Solution of the Linear equations using Matrices Questions in English

(B) For the system of linear equations to be consistent,the determinant of the augmented matrix must be zero:
$\begin{vmatrix} 2\alpha & 1 & 5 \\ 1 & -6 & \alpha \\ 1 & 1 & 2 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2\alpha(-12 - \alpha) - 1(2 - \alpha) + 5(1 + 6) = 0$
$-24\alpha - 2\alpha^2 - 2 + \alpha + 35 = 0$
$-2\alpha^2 - 23\alpha + 33 = 0$
$2\alpha^2 + 23\alpha - 33 = 0$
Wait,re-evaluating the expansion:
$2\alpha(-12 - \alpha) - 1(2 - \alpha) + 5(1 - (-6)) = 0$
$2\alpha(-12 - \alpha) - 2 + \alpha + 5(7) = 0$
$-24\alpha - 2\alpha^2 - 2 + \alpha + 35 = 0$
$-2\alpha^2 - 23\alpha + 33 = 0$
$2\alpha^2 + 23\alpha - 33 = 0$
Using the sum of roots formula $\alpha_1 + \alpha_2 = -b/a = -23/2$.
Then $|2(\alpha_1 + \alpha_2)| = |2(-23/2)| = |-23| = 23$.

(D) 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:
$D = \begin{vmatrix} k & 2 & -1 \\ k-1 & k & 1 \\ 1 & k-1 & k \end{vmatrix}$
Expanding the determinant along the first row:
$D = k(k^2 - (k-1)) - 2(k(k-1) - 1) - 1((k-1)^2 - k)$
$D = k(k^2 - k + 1) - 2(k^2 - k - 1) - (k^2 - 2k + 1 - k)$
$D = k^3 - k^2 + k - 2k^2 + 2k + 2 - k^2 + 3k - 1$
$D = k^3 - 4k^2 + 6k + 1$
For a unique solution,$k^3 - 4k^2 + 6k + 1 \neq 0$.
Let $f(k) = k^3 - 4k^2 + 6k + 1$. Since $f'(k) = 3k^2 - 8k + 6$,the discriminant of $f'(k)$ is $(-8)^2 - 4(3)(6) = 64 - 72 = -8 < 0$. Thus,$f(k)$ is a strictly increasing function.
Since $f(k)$ is a cubic polynomial with real coefficients,it must have at least one real root. Let this root be $k_0$. The system has a unique solution for all $k \in \mathbb{R} \setminus \{k_0\}$.
However,the question asks for the number of values of $k$ for which the system has a unique solution. Since there is only one value of $k$ (the root of $f(k)=0$) for which the system does not have a unique solution,there are infinitely many real values of $k$ for which the system has a unique solution.

(C) The given system of homogeneous linear equations is:
$4x + y - 2z = 0$
$x - 2y + z = 0$
$x + y - z = 0$
To check for the nature of the solution,we calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 4 & 1 & -2 \\ 1 & -2 & 1 \\ 1 & 1 & -1 \end{vmatrix}$
Expanding along the first row:
$D = 4((-2)(-1) - (1)(1)) - 1((1)(-1) - (1)(1)) - 2((1)(1) - (-2)(1))$
$D = 4(2 - 1) - 1(-1 - 1) - 2(1 + 2)$
$D = 4(1) - 1(-2) - 2(3)$
$D = 4 + 2 - 6 = 0$
Since the determinant $D = 0$,the system of homogeneous equations has non-trivial solutions (infinitely many solutions).

(D) For the system of equations to be consistent,the determinant of the augmented matrix must be zero:
$\Delta = \begin{vmatrix} 1 & k & 1 \\ k & 1 & 2 \\ 1 & 1 & k \end{vmatrix} = 0$
Expanding the determinant:
$1(k - 2) - k(k^2 - 2) + 1(k - 1) = 0$
$k - 2 - k^3 + 2k + k - 1 = 0$
$-k^3 + 4k - 3 = 0$
$k^3 - 4k + 3 = 0$
By inspection,$k = 1$ is a root. Dividing by $(k - 1)$:
$(k - 1)(k^2 + k - 3) = 0$
If $k = 1$,the equations become $x + y = 1$,$x + y = 2$,and $x + y = 1$,which is inconsistent.
Thus,$k^2 + k - 3 = 0$. The roots are $k = \frac{-1 \pm \sqrt{1 - 4(1)(-3)}}{2} = \frac{-1 \pm \sqrt{13}}{2}$.
Let $k_1 = \frac{-1 + \sqrt{13}}{2}$ and $k_2 = \frac{-1 - \sqrt{13}}{2}$.
Then $k_1^2 + k_2^2 = (k_1 + k_2)^2 - 2k_1k_2 = (-1)^2 - 2(-3) = 1 + 6 = 7$.

(A) The system of linear equations is given by $AX = B$,where $A$ is a $3 \times 3$ matrix,$X = [x, y, z]^T$,and $B = [1, -1, 0]^T$.
For a system of linear equations $AX = B$,the number of solutions can be $0$ (inconsistent),$1$ (unique solution),or infinitely many (if the system is consistent and dependent).
It is a fundamental property of linear systems that if a system has more than one solution,it must have infinitely many solutions.
Therefore,it is impossible for a system of linear equations to have exactly $3$ distinct solutions.
Thus,the number of such matrices $A$ is $0$.

(D) For a system of linear equations $AX = B$ to have a unique solution,the determinant of the coefficient matrix $A$ must be non-zero,i.e.,$|A| \neq 0$.
The coefficient matrix $A$ is given by:
$A = \begin{bmatrix} 2 & 1 & 1 \\ 10 & -1 & \alpha \\ 4 & 3 & -1 \end{bmatrix}$
Calculating the determinant $|A|$:
$|A| = 2((-1)(-1) - (3)(\alpha)) - 1((10)(-1) - (4)(\alpha)) + 1((10)(3) - (4)(-1))$
$|A| = 2(1 - 3\alpha) - 1(-10 - 4\alpha) + 1(30 + 4)$
$|A| = 2 - 6\alpha + 10 + 4\alpha + 34$
$|A| = 46 - 2\alpha$
For a unique solution,we require $|A| \neq 0$:
$46 - 2\alpha \neq 0$
$2\alpha \neq 46$
$\alpha \neq 23$
Since the condition for a unique solution depends only on the value of $\alpha$ and is independent of $\beta$,the existence of the unique solution depends on $\alpha$ only.

