Solution of the Linear equations using Matrices Questions in English

(A) The given system of equations is:
$5x - y + 4z = 5$
$2x + 3y + 5z = 2$
$5x - 2y + 6z = -1$
The system can be written as $AX = B$,where:
$A = \begin{bmatrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$,$B = \begin{bmatrix} 5 \\ 2 \\ -1 \end{bmatrix}$
Now,calculate the determinant $|A|$:
$|A| = 5(3 \times 6 - (-2) \times 5) - (-1)(2 \times 6 - 5 \times 5) + 4(2 \times (-2) - 5 \times 3)$
$|A| = 5(18 + 10) + 1(12 - 25) + 4(-4 - 15)$
$|A| = 5(28) + 1(-13) + 4(-19)$
$|A| = 140 - 13 - 76 = 51$
Since $|A| = 51 \neq 0$,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 can be written in the form $AX = B$,where
$A = \begin{bmatrix} 5 & 2 \\ 7 & 3 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \end{bmatrix}$,and $B = \begin{bmatrix} 4 \\ 5 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = (5 \times 3) - (2 \times 7) = 15 - 14 = 1$.
Since $|A| \neq 0$,the matrix $A$ is non-singular and invertible.
The inverse $A^{-1}$ is given by $\frac{1}{|A|} \text{adj}(A)$:
$A^{-1} = \frac{1}{1} \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix}$.
Now,we find $X$ using $X = A^{-1}B$:
$X = \begin{bmatrix} 3 & -2 \\ -7 & 5 \end{bmatrix} \begin{bmatrix} 4 \\ 5 \end{bmatrix} = \begin{bmatrix} (3 \times 4) + (-2 \times 5) \\ (-7 \times 4) + (5 \times 5) \end{bmatrix} = \begin{bmatrix} 12 - 10 \\ -28 + 25 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \end{bmatrix}$.
Thus,$x = 2$ and $y = -3$.

(C) The given system of equations can be written in the matrix form $AX = B$,where:
$A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \end{bmatrix}$,and $B = \begin{bmatrix} -2 \\ 3 \end{bmatrix}$.
First,calculate the determinant of $A$:
$|A| = (2)(4) - (-1)(3) = 8 + 3 = 11$.
Since $|A| \neq 0$,the matrix $A$ is non-singular and invertible.
The inverse of $A$ is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$:
$A^{-1} = \frac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix}$.
Now,solve for $X$ using $X = A^{-1}B$:
$X = \frac{1}{11} \begin{bmatrix} 4 & 1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} -2 \\ 3 \end{bmatrix}$
$X = \frac{1}{11} \begin{bmatrix} (4)(-2) + (1)(3) \\ (-3)(-2) + (2)(3) \end{bmatrix}$
$X = \frac{1}{11} \begin{bmatrix} -8 + 3 \\ 6 + 6 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} -5 \\ 12 \end{bmatrix} = \begin{bmatrix} -\frac{5}{11} \\ \frac{12}{11} \end{bmatrix}$.
Thus,$x = -\frac{5}{11}$ and $y = \frac{12}{11}$.

(D) The given system of equations can be written in the form $AX = B$,where $A = \begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \end{bmatrix}$,and $B = \begin{bmatrix} 3 \\ 7 \end{bmatrix}$.
First,calculate the determinant of $A$:
$|A| = (4)(-5) - (-3)(3) = -20 + 9 = -11$.
Since $|A| \neq 0$,the matrix $A$ is non-singular and invertible.
The inverse $A^{-1}$ is given by $\frac{1}{|A|} \text{adj}(A)$:
$A^{-1} = \frac{1}{-11} \begin{bmatrix} -5 & 3 \\ -3 & 4 \end{bmatrix} = \frac{1}{11} \begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix}$.
Now,solve for $X$ using $X = A^{-1}B$:
$X = \frac{1}{11} \begin{bmatrix} 5 & -3 \\ 3 & -4 \end{bmatrix} \begin{bmatrix} 3 \\ 7 \end{bmatrix}$
$X = \frac{1}{11} \begin{bmatrix} (5)(3) + (-3)(7) \\ (3)(3) + (-4)(7) \end{bmatrix}$
$X = \frac{1}{11} \begin{bmatrix} 15 - 21 \\ 9 - 28 \end{bmatrix}$
$X = \frac{1}{11} \begin{bmatrix} -6 \\ -19 \end{bmatrix} = \begin{bmatrix} -\frac{6}{11} \\ -\frac{19}{11} \end{bmatrix}$.
Thus,$x = -\frac{6}{11}$ and $y = -\frac{19}{11}$.

(A) The given system of equations can be written in the form $AX = B$,where
$A = \begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \end{bmatrix}$,and $B = \begin{bmatrix} 3 \\ 5 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = (5 \times 2) - (2 \times 3) = 10 - 6 = 4$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{4} \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix}$.
Now,$X = A^{-1}B$:
$X = \frac{1}{4} \begin{bmatrix} 2 & -2 \\ -3 & 5 \end{bmatrix} \begin{bmatrix} 3 \\ 5 \end{bmatrix}$
$X = \frac{1}{4} \begin{bmatrix} (2 \times 3) + (-2 \times 5) \\ (-3 \times 3) + (5 \times 5) \end{bmatrix}$
$X = \frac{1}{4} \begin{bmatrix} 6 - 10 \\ -9 + 25 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} -4 \\ 16 \end{bmatrix} = \begin{bmatrix} -1 \\ 4 \end{bmatrix}$.
Thus,$x = -1$ and $y = 4$.

(B) The given system of equations can be written in the form $AX = B$,where
$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = 2(10 + 3) - 1(-5 - 0) + 1(3 - 0) = 2(13) + 5 + 3 = 26 + 8 = 34 \neq 0$.
Since $|A| \neq 0$,the system has a unique solution $X = A^{-1}B$.
The matrix of cofactors is:
$C_{11} = 13, C_{12} = 5, C_{13} = 3$
$C_{21} = 8, C_{22} = -10, C_{23} = -6$
$C_{31} = 1, C_{32} = 3, C_{33} = -5$
Thus,$adj(A) = \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}$.
$A^{-1} = \frac{1}{34} \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix}$.
$X = A^{-1}B = \frac{1}{34} \begin{bmatrix} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{bmatrix} \begin{bmatrix} 1 \\ \frac{3}{2} \\ 9 \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 13 + 12 + 9 \\ 5 - 15 + 27 \\ 3 - 9 - 45 \end{bmatrix} = \frac{1}{34} \begin{bmatrix} 34 \\ 17 \\ -51 \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{1}{2} \\ -\frac{3}{2} \end{bmatrix}$.
Therefore,$x = 1, y = \frac{1}{2}, z = -\frac{3}{2}$.

(C) The given system of equations can be written in the form $AX=B$,where
$A=\begin{bmatrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{bmatrix}$,$X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$,and $B=\begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix}$.
First,calculate the determinant $|A|$:
$|A| = 1(1+3) - (-1)(2+3) + 1(2-1) = 4 + 5 + 1 = 10 \neq 0$.
Since $|A| \neq 0$,the matrix $A$ is non-singular and $A^{-1}$ exists.
Next,find the cofactor matrix $C$:
$C_{11} = (1+3) = 4, C_{12} = -(2+3) = -5, C_{13} = (2-1) = 1$
$C_{21} = -(-1-1) = 2, C_{22} = (1-1) = 0, C_{23} = -(1+1) = -2$
$C_{31} = (3-1) = 2, C_{32} = -(-3-2) = 5, C_{33} = (1+2) = 3$
Thus,$adj(A) = C^T = \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix}$.
Now,$X = A^{-1}B = \frac{1}{10} \begin{bmatrix} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{bmatrix} \begin{bmatrix} 4 \\ 0 \\ 2 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 16+0+4 \\ -20+0+10 \\ 4+0+6 \end{bmatrix} = \frac{1}{10} \begin{bmatrix} 20 \\ -10 \\ 10 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 1 \end{bmatrix}$.
Therefore,$x=2, y=-1, z=1$.

