zero.
Therefore,the system has infinitely many solutions.

(A) The given system of equations is:
$x + 2y - 3z = 1$
$0x + 0y + (k + 3)z = 3$
$(2k + 1)x + 0y + z = 0$
The system is represented by the matrix equation $AX = B$,where $A = \begin{bmatrix} 1 & 2 & -3 \\ 0 & 0 & k+3 \\ 2k+1 & 0 & 1 \end{bmatrix}$.
For the system to be inconsistent,the determinant $D = |A|$ must be $0$,and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ must be non-zero.
First,calculate $D$:
$D = 1(0 - 0) - 2(0 - (k+3)(2k+1)) - 3(0 - 0)$
$D = 2(k+3)(2k+1)$
Setting $D = 0$ gives $k = -3$ or $k = -1/2$.
If $k = -3$,the second equation becomes $0 = 3$,which is a contradiction. Thus,the system is inconsistent for $k = -3$.
If $k = -1/2$,the third equation becomes $0x + 0y + z = 0$,and the second equation becomes $2.5z = 3$,which is consistent.

(B) The given system of equations is:
$a + b - 2c = 0$
$2a - 3b + c = 0$
$a - 5b + 4c = \alpha$
For a system of linear equations to be consistent,the determinant of the coefficient matrix $D$ must be zero,and the augmented determinants $D_1, D_2, D_3$ must also be zero.
First,calculate the determinant of the coefficient matrix $D$:
$D = \begin{vmatrix} 1 & 1 & -2 \\ 2 & -3 & 1 \\ 1 & -5 & 4 \end{vmatrix}$
$D = 1((-3)(4) - (1)(-5)) - 1((2)(4) - (1)(1)) - 2((2)(-5) - (-3)(1))$
$D = 1(-12 + 5) - 1(8 - 1) - 2(-10 + 3)$
$D = 1(-7) - 1(7) - 2(-7) = -7 - 7 + 14 = 0$
Since $D = 0$,the system is consistent if the determinant $D_1$ (replacing the first column with the constants) is zero:
$D_1 = \begin{vmatrix} 0 & 1 & -2 \\ 0 & -3 & 1 \\ \alpha & -5 & 4 \end{vmatrix} = 0$
Expanding along the first column:
$D_1 = \alpha \begin{vmatrix} 1 & -2 \\ -3 & 1 \end{vmatrix} = \alpha(1 - 6) = -5\alpha$
Setting $D_1 = 0$,we get $-5\alpha = 0$,which implies $\alpha = 0$.

(C) To determine the nature of the solution,we calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{vmatrix}$
Expanding along the first row:
$D = 1(3 \times 2 - 1 \times 1) - 2(2 \times 2 - 3 \times 1) + 3(2 \times 1 - 3 \times 3)$
$D = 1(6 - 1) - 2(4 - 3) + 3(2 - 9)$
$D = 1(5) - 2(1) + 3(-7)$
$D = 5 - 2 - 21 = -18$
Since $D \neq 0$,the system of linear equations has a unique solution for any values of $a, b, c$.

(D) For a system of homogeneous linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be equal to zero.
Given the system:
$3x - 2y + z = 0$
$\lambda x - 14y + 15z = 0$
$x + 2y + 3z = 0$
The determinant of the coefficient matrix is:
$\Delta = \begin{vmatrix} 3 & -2 & 1 \\ \lambda & -14 & 15 \\ 1 & 2 & 3 \end{vmatrix} = 0$
Expanding along the first row:
$3((-14)(3) - (15)(2)) - (-2)((\lambda)(3) - (15)(1)) + 1((\lambda)(2) - (-14)(1)) = 0$
$3(-42 - 30) + 2(3\lambda - 15) + (2\lambda + 14) = 0$
$3(-72) + 6\lambda - 30 + 2\lambda + 14 = 0$
$-216 + 8\lambda - 16 = 0$
$8\lambda - 232 = 0$
$8\lambda = 232$
$\lambda = \frac{232}{8} = 29$
Thus,the value of $\lambda$ is $29$.

(A) system of linear equations $AX = B$ has a unique solution if and only if the determinant of the coefficient matrix $A$ is non-zero,i.e.,$|A| \ne 0$.
The coefficient matrix $A$ is given by:
$A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{bmatrix}$
We calculate the determinant $|A|$:
$|A| = 1(1 \cdot k - (-1) \cdot 2) - 1(2 \cdot k - (-1) \cdot 3) + 1(2 \cdot 2 - 1 \cdot 3)$
$|A| = 1(k + 2) - 1(2k + 3) + 1(4 - 3)$
$|A| = k + 2 - 2k - 3 + 1$
$|A| = -k$
For a unique solution,we require $|A| \ne 0$,which implies $-k \ne 0$,or $k \ne 0$.

(C) The system of equations is given by:
$x_1 - x_2 + x_3 = 2$
$3x_1 - x_2 + 2x_3 = -6$
$3x_1 + x_2 + x_3 = -18$
First,we calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 1 & -1 & 1 \\ 3 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} = 1(-1 - 2) - (-1)(3 - 6) + 1(3 + 3) = 1(-3) + 1(-3) + 1(6) = -3 - 3 + 6 = 0$
Next,we calculate $D_1$ by replacing the first column with the constants:
$D_1 = \begin{vmatrix} 2 & -1 & 1 \\ -6 & -1 & 2 \\ -18 & 1 & 1 \end{vmatrix} = 2(-1 - 2) - (-1)(-6 + 36) + 1(-6 - 18) = 2(-3) + 1(30) - 24 = -6 + 30 - 24 = 0$
Similarly,calculating $D_2$ and $D_3$:
$D_2 = \begin{vmatrix} 1 & 2 & 1 \\ 3 & -6 & 2 \\ 3 & -18 & 1 \end{vmatrix} = 1(-6 + 36) - 2(3 - 6) + 1(-54 + 18) = 30 - 2(-3) - 36 = 30 + 6 - 36 = 0$
$D_3 = \begin{vmatrix} 1 & -1 & 2 \\ 3 & -1 & -6 \\ 3 & 1 & -18 \end{vmatrix} = 1(18 + 6) - (-1)(-54 + 18) + 2(3 + 3) = 24 - 36 + 12 = 0$
Since $D = D_1 = D_2 = D_3 = 0$,the system is consistent and has infinitely many solutions.