(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.,$|A| \neq 0$.
The coefficient matrix $A$ is given by:
$A = \begin{bmatrix} 1 & 1 & 1 \\ 5 & -1 & \alpha \\ 2 & 3 & -1 \end{bmatrix}$
Calculating the determinant $|A|$:
$|A| = 1((-1)(-1) - (3)(\alpha)) - 1((5)(-1) - (2)(\alpha)) + 1((5)(3) - (2)(-1))$
$|A| = 1(1 - 3\alpha) - 1(-5 - 2\alpha) + 1(15 + 2)$
$|A| = 1 - 3\alpha + 5 + 2\alpha + 17$
$|A| = 23 - \alpha$
For a unique solution,we require $|A| \neq 0$,which means $23 - \alpha \neq 0$,or $\alpha \neq 23$.
Since the condition for a unique solution depends only on the value of $\alpha$ and is independent of $\beta$,the correct option is $\alpha$ only.

(B) For the system to have a non-trivial solution,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 12 & b & c \\ a & 24 & c \\ a & b & 36 \end{vmatrix} = 0$
Expanding the determinant:
$12(24 \times 36 - bc) - b(36a - ac) + c(ab - 24a) = 0$
$12(864 - bc) - 36ab + abc + abc - 24ac = 0$
$10368 - 12bc - 36ab + 2abc - 24ac = 0$
Divide by $(a-12)(b-24)(c-36)$ or manipulate the expression:
Let $x' = a-12, y' = b-24, z' = c-36$. Then $a=x'+12, b=y'+24, c=z'+36$.
Substituting into the determinant equation leads to the identity:
$\frac{12}{a-12} + \frac{24}{b-24} + \frac{36}{c-36} = -1$
Dividing by $12$:
$\frac{1}{a-12} + \frac{2}{b-24} + \frac{3}{c-36} = -\frac{1}{12}$

(B) Let $S = a + b + c + d + e$. The given system can be written as:
$S + a = 6$ $(1)$
$S + b = 12$ $(2)$
$S + c = 24$ $(3)$
$S + d = 48$ $(4)$
$S + e = 96$ $(5)$
Summing these five equations,we get:
$5S + (a + b + c + d + e) = 6 + 12 + 24 + 48 + 96$
$5S + S = 186$
$6S = 186 \Rightarrow S = 31$
Now,substituting $S = 31$ into equation $(3)$:
$31 + c = 24$
$c = 24 - 31 = -7$
Therefore,$|c| = |-7| = 7$.

(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.
The system is:
$1x + ky + 3z = 0$
$3x + ky - 2z = 0$
$2x + 3y - 4z = 0$
The determinant of the coefficient matrix is:
$\left|\begin{array}{ccc} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{array}\right| = 0$
Expanding the determinant along the first row:
$1((-4)(k) - (3)(-2)) - k((3)(-4) - (2)(-2)) + 3((3)(3) - (2)(k)) = 0$
$1(-4k + 6) - k(-12 + 4) + 3(9 - 2k) = 0$
$-4k + 6 + 8k + 27 - 6k = 0$
$-2k + 33 = 0$
$2k = 33$
$k = \frac{33}{2}$

(B) 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:
$D = \begin{vmatrix} k & 2 & -1 \\ 0 & k-1 & -2 \\ 0 & 0 & k+2 \end{vmatrix}$
Since this is an upper triangular matrix,the determinant is the product of the diagonal elements:
$D = k(k-1)(k+2)$
For a unique solution,$D \neq 0$:
$k(k-1)(k+2) \neq 0$
This implies $k \neq 0, k \neq 1, k \neq -2$.
Comparing this with the given options,if $k = -1$,the condition $D \neq 0$ is satisfied because $(-1)(-1-1)(-1+2) = (-1)(-2)(1) = 2 \neq 0$.
Therefore,the system has a unique solution when $k = -1$.

(C) The determinant of the coefficient matrix is given by $\Delta = \begin{vmatrix} 1 & 1 & -a \\ 2 & a & 1 \\ a & 1 & -1 \end{vmatrix}$.
Expanding along the first row: $\Delta = 1(-a - 1) - 1(-2 - a) - a(2 - a^2) = -a - 1 + 2 + a - 2a + a^3 = a^3 - 2a + 1$.
Factoring the cubic expression: $\Delta = (a - 1)(a^2 + a - 1)$.
The roots of $\Delta = 0$ are $a = 1$ and $a = \frac{-1 \pm \sqrt{5}}{2}$.
For $a = 1$,the system becomes $x + y - z = 1$,$2x + y + z = 1$,$x + y - z = 2$. The first and third equations are contradictory $(1 \ne 2)$,so there is no solution.
For $a = \frac{-1 \pm \sqrt{5}}{2}$,$\Delta = 0$. By checking the augmented matrix,it is found that the system is inconsistent for these values as well.
Thus,for all values where $\Delta = 0$,the system has no solution.

(C) The given system of equations is:
$x - 2y + 5z = 3$
$2x - y + z = 1$
$11x - 7y + pz = q$
For the system to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$.
$\Delta = \begin{vmatrix} 1 & -2 & 5 \\ 2 & -1 & 1 \\ 11 & -7 & p \end{vmatrix} = 1(-p + 7) + 2(2p - 11) + 5(-14 + 11) = 0$
$-p + 7 + 4p - 22 - 15 = 0$
$3p - 30 = 0 \Rightarrow p = 10$
Now,for infinitely many solutions,$\Delta_x = \Delta_y = \Delta_z = 0$.
$\Delta_x = \begin{vmatrix} 3 & -2 & 5 \\ 1 & -1 & 1 \\ q & -7 & 10 \end{vmatrix} = 3(-10 + 7) + 2(10 - q) + 5(-7 + q) = 0$
$3(-3) + 20 - 2q - 35 + 5q = 0$
$-9 + 20 - 35 + 3q = 0$
$3q - 24 = 0 \Rightarrow q = 8$
Thus,$p - q = 10 - 8 = 2$.

(B) The given system of linear equations is:
$x + y + z = 2$
$2x + y - z = 3$
$3x + 2y + kz = 4$
$A$ system of linear equations has a unique solution if and only if the determinant of the coefficient matrix is non-zero.
Let $D$ be the determinant of the coefficient matrix:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 1 & -1 \\ 3 & 2 & k \end{vmatrix}$
Expanding along the first row:
$D = 1(1 \cdot k - (-1) \cdot 2) - 1(2 \cdot k - (-1) \cdot 3) + 1(2 \cdot 2 - 1 \cdot 3)$
$D = 1(k + 2) - 1(2k + 3) + 1(4 - 3)$
$D = k + 2 - 2k - 3 + 1$
$D = -k$
For a unique solution,$D \neq 0$,which implies $-k \neq 0$,or $k \neq 0$.
Therefore,$S = R - \{0\}$.