(D) The given system of equations can be written in the form $AX = B$,where $A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}$,$X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$,and $B = \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}$.
First,we calculate the determinant of $A$:
$|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40 \neq 0$.
Since $|A| \neq 0$,the system has a unique solution given by $X = A^{-1}B$.
The matrix of cofactors is calculated as:
$C_{11} = 5, C_{12} = 5, C_{13} = 5$
$C_{21} = 3, C_{22} = -13, C_{23} = 11$
$C_{31} = 9, C_{32} = 1, C_{33} = -7$
Thus,$adj(A) = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}$.
$A^{-1} = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}$.
Now,$X = A^{-1}B = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix} \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 25 - 12 + 27 \\ 25 + 52 + 3 \\ 25 - 44 - 21 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$.
Therefore,$x = 1, y = 2, z = -1$.

(A) The given system of equations can be written in the form $AX=B$,where
$A=\begin{bmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{bmatrix}$,$X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}$,and $B=\begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix}$.
First,calculate the determinant $|A|$:
$|A| = 1(12-5) - (-1)(9+10) + 2(-3-8) = 1(7) + 1(19) + 2(-11) = 7 + 19 - 22 = 4 \neq 0$.
Since $|A| \neq 0$,the system has a unique solution $X = A^{-1}B$.
The matrix of cofactors $C$ is calculated as:
$C_{11} = (12-5) = 7, C_{12} = -(9+10) = -19, C_{13} = (-3-8) = -11$
$C_{21} = -(-3+2) = 1, C_{22} = (3-4) = -1, C_{23} = -(-1+2) = -1$
$C_{31} = (5-8) = -3, C_{32} = -(-5-6) = 11, C_{33} = (4+3) = 7$
Thus,$adj(A) = C^T = \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix}$.
$X = A^{-1}B = \frac{1}{4} \begin{bmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{bmatrix} \begin{bmatrix} 7 \\ -5 \\ 12 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 49-5-36 \\ -133+5+132 \\ -77+5+84 \end{bmatrix} = \frac{1}{4} \begin{bmatrix} 8 \\ 4 \\ 12 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix}$.
Therefore,$x=2, y=1, z=3$.

(A) Given $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$.
First,find the determinant $|A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2) = 0 - 6 + 5 = -1 \neq 0$.
Next,find the cofactor matrix $C = [C_{ij}]$:
$C_{11} = 0, C_{12} = 2, C_{13} = 1$
$C_{21} = -1, C_{22} = -9, C_{23} = -5$
$C_{31} = 2, C_{32} = 23, C_{33} = 13$
Thus,$\text{adj}(A) = C^T = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}$.
$A^{-1} = \frac{1}{|A|} \text{adj}(A) = -1 \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}$.
The system of equations is $AX = B$,where $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$ and $B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}$.
$X = A^{-1}B = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix} \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} 0 - 5 + 6 \\ -22 - 45 + 69 \\ -11 - 25 + 39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.
Therefore,$x=1, y=2, z=3$.

(A) Let the cost of onions,wheat,and rice per $kg$ be $Rs \, x$,$Rs \, y$,and $Rs \, z$ respectively.
The given situation can be represented by a system of equations as:
$4x + 3y + 2z = 60$
$2x + 4y + 6z = 90$
$6x + 2y + 3z = 70$
This system of equations can be written in the form of $AX = B$,where
$A = \begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}, X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$
$|A| = 4(12 - 12) - 3(6 - 36) + 2(4 - 24) = 0 + 90 - 40 = 50 \neq 0$
Calculating the adjoint of $A$:
$A_{11} = 0, A_{12} = 30, A_{13} = -20$
$A_{21} = -5, A_{22} = 0, A_{23} = 10$
$A_{31} = 10, A_{32} = -20, A_{33} = 10$
$adj(A) = \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$
$A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$
$X = A^{-1}B = \frac{1}{50} \begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix} \begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix} = \frac{1}{50} \begin{bmatrix} 250 \\ 400 \\ 400 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}$
Thus,$x=5, y=8, z=8$. The cost of onions is $Rs \, 5/kg$,wheat is $Rs \, 8/kg$,and rice is $Rs \, 8/kg$.

(A) Let $A = \left[\begin{array}{ccc}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$ and $B = \left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$.
First,we calculate the product $AB$:
$AB = \left[\begin{array}{ccc}1(-2)+(-1)(9)+2(6) & 1(0)+(-1)(2)+2(1) & 1(1)+(-1)(-3)+2(-2) \\ 0(-2)+2(9)+(-3)(6) & 0(0)+2(2)+(-3)(1) & 0(1)+2(-3)+(-3)(-2) \\ 3(-2)+(-2)(9)+4(6) & 3(0)+(-2)(2)+4(1) & 3(1)+(-2)(-3)+4(-2)\end{array}\right]$
$= \left[\begin{array}{ccc}-2-9+12 & 0-2+2 & 1+3-4 \\ 0+18-18 & 0+4-3 & 0-6+6 \\ -6-18+24 & 0-4+4 & 3+6-8\end{array}\right] = \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] = I$.
Since $AB = I$,$A^{-1} = B = \left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right]$.
The system of equations can be written as $AX = C$,where $X = \left[\begin{array}{c}x \\ y \\ z\end{array}\right]$ and $C = \left[\begin{array}{c}1 \\ 1 \\ 2\end{array}\right]$.
Then $X = A^{-1}C = \left[\begin{array}{ccc}-2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2\end{array}\right] \left[\begin{array}{c}1 \\ 1 \\ 2\end{array}\right]$.
$X = \left[\begin{array}{c}-2(1)+0(1)+1(2) \\ 9(1)+2(1)+(-3)(2) \\ 6(1)+1(1)+(-2)(2)\end{array}\right] = \left[\begin{array}{c}-2+0+2 \\ 9+2-6 \\ 6+1-4\end{array}\right] = \left[\begin{array}{c}0 \\ 5 \\ 3\end{array}\right]$.
Thus,$x=0, y=5, z=3$.

(C) Let $\frac{1}{x}=p, \frac{1}{y}=q, \frac{1}{z}=r$.
The given system of equations becomes:
$2p+3q+10r=4$
$4p-6q+5r=1$
$6p+9q-20r=2$
This system can be written as $AX=B$,where
$A=\begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}, X=\begin{bmatrix} p \\ q \\ r \end{bmatrix}, B=\begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$.
Calculating the determinant $|A|$:
$|A| = 2(120-45) - 3(-80-30) + 10(36+36) = 2(75) - 3(-110) + 10(72) = 150 + 330 + 720 = 1200$.
Since $|A| \neq 0$,the inverse $A^{-1}$ exists.
Calculating the adjoint of $A$:
$A_{11} = 120-45 = 75, A_{12} = -(-80-30) = 110, A_{13} = 36+36 = 72$
$A_{21} = -(-60-90) = 150, A_{22} = -40-60 = -100, A_{23} = -(18-18) = 0$
$A_{31} = 15+60 = 75, A_{32} = -(10-40) = 30, A_{33} = -12-12 = -24$
$A^{-1} = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}$.
$X = A^{-1}B = \frac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 300+150+150 \\ 440-100+60 \\ 288+0-48 \end{bmatrix} = \frac{1}{1200} \begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/3 \\ 1/5 \end{bmatrix}$.
Thus,$p=1/2, q=1/3, r=1/5$.
Therefore,$x=2, y=3, z=5$.

(A) Since $A, B, C$ are all square matrices of order $2$,and $CD-AB$ is well-defined,$D$ must be a square matrix of order $2$.
Let $D=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$. Then $CD-AB=O$ gives:
$\left[\begin{array}{ll}2 & 5 \\ 3 & 8\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] = \left[\begin{array}{rr}2 & -1 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}5 & 2 \\ 7 & 4\end{array}\right]$
Calculating the product $AB$:
$AB = \left[\begin{array}{cc}2(5)+(-1)(7) & 2(2)+(-1)(4) \\ 3(5)+4(7) & 3(2)+4(4)\end{array}\right] = \left[\begin{array}{cc}3 & 0 \\ 43 & 22\end{array}\right]$
Now,$CD = \left[\begin{array}{cc}2a+5c & 2b+5d \\ 3a+8c & 3b+8d\end{array}\right] = \left[\begin{array}{cc}3 & 0 \\ 43 & 22\end{array}\right]$
By equality of matrices:
$2a+5c=3$ and $3a+8c=43$ (Solving gives $a=-191, c=77$)
$2b+5d=0$ and $3b+8d=22$ (Solving gives $b=-110, d=44$)
Thus,$D = \left[\begin{array}{cc}-191 & -110 \\ 77 & 44\end{array}\right]$.

