Solution of the Linear equations using Matrices Questions in English

(C) For a system of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.
Given equations are $2x - 3y = \gamma + 5$ and $\alpha