(A) For a system of linear equations to have a unique solution,the determinant of the coefficient matrix $D$ must be non-zero $(D \neq 0)$.
The coefficient matrix is given by:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 5 & -1 & \mu \\ 2 & 3 & -1 \end{vmatrix}$
Expanding the determinant along the first row:
$D = 1((-1)(-1) - (3)(\mu)) - 1((5)(-1) - (2)(\mu)) + 1((5)(3) - (2)(-1))$
$D = 1(1 - 3\mu) - 1(-5 - 2\mu) + 1(15 + 2)$
$D = 1 - 3\mu + 5 + 2\mu + 17$
$D = 23 - \mu$
For a unique solution,$D \neq 0$,which implies $23 - \mu \neq 0$,or $\mu \neq 23$.
Since the condition $D \neq 0$ depends only on the value of $\mu$ and is independent of $\lambda$,the existence of a unique solution depends on $\mu$ only.

(A) The given system of equations can be represented in matrix form $AX = B$ as:
$\begin{bmatrix} 1 & 1 & 1 \\ 3 & -1 & 2 \\ 3 & 1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 2 \\ 6 \\ -18 \end{bmatrix}$
To determine the nature of the solution,we calculate the determinant of the coefficient matrix $A$:
$|A| = 1((-1)(1) - (2)(1)) - 1((3)(1) - (2)(3)) + 1((3)(1) - (-1)(3))$
$|A| = 1(-1 - 2) - 1(3 - 6) + 1(3 + 3)$
$|A| = 1(-3) - 1(-3) + 1(6) = -3 + 3 + 6 = 6$
Since $|A| \neq 0$,the matrix $A$ is non-singular and invertible.
Therefore,the system of equations has a unique solution.

(D) The system of equations is given by:
$x + y + z = 6$
$x + 2y + 3z = 10$
$x + 2y + \lambda z = 12$
$A$ system of linear equations $AX = B$ is inconsistent if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.
First,calculate the determinant $D$:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix}$
Expanding along the first row:
$D = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2)$
$D = 2\lambda - 6 - \lambda + 3 + 0$
$D = \lambda - 3$
For the system to be inconsistent,we set $D = 0$:
$\lambda - 3 = 0 \Rightarrow \lambda = 3$
Now,check for consistency at $\lambda = 3$ by calculating $D_z$:
$D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 2 & 12 \end{vmatrix}$
$D_z = 1(24 - 20) - 1(12 - 10) + 6(2 - 2)$
$D_z = 1(4) - 1(2) + 6(0) = 4 - 2 = 2$
Since $D = 0$ and $D_z \neq 0$,the system is inconsistent when $\lambda = 3$.

(A) The given system of equations is homogeneous:
$x + ay + 0z = 0$
$0x + y + az = 0$
$ax + 0y + z = 0$
For a homogeneous system to have infinite solutions (non-trivial solutions),the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(1 \times 1 - a \times 0) - a(0 \times 1 - a \times a) + 0(0 \times 0 - 1 \times a) = 0$
$1(1) - a(-a^2) + 0 = 0$
$1 + a^3 = 0$
$a^3 = -1$
Taking the cube root on both sides,we get $a = -1$.

(D) Given the system of equations:
$1) 3x + y + 2z = 3$
$2) 2x - 3y - z = -3$
$3) x + 2y + z = 4$
We can solve this using the substitution method or by checking the given options.
Let us check option $(d)$ where $x = 1, y = 2, z = -1$:
For equation $(1): 3(1) + (2) + 2(-1) = 3 + 2 - 2 = 3$ (Satisfied)
For equation $(2): 2(1) - 3(2) - (-1) = 2 - 6 + 1 = -3$ (Satisfied)
For equation $(3): (1) + 2(2) + (-1) = 1 + 4 - 1 = 4$ (Satisfied)
Since all equations are satisfied,the correct values are $x = 1, y = 2, z = -1$.

(D) The system of equations has no solution if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.
First, calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{vmatrix}$
$D = 2(-2\lambda - 1) - (-1)(\lambda - 1) + (-1)(1 - (-2))$
$D = -4\lambda - 2 + \lambda - 1 - 3$
$D = -3\lambda - 6$
For the system to have no solution or infinitely many solutions, we set $D = 0$:
$-3\lambda - 6 = 0 \Rightarrow \lambda = -2$
Now, check for $\lambda = -2$ using $D_x$:
$D_x = \begin{vmatrix} 12 & -1 & -1 \\ -4 & -2 & 1 \\ 4 & 1 & -2 \end{vmatrix} = 12(4 - 1) - (-1)(8 - 4) + (-1)(-4 + 8)$
$D_x = 12(3) + 1(4) - 1(4) = 36 \neq 0$
Since $D = 0$ and $D_x \neq 0$, the system has no solution for $\lambda = -2$.