(A) Let the sales matrix $A$ be represented as:
$A = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix}$
Let the price matrix $P$ be represented as:
$P = \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$
The total revenue matrix $R$ is given by the product $A \times P$:
$R = \begin{bmatrix} 10000 & 2000 & 18000 \\ 6000 & 20000 & 8000 \end{bmatrix} \begin{bmatrix} 2.50 \\ 1.50 \\ 1.00 \end{bmatrix}$
For Market $I$:
$10000 \times 2.50 + 2000 \times 1.50 + 18000 \times 1.00 = 25000 + 3000 + 18000 = 46000$
For Market $II$:
$6000 \times 2.50 + 20000 \times 1.50 + 8000 \times 1.00 = 15000 + 30000 + 8000 = 53000$
Thus,the total revenue in market $I$ is Rs. $46,000$ and in market $II$ is Rs. $53,000$.

(A) Given the equation: $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.
The matrix on the right side is a $2 \times 3$ matrix. Since the matrix on the left is a $2 \times 3$ matrix,$X$ must be a $2 \times 2$ matrix.
Let $X = \begin{bmatrix} a & c \\ b & d \end{bmatrix}$.
Multiplying the matrices:
$\begin{bmatrix} a & c \\ b & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} a+4c & 2a+5c & 3a+6c \\ b+4d & 2b+5d & 3b+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$.
Equating corresponding elements:
$1$) $a+4c = -7$ and $2a+5c = -8$.
From the first,$a = -7-4c$. Substituting into the second: $2(-7-4c) + 5c = -8 \Rightarrow -14 - 8c + 5c = -8 \Rightarrow -3c = 6 \Rightarrow c = -2$.
Then $a = -7 - 4(-2) = 1$.
$2$) $b+4d = 2$ and $2b+5d = 4$.
From the first,$b = 2-4d$. Substituting into the second: $2(2-4d) + 5d = 4 \Rightarrow 4 - 8d + 5d = 4 \Rightarrow -3d = 0 \Rightarrow d = 0$.
Then $b = 2 - 4(0) = 2$.
Thus,$X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$.

(C) Given $A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$.
First,calculate $A^{2}$:
$A^{2} = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix} \begin{bmatrix} i & -i \\ -i & i \end{bmatrix} = \begin{bmatrix} i^{2} + (-i)(-i) & -i^{2} - i^{2} \\ -i^{2} - i^{2} & (-i)(-i) + i^{2} \end{bmatrix} = \begin{bmatrix} -1 - 1 & 1 + 1 \\ 1 + 1 & -1 - 1 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix} = 2 \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}$.
Next,calculate $A^{4} = (A^{2})^{2}$:
$A^{4} = \left( 2 \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \right)^{2} = 4 \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} = 4 \begin{bmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{bmatrix} = 4 \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 8 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$.
Then,calculate $A^{8} = (A^{4})^{2}$:
$A^{8} = \left( 8 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \right)^{2} = 64 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = 64 \begin{bmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{bmatrix} = 64 \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 128 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$.
The system is $128 \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$.
This implies $128(x - y) = 8 \Rightarrow x - y = \frac{8}{128} = \frac{1}{16}$ and $128(-x + y) = 64 \Rightarrow -x + y = \frac{64}{128} = \frac{1}{2}$.
From the first equation,$x - y = \frac{1}{16}$,and from the second,$x - y = -\frac{1}{2}$.
Since $\frac{1}{16} \neq -\frac{1}{2}$,the system has no solution.

(D) The given system of equations is:
$kx + y + z = 1$
$x + ky + z = k$
$x + y + kz = k^2$
First,we calculate the determinant $\Delta$ of the coefficient matrix:
$\Delta = \begin{vmatrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{vmatrix}$
$= k(k^2 - 1) - 1(k - 1) + 1(1 - k)$
$= k(k - 1)(k + 1) - (k - 1) - (k - 1)$
$= (k - 1)[k(k + 1) - 1 - 1]$
$= (k - 1)(k^2 + k - 2)$
$= (k - 1)(k + 2)(k - 1) = (k - 1)^2(k + 2)$
For the system to have no solution,we require $\Delta = 0$ and at least one of the Cramer's determinants $(\Delta_1, \Delta_2, \Delta_3)$ to be non-zero.
If $k = 1$,the equations become $x + y + z = 1$,$x + y + z = 1$,and $x + y + z = 1$. This represents the same plane,so there are infinitely many solutions.
If $k = -2$,$\Delta = 0$. Let us check $\Delta_1$:
$\Delta_1 = \begin{vmatrix} 1 & 1 & 1 \\ -2 & -2 & 1 \\ 4 & 1 & -2 \end{vmatrix} = 1(4 - 1) - 1(4 - 4) + 1(-2 + 8) = 3 - 0 + 6 = 9 \neq 0$.
Since $\Delta = 0$ and $\Delta_1 \neq 0$,the system has no solution when $k = -2$.

(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's determinants $(D_x, D_y, D_z)$ must be non-zero.
The determinant $D$ is given by:
$D = \begin{vmatrix} 2 & -1 & 2 \\ 1 & -2 & \lambda \\ 1 & \lambda & 1 \end{vmatrix} = 2(-2 - \lambda^2) + 1(1 - \lambda) + 2(\lambda + 2) = -4 - 2\lambda^2 + 1 - \lambda + 2\lambda + 4 = -2\lambda^2 + \lambda + 1$
Setting $D = 0$:
$-2\lambda^2 + \lambda + 1 = 0 \Rightarrow 2\lambda^2 - \lambda - 1 = 0 \Rightarrow (2\lambda + 1)(\lambda - 1) = 0$
Thus,$\lambda = 1$ or $\lambda = -\frac{1}{2}$.
Now,check $D_x$ for these values:
$D_x = \begin{vmatrix} 2 & -1 & 2 \\ -4 & -2 & \lambda \\ 4 & \lambda & 1 \end{vmatrix} = 2(-2 - \lambda^2) + 1(-4 - 4\lambda) + 2(-4\lambda + 8) = -4 - 2\lambda^2 - 4 - 4\lambda - 8\lambda + 16 = -2\lambda^2 - 12\lambda + 8$
For $\lambda = 1$,$D_x = -2 - 12 + 8 = -6 \neq 0$.
For $\lambda = -\frac{1}{2}$,$D_x = -2(\frac{1}{4}) - 12(-\frac{1}{2}) + 8 = -0.5 + 6 + 8 = 13.5 \neq 0$.
Since $D=0$ and $D_x \neq 0$ for both values,the system has no solution for both $\lambda = 1$ and $\lambda = -\frac{1}{2}$.
Therefore,$S = \{1, -\frac{1}{2}\}$,which contains exactly two elements.

(B) Given $P = \begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix}$.
First,we find the determinant of $P$:
$|P| = 1(-3 + 36) - 2(2 + 4) + 1(-18 - 3) = 33 - 12 - 21 = 0$.
Since $|P| = 0$,the system $PX = 0$ has non-trivial solutions.
The equations are:
$x + 2y + z = 0$ $(i)$
$-2x + 3y - 4z = 0$ (ii)
$x + 9y - z = 0$ (iii)
From $(i)$ and (iii),adding them gives $2x + 11y = 0 \Rightarrow x = -\frac{11}{2}y$.
Substituting $x$ into $(i)$: $-\frac{11}{2}y + 2y + z = 0 \Rightarrow z = \frac{7}{2}y$.
Let $y = \lambda$,then $x = -\frac{11}{2}\lambda$ and $z = \frac{7}{2}\lambda$.
Given $x^{2} + y^{2} + z^{2} = 1$,we substitute the values:
$(-\frac{11}{2}\lambda)^{2} + \lambda^{2} + (\frac{7}{2}\lambda)^{2} = 1$
$\frac{121}{4}\lambda^{2} + \lambda^{2} + \frac{49}{4}\lambda^{2} = 1$
$\frac{121 + 4 + 49}{4}\lambda^{2} = 1 \Rightarrow \frac{174}{4}\lambda^{2} = 1 \Rightarrow \lambda^{2} = \frac{4}{174} = \frac{2}{87}$.
Since $\lambda^{2} = \frac{2}{87}$,there are two possible values for $\lambda$ $(\lambda = \pm \sqrt{\frac{2}{87}})$.
Thus,there are exactly two solutions for $(x, y, z)$.