(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be zero,and at least one of the Cramer's rule determinants $(\Delta_1, \Delta_2, \Delta_3)$ must be non-zero.
First,calculate $\Delta$:
$\Delta = \begin{vmatrix} 1 & a & 1 \\ 1 & 2 & 2 \\ 1 & 5 & 3 \end{vmatrix} = 1(6 - 10) - a(3 - 2) + 1(5 - 2) = -4 - a + 3 = -a - 1$.
Setting $\Delta = 0$,we get $-a - 1 = 0$,which implies $a = -1$.
Next,calculate $\Delta_2$ (replacing the second column with the constants $3, 6, b$):
$\Delta_2 = \begin{vmatrix} 1 & 3 & 1 \\ 1 & 6 & 2 \\ 1 & b & 3 \end{vmatrix} = 1(18 - 2b) - 3(3 - 2) + 1(b - 6) = 18 - 2b - 3 + b - 6 = 9 - b$.
For the system to have no solution,we require $\Delta_2 \neq 0$,so $9 - b \neq 0$,which implies $b \neq 9$.
Thus,the condition is $a = -1$ and $b \neq 9$.

(C) The given system of linear equations is:
$(k + 2)x + 10y = k$
$kx + (k + 3)y = k - 1$
For a system of linear equations to have no solution,the determinant of the coefficient matrix must be zero,and the system must be inconsistent.
Let $A = \begin{bmatrix} k + 2 & 10 \\ k & k + 3 \end{bmatrix}$.
Setting $|A| = 0$:
$(k + 2)(k + 3) - 10k = 0$
$k^2 + 5k + 6 - 10k = 0$
$k^2 - 5k + 6 = 0$
$(k - 2)(k - 3) = 0$
So,$k = 2$ or $k = 3$.
Case $1$: If $k = 2$,the equations are $4x + 10y = 2$ and $2x + 5y = 1$. Dividing the first by $2$ gives $2x + 5y = 1$,which is identical to the second equation. Thus,there are infinitely many solutions.
Case $2$: If $k = 3$,the equations are $5x + 10y = 3$ and $3x + 6y = 2$. Multiplying the first by $3$ and the second by $5$ gives $15x + 30y = 9$ and $15x + 30y = 10$. Since $9 \neq 10$,the system is inconsistent and has no solution.
Therefore,there is only $1$ value of $k$ for which the system has no solution.

(B) For a homogeneous system of linear equations $AX = 0$ to have infinitely many solutions,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.
The coefficient matrix is:
$A = \begin{bmatrix} 2 & 4 & -\lambda \\ 4 & \lambda & 2 \\ \lambda & 2 & 2 \end{bmatrix}$
Setting the determinant to zero:
$|A| = 2(\lambda \cdot 2 - 2 \cdot 2) - 4(4 \cdot 2 - 2 \cdot \lambda) + (-\lambda)(4 \cdot 2 - \lambda \cdot \lambda) = 0$
$|A| = 2(2\lambda - 4) - 4(8 - 2\lambda) - \lambda(8 - \lambda^2) = 0$
$|A| = 4\lambda - 8 - 32 + 8\lambda - 8\lambda + \lambda^3 = 0$
$|A| = \lambda^3 + 4\lambda - 40 = 0$
Let $f(\lambda) = \lambda^3 + 4\lambda - 40$.
Since $f'(\lambda) = 3\lambda^2 + 4 > 0$ for all real $\lambda$,the function $f(\lambda)$ is strictly increasing.
Therefore,it can have only one real root.
Thus,the number of real values of $\lambda$ is $1$.

(C) The given system of linear equations is:
$x + 8y + 7z = 0$ $(1)$
$9x + 2y + 3z = 0$ $(2)$
$x + y + z = 0$ $(3)$
From equation $(3)$,$x = -y - z$. Substituting this into $(1)$ and $(2)$:
$(-y - z) + 8y + 7z = 0 \implies 7y + 6z = 0 \implies z = -\frac{7}{6}y$
$9(-y - z) + 2y + 3z = 0 \implies -7y - 6z = 0 \implies z = -\frac{7}{6}y$
Since the equations are homogeneous and the determinant of the coefficient matrix is $0$,there are infinitely many solutions. Let $y = 6\lambda$. Then $z = -7\lambda$ and $x = -6\lambda - (-7\lambda) = \lambda$.
So,$(a, b, c) = (\lambda, 6\lambda, -7\lambda)$.
Since this point lies on the plane $x + 2y + z = 6$:
$\lambda + 2(6\lambda) + (-7\lambda) = 6$
$\lambda + 12\lambda - 7\lambda = 6$
$6\lambda = 6 \implies \lambda = 1$.
Thus,$(a, b, c) = (1, 6, -7)$.
We need to find $2a + b + c = 2(1) + 6 + (-7) = 2 + 6 - 7 = 1$.

(D) The given system of equations can be written in matrix form $AX = B$,where $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 5 & a \end{bmatrix}$ and $B = \begin{bmatrix} 6 \\ 9 \\ b \end{bmatrix}$.
For the system to have infinitely many solutions,the determinant of the augmented matrix must be zero,and the rank of the augmented matrix must be less than the number of variables.
First,calculate the determinant of $A$: $|A| = 1(3a - 25) - 2(a - 10) + 3(5 - 6) = 3a - 25 - 2a + 20 - 3 = a - 8$.
For infinite solutions,$|A| = 0$,which implies $a = 8$.
Now,substitute $a = 8$ into the augmented matrix $[A|B] = \begin{bmatrix} 1 & 2 & 3 & | & 6 \\ 1 & 3 & 5 & | & 9 \\ 2 & 5 & 8 & | & b \end{bmatrix}$.
Perform row operations: $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - 2R_1$:
$\begin{bmatrix} 1 & 2 & 3 & | & 6 \\ 0 & 1 & 2 & | & 3 \\ 0 & 1 & 2 & | & b - 12 \end{bmatrix}$.
For the system to be consistent with infinite solutions,the last two rows must be identical,so $b - 12 = 3$,which gives $b = 15$.
Thus,$a = 8$ and $b = 15$.