(A) 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 $x = \frac{D_1}{D}$.
Here,$D$ is the determinant of the coefficient matrix:
$D = \begin{vmatrix} 2 & 3 & 4 \\ 4 & 9 & 3 \\ 5 & 10 & 5 \end{vmatrix}$.
$D_1$ is the determinant obtained by replacing the first column of $D$ with the constants $d_1, d_2, d_3$:
$D_1 = \begin{vmatrix} 9 & 3 & 4 \\ 10 & 9 & 3 \\ 11 & 10 & 5 \end{vmatrix}$.
Thus,$x = \frac{D_1}{D} = \frac{\begin{vmatrix} 9 & 3 & 4 \\ 10 & 9 & 3 \\ 11 & 10 & 5 \end{vmatrix}}{\begin{vmatrix} 2 & 3 & 4 \\ 4 & 9 & 3 \\ 5 & 10 & 5 \end{vmatrix}}$.
Therefore,option $A$ is the correct answer.

(B) The system of equations is given by:
$x + y + z = 6$
$x + 2y + 3z = 10$
$x + 2y + \lambda z = \mu$
The coefficient matrix $A$ is:
$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{bmatrix}$
The determinant $|A| = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2) = 2\lambda - 6 - \lambda + 3 = \lambda - 3$.
For the system to have no solution or infinitely many solutions,we must have $|A| = 0$,which implies $\lambda = 3$.
When $\lambda = 3$,the system becomes:
$x + y + z = 6$
$x + 2y + 3z = 10$
$x + 2y + 3z = \mu$
Subtracting the second equation from the third,we get $0 = \mu - 10$.
If $\mu \ne 10$,this leads to a contradiction $(0 = \text{non-zero constant})$,meaning the system has no solution.
Thus,the condition for no solution is $\lambda = 3$ and $\mu \ne 10$.

(C) For the system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.
$\Delta = \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$
Applying column operation $C_2 \to C_2 - 2C_3$:
$\begin{vmatrix} 1 & 0 & a \\ 1 & b & b \\ 1 & 2c & c \end{vmatrix} = 0$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\begin{vmatrix} 1 & 0 & a \\ 0 & b & b-a \\ 0 & 2c-b & c-b \end{vmatrix} = 0$
Expanding along the first column:
$1 \cdot [b(c - b) - (b - a)(2c - b)] = 0$
$bc - b^2 - (2bc - b^2 - 2ac + ab) = 0$
$bc - b^2 - 2bc + b^2 + 2ac - ab = 0$
$2ac - ab - bc = 0$
$2ac = ab + bc$
Dividing both sides by $abc$:
$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$
This is the condition for $a, b, c$ to be in Harmonic Progression ($H$.$P$.).

(C) For a system of linear equations to have no solution or infinitely many solutions,the determinant of the coefficient matrix $D$ must be zero.
$D = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} = 0$
Expanding the determinant:
$\alpha(\alpha^2 - 1) - 1(\alpha - 1) + 1(1 - \alpha) = 0$
$\alpha(\alpha - 1)(\alpha + 1) - 1(\alpha - 1) - 1(\alpha - 1) = 0$
$(\alpha - 1)[\alpha(\alpha + 1) - 2] = 0$
$(\alpha - 1)(\alpha^2 + \alpha - 2) = 0$
$(\alpha - 1)(\alpha + 2)(\alpha - 1) = 0$
$(\alpha - 1)^2(\alpha + 2) = 0$
So,$\alpha = 1$ or $\alpha = -2$.
If $\alpha = 1$,the equations become $x + y + z = 0$,which has infinitely many solutions.
If $\alpha = -2$,the equations become:
$-2x + y + z = -3$
$x - 2y + z = -3$
$x + y - 2z = -3$
Adding these three equations gives $0 = -9$,which is a contradiction. Thus,there is no solution when $\alpha = -2$.

(A) Let $A = \begin{bmatrix} 3 & 1 \\ 4 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}$. We have $AX = B$,which implies $X = A^{-1}B$.
First,find the determinant of $A$: $|A| = (3)(1) - (1)(4) = 3 - 4 = -1$.
Next,find the adjoint of $A$: $\text{adj}(A) = \begin{bmatrix} 1 & -1 \\ -4 & 3 \end{bmatrix}$.
Then,$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-1} \begin{bmatrix} 1 & -1 \\ -4 & 3 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 4 & -3 \end{bmatrix}$.
Now,calculate $X = A^{-1}B = \begin{bmatrix} -1 & 1 \\ 4 & -3 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ 2 & 3 \end{bmatrix}$.
$X = \begin{bmatrix} (-1)(5) + (1)(2) & (-1)(-1) + (1)(3) \\ (4)(5) + (-3)(2) & (4)(-1) + (-3)(3) \end{bmatrix} = \begin{bmatrix} -5+2 & 1+3 \\ 20-6 & -4-9 \end{bmatrix} = \begin{bmatrix} -3 & 4 \\ 14 & -13 \end{bmatrix}$.

(B) Given $A = \begin{bmatrix} -1 & 2 \\ 2 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$.
From $AX = B$,we have $X = A^{-1}B$.
First,find the determinant $|A| = (-1)(-1) - (2)(2) = 1 - 4 = -3$.
The adjoint of $A$ is $adj(A) = \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix}$.
Thus,$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-3} \begin{bmatrix} -1 & -2 \\ -2 & -1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$.
Now,calculate $X = A^{-1}B = \frac{1}{3} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 1 \end{bmatrix}$.
$X = \frac{1}{3} \begin{bmatrix} (1)(3) + (2)(1) \\ (2)(3) + (1)(1) \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 3 + 2 \\ 6 + 1 \end{bmatrix} = \frac{1}{3} \begin{bmatrix} 5 \\ 7 \end{bmatrix}$.