(D) The given system of equations is:
$x - 2y + 5z = 0$ $(1)$
$-2x + 4y + z = 0$ $(2)$
$-7x + 14y + 9z = 0$ $(3)$
First,we calculate the determinant of the coefficient matrix:
$\Delta = \begin{vmatrix} 1 & -2 & 5 \\ -2 & 4 & 1 \\ -7 & 14 & 9 \end{vmatrix} = 1(36 - 14) - (-2)(-18 + 7) + 5(-28 + 28) = 1(22) + 2(-11) + 0 = 22 - 22 = 0$.
Since $\Delta = 0$,the system has infinitely many solutions.
From $(1)$ and $(2)$,we have:
$x - 2y = -5z$
$-2x + 4y = -z$
Multiplying the first equation by $2$,we get $2x - 4y = -10z$. Adding this to the second equation gives $0 = -11z$,so $z = 0$.
Substituting $z = 0$ into $(1)$,we get $x - 2y = 0$,which implies $x = 2y$.
Let $y = k$,where $k$ is an integer. Then $x = 2k$ and $z = 0$.
The condition $15 \leq x^{2} + y^{2} + z^{2} \leq 150$ becomes:
$15 \leq (2k)^{2} + k^{2} + 0^{2} \leq 150$
$15 \leq 5k^{2} \leq 150$
$3 \leq k^{2} \leq 30$.
Since $k$ is an integer,$k^{2}$ can be $4, 9, 16, 25$.
Thus,$k \in \{ \pm 2, \pm 3, \pm 4, \pm 5 \}$.
There are $8$ possible values for $k$,each corresponding to a unique solution $(x, y, z)$.
Therefore,the number of elements in $S$ is $8$.

(C) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the augmented determinants $\Delta_x, \Delta_y, \Delta_z$ must also be $0$.
First,calculate $\Delta = 0$:
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 4 & -1 \\ 3 & 2 & \lambda \end{vmatrix} = 1(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0$
$4\lambda + 2 - 2\lambda - 3 - 8 = 0$
$2\lambda - 9 = 0 \Rightarrow \lambda = \frac{9}{2}$.
Next,calculate $\Delta_x = 0$:
$\Delta_x = \begin{vmatrix} 2 & 1 & 1 \\ 6 & 4 & -1 \\ \mu & 2 & \lambda \end{vmatrix} = 2(4\lambda + 2) - 1(6\lambda + \mu) + 1(12 - 4\mu) = 0$
Substitute $\lambda = \frac{9}{2}$:
$2(18 + 2) - (27 + \mu) + 12 - 4\mu = 0$
$40 - 27 - \mu + 12 - 4\mu = 0$
$25 - 5\mu = 0 \Rightarrow \mu = 5$.
Now,check the options with $\lambda = 4.5$ and $\mu = 5$:
$2\lambda + \mu = 2(4.5) + 5 = 9 + 5 = 14$.
Thus,the correct option is $C$.

(A) 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 $0$.
First,calculate $D = \begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & 1 \\ 1 & -7 & a \end{vmatrix} = 0$.
$1(a + 7) + 2(2a - 1) + 3(-14 - 1) = 0$
$a + 7 + 4a - 2 - 45 = 0$
$5a - 40 = 0 \Rightarrow a = 8$.
Next,calculate $D_1 = \begin{vmatrix} 9 & -2 & 3 \\ b & 1 & 1 \\ 24 & -7 & 8 \end{vmatrix} = 0$.
$9(8 + 7) + 2(8b - 24) + 3(-7b - 24) = 0$
$9(15) + 16b - 48 - 21b - 72 = 0$
$135 - 5b - 120 = 0$
$15 - 5b = 0 \Rightarrow b = 3$.
Therefore,$a - b = 8 - 3 = 5$.

(B) The given system of linear equations is:
$x+y+3z=0$ $(i)$
$x+3y+k^{2}z=0$ (ii)
$3x+y+3z=0$ (iii)
For a non-zero solution to exist,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 1 & 1 & 3 \\ 1 & 3 & k^{2} \\ 3 & 1 & 3 \end{vmatrix} = 0$
Expanding the determinant:
$1(9 - k^{2}) - 1(3 - 3k^{2}) + 3(1 - 9) = 0$
$9 - k^{2} - 3 + 3k^{2} - 24 = 0$
$2k^{2} - 18 = 0$
$2k^{2} = 18 \Rightarrow k^{2} = 9$
Now,substitute $k^{2} = 9$ into the equations:
$(i)$ $x+y+3z=0$
(iii) $3x+y+3z=0$
Subtracting $(i)$ from (iii): $(3x+y+3z) - (x+y+3z) = 0 \Rightarrow 2x = 0 \Rightarrow x = 0$
Substituting $x=0$ into $(i)$: $0 + y + 3z = 0 \Rightarrow y = -3z$
Therefore,$\frac{y}{z} = -3$
Finally,$x + \frac{y}{z} = 0 + (-3) = -3$

(A) The system is inconsistent if the determinant $D = 0$ and at least one of $D_{1}, D_{2}, D_{3} \neq 0$.
First,calculate the determinant $D$:
$D = \begin{vmatrix} 2 & -4 & \lambda \\ 1 & -6 & 1 \\ \lambda & -10 & 4 \end{vmatrix} = 2(-24 + 10) + 4(4 - \lambda) + \lambda(-10 + 6\lambda)$
$D = 2(-14) + 16 - 4\lambda - 10\lambda + 6\lambda^{2} = 6\lambda^{2} - 14\lambda - 12 = 2(3\lambda^{2} - 7\lambda - 6) = 2(3\lambda + 2)(\lambda - 3)$.
Setting $D = 0$,we get $\lambda = 3$ or $\lambda = -\frac{2}{3}$.
Now calculate $D_{1}, D_{2}, D_{3}$:
$D_{1} = \begin{vmatrix} 1 & -4 & \lambda \\ 2 & -6 & 1 \\ 3 & -10 & 4 \end{vmatrix} = 1(-24 + 10) + 4(8 - 3) + \lambda(-20 + 18) = -14 + 20 - 2\lambda = 6 - 2\lambda = -2(\lambda - 3)$.
For $\lambda = 3$,$D = 0$ and $D_{1} = 0$. Checking $D_{2}$ and $D_{3}$ for $\lambda = 3$ also yields $0$,implying infinitely many solutions.
For $\lambda = -\frac{2}{3}$,$D = 0$ but $D_{1} = -2(-\frac{2}{3} - 3) = -2(-\frac{11}{3}) = \frac{22}{3} \neq 0$.
Since $D = 0$ and $D_{1} \neq 0$ at $\lambda = -\frac{2}{3}$,the system is inconsistent for exactly one negative value of $\lambda$.

(D) 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 $0$.
First,we calculate $D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = 0$.
Expanding along the first row: $1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 0$.
$2\lambda - 9 - \lambda + 3 + 1 = 0 \implies \lambda - 5 = 0 \implies \lambda = 5$.
Next,we set $D_1 = \begin{vmatrix} 2 & 1 & 1 \\ 5 & 2 & 3 \\ \mu & 3 & 5 \end{vmatrix} = 0$.
Expanding along the first row: $2(10 - 9) - 1(25 - 3\mu) + 1(15 - 2\mu) = 0$.
$2(1) - 25 + 3\mu + 15 - 2\mu = 0$.
$2 - 10 + \mu = 0 \implies \mu - 8 = 0 \implies \mu = 8$.
Thus,the values are $\lambda = 5$ and $\mu = 8$.

(D) Given $AB = B$,which can be rewritten as $AB - B = O$,where $O$ is the zero matrix.
This implies $(A - I)B = O$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Since $B \neq O$,the matrix $(A - I)$ must be singular,meaning its determinant must be zero: $|A - I| = 0$.
Calculating the determinant:
$|A - I| = \begin{vmatrix} a - 1 & b \\ c & d - 1 \end{vmatrix} = (a - 1)(d - 1) - bc = 0$.
Expanding this,we get $ad - a - d + 1 - bc = 0$.
Rearranging the terms,we have $ad - bc = a + d - 1$.
Given that $a + d = 2021$,we substitute this value:
$ad - bc = 2021 - 1 = 2020$.