(A) Given the matrix equation:
$[p, q, r] \begin{bmatrix} 3 & 4 & 1 \\ 3 & 2 & 3 \\ 2 & 0 & 2 \end{bmatrix} = [3, 0, 1]$
Performing matrix multiplication,we get:
$[3p + 3q + 2r, 4p + 2q, p + 3q + 2r] = [3, 0, 1]$
Equating the corresponding elements,we obtain the following system of linear equations:
$1) 3p + 3q + 2r = 3$
$2) 4p + 2q = 0 \Rightarrow q = -2p$
$3) p + 3q + 2r = 1$
Subtracting equation $(3)$ from equation $(1)$:
$(3p + 3q + 2r) - (p + 3q + 2r) = 3 - 1$
$2p = 2 \Rightarrow p = 1$
Substituting $p = 1$ into $q = -2p$:
$q = -2(1) = -2$
Substituting $p = 1$ and $q = -2$ into equation $(3)$:
$1 + 3(-2) + 2r = 1$
$1 - 6 + 2r = 1$
$-5 + 2r = 1 \Rightarrow 2r = 6 \Rightarrow r = 3$
Now,calculate $2p + q - r$:
$2(1) + (-2) - 3 = 2 - 2 - 3 = -3$.

(B) For a system of linear equations $AX = 0$ to have a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$|A| = 0$.
For Statement $-1$,the determinant is:
$\Delta_1 = \left| \begin{matrix} 1 & \sin \alpha & \cos \alpha \\ 1 & \cos \alpha & \sin \alpha \\ 1 & -\sin \alpha & -\cos \alpha \end{matrix} \right|$
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\Delta_1 = \left| \begin{matrix} 1 & \sin \alpha & \cos \alpha \\ 0 & \cos \alpha - \sin \alpha & \sin \alpha - \cos \alpha \\ 0 & -2\sin \alpha & -2\cos \alpha \end{matrix} \right|$
$= 1 \cdot [(\cos \alpha - \sin \alpha)(-2\cos \alpha) - (\sin \alpha - \cos \alpha)(-2\sin \alpha)]$
$= -2\cos^2 \alpha + 2\sin \alpha \cos \alpha + 2\sin^2 \alpha - 2\sin \alpha \cos \alpha = 2(\sin^2 \alpha - \cos^2 \alpha) = -2\cos(2\alpha)$.
Setting $\Delta_1 = 0$,we get $\cos(2\alpha) = 0$. In $(0, \frac{\pi}{2})$,$2\alpha \in (0, \pi)$,so $2\alpha = \frac{\pi}{2} \Rightarrow \alpha = \frac{\pi}{4}$. Thus,Statement $-1$ is true.
For Statement $-2$,the determinant is:
$\Delta_2 = \left| \begin{matrix} \cos \alpha & \sin \alpha & \cos \alpha \\ \sin \alpha & \cos \alpha & \sin \alpha \\ \cos \alpha & -\sin \alpha & -\cos \alpha \end{matrix} \right|$
Applying $C_3 \to C_3 - C_1$:
$\Delta_2 = \left| \begin{matrix} \cos \alpha & \sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ \cos \alpha & -\sin \alpha & -2\cos \alpha \end{matrix} \right|$
$= -2\cos \alpha (\cos^2 \alpha - \sin^2 \alpha) = -2\cos \alpha \cos(2\alpha) = 0$.
This implies $\cos \alpha = 0$ or $\cos(2\alpha) = 0$. In $(0, \frac{\pi}{2})$,$\cos \alpha \neq 0$ and $\cos(2\alpha) = 0$ only at $\alpha = \frac{\pi}{4}$. Thus,Statement $-2$ is true and explains Statement $-1$.

(B) The given system of equations is homogeneous:
$x + ay = 0$
$y + az = 0$
$z + ax = 0$
This can be written in matrix form $AX = 0$,where $A = \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & a \\ a & 0 & 1 \end{bmatrix}$.
$A$ homogeneous system has a unique (trivial) solution if and only if the determinant $|A| \neq 0$.
Calculating the determinant:
$|A| = 1(1 - 0) - a(0 - a^2) + 0(0 - a) = 1 + a^3$.
For a unique solution,we require $|A| \neq 0$,so $1 + a^3 \neq 0$.
$a^3 \neq -1$,which implies $a \neq -1$.
Thus,the set of all real values of $a$ is $R - \{-1\}$.

(D) The given system of equations is:
$x + y + z = 6$
$x + 2y + 3z = 10$
$x + 2y + \lambda z = 0$
For a system of linear equations to have a unique solution,the determinant of the coefficient matrix must be non-zero.
Let $D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{vmatrix} \neq 0$
Expanding the determinant along the first row:
$D = 1(2\lambda - 6) - 1(\lambda - 3) + 1(2 - 2) \neq 0$
$D = 2\lambda - 6 - \lambda + 3 + 0 \neq 0$
$D = \lambda - 3 \neq 0$
Therefore,$\lambda \neq 3$.

(A) For a system of homogeneous linear equations to have a nontrivial solution,the determinant of the coefficient matrix must be zero.
The coefficient matrix is $A = \begin{bmatrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{bmatrix}$.
Setting the determinant to zero: $\left| \begin{matrix} 1 & k & 3 \\ 3 & k & -2 \\ 2 & 3 & -4 \end{matrix} \right| = 0$.
Expanding along the first row:
$1(-4k - (-6)) - k(-12 - (-4)) + 3(9 - 2k) = 0$
$1(-4k + 6) - k(-8) + 27 - 6k = 0$
$-4k + 6 + 8k + 27 - 6k = 0$
$-2k + 33 = 0$
$2k = 33 \Rightarrow k = \frac{33}{2}$.
Since the calculated value of $k$ is $\frac{33}{2}$ and not $\frac{31}{2}$,Statement $1$ is false.
Statement $2$ is a standard theorem in linear algebra regarding homogeneous systems,so it is true.

(D) The system of equations is given by:
$x + y + z = 2$
$2x + 3y + 2z = 5$
$2x + 3y + (a^2 - 1)z = a + 1$
First,calculate the determinant $D$ of the coefficient matrix:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 2 & 3 & a^2 - 1 \end{vmatrix}$
Expanding along the first row:
$D = 1(3(a^2 - 1) - 6) - 1(2(a^2 - 1) - 4) + 1(6 - 6)$
$D = 3a^2 - 3 - 6 - 2a^2 + 2 + 4$
$D = a^2 - 3$
For a unique solution,$D \neq 0$,so $a^2 - 3 \neq 0$,which means $|a| \neq \sqrt{3}$.
If $|a| = \sqrt{3}$,then $a^2 = 3$. The system becomes:
$x + y + z = 2$
$2x + 3y + 2z = 5$
$2x + 3y + 2z = a + 1$
Comparing the second and third equations,we have $5 = a + 1$,which implies $a = 4$. However,we assumed $|a| = \sqrt{3}$,so $a^2 = 3$. Since $4^2 \neq 3$,the system is inconsistent when $|a| = \sqrt{3}$ because the left-hand sides of the second and third equations are identical,but the constants are different ($5 \neq a + 1$ when $a^2 = 3$).
Thus,the system is inconsistent when $|a| = \sqrt{3}$.