(A) First,we perform the matrix multiplication on the right side of the equation:
$\begin{bmatrix} 4 & -2 \\ 0 & -6 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 4(2) + (-2)(1) \\ 0(2) + (-6)(1) \\ -1(2) + 2(1) \end{bmatrix} = \begin{bmatrix} 8 - 2 \\ 0 - 6 \\ -2 + 2 \end{bmatrix} = \begin{bmatrix} 6 \\ -6 \\ 0 \end{bmatrix}$
Now,we equate this to the left side:
$\begin{bmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 6 \\ -6 \\ 0 \end{bmatrix}$
This gives us the system of linear equations:
$1) \ x + 2y + 3z = 6$
$2) \ 3x + y + 2z = -6$
$3) \ 2x + 3y + z = 0$
Adding the three equations:
$(x + 3x + 2x) + (2y + y + 3y) + (3z + 2z + z) = 6 - 6 + 0$
$6x + 6y + 6z = 0 \implies x + y + z = 0$
From equation $(3)$,$z = -2x - 3y$. Substituting into $(1)$:
$x + 2y + 3(-2x - 3y) = 6 \implies x + 2y - 6x - 9y = 6 \implies -5x - 7y = 6$
Solving the system,we find $x = -4, y = 2, z = 2$.

(D) Given the matrix equation $AX = B$,where $A = \begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$,and $B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$.
By performing matrix multiplication,we get the system of linear equations:
$x + z = 1$ $(1)$
$-x + y = 1$ $(2)$
$-y + z = 2$ $(3)$
From equation $(2)$,we have $y = x + 1$.
From equation $(1)$,we have $z = 1 - x$.
Substitute $y$ and $z$ into equation $(3)$:
$(1 - x) - (x + 1) = 2$
$1 - x - x - 1 = 2$
$-2x = 2$
$x = -1$.
Now,find $y$ and $z$:
$y = -1 + 1 = 0$
$z = 1 - (-1) = 2$.
Thus,$(x, y, z) = (-1, 0, 2)$.

(A) The given system of equations is:
$x + 2y + 3z = 1$
$2x + y + 3z = 2$
$5x + 5y + 9z = 4$
We can represent this as $AX = B$,where $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 5 & 5 & 9 \end{bmatrix}$.
Now,calculate the determinant $|A|$:
$|A| = 1(1 \times 9 - 3 \times 5) - 2(2 \times 9 - 3 \times 5) + 3(2 \times 5 - 1 \times 5)$
$|A| = 1(9 - 15) - 2(18 - 15) + 3(10 - 5)$
$|A| = 1(-6) - 2(3) + 3(5)$
$|A| = -6 - 6 + 15 = 3$
Since $|A| = 3 \neq 0$,the system of equations has a unique solution.

(A) The given system of linear equations is:
$0x_1 + 1x_2 - 1x_3 = 1$
$-1x_1 + 0x_2 + 2x_3 = -2$
$1x_1 - 2x_2 + 0x_3 = 3$
First,we calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 0 & 1 & -1 \\ -1 & 0 & 2 \\ 1 & -2 & 0 \end{vmatrix} = 0(0 - (-4)) - 1(0 - 2) + (-1)(2 - 0) = 0 + 2 - 2 = 0$
Since $D = 0$,the system is either inconsistent or has infinitely many solutions. We calculate $D_1$ by replacing the first column with the constants:
$D_1 = \begin{vmatrix} 1 & 1 & -1 \\ -2 & 0 & 2 \\ 3 & -2 & 0 \end{vmatrix} = 1(0 - (-4)) - 1(0 - 6) + (-1)(4 - 0) = 4 + 6 - 4 = 6$
Since $D = 0$ and $D_1 \neq 0$,the system of equations is inconsistent and has no solution.

(D) Given the matrix equation: $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -2 & -2 \\ 1 & 3 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \\ 4 \end{bmatrix}$.
This corresponds to the system of linear equations:
$1) x + y + z = 0$
$2) x - 2y - 2z = 3$
$3) x + 3y + z = 4$
Subtracting equation $(1)$ from equation $(3)$:
$(x + 3y + z) - (x + y + z) = 4 - 0$
$2y = 4 \implies y = 2$.
Substitute $y = 2$ into equations $(1)$ and $(2)$:
$x + 2 + z = 0 \implies x + z = -2$ $(4)$
$x - 2(2) - 2z = 3 \implies x - 2z = 7$ $(5)$
Subtracting equation $(5)$ from equation $(4)$:
$(x + z) - (x - 2z) = -2 - 7$
$3z = -9 \implies z = -3$.
Substitute $z = -3$ into equation $(4)$:
$x - 3 = -2 \implies x = 1$.
Thus,the solution is $\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -3 \end{bmatrix}$.

(B) Let $\frac{x^2}{a^2} = X, \frac{y^2}{b^2} = Y$ and $\frac{z^2}{c^2} = Z$.
The system of equations becomes:
$X + Y - Z = 1$
$X - Y + Z = 1$
$-X + Y + Z = 1$
The coefficient matrix $A$ is given by:
$A = \begin{bmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ -1 & 1 & 1 \end{bmatrix}$
Calculating the determinant $|A|$:
$|A| = 1((-1)(1) - (1)(1)) - 1((1)(1) - (1)(-1)) + (-1)((1)(1) - (-1)(-1))$
$|A| = 1(-2) - 1(2) - 1(0) = -2 - 2 = -4$
Since $|A| = -4 \neq 0$,the system has a unique solution for $(X, Y, Z)$.
Solving the system:
Adding the equations: $(X+Y-Z) + (X-Y+Z) = 1+1 \implies 2X = 2 \implies X = 1$.
Similarly,$Y = 1$ and $Z = 1$.
Thus,$\frac{x^2}{a^2} = 1, \frac{y^2}{b^2} = 1, \frac{z^2}{c^2} = 1$,which gives $x = \pm a, y = \pm b, z = \pm c$. Since there are finite values for $x, y, z$,the system has a unique solution for $(X, Y, Z)$ and finitely many solutions for $(x, y, z)$. However,in the context of the matrix system for $(X, Y, Z)$,it is a unique solution.