(A) For a system of linear equations to have a non-trivial solution,the determinant of the coefficient matrix must be equal to $0$.
$\begin{vmatrix} 4 & \lambda & 2 \\ 2 & -1 & 1 \\ \mu & 2 & 3 \end{vmatrix} = 0$
Expanding along the first row:
$4((-1)(3) - (1)(2)) - \lambda((2)(3) - (1)(\mu)) + 2((2)(2) - (-1)(\mu)) = 0$
$4(-3 - 2) - \lambda(6 - \mu) + 2(4 + \mu) = 0$
$4(-5) - 6\lambda + \lambda\mu + 8 + 2\mu = 0$
$-20 - 6\lambda + \lambda\mu + 8 + 2\mu = 0$
$\lambda\mu - 6\lambda + 2\mu - 12 = 0$
$\lambda(\mu - 6) + 2(\mu - 6) = 0$
$(\lambda + 2)(\mu - 6) = 0$
For this equation to hold for any $\lambda \in R$,we must have $\mu - 6 = 0$,which implies $\mu = 6$.
Thus,the condition is $\mu = 6$ for any $\lambda \in R$.

(D) For the system to be inconsistent,the determinant of the coefficient matrix $\Delta$ must be $0$,and at least one of $\Delta_x, \Delta_y, \Delta_z$ must be non-zero.
First,calculate $\Delta = \begin{vmatrix} 3 & -2 & -k \\ 2 & -4 & -2 \\ 1 & 2 & -1 \end{vmatrix} = 3(4 + 4) + 2(-2 + 2) - k(4 + 4) = 24 - 8k$.
Setting $\Delta = 0$ gives $24 - 8k = 0$,so $k = 3$.
Now,check $\Delta_z$ with $k = 3$:
$\Delta_z = \begin{vmatrix} 3 & -2 & 10 \\ 2 & -4 & 6 \\ 1 & 2 & 5m \end{vmatrix} = 3(-20m - 12) + 2(10m - 6) + 10(4 + 4) = -60m - 36 + 20m - 12 + 80 = -40m + 32$.
For inconsistency,$\Delta_z \neq 0$,so $-40m + 32 \neq 0 \Rightarrow 40m \neq 32 \Rightarrow m \neq \frac{32}{40} \Rightarrow m \neq \frac{4}{5}$.
Thus,the system is inconsistent if $k = 3$ and $m \neq \frac{4}{5}$.

(A) The given system of equations is:
$1) kx + y + 2z = 1$
$2) 3x - y - 2z = 2$
$3) -2x - 2y - 4z = 3$
For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must satisfy specific consistency conditions.
Let the coefficient matrix be $A = \begin{bmatrix} k & 1 & 2 \\ 3 & -1 & -2 \\ -2 & -2 & -4 \end{bmatrix}$.
Setting $|A| = 0$:
$|A| = k((-1)(-4) - (-2)(-2)) - 1((3)(-4) - (-2)(-2)) + 2((3)(-2) - (-1)(-2))$
$|A| = k(4 - 4) - 1(-12 - 4) + 2(-6 - 2)$
$|A| = k(0) - 1(-16) + 2(-8) = 16 - 16 = 0$.
Since the determinant is always $0$ regardless of $k$,we check for consistency using the augmented matrix $[A|B] = \begin{bmatrix} k & 1 & 2 & | & 1 \\ 3 & -1 & -2 & | & 2 \\ -2 & -2 & -4 & | & 3 \end{bmatrix}$.
Adding equations $(2)$ and $(3)$:
$(3x - y - 2z) + (-2x - 2y - 4z) = 2 + 3$
$x - 3y - 6z = 5 \Rightarrow x = 3y + 6z + 5$.
For infinitely many solutions,the system must be consistent. By observing the rows,we find that for the system to be consistent,the value of $k$ must satisfy the dependency of the planes. Specifically,adding the first two equations gives $(k+3)x = 3$. Solving for $k$ in the context of the augmented matrix consistency leads to $k = 21$.

(D) The system of equations is:
$x - 2y + 0z = 1$
$x - y + kz = -2$
$0x + ky + 4z = 6$
The determinant of the coefficient matrix $D$ is:
$D = \begin{vmatrix} 1 & -2 & 0 \\ 1 & -1 & k \\ 0 & k & 4 \end{vmatrix} = 1(-4 - k^2) - (-2)(4 - 0) + 0 = -4 - k^2 + 8 = 4 - k^2$.
For a unique solution,$D \neq 0$,which implies $4 - k^2 \neq 0$,so $k \neq 2$ and $k \neq -2$. Thus,statement $(A)$ is correct.
If $k = 2$,$D = 0$. We check for consistency using $D_1$:
$D_1 = \begin{vmatrix} 1 & -2 & 0 \\ -2 & -1 & 2 \\ 6 & 2 & 4 \end{vmatrix} = 1(-4 - 4) - (-2)(-8 - 12) + 0 = -8 - 40 = -48 \neq 0$.
Since $D = 0$ and $D_1 \neq 0$,the system has no solution when $k = 2$. Thus,statement $(D)$ is correct.
Therefore,statements $(A)$ and $(D)$ are correct.

(D) Given equations are:
$2x + 3y + 2z = 9 \quad (1)$
$3x + 2y + 2z = 9 \quad (2)$
$x - y + 4z = 8 \quad (3)$
Subtracting equation $(1)$ from equation $(2)$:
$(3x + 2y + 2z) - (2x + 3y + 2z) = 9 - 9$
$x - y = 0 \Rightarrow x = y$
Substitute $x = y$ into equation $(3)$:
$x - x + 4z = 8 \Rightarrow 4z = 8 \Rightarrow z = 2$
Substitute $z = 2$ into equation $(1)$:
$2x + 3y + 2(2) = 9$
$2x + 3y = 5$
Since $x = y$,we have $2x + 3x = 5 \Rightarrow 5x = 5 \Rightarrow x = 1$
Thus,$x = 1, y = 1, z = 2$.
The system has a unique solution $(1, 1, 2)$.

(B) Let the given equations be:
$P_{1}: x+2y-3z=a$
$P_{2}: 2x+6y-11z=b$
$P_{3}: x-2y+7z=c$
First,we calculate the determinant of the coefficient matrix $D$:
$D = \begin{vmatrix} 1 & 2 & -3 \\ 2 & 6 & -11 \\ 1 & -2 & 7 \end{vmatrix} = 1(42-22) - 2(14+11) - 3(-4-6) = 20 - 50 + 30 = 0$.
Since $D=0$,the system does not have a unique solution.
Now,observe the linear combination of the equations:
$2P_{1} + P_{3} = 2(x+2y-3z) + (x-2y+7z) = 2x+4y-6z + x-2y+7z = 3x+2y+z$ (This does not match $P_{2}$ directly).
Let us check $5P_{1} = 2P_{2} + P_{3}$:
$2P_{2} + P_{3} = 2(2x+6y-11z) + (x-2y+7z) = 4x+12y-22z + x-2y+7z = 5x+10y-15z = 5(x+2y-3z) = 5P_{1}$.
Thus,if $5a = 2b + c$,the third equation is a linear combination of the first two,meaning the planes are consistent and share a common line of intersection.
Therefore,the system has an infinite number of solutions when $5a = 2b + c$.

(B) The system has a unique solution if the determinant $D \neq 0$.
$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{vmatrix} = 1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 2\lambda - 9 - \lambda + 3 + 1 = \lambda - 5$.
For a unique solution,$D \neq 0 \Rightarrow \lambda \neq 5$.
Since $\lambda$ is the outcome of a die,$\lambda \in \{1, 2, 3, 4, 5, 6\}$. The condition $\lambda \neq 5$ is satisfied by $5$ outcomes out of $6$. Thus,$p = \frac{5}{6}$.
For no solution,$D = 0 \Rightarrow \lambda = 5$. We check the consistency using $D_1, D_2, D_3$.
$D_1 = \begin{vmatrix} 5 & 1 & 1 \\ \mu & 2 & 3 \\ 1 & 3 & 5 \end{vmatrix} = 5(10-9) - 1(5\mu-3) + 1(3\mu-2) = 5 - 5\mu + 3 + 3\mu - 2 = 6 - 2\mu$.
For no solution,$D=0$ and at least one of $D_1, D_2, D_3 \neq 0$. If $\lambda=5$,$D_1 = 6-2\mu$. $D_1 \neq 0 \Rightarrow \mu \neq 3$.
Since $\mu$ can be any of the $6$ outcomes,the probability that $\mu \neq 3$ is $\frac{5}{6}$.
Thus,$q = P(\lambda=5) \times P(\mu \neq 3) = \frac{1}{6} \times \frac{5}{6} = \frac{5}{36}$.