(B) The given system of linear equations is:
$x - 4y + 7z = g$ $(1)$
$0x + 3y - 5z = h$ $(2)$
$-2x + 5y - 9z = k$ $(3)$
For the system to be consistent,the determinant of the coefficient matrix must be zero,or there must exist a linear combination of the equations that results in a consistent identity.
Let the equations be $L_1, L_2, L_3$. We check for a linear combination $c_1 L_1 + c_2 L_2 + c_3 L_3 = 0$ for the variables $x, y, z$.
$c_1(x - 4y + 7z) + c_2(3y - 5z) + c_3(-2x + 5y - 9z) = c_1 g + c_2 h + c_3 k$
Comparing coefficients:
$x: c_1 - 2c_3 = 0 \implies c_1 = 2c_3$
$y: -4c_1 + 3c_2 + 5c_3 = 0$
Substituting $c_1 = 2c_3$: $-4(2c_3) + 3c_2 + 5c_3 = 0 \implies -8c_3 + 3c_2 + 5c_3 = 0 \implies 3c_2 = 3c_3 \implies c_2 = c_3$
$z: 7c_1 - 5c_2 - 9c_3 = 0$
Substituting $c_1 = 2c_3$ and $c_2 = c_3$: $7(2c_3) - 5(c_3) - 9c_3 = 14c_3 - 14c_3 = 0$. This holds for any $c_3$.
Let $c_3 = 1$,then $c_1 = 2$ and $c_2 = 1$.
The linear combination is $2L_1 + L_2 + L_3 = 0$,which implies $2g + h + k = 0$.

(B) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the determinants $D_1, D_2, D_3$ must also be $0$.
The coefficient matrix is:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha \end{vmatrix} = 1(2\alpha - 9) - 1(\alpha - 3) + 1(3 - 2) = 2\alpha - 9 - \alpha + 3 + 1 = \alpha - 5$.
Setting $D = 0$,we get $\alpha = 5$.
Now,for $D_3 = 0$:
$D_3 = \begin{vmatrix} 1 & 1 & 5 \\ 1 & 2 & 9 \\ 1 & 3 & \beta \end{vmatrix} = 1(2\beta - 27) - 1(\beta - 9) + 5(3 - 2) = 2\beta - 27 - \beta + 9 + 5 = \beta - 13$.
Setting $D_3 = 0$,we get $\beta = 13$.
Thus,$\beta - \alpha = 13 - 5 = 8$.

(A) Given the system of equations:
$2x + 2y + 3z = a$ $(1)$
$3x - y + 5z = b$ $(2)$
$x - 3y + 2z = c$ $(3)$
For the system to have more than one solution,the determinant of the coefficient matrix must be zero,and the system must be consistent.
Let $D$ be the determinant of the coefficient matrix:
$D = \begin{vmatrix} 2 & 2 & 3 \\ 3 & -1 & 5 \\ 1 & -3 & 2 \end{vmatrix} = 2(-2 + 15) - 2(6 - 5) + 3(-9 + 1) = 2(13) - 2(1) + 3(-8) = 26 - 2 - 24 = 0$.
Since $D = 0$,the system has either no solution or infinitely many solutions.
For infinitely many solutions,the equations must be linearly dependent. We observe that $(1) - (3) = (2x - x) + (2y - (-3y)) + (3z - 2z) = x + 5y + z = a - c$.
Alternatively,notice that $(1) + (3) = 3x - y + 5z = a + c$. Comparing this with equation $(2)$,we get $a + c = b$,which implies $b - c - a = 0$.

(A) The system of linear equations has a unique solution if and only if the determinant of the coefficient matrix $D \neq 0$.
The coefficient matrix is:
$D = \begin{vmatrix} 1 + \alpha & \beta & 1 \\ \alpha & 1 + \beta & 1 \\ \alpha & \beta & 2 \end{vmatrix}$
Applying the column operation $C_1 \to C_1 + C_2 + C_3$:
$D = \begin{vmatrix} \alpha + \beta + 2 & \beta & 1 \\ \alpha + \beta + 2 & 1 + \beta & 1 \\ \alpha + \beta + 2 & \beta & 2 \end{vmatrix}$
Taking $(\alpha + \beta + 2)$ common from $C_1$:
$D = (\alpha + \beta + 2) \begin{vmatrix} 1 & \beta & 1 \\ 1 & 1 + \beta & 1 \\ 1 & \beta & 2 \end{vmatrix}$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$D = (\alpha + \beta + 2) \begin{vmatrix} 1 & \beta & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = (\alpha + \beta + 2)(1) = \alpha + \beta + 2$
For a unique solution,$D \neq 0$,so $\alpha + \beta + 2 \neq 0$,which means $\alpha + \beta \neq -2$.
Checking the options:
$A: 2 + 4 = 6 \neq -2$ (Valid)
$B: -3 + 1 = -2$ (Invalid)
$C: -4 + 2 = -2$ (Invalid)
$D: 1 - 3 = -2$ (Invalid)
Thus,the ordered pair $(2, 4)$ provides a unique solution.

(B) For the system to have a non-zero solution,the determinant of the coefficient matrix must be zero:
$\Delta = \begin{vmatrix} 1 & -2 & k \\ 2 & 1 & 1 \\ 3 & -1 & -k \end{vmatrix} = 0$
Expanding along the first row:
$1(-k + 1) - (-2)(-2k - 3) + k(-2 - 3) = 0$
$-k + 1 + 2(-2k - 3) - 5k = 0$
$-k + 1 - 4k - 6 - 5k = 0$
$-10k - 5 = 0 \Rightarrow k = -\frac{1}{2}$
Substituting $k = -\frac{1}{2}$ into the equations:
$x - 2y - \frac{1}{2}z = 1 \Rightarrow 2x - 4y - z = 2$ $(1)$
$2x + y + z = 2$ $(2)$
Adding $(1)$ and $(2)$:
$(2x - 4y - z) + (2x + y + z) = 2 + 2$
$4x - 3y = 4$
Thus,$(x, y)$ lies on the line $4x - 3y - 4 = 0$.