(D) Given the equations:
${x^a}{y^b} = {e^m}$ and ${x^c}{y^d} = {e^n}$
Taking the natural logarithm on both sides of both equations:
$a \ln x + b \ln y = m$
$c \ln x + d \ln y = n$
This is a system of linear equations in terms of $\ln x$ and $\ln y$.
Using Cramer's rule,we define the determinants:
${\Delta _3} = \left| {\begin{array}{*{20}{c}} a & b \\ c & d \end{array}} \right|$
${\Delta _1} = \left| {\begin{array}{*{20}{c}} m & b \\ n & d \end{array}} \right|$
${\Delta _2} = \left| {\begin{array}{*{20}{c}} a & m \\ c & n \end{array}} \right|$
According to Cramer's rule:
$\ln x = \frac{{{\Delta _1}}}{{{\Delta _3}}}$ and $\ln y = \frac{{{\Delta _2}}}{{{\Delta _3}}}$
Therefore,solving for $x$ and $y$:
$x = {e^{{\Delta _1}/{\Delta _3}}}$ and $y = {e^{{\Delta _2}/{\Delta _3}}}$
Thus,the correct option is $(d)$.

(C) Given the system of equations:
$3X + 2Y = I$ $(i)$
$2X - Y = O$ $(ii)$
Multiply equation $(ii)$ by $2$:
$4X - 2Y = 2O = O$ $(iii)$
Adding equation $(i)$ and $(iii)$:
$(3X + 2Y) + (4X - 2Y) = I + O$
$7X = I$
$X = \frac{1}{7}I$
Substitute $X$ into equation $(ii)$:
$2(\frac{1}{7}I) - Y = O$
$Y = \frac{2}{7}I$
Thus,$X = \frac{1}{7}I$ and $Y = \frac{2}{7}I$.

(A) Given equations are:
$x + y = 10$ $(i)$
$2x + y = 18$ $(ii)$
$4x - 3y = 26$ $(iii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(2x + y) - (x + y) = 18 - 10$
$x = 8$
Substituting $x = 8$ in equation $(i)$:
$8 + y = 10$
$y = 2$
Now,check if the point $(8, 2)$ satisfies equation $(iii)$:
$L.H.S. = 4(8) - 3(2) = 32 - 6 = 26$
$R.H.S. = 26$
Since $L.H.S. = R.H.S.$,the point $(8, 2)$ satisfies all three equations.
Therefore,the system of equations has exactly one common solution.

(B) For a system of linear equations to have no solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and at least one of the determinants $\Delta_1$ or $\Delta_2$ must be non-zero.
$\Delta = \begin{vmatrix} k+1 & 8 \\ k & k+3 \end{vmatrix} = (k+1)(k+3) - 8k = k^2 + 4k + 3 - 8k = k^2 - 4k + 3 = (k-3)(k-1)$.
Setting $\Delta = 0$,we get $k = 1$ or $k = 3$.
Now,calculate $\Delta_1$ and $\Delta_2$ for these values:
$\Delta_1 = \begin{vmatrix} 4k & 8 \\ 3k-1 & k+3 \end{vmatrix} = 4k(k+3) - 8(3k-1) = 4k^2 + 12k - 24k + 8 = 4k^2 - 12k + 8 = 4(k-1)(k-2)$.
$\Delta_2 = \begin{vmatrix} k+1 & 4k \\ k & 3k-1 \end{vmatrix} = (k+1)(3k-1) - 4k^2 = 3k^2 + 2k - 1 - 4k^2 = -k^2 + 2k - 1 = -(k-1)^2$.
Case $k=1$: $\Delta = 0, \Delta_1 = 0, \Delta_2 = 0$. This leads to infinitely many solutions (or consistency).
Case $k=3$: $\Delta = 0, \Delta_1 = 4(3-1)(3-2) = 8 \neq 0, \Delta_2 = -(3-1)^2 = -4 \neq 0$.
Since $\Delta = 0$ and $\Delta_1, \Delta_2 \neq 0$ for $k=3$,the system has no solutions.
Thus,there is only $1$ value of $k$.

(A) The system of equations has no solution if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.
The determinant $D$ is given by:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & a & 1 \\ a & b & 1 \end{vmatrix} = 1(a - b) - 1(1 - a) + 1(b - a^2) = a - b - 1 + a + b - a^2 = -a^2 + 2a - 1 = -(a - 1)^2$.
For the system to have no solution or infinitely many solutions,we must have $D = 0$,which implies $-(a - 1)^2 = 0$,so $a = 1$.
Substituting $a = 1$ into the system:
$x + y + z = 1$
$x + y + z = 1$
$x + by + z = 0$
For the system to have no solution,the third equation must be inconsistent with the first two. Since the first two equations represent the same plane $x + y + z = 1$,the third equation $x + by + z = 0$ must represent a plane parallel to the first one but not identical to it.
Comparing $x + y + z = 1$ and $x + by + z = 0$,we see that for the planes to be parallel,the coefficients of $x, y, z$ must be proportional. Thus,$1/1 = b/1 = 1/1$,which implies $b = 1$.
However,if $b = 1$,the third equation becomes $x + y + z = 0$,which is parallel to $x + y + z = 1$ but distinct (since $0 \neq 1$). Thus,the system has no solution when $b = 1$.
Therefore,$S = \{1\}$,which is a singleton set.