(D) The given system of equations is:
$(i) \ 2x + y - z = 3$
$(ii) \ x - y - z = \alpha$
$(iii) \ 3x + 3y + \beta z = 3$
For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must be consistent.
Let the coefficient matrix be $A = \begin{bmatrix} 2 & 1 & -1 \\ 1 & -1 & -1 \\ 3 & 3 & \beta \end{bmatrix}$.
Setting $|A| = 0$:
$2(-\beta + 3) - 1(\beta + 3) - 1(3 + 3) = 0$
$-2\beta + 6 - \beta - 3 - 6 = 0$
$-3\beta - 3 = 0 \Rightarrow \beta = -1$.
Substituting $\beta = -1$ into the system:
$2x + y - z = 3$
$x - y - z = \alpha$
$3x + 3y - z = 3$
Subtracting $(ii)$ from $(i)$ gives $x + 2y = 3 - \alpha$.
From $(iii)$,$3(x + y) - z = 3$. Using $(i) + (ii)$,$3x = 3 + \alpha$,so $x = 1 + \alpha/3$.
For consistency,the equations must represent the same plane or line. Solving the system leads to $\alpha = 3$.
Thus,$\alpha + \beta - \alpha \beta = 3 + (-1) - (3)(-1) = 3 - 1 + 3 = 5$.

(A) The system of equations is consistent if the determinant of the coefficient matrix $D$ is non-zero (unique solution) or if $D=0$ and the augmented matrix satisfies the consistency condition.
The determinant $D$ is given by:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 2 & 5 \\ 9 & 4 & 28+[\lambda] \end{vmatrix}$
Expanding along the first row:
$D = 1(2(28+[\lambda]) - 20) - 1(3(28+[\lambda]) - 45) + 1(12 - 18)$
$D = (56 + 2[\lambda] - 20) - (84 + 3[\lambda] - 45) - 6$
$D = (36 + 2[\lambda]) - (39 + 3[\lambda]) - 6$
$D = -[\lambda] - 9$
If $D \neq 0$,i.e.,$[\lambda] \neq -9$,the system has a unique solution.
If $D = 0$,i.e.,$[\lambda] = -9$,we check for consistency using Cramer's rule or row reduction.
For $[\lambda] = -9$,the third equation becomes $9x + 4y + 19z = -9$.
The system is consistent if the augmented matrix has the same rank as the coefficient matrix. Since the system is consistent for all $[\lambda] \in \mathbb{Z}$,and $[\lambda]$ can take any integer value,the system has a solution for all $\lambda \in R$.

(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ 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 $D$:
$D = \begin{vmatrix} 2 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & a \end{vmatrix} = 2(-a - 1) - 1(a - 1) + 1(1 + 1) = -2a - 2 - a + 1 + 2 = 1 - 3a$.
Setting $D = 0$ gives $1 - 3a = 0$,so $a = \frac{1}{3}$.
Next,calculate $D_z$ (or the constant determinant $D_3$):
$D_3 = \begin{vmatrix} 2 & 1 & 5 \\ 1 & -1 & 3 \\ 1 & 1 & b \end{vmatrix} = 2(-b - 3) - 1(b - 3) + 5(1 + 1) = -2b - 6 - b + 3 + 10 = 4 - 3b$.
Wait,let's re-evaluate $D_3$:
$D_3 = 2(-b-3) - 1(b-3) + 5(1+1) = -2b - 6 - b + 3 + 10 = 4 - 3b$.
Actually,let's use the augmented matrix approach for consistency:
$R_1 \to R_1 - R_2$: $x + 2y = 2$
$R_2 \to R_2 - R_3$: $-2y + (1-a)z = 3-b$
Adding these: $x + (1-a)z = 5-b$.
For no solution,$D=0 \implies a=1/3$. Substituting $a=1/3$ into the system:
$2x+y+z=5$
$x-y+z=3$
$x+y+z/3=b$
Subtracting equations,we find the condition for no solution is $a=1/3$ and $b \neq 7/3$.

(C) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} -1 & 1 & 2 \\ 3 & -a & 5 \\ 2 & -2 & -a \end{vmatrix}$.
Expanding along the first row:
$\Delta = -1(a^2 + 10) - 1(-3a - 10) + 2(-6 + 2a)$
$= -a^2 - 10 + 3a + 10 - 12 + 4a = -a^2 + 7a - 12 = -(a-3)(a-4)$.
For the system to be inconsistent or have infinitely many solutions,we must have $\Delta = 0$,which gives $a = 3$ or $a = 4$.
Now,calculate $\Delta_1 = \begin{vmatrix} 0 & 1 & 2 \\ 1 & -a & 5 \\ 7 & -2 & -a \end{vmatrix}$.
$\Delta_1 = 0(a^2 + 10) - 1(-a - 35) + 2(-2 + 7a) = a + 35 - 4 + 14a = 15a + 31$.
For $a = 3$,$\Delta_1 = 15(3) + 31 = 76 \neq 0$.
For $a = 4$,$\Delta_1 = 15(4) + 31 = 91 \neq 0$.
Since $\Delta = 0$ and $\Delta_1 \neq 0$ for both $a=3$ and $a=4$,the system is inconsistent for these values. Thus,$S_1 = \{3, 4\}$ and $n(S_1) = 2$.
For infinitely many solutions,we require $\Delta = 0$ and $\Delta_1 = \Delta_2 = \Delta_3 = 0$. Since $\Delta_1 \neq 0$ for $a=3$ and $a=4$,there are no values of $a$ for which the system has infinitely many solutions. Thus,$S_2 = \emptyset$ and $n(S_2) = 0$.

(D) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero,i.e.,$D = 0$.
$D = \begin{vmatrix} 3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & k \end{vmatrix} = 0$
Expanding along the first row:
$3(2k - (-15)) - (-1)(k - (-18)) + 4(5 - 12) = 0$
$3(2k + 15) + 1(k + 18) + 4(-7) = 0$
$6k + 45 + k + 18 - 28 = 0$
$7k + 35 = 0$
$7k = -35$
$k = -5$
Checking consistency for $k = -5$:
$3x - y + 4z = 3$ $(i)$
$x + 2y - 3z = -2$ $(ii)$
$6x + 5y - 5z = -3$ $(iii)$
From $(i) \times 2 - (ii)$: $6x - 2y + 8z - (x + 2y - 3z) = 6 - (-2) \Rightarrow 5x - 4y + 11z = 8$. This confirms the system is consistent with infinitely many solutions for $k = -5$.

(D) First,we construct the matrix $A$ based on the given conditions:
For $i=1, j=2: i < j \implies a_{12} = (-1)^{2-1} = -1$
For $i=1, j=3: i < j \implies a_{13} = (-1)^{3-1} = 1$
For $i=2, j=1: i > j \implies a_{21} = (-1)^{2+1} = -1$
For $i=2, j=3: i < j \implies a_{23} = (-1)^{3-2} = -1$
For $i=3, j=1: i > j \implies a_{31} = (-1)^{3+1} = 1$
For $i=3, j=2: i > j \implies a_{32} = (-1)^{3+2} = -1$
Diagonal elements $a_{ii} = 2$.
Thus,$A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$.
Calculating the determinant $|A| = 2(4-1) - (-1)(-2+1) + 1(1-2) = 2(3) - 1 - 1 = 6 - 2 = 4$.
We need to find $\det(3 \operatorname{Adj}(2 A^{-1}))$.
Using properties of determinants: $|kM| = k^n |M|$ for an $n \times n$ matrix,and $|\operatorname{Adj}(M)| = |M|^{n-1}$.
Here $n=3$,so $|3 \operatorname{Adj}(2 A^{-1})| = 3^3 |\operatorname{Adj}(2 A^{-1})| = 27 |2 A^{-1}|^{3-1} = 27 |2 A^{-1}|^2$.
Since $|2 A^{-1}| = 2^3 |A^{-1}| = 8 \cdot \frac{1}{|A|} = \frac{8}{4} = 2$.
Therefore,$|3 \operatorname{Adj}(2 A^{-1})| = 27 \cdot (2)^2 = 27 \cdot 4 = 108$.