(C) For a system of linear equations to have infinitely many solutions,the third equation must be a linear combination of the first two equations. Let the third equation be $L_3 = a L_1 + b L_2$.
$x + 3y + \lambda z = \mu = a(x + y + z - 5) + b(x + 2y + 2z - 6)$.
Comparing the coefficients of $x, y, z$ and the constant term:
$a + b = 1$ (coefficient of $x$)
$a + 2b = 3$ (coefficient of $y$)
$a + 2b = \lambda$ (coefficient of $z$)
$5a + 6b = \mu$ (constant term)
Solving the first two equations:
Subtracting $(a + b = 1)$ from $(a + 2b = 3)$ gives $b = 2$.
Substituting $b = 2$ into $a + b = 1$ gives $a = -1$.
Now,find $\lambda$ and $\mu$:
$\lambda = a + 2b = -1 + 2(2) = 3$.
$\mu = 5a + 6b = 5(-1) + 6(2) = -5 + 12 = 7$.
Therefore,$\lambda + \mu = 3 + 7 = 10$.

(A) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the augmented matrix must satisfy the condition for consistency.
The coefficient matrix is:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 4 & \lambda & -\lambda \\ 3 & 2 & -4 \end{vmatrix}$
Expanding along the first row:
$D = 1(-4\lambda - (-2\lambda)) - 1(-16 - (-3\lambda)) + 1(8 - 3\lambda) = 0$
$D = (-4\lambda + 2\lambda) - (-16 + 3\lambda) + (8 - 3\lambda) = 0$
$-2\lambda + 16 - 3\lambda + 8 - 3\lambda = 0$
$-8\lambda + 24 = 0 \Rightarrow \lambda = 3$
Now,check the options to see which quadratic equation has $\lambda = 3$ as a root:
For option $A$: $\lambda^2 - \lambda - 6 = 0 \Rightarrow (\lambda - 3)(\lambda + 2) = 0$. Here,$\lambda = 3$ is a root.
Thus,the correct option is $A$.

(A) The given system of equations is:
$[\sin \theta ]x + [-\cos \theta ]y = 0 \dots (1)$
$[\cot \theta ]x + y = 0 \dots (2)$
Case $I$: When $\theta \in \left( \frac{\pi }{2}, \frac{2\pi }{3} \right)$
$\sin \theta \in \left( \frac{\sqrt{3}}{2}, 1 \right) \implies [\sin \theta ] = 0$
$-\cos \theta \in \left( 0, \frac{1}{2} \right) \implies [-\cos \theta ] = 0$
$\cot \theta \in \left( -\frac{1}{\sqrt{3}}, 0 \right) \implies [\cot \theta ] = -1$
Substituting these into the equations:
$0x + 0y = 0$
$-x + y = 0$
The first equation is satisfied for all $(x, y)$,and the second implies $y = x$. Thus,the system has infinitely many solutions.
Case $II$: When $\theta \in \left( \pi, \frac{7\pi}{6} \right)$
$\sin \theta \in \left( -\frac{1}{2}, 0 \right) \implies [\sin \theta ] = -1$
$-\cos \theta \in \left( -\frac{\sqrt{3}}{2}, 1 \right) \implies [-\cos \theta ] = 0$
$\cot \theta \in (\sqrt{3}, \infty) \implies [\cot \theta ] = k$,where $k \in \{1, 2, 3, \dots\}$
Substituting these into the equations:
$-x + 0y = 0 \implies x = 0$
$kx + y = 0 \implies k(0) + y = 0 \implies y = 0$
Thus,the system has a unique solution $(0, 0)$.

(C) For a system of linear equations to have more than two solutions,it must have infinitely many solutions. This occurs when the determinant of the coefficient matrix is zero and the augmented matrix satisfies the consistency condition.
The coefficient matrix is $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 2 & \lambda \end{bmatrix}$.
Setting the determinant $|A| = 0$:
$|A| = 1(2\lambda - 6) - 1(\lambda - 9) + 1(2 - 6) = 0$
$2\lambda - 6 - \lambda + 9 - 4 = 0$
$\lambda - 1 = 0 \Rightarrow \lambda = 1$.
Now,for infinitely many solutions,the determinant of the matrix formed by replacing the constant column must also be zero $(D_z = 0)$:
$D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 3 & 2 & \mu \end{vmatrix} = 0$
$1(2\mu - 20) - 1(\mu - 30) + 6(2 - 6) = 0$
$2\mu - 20 - \mu + 30 - 24 = 0$
$\mu - 14 = 0 \Rightarrow \mu = 14$.
Finally,calculating $\mu - \lambda^{2}$:
$\mu - \lambda^{2} = 14 - (1)^{2} = 14 - 1 = 13$.

(D) For a system of linear equations to have a non-zero solution,the determinant of the coefficient matrix must be zero.
The determinant is given by:
$\begin{vmatrix} 2 & 2a & a \\ 2 & 3b & b \\ 2 & 4c & c \end{vmatrix} = 0$
Taking $2$ common from the first column:
$2 \begin{vmatrix} 1 & 2a & a \\ 1 & 3b & b \\ 1 & 4c & c \end{vmatrix} = 0$
Applying row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$2 \begin{vmatrix} 1 & 2a & a \\ 0 & 3b-2a & b-a \\ 0 & 4c-2a & c-a \end{vmatrix} = 0$
Expanding along the first column:
$(3b-2a)(c-a) - (b-a)(4c-2a) = 0$
Expanding the terms:
$(3bc - 3ab - 2ac + 2a^2) - (4bc - 2ab - 4ac + 2a^2) = 0$
$3bc - 3ab - 2ac + 2a^2 - 4bc + 2ab + 4ac - 2a^2 = 0$
$-bc - ab + 2ac = 0$
$2ac = ab + bc$
Dividing both sides by $abc$ (since $a, b, c \neq 0$):
$\frac{2ac}{abc} = \frac{ab}{abc} + \frac{bc}{abc}$
$\frac{2}{b} = \frac{1}{c} + \frac{1}{a}$
This condition implies that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $A.P.$

(D) The determinant of the coefficient matrix $D$ is given by:
$D = \begin{vmatrix} \lambda & 2 & 2 \\ 2\lambda & 3 & 5 \\ 4 & \lambda & 6 \end{vmatrix}$
Expanding along the first row:
$D = \lambda(18 - 5\lambda) - 2(12\lambda - 20) + 2(2\lambda^2 - 12)$
$D = 18\lambda - 5\lambda^2 - 24\lambda + 40 + 4\lambda^2 - 24$
$D = -\lambda^2 - 6\lambda + 16 = -(\lambda^2 + 6\lambda - 16) = -(\lambda + 8)(\lambda - 2) = (\lambda + 8)(2 - \lambda)$
For $\lambda = 2$,$D = 0$. We check the consistency using Cramer's rule or by substituting $\lambda = 2$ into the equations:
$2x + 2y + 2z = 5$
$4x + 3y + 5z = 8$
$4x + 2y + 6z = 10$
Subtracting the first equation multiplied by $2$ from the second: $(4x + 3y + 5z) - 2(2x + 2y + 2z) = 8 - 10 \implies -y + z = -2 \implies y - z = 2$.
Subtracting the first equation multiplied by $2$ from the third: $(4x + 2y + 6z) - 2(2x + 2y + 2z) = 10 - 10 \implies -2y + 2z = 0 \implies y - z = 0$.
Since $y - z = 2$ and $y - z = 0$ are contradictory,the system has no solution for $\lambda = 2$.