(A) For a system of homogeneous linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 4 & -3 \end{vmatrix} = 0$
Expanding along the first row:
$1(-3k + 8) - k(-9 + 4) + 3(12 - 2k) = 0$
$-3k + 8 + 5k + 36 - 6k = 0$
$-4k + 44 = 0 \Rightarrow k = 11$
Substituting $k = 11$ into the equations:
$x + 11y + 3z = 0$ $(1)$
$3x + 11y - 2z = 0$ $(2)$
$2x + 4y - 3z = 0$ $(3)$
Subtracting $(1)$ from $(2)$:
$(3x - x) + (11y - 11y) + (-2z - 3z) = 0 \Rightarrow 2x - 5z = 0 \Rightarrow x = \frac{5}{2}z$
Substituting $x = \frac{5}{2}z$ into $(1)$:
$\frac{5}{2}z + 11y + 3z = 0 \Rightarrow 11y = -\frac{11}{2}z \Rightarrow y = -\frac{1}{2}z$
Now,calculate $\frac{xz}{y^2}$:
$\frac{xz}{y^2} = \frac{(\frac{5}{2}z)(z)}{(-\frac{1}{2}z)^2} = \frac{\frac{5}{2}z^2}{\frac{1}{4}z^2} = \frac{5}{2} \times 4 = 10$

(A) Given the matrix equation $AX = B$.
Multiplying both sides by $A^{-1}$ on the left,we get $A^{-1}(AX) = A^{-1}B$.
Since $A^{-1}A = I$,we have $IX = A^{-1}B$,which simplifies to $X = A^{-1}B$.
Now,substitute the given matrices:
$X = \begin{bmatrix} 3 & -\frac{1}{2} & -\frac{1}{2} \\ -4 & \frac{3}{4} & \frac{5}{4} \\ 2 & -\frac{1}{4} & -\frac{3}{4} \end{bmatrix} \begin{bmatrix} 9 \\ 52 \\ 0 \end{bmatrix}$
Calculating the product:
$X = \begin{bmatrix} (3 \times 9) + (-\frac{1}{2} \times 52) + (-\frac{1}{2} \times 0) \\ (-4 \times 9) + (\frac{3}{4} \times 52) + (\frac{5}{4} \times 0) \\ (2 \times 9) + (-\frac{1}{4} \times 52) + (-\frac{3}{4} \times 0) \end{bmatrix}$
$X = \begin{bmatrix} 27 - 26 + 0 \\ -36 + 39 + 0 \\ 18 - 13 + 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix}$.

(B) The given system of linear equations is:
$a_1x + b_1y + c_1z = -d_1$
$a_2x + b_2y + c_2z = -d_2$
$a_3x + b_3y + c_3z = -d_3$
According to Cramer's rule,the value of $x$ is given by $x = \frac{D_x}{D}$,where $D = \Delta (a,b,c) = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}$ and $D_x = \begin{vmatrix} -d_1 & b_1 & c_1 \\ -d_2 & b_2 & c_2 \\ -d_3 & b_3 & c_3 \end{vmatrix}$.
Taking $-1$ common from the first column of $D_x$,we get:
$D_x = -1 \times \begin{vmatrix} d_1 & b_1 & c_1 \\ d_2 & b_2 & c_2 \\ d_3 & b_3 & c_3 \end{vmatrix} = -\Delta (d,b,c) = -\Delta (b,c,d)$.
Thus,$x = \frac{-\Delta (bcd)}{\Delta (abc)}$.

(B) For a system of $n$ linear equations $AX = B$ to have a unique solution,the determinant of the coefficient matrix $A$ must be non-zero $(|A| \neq 0)$. This means the matrix $A$ is non-singular. Thus,statement $(B)$ is correct.
Regarding $(A)$,it is incorrect because a unique solution requires a non-singular matrix.
Regarding $(C)$,if $A^{-1}$ exists,then $|A| \neq 0$. Since $|adj A| = |A|^{n-1}$,it follows that $|adj A| \neq 0$,so $(adj A)^{-1}$ must exist.
Regarding $(D)$,using the property of the given matrix,$F(x) \cdot F(y) = F(x + y)$,not $F(x - y)$.

(D) For a system of linear equations to have an infinite number of solutions,the determinant of the coefficient matrix $D$ must be $0$,and $D_x = D_y = D_z = 0$.
The coefficient matrix is $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & p & 2 \\ 1 & 4 & \mu \end{bmatrix}$.
Setting $D = 0$:
$1(p\mu - 8) - 2(\mu - 2) + 3(4 - p) = 0$
$p\mu - 8 - 2\mu + 4 + 12 - 3p = 0$
$p\mu - 3p - 2\mu + 8 = 0$
$(p - 2)(\mu - 3) = 2$ (This does not immediately yield a simple integer solution).
Alternatively,using the augmented matrix $\begin{bmatrix} 1 & 2 & 3 & | & 4 \\ 1 & p & 2 & | & 3 \\ 1 & 4 & \mu & | & 3 \end{bmatrix}$.
Subtracting row $1$ from row $2$ and row $3$:
$R_2 \to R_2 - R_1: (0, p-2, -1, | -1)$
$R_3 \to R_3 - R_1: (0, 2, \mu-3, | -1)$
For infinite solutions,the rows must be proportional: $\frac{p-2}{2} = \frac{-1}{\mu-3} = \frac{-1}{-1} = 1$.
Thus,$p-2 = 2 \implies p = 4$ and $\mu-3 = -1 \implies \mu = 2$.
Checking the options,none match $p=4, \mu=2$.