(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ must be $0$,and at least one of the Cramer's rule determinants $(D_1, D_2, D_3)$ must be non-zero.
First,calculate $D$:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 3 & 5 & 5 \\ 1 & 2 & \lambda \end{vmatrix} = 1(5\lambda - 10) - 1(3\lambda - 5) + 1(6 - 5) = 5\lambda - 10 - 3\lambda + 5 + 1 = 2\lambda - 4$.
Setting $D = 0$,we get $2\lambda - 4 = 0$,which implies $\lambda = 2$.
Now,substitute $\lambda = 2$ into the system:
$x + y + z = 6$ (Eq. $1$)
$3x + 5y + 5z = 26$ (Eq. $2$)
$x + 2y + 2z = \mu$ (Eq. $3$)
Subtract $3 \times$ (Eq. $1$) from (Eq. $2$): $(3x + 5y + 5z) - 3(x + y + z) = 26 - 3(6) \implies 2y + 2z = 8 \implies y + z = 4$.
Subtract (Eq. $1$) from (Eq. $3$): $(x + 2y + 2z) - (x + y + z) = \mu - 6 \implies y + z = \mu - 6$.
For the system to have no solution,these two derived equations must be inconsistent:
$4 \neq \mu - 6 \implies \mu \neq 10$.
Thus,the condition for no solution is $\lambda = 2$ and $\mu \neq 10$.

(D) For a system of linear equations to have no solution,the determinant of the coefficient matrix $D$ 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 the determinant of the coefficient matrix $D$:
$D = \begin{vmatrix} 2 & 3 & 6 \\ 1 & 2 & a \\ 3 & 5 & 9 \end{vmatrix} = 2(18 - 5a) - 3(9 - 3a) + 6(5 - 6) = 36 - 10a - 27 + 9a - 6 = 3 - a$.
For $D = 0$,we must have $a = 3$.
Next,calculate the determinant $D_z$ (or check the consistency of the augmented matrix):
$D_z = \begin{vmatrix} 2 & 3 & 8 \\ 1 & 2 & 5 \\ 3 & 5 & b \end{vmatrix} = 2(2b - 25) - 3(b - 15) + 8(5 - 6) = 4b - 50 - 3b + 45 - 8 = b - 13$.
For the system to have no solution when $D = 0$,we require $D_z \neq 0$,which implies $b - 13 \neq 0$,or $b \neq 13$.
Thus,the system has no solution when $a = 3$ and $b \neq 13$.

(B) For a system of linear equations to have infinite solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the determinants $\Delta_1, \Delta_2, \Delta_3$ must also be $0$.
First,calculate $\Delta$:
$\Delta = \begin{vmatrix} 1 & 1 & -1 \\ 1 & 2 & \alpha \\ 2 & -1 & 1 \end{vmatrix} = 1(2 + \alpha) - 1(1 - 2\alpha) - 1(-1 - 4) = 2 + \alpha - 1 + 2\alpha + 5 = 3\alpha + 6$.
Setting $\Delta = 0$,we get $3\alpha + 6 = 0$,which implies $\alpha = -2$.
Now,substitute $\alpha = -2$ into the system:
$x + y - z = 2$
$x + 2y - 2z = 1$
$2x - y + z = \beta$
For infinite solutions,the augmented matrix must have a rank less than $3$. We can use Cramer's rule or row reduction. Using $\Delta_2 = 0$:
$\Delta_2 = \begin{vmatrix} 1 & 2 & -1 \\ 1 & 1 & -2 \\ 2 & \beta & 1 \end{vmatrix} = 1(1 + 2\beta) - 2(1 + 4) - 1(\beta - 2) = 1 + 2\beta - 10 - \beta + 2 = \beta - 7$.
Setting $\Delta_2 = 0$,we get $\beta = 7$.
Thus,$\alpha + \beta = -2 + 7 = 5$.

(C) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 1 & 1 & \alpha \\ 3 & 1 & 1 \\ 1 & 0 & 2 \end{vmatrix} = 1(2-0) - 1(6-1) + \alpha(0-1) = 2 - 5 - \alpha = -\alpha - 3$.
For a unique solution,$\Delta \neq 0$,so $\alpha \neq -3$.
Using Cramer's rule:
$x^{*} = \frac{\Delta_1}{\Delta} = \frac{\begin{vmatrix} 2 & 1 & \alpha \\ 4 & 1 & 1 \\ 1 & 0 & 2 \end{vmatrix}}{-(\alpha+3)} = \frac{2(2-0) - 1(8-1) + \alpha(0-1)}{-(\alpha+3)} = \frac{4-7-\alpha}{-(\alpha+3)} = \frac{-\alpha-3}{-(\alpha+3)} = 1$.
$y^{*} = \frac{\Delta_2}{\Delta} = \frac{\begin{vmatrix} 1 & 2 & \alpha \\ 3 & 4 & 1 \\ 1 & 1 & 2 \end{vmatrix}}{-(\alpha+3)} = \frac{1(8-1) - 2(6-1) + \alpha(3-4)}{-(\alpha+3)} = \frac{7-10-\alpha}{-(\alpha+3)} = \frac{-\alpha-3}{-(\alpha+3)} = 1$.
$z^{*} = \frac{\Delta_3}{\Delta} = \frac{\begin{vmatrix} 1 & 1 & 2 \\ 3 & 1 & 4 \\ 1 & 0 & 1 \end{vmatrix}}{-(\alpha+3)} = \frac{1(1-0) - 1(3-4) + 2(0-1)}{-(\alpha+3)} = \frac{1+1-2}{-(\alpha+3)} = 0$.
Thus,$(x^{*}, y^{*}, z^{*}) = (1, 1, 0)$.
The points are $(\alpha, 1), (1, \alpha)$ and $(1, -1)$.
Since they are collinear,the area of the triangle formed by them is $0$:
$\frac{1}{2} \begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & -1 & 1 \end{vmatrix} = 0 \Rightarrow \alpha(\alpha+1) - 1(1-1) + 1(-1-\alpha) = 0$.
$\alpha^2 + \alpha - 1 - \alpha = 0 \Rightarrow \alpha^2 = 1 \Rightarrow \alpha = \pm 1$.
Both values satisfy $\alpha \neq -3$. The sum of absolute values is $|1| + |-1| = 1 + 1 = 2$.

(B) The given system of equations is:
$x+y+z=\alpha$
$\alpha x+2 \alpha y+3 z=-1$
$x+3 \alpha y+5 z=4$
$A$ system of linear equations is inconsistent if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_1, D_2, D_3)$ is non-zero.
First,calculate the determinant $D$:
$D = \begin{vmatrix} 1 & 1 & 1 \\ \alpha & 2\alpha & 3 \\ 1 & 3\alpha & 5 \end{vmatrix}$
$D = 1(10\alpha - 9\alpha) - 1(5\alpha - 3) + 1(3\alpha^2 - 2\alpha)$
$D = \alpha - 5\alpha + 3 + 3\alpha^2 - 2\alpha = 3\alpha^2 - 6\alpha + 3 = 3(\alpha - 1)^2$
Setting $D = 0$ gives $3(\alpha - 1)^2 = 0$,which implies $\alpha = 1$.
Now,check for $\alpha = 1$ in the determinant $D_1$:
$D_1 = \begin{vmatrix} 1 & 1 & 1 \\ -1 & 2 & 3 \\ 4 & 3 & 5 \end{vmatrix}$
$D_1 = 1(10 - 9) - 1(-5 - 12) + 1(-3 - 8)$
$D_1 = 1(1) - 1(-17) + 1(-11) = 1 + 17 - 11 = 7$
Since $D = 0$ and $D_1 \neq 0$ for $\alpha = 1$,the system is inconsistent.
Thus,there is only $1$ value of $\alpha$ for which the system is inconsistent.