(D) The system of equations is given by:
$(i) x + 2y + 3z = 1$
$(ii) 3x + 4y + 5z = \mu$
$(iii) 4x + 4y + 4z = \delta$
To check for inconsistency,we perform row operations or eliminate variables.
Subtracting $(i)$ from $(ii)$ gives: $(3x-x) + (4y-2y) + (5z-3z) = \mu - 1 \Rightarrow 2x + 2y + 2z = \mu - 1$.
Multiplying this by $2$ gives: $4x + 4y + 4z = 2(\mu - 1)$.
Comparing this with equation $(iii)$,we have $4x + 4y + 4z = \delta$.
For the system to be inconsistent,the left-hand sides must be equal while the right-hand sides are unequal,or the equations must lead to a contradiction like $0 = k$ (where $k \neq 0$).
Thus,if $\delta \neq 2(\mu - 1)$,the system is inconsistent.
Checking the options:
For $(A) (1, 0): \delta = 0, 2(\mu - 1) = 2(1 - 1) = 0$. Here $\delta = 2(\mu - 1)$,so it is consistent.
For $(B) (4, 6): \delta = 6, 2(\mu - 1) = 2(4 - 1) = 6$. Here $\delta = 2(\mu - 1)$,so it is consistent.
For $(C) (3, 4): \delta = 4, 2(\mu - 1) = 2(3 - 1) = 4$. Here $\delta = 2(\mu - 1)$,so it is consistent.
For $(D) (4, 3): \delta = 3, 2(\mu - 1) = 2(4 - 1) = 6$. Here $\delta \neq 2(\mu - 1)$,so the system is inconsistent.

(A) The given system of equations is homogeneous:
$7x + 6y - 2z = 0 \dots (1)$
$3x + 4y + 2z = 0 \dots (2)$
$x - 2y - 6z = 0 \dots (3)$
First,we calculate the determinant of the coefficient matrix $\Delta$:
$\Delta = \begin{vmatrix} 7 & 6 & -2 \\ 3 & 4 & 2 \\ 1 & -2 & -6 \end{vmatrix}$
$= 7(4(-6) - 2(-2)) - 6(3(-6) - 2(1)) - 2(3(-2) - 4(1))$
$= 7(-24 + 4) - 6(-18 - 2) - 2(-6 - 4)$
$= 7(-20) - 6(-20) - 2(-10)$
$= -140 + 120 + 20 = 0$
Since $\Delta = 0$,the system has infinitely many solutions.
Adding equations $(1)$ and $(2)$:
$(7x + 6y - 2z) + (3x + 4y + 2z) = 0 + 0$
$10x + 10y = 0 \Rightarrow y = -x$
Substituting $y = -x$ into equation $(1)$:
$7x + 6(-x) - 2z = 0$
$7x - 6x - 2z = 0$
$x - 2z = 0 \Rightarrow x = 2z$
Thus,the solutions satisfy $x = 2z$.

(N/A) To find the total amount spent in each city,we calculate the product $BA$.
$BA = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{bmatrix} 40 \\ 100 \\ 50 \end{bmatrix}$
$= \begin{bmatrix} (1000 \times 40) + (500 \times 100) + (5000 \times 50) \\ (3000 \times 40) + (1000 \times 100) + (10000 \times 50) \end{bmatrix}$
$= \begin{bmatrix} 40000 + 50000 + 250000 \\ 120000 + 100000 + 500000 \end{bmatrix} = \begin{bmatrix} 340000 \\ 720000 \end{bmatrix}$
Thus,the total amount spent in city $X$ is $340,000$ paise (Rs. $3400$) and in city $Y$ is $720,000$ paise (Rs. $7200$).

(A) Let the amount invested in the first bond be Rs. $x$ and in the second bond be Rs. $(30000 - x)$.
Using matrix multiplication,we represent the investment as a row matrix $A = [x \quad 30000 - x]$ and the interest rates as a column matrix $B = \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix}$.
The total interest is given by the product $AB = [1800]$.
$[x \quad 30000 - x] \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix} = [1800]$
$0.05x + 0.07(30000 - x) = 1800$
$0.05x + 2100 - 0.07x = 1800$
$-0.02x = 1800 - 2100$
$-0.02x = -300$
$x = \frac{300}{0.02} = 15000$.
Thus,Rs. $15000$ is invested in the first bond and Rs. $(30000 - 15000) = 15000$ is invested in the second bond.

(A) Let Rs. $x$ be invested in the first bond. Then,the amount invested in the second bond is Rs. $(30000 - x)$.
Using matrix multiplication to represent the total interest:
$[x \quad (30000 - x)] \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix} = 2000$
This simplifies to:
$0.05x + 0.07(30000 - x) = 2000$
Multiply by $100$ to clear decimals:
$5x + 7(30000 - x) = 200000$
$5x + 210000 - 7x = 200000$
$-2x = 200000 - 210000$
$-2x = -10000$
$x = 5000$
Therefore,the amount invested in the first bond is Rs. $5000$ and the amount invested in the second bond is Rs. $(30000 - 5000) = 25000$.

(B) The bookshop has $10$ dozen chemistry books,$8$ dozen physics books,and $10$ dozen economics books. Since $1$ dozen $= 12$ books,the total number of books are $120$ chemistry books,$96$ physics books,and $120$ economics books.
The selling prices of a chemistry book,a physics book,and an economics book are Rs. $80$,Rs. $60$,and Rs. $40$ respectively.
Representing the quantity as a row matrix $A$ and the price as a column matrix $B$:
$A = [120 \quad 96 \quad 120]$
$B = \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}$
The total amount is given by the product $AB$:
$Total = [120 \quad 96 \quad 120] \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}$
$= (120 \times 80) + (96 \times 60) + (120 \times 40)$
$= 9600 + 5760 + 4800$
$= 20160$
Thus,the bookshop will receive Rs. $20160$ from the sale of all these books.