(B) matrix is singular if its determinant is equal to $0$.
Given $A = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Then $A - \lambda I = \begin{bmatrix} 1 & 3 \\ 2 & 2 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} = \begin{bmatrix} 1 - \lambda & 3 \\ 2 & 2 - \lambda \end{bmatrix}$.
For $A - \lambda I$ to be a singular matrix,$\det(A - \lambda I) = 0$.
$\det(A - \lambda I) = (1 - \lambda)(2 - \lambda) - (3)(2) = 0$.
$2 - \lambda - 2\lambda + \lambda^2 - 6 = 0$.
$\lambda^2 - 3\lambda - 4 = 0$.
Thus,the correct option is $B$.

(A) Given the system of equations:
$a^2 x - ay = 1 - a$ $(1)$
$bx + (3 - 2b) y = 3 + a$ $(2)$
Since $(x, y) = (1, 1)$ is a solution,substitute these values into both equations:
For equation $(1)$: $a^2(1) - a(1) = 1 - a \implies a^2 - a = 1 - a \implies a^2 = 1 \implies a = \pm 1$.
For equation $(2)$: $b(1) + (3 - 2b)(1) = 3 + a \implies b + 3 - 2b = 3 + a \implies 3 - b = 3 + a \implies b = -a$.
Case $I$: If $a = 1$,then $b = -1$. Check for uniqueness: The determinant of the coefficient matrix is $D = \begin{vmatrix} a^2 & -a \\ b & 3 - 2b \end{vmatrix} = \begin{vmatrix} 1 & -1 \\ -1 & 5 \end{vmatrix} = 5 - 1 = 4 \neq 0$. Thus,$a = 1, b = -1$ gives a unique solution.
Case $II$: If $a = -1$,then $b = 1$. Check for uniqueness: $D = \begin{vmatrix} (-1)^2 & -(-1) \\ 1 & 3 - 2(1) \end{vmatrix} = \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = 1 - 1 = 0$. Since $D = 0$,the system does not have a unique solution for $a = -1, b = 1$.
Therefore,the only valid solution is $a = 1, b = -1$.

(B) Given the system of equations:
$ax - by = 2a - b$ $(1)$
$(c + 1)x + cy = 10 - a + 3b$ $(2)$
Since $(x = 1, y = 3)$ is a solution,substitute these values into the equations:
From $(1)$: $a(1) - b(3) = 2a - b \implies a - 3b = 2a - b \implies a = -2b$.
From $(2)$: $(c + 1)(1) + c(3) = 10 - a + 3b \implies c + 1 + 3c = 10 - a + 3b \implies 4c + 1 = 10 - a + 3b$.
Substitute $a = -2b$ into the second result: $4c + 1 = 10 - (-2b) + 3b \implies 4c + 1 = 10 + 5b \implies 4c = 9 + 5b \implies c = \frac{9 + 5b}{4}$.
For infinitely many solutions,the ratio of coefficients must be equal:
$\frac{a}{c + 1} = \frac{-b}{c} = \frac{2a - b}{10 - a + 3b}$.
Using $\frac{a}{c + 1} = \frac{-b}{c}$ and substituting $a = -2b$: $\frac{-2b}{c + 1} = \frac{-b}{c}$.
This gives $b = 0$ or $c = 1$.
Case $1$: If $b = 0$,then $a = -2(0) = 0$. Substituting into $c = \frac{9 + 5(0)}{4}$,we get $c = \frac{9}{4}$. This gives the triplet $(0, 0, 9/4)$.
Case $2$: If $c = 1$,then $1 = \frac{9 + 5b}{4} \implies 4 = 9 + 5b \implies 5b = -5 \implies b = -1$. Then $a = -2(-1) = 2$. This gives the triplet $(2, -1, 1)$.
Thus,there are exactly two such triplets.

(B) To determine the nature of the system,we write it in the matrix form $AX = B$,where $A = \begin{bmatrix} 3 & -7 & 5 \\ 3 & 1 & 5 \\ 2 & 3 & 5 \end{bmatrix}$.
We calculate the determinant $|A|$:
$|A| = 3(1 \times 5 - 5 \times 3) - (-7)(3 \times 5 - 5 \times 2) + 5(3 \times 3 - 1 \times 2)$
$|A| = 3(5 - 15) + 7(15 - 10) + 5(9 - 2)$
$|A| = 3(-10) + 7(5) + 5(7)$
$|A| = -30 + 35 + 35 = 40$.
Since $|A| \neq 0$,the system has a unique solution.
By Cramer's rule or matrix inversion,we can find the values of $x, y, z$. Since the constants on the right side are non-zero,the solution is non-trivial.
Thus,the system is consistent with a unique non-trivial solution.