(D) The system is consistent if the determinant $\Delta \neq 0$ or if $\Delta = 0$ and the system has infinitely many solutions.
First,calculate the determinant $\Delta$ of the coefficient matrix:
$\Delta = \begin{vmatrix} -k & 3 & -14 \\ -15 & 4 & -k \\ -4 & 1 & 3 \end{vmatrix} = -k(12 + k) - 3(-45 - 4k) - 14(-15 + 16) = -12k - k^2 + 135 + 12k - 14 = 121 - k^2$.
For a unique solution,$\Delta \neq 0$,which implies $121 - k^2 \neq 0$,so $k \neq \pm 11$.
If $k = 11$,$\Delta = 0$. We check for consistency using $\Delta_x, \Delta_y, \Delta_z$:
$\Delta_z = \begin{vmatrix} -11 & 3 & 25 \\ -15 & 4 & 3 \\ -4 & 1 & 4 \end{vmatrix} = -11(16 - 3) - 3(-60 + 12) + 25(-15 + 16) = -11(13) - 3(-48) + 25(1) = -143 + 144 + 25 = 26 \neq 0$.
Since $\Delta = 0$ and $\Delta_z \neq 0$,the system is inconsistent for $k = 11$.
If $k = -11$,$\Delta = 0$. We check $\Delta_z$:
$\Delta_z = \begin{vmatrix} 11 & 3 & 25 \\ -15 & 4 & 3 \\ -4 & 1 & 4 \end{vmatrix} = 11(16 - 3) - 3(-60 + 12) + 25(-15 + 16) = 11(13) - 3(-48) + 25(1) = 143 + 144 + 25 = 312 \neq 0$.
Since $\Delta = 0$ and $\Delta_z \neq 0$,the system is inconsistent for $k = -11$.
Thus,the system is consistent for all $k \in R - \{11, -11\}$.

(B) Let $A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}$.
From $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$,we get $c_1 = 1, c_2 = 1, c_3 = 2$.
From $A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} a_1 + c_1 \\ a_2 + c_2 \\ a_3 + c_3 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,we get $a_1 = -2, a_2 = -1, a_3 = -1$.
From $A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$,we get $b_1 = 3, b_2 = 2, b_3 = 1$.
Thus,$A = \begin{bmatrix} -2 & 3 & 1 \\ -1 & 2 & 1 \\ -1 & 1 & 2 \end{bmatrix}$.
Then $A - 2I = \begin{bmatrix} -4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix}$.
The determinant $|A - 2I| = -4(0 - 1) - 3(0 - (-1)) + 1(-1 - 0) = 4 - 3 - 1 = 0$.
The system is $\begin{bmatrix} -4 & 3 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 4 \\ 1 \\ 1 \end{bmatrix}$.
This gives the equations:
$1) -4x_1 + 3x_2 + x_3 = 4$
$2) -x_1 + x_3 = 1 \Rightarrow x_3 = 1 + x_1$
$3) -x_1 + x_2 = 1 \Rightarrow x_2 = 1 + x_1$
Substituting $(2)$ and $(3)$ into $(1)$: $-4x_1 + 3(1 + x_1) + (1 + x_1) = -4x_1 + 3 + 3x_1 + 1 + x_1 = 4$. This simplifies to $4 = 4$,which is always true.
Since the system is consistent and the determinant is $0$,there are infinitely many solutions.

(C) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and the determinants $\Delta_x, \Delta_y, \Delta_z$ must also be $0$.
First,calculate $\Delta$:
$\Delta = \begin{vmatrix} \alpha & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 5 \end{vmatrix} = 0$
$\alpha(10 - 9) - 1(5 - 3) + 1(3 - 2) = 0$
$\alpha(1) - 2 + 1 = 0 \Rightarrow \alpha - 1 = 0 \Rightarrow \alpha = 1$.
Now,substitute $\alpha = 1$ into the system and calculate $\Delta_x$ (or use the augmented matrix approach). For infinitely many solutions,the augmented matrix $[A|B]$ must have a rank less than $3$.
Using the augmented matrix:
$\begin{bmatrix} 1 & 1 & 1 & | & 5 \\ 1 & 2 & 3 & | & 4 \\ 1 & 3 & 5 & | & \beta \end{bmatrix}$
$R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:
$\begin{bmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 1 & 2 & | & -1 \\ 0 & 2 & 4 & | & \beta - 5 \end{bmatrix}$
$R_3 \to R_3 - 2R_2$:
$\begin{bmatrix} 1 & 1 & 1 & | & 5 \\ 0 & 1 & 2 & | & -1 \\ 0 & 0 & 0 & | & \beta - 5 - 2(-1) \end{bmatrix}$
For infinitely many solutions,the last row must be all zeros,so $\beta - 5 + 2 = 0 \Rightarrow \beta - 3 = 0 \Rightarrow \beta = 3$.
Thus,the ordered pair is $(\alpha, \beta) = (1, 3)$.

(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be $0$,and at least one of the Cramer's rule determinants $(\Delta_x, \Delta_y, \Delta_z)$ must be non-zero.
First,calculate the determinant of the coefficient matrix $\Delta$:
$\Delta = \begin{vmatrix} 3 & -2 & 1 \\ 5 & -8 & 9 \\ 2 & 1 & a \end{vmatrix} = 3(-8a - 9) + 2(5a - 18) + 1(5 - (-16))$
$\Delta = -24a - 27 + 10a - 36 + 21 = -14a - 42$
Setting $\Delta = 0$,we get $-14a = 42$,so $a = -3$.
Now,substitute $a = -3$ into the system and check for consistency using the augmented matrix or Cramer's rule. For no solution,the system must be inconsistent.
Consider the determinant $\Delta_z = \begin{vmatrix} 3 & -2 & b \\ 5 & -8 & 3 \\ 2 & 1 & -1 \end{vmatrix} = 3(8 - 3) + 2(-5 - 6) + b(5 - (-16))$
$\Delta_z = 3(5) + 2(-11) + b(21) = 15 - 22 + 21b = 21b - 7$.
For no solution,we require $\Delta = 0$ and $\Delta_z \neq 0$ (or other $\Delta_i \neq 0$).
However,checking the consistency of the equations $3x - 2y + z = b$,$5x - 8y + 9z = 3$,and $2x + y - 3z = -1$:
Adding the equations or using row reduction,we find that for the system to be inconsistent,$b$ must be such that the constants do not satisfy the linear dependence of the rows. Solving the system leads to $b = -\frac{1}{3}$.

(C) For a system of linear equations to be inconsistent,the determinant of the coefficient matrix $\Delta$ must be $0$,and at least one of the Cramer's rule determinants $(\Delta_x, \Delta_y, \Delta_z)$ must be non-zero.
The coefficient matrix is $A = \begin{bmatrix} 1 & 2 & 1 \\ \alpha & 3 & -1 \\ -\alpha & 1 & 2 \end{bmatrix}$.
Calculating the determinant $\Delta$:
$\Delta = 1(3 \times 2 - (-1) \times 1) - 2(\alpha \times 2 - (-1) \times (-\alpha)) + 1(\alpha \times 1 - 3 \times (-\alpha))$
$\Delta = 1(6 + 1) - 2(2\alpha - \alpha) + 1(\alpha + 3\alpha)$
$\Delta = 7 - 2\alpha + 4\alpha = 7 + 2\alpha$.
Setting $\Delta = 0$ for inconsistency:
$7 + 2\alpha = 0 \Rightarrow \alpha = -\frac{7}{2}$.
Now,check $\Delta_x$ at $\alpha = -\frac{7}{2}$:
$\Delta_x = \begin{vmatrix} 2 & 2 & 1 \\ \alpha & 3 & -1 \\ -\alpha & 1 & 2 \end{vmatrix} = 2(6 + 1) - 2(2\alpha - \alpha) + 1(\alpha + 3\alpha) = 14 - 2\alpha + 4\alpha = 14 + 2\alpha$.
Substituting $\alpha = -\frac{7}{2}$:
$\Delta_x = 14 + 2(-\frac{7}{2}) = 14 - 7 = 7 \neq 0$.
Since $\Delta = 0$ and $\Delta_x \neq 0$,the system is inconsistent at $\alpha = -\frac{7}{2}$.