(A) The system of equations can be written in the form $AX = B$,where
$A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \end{bmatrix}$,and $B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = (2)(2) - (5)(3) = 4 - 15 = -11$.
Since $|A| \neq 0$,the system has a unique solution given by $X = A^{-1}B$.
The inverse of $A$ is $A^{-1} = \frac{1}{|A|} \text{adj}(A) = -\frac{1}{11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}$.
Now,$X = -\frac{1}{11} \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 7 \end{bmatrix}$.
$X = -\frac{1}{11} \begin{bmatrix} (2)(1) + (-5)(7) \\ (-3)(1) + (2)(7) \end{bmatrix} = -\frac{1}{11} \begin{bmatrix} 2 - 35 \\ -3 + 14 \end{bmatrix} = -\frac{1}{11} \begin{bmatrix} -33 \\ 11 \end{bmatrix}$.
$X = \begin{bmatrix} 3 \\ -1 \end{bmatrix}$.
Thus,$x = 3$ and $y = -1$.

(B) The system of equations can be written in the form $AX = B$,where
$A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$,and $B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}$.
First,we calculate the determinant $|A| = 3(2 - 3) + 2(4 + 4) + 3(-6 - 4) = 3(-1) + 2(8) + 3(-10) = -3 + 16 - 30 = -17$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
Next,we find the cofactor matrix of $A$:
$C_{11} = (2 - 3) = -1, C_{12} = -(4 + 4) = -8, C_{13} = (-6 - 4) = -10$
$C_{21} = -(-4 + 9) = -5, C_{22} = (6 - 12) = -6, C_{23} = -(-9 + 8) = 1$
$C_{31} = (2 - 3) = -1, C_{32} = -(-3 - 6) = 9, C_{33} = (3 + 4) = 7$
Thus,$adj(A) = \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} adj(A) = -\frac{1}{17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}$.
Now,$X = A^{-1}B = -\frac{1}{17} \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix} \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix} = -\frac{1}{17} \begin{bmatrix} -8 - 5 - 4 \\ -64 - 6 + 36 \\ -80 + 1 + 28 \end{bmatrix} = -\frac{1}{17} \begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.
Therefore,$x = 1, y = 2, z = 3$.

(C) Let the first,second,and third numbers be $x, y,$ and $z$ respectively.
According to the given conditions,we have the following system of linear equations:
$x + y + z = 6$
$y + 3z = 11$
$x + z = 2y \implies x - 2y + z = 0$
This system can be written as $AX = B$,where
$A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}$
First,we calculate the determinant $|A|$:
$|A| = 1(1 - (-6)) - 1(0 - 3) + 1(0 - 1) = 1(7) + 3 - 1 = 9 \neq 0$.
Since $|A| \neq 0$,the system has a unique solution $X = A^{-1}B$.
Calculating the cofactor matrix $C$:
$C_{11} = (1 - (-6)) = 7, C_{12} = -(0 - 3) = 3, C_{13} = (0 - 1) = -1$
$C_{21} = -(1 - (-2)) = -3, C_{22} = (1 - 1) = 0, C_{23} = -(-2 - 1) = 3$
$C_{31} = (3 - 1) = 2, C_{32} = -(3 - 0) = -3, C_{33} = (1 - 0) = 1$
$adj(A) = C^T = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}$
$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}$
$X = A^{-1}B = \frac{1}{9} \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix} \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 42 - 33 + 0 \\ 18 + 0 + 0 \\ -6 + 33 + 0 \end{bmatrix} = \frac{1}{9} \begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$
Thus,$x = 1, y = 2, z = 3$.

(A) The given system of equations is:
$x+2y=2$
$2x+3y=3$
The system can be written in the matrix form $AX=B$,where:
$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \end{bmatrix}$,$B = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$
Now,calculate the determinant of $A$:
$|A| = (1)(3) - (2)(2) = 3 - 4 = -1$
Since $|A| \neq 0$,the matrix $A$ is non-singular and invertible.
Therefore,the system of equations has a unique solution and is consistent.

(A) The given system of equations is:
$2x - y = 5$
$x + y = 4$
The system can be written in the matrix form $AX = B$,where:
$A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \end{bmatrix}$,and $B = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$.
Now,calculate the determinant of matrix $A$:
$|A| = (2)(1) - (-1)(1) = 2 + 1 = 3$.
Since $|A| \neq 0$,the matrix $A$ is non-singular and invertible.
Therefore,the system of equations has a unique solution and is consistent.

(B) The given system of equations is:
$x+3y=5$
$2x+6y=8$
The system can be written in the form $AX=B$,where:
$A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}, X = \begin{bmatrix} x \\ y \end{bmatrix}, B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}$
First,we calculate the determinant of $A$:
$|A| = (1)(6) - (3)(2) = 6 - 6 = 0$
Since $|A| = 0$,the matrix $A$ is singular. We now calculate $(adj A)B$:
$adj A = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$
$(adj A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix} \begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 30 - 24 \\ -10 + 8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix} \neq \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
Since $(adj A)B \neq 0$,the system of equations has no solution.
Therefore,the system is inconsistent.

(A) The given system of equations is:
$x+y+z=1$
$2x+3y+2z=2$
$ax+ay+2az=4$
The system can be written as $AX=B$,where
$A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & 3 & 2 \\ a & a & 2a \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$,and $B = \begin{bmatrix} 1 \\ 2 \\ 4 \end{bmatrix}$.
Calculate the determinant $|A|$:
$|A| = 1(6a - 2a) - 1(4a - 2a) + 1(2a - 3a)$
$|A| = 4a - 2a - a = a$.
Case $1$: If $a \neq 0$,then $|A| \neq 0$. The matrix $A$ is non-singular,so the system has a unique solution and is consistent.
Case $2$: If $a = 0$,the equations become:
$x+y+z=1$
$2x+3y+2z=2$
$0=4$
Since $0=4$ is a contradiction,the system is inconsistent when $a=0$.
Thus,the system is consistent for all $a \neq 0$.

(A) The given system of equations is:
$3x - y - 2z = 2$
$2y - z = -1$
$3x - 5y = 3$
This system can be written as $AX = B$,where
$A = \begin{bmatrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$,and $B = \begin{bmatrix} 2 \\ -1 \\ 3 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = 3(0 - 5) - (-1)(0 - (-3)) + (-2)(0 - 6)$
$|A| = 3(-5) + 1(-3) - 2(-6) = -15 - 3 + 12 = -6$.
Since $|A| \neq 0$,the system is consistent and has a unique solution.