(A) For a system of homogeneous linear equations $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.
The coefficient matrix is:
$A = \begin{bmatrix} \lambda & -1 & \cos\theta \\ 3 & 1 & 2 \\ \cos\theta & 1 & 2 \end{bmatrix}$
Calculating the determinant $|A|$:
$|A| = \lambda(1 \times 2 - 1 \times 2) - (-1)(3 \times 2 - \cos\theta \times 2) + \cos\theta(3 \times 1 - \cos\theta \times 1)$
$|A| = \lambda(0) + (6 - 2\cos\theta) + \cos\theta(3 - \cos\theta)$
$|A| = 6 - 2\cos\theta + 3\cos\theta - \cos^2\theta$
$|A| = 6 + \cos\theta - \cos^2\theta$
For a non-trivial solution,$|A| = 0$:
$-\cos^2\theta + \cos\theta + 6 = 0$
$\cos^2\theta - \cos\theta - 6 = 0$
Let $t = \cos\theta$. Then $t^2 - t - 6 = 0$.
$(t - 3)(t + 2) = 0$
So,$\cos\theta = 3$ or $\cos\theta = -2$.
Since the range of $\cos\theta$ is $[-1, 1]$,neither $\cos\theta = 3$ nor $\cos\theta = -2$ is possible for any real $\theta$.
Therefore,$|A|$ is never zero for any $\theta \in (0, 2\pi)$.
Thus,the system has only the trivial solution $(x=0, y=0, z=0)$ for all values of $\lambda$ and $\theta$.

(B) The system of equations can be represented in matrix form $AX = B$,where $A = \begin{bmatrix} \sin\theta & 0 & 2 \\ \cos\theta & \sin\theta & 0 \\ 0 & \cos\theta & 2 \end{bmatrix}$.
To check for a unique solution,we calculate the determinant $D = |A|$.
$D = \sin\theta (\sin\theta \cdot 2 - 0 \cdot \cos\theta) - 0 + 2 (\cos\theta \cdot \cos\theta - 0 \cdot \sin\theta)$
$D = 2\sin^2\theta + 2\cos^2\theta = 2(\sin^2\theta + \cos^2\theta) = 2(1) = 2$.
Since $D = 2 \neq 0$,the system has a unique solution.
Using Cramer's Rule,$x = \frac{D_x}{D}$,$y = \frac{D_y}{D}$,and $z = \frac{D_z}{D}$.
$D_x = \begin{vmatrix} 0 & 0 & 2 \\ 0 & \sin\theta & 0 \\ a & \cos\theta & 2 \end{vmatrix} = 2(0 - a\sin\theta) = -2a\sin\theta$.
$D_y = \begin{vmatrix} \sin\theta & 0 & 2 \\ \cos\theta & 0 & 0 \\ 0 & a & 2 \end{vmatrix} = -0 + 0 - 2(a\cos\theta) = -2a\cos\theta$.
$D_z = \begin{vmatrix} \sin\theta & 0 & 0 \\ \cos\theta & \sin\theta & 0 \\ 0 & \cos\theta & a \end{vmatrix} = \sin\theta(a\sin\theta - 0) = a\sin^2\theta$.
Thus,$x = -a\sin\theta$,$y = -a\cos\theta$,and $z = \frac{a\sin^2\theta}{2}$.
Since the solution depends on both $a$ and $\theta$,the correct option is $B$.

(A) To determine the nature of the solutions for the system $Ax = b$,we first calculate the determinant of matrix $A$,denoted as $|A|$.
$|A| = 1(0 \times 1 - 5 \times 2) - 2(2 \times 1 - 5 \times 0) + 3(2 \times 2 - 0 \times 0)$
$|A| = 1(0 - 10) - 2(2 - 0) + 3(4 - 0)$
$|A| = -10 - 4 + 12 = -2$
Since $|A| = -2$,which is not equal to $0$,the matrix $A$ is non-singular (invertible).
$A$ system of linear equations $Ax = b$ has a unique solution if and only if the matrix $A$ is invertible $(|A| \neq 0)$.
Therefore,the system $Ax = b$ has a unique solution.

(B) Statement $-1$: If two lines are neither parallel nor coincident,they must intersect at exactly one point. Thus,the system has a unique solution. This statement is $T$.
Statement $-2$: The system $ax + by = 0$ and $cx + dy = 0$ is a homogeneous system. If it has a non-zero solution,the determinant $ad - bc$ must be $0$. This implies the lines are coincident,leading to infinitely many solutions. This statement is $T$.
Statement $-3$: Substitute $x = y$ and $y = 1 + z$ into $x + y + z = 1$. We get $(1 + z) + (1 + z) + z = 1$,which simplifies to $3z + 2 = 1$,so $3z = -1$ or $z = -1/3$. This gives $y = 2/3$ and $x = 2/3$. Since a solution exists,the system is consistent. This statement is $F$.
Statement $-4$: If two equations in a system are inconsistent (e.g.,$x + y = 1$ and $x + y = 2$),they cannot be satisfied simultaneously,making the entire system inconsistent. This statement is $T$.
The correct order is $TTFT$.

(D) To find the nature of the solutions,we calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 1 & -1 & 3 \\ 2 & -1 & 1 \\ 1 & -2 & \alpha \end{vmatrix}$
$D = 1(-\alpha + 2) - (-1)(2\alpha - 1) + 3(-4 + 1)$
$D = -\alpha + 2 + 2\alpha - 1 - 9 = \alpha - 8$
For a unique solution,$D \neq 0$,which implies $\alpha \neq 8$. Thus,option $b$ is correct.
If $\alpha = 8$,then $D = 0$. We check for consistency by calculating $D_1, D_2, D_3$:
$D_1 = \begin{vmatrix} 2 & -1 & 3 \\ 4 & -1 & 1 \\ 3 & -2 & 8 \end{vmatrix} = 2(-8 + 2) + 1(32 - 3) + 3(-8 + 3) = -12 + 29 - 15 = 2 \neq 0$
Since $D = 0$ and $D_1 \neq 0$,the system has no solution for $\alpha = 8$. Thus,option $c$ is also correct.
Therefore,both $b$ and $c$ are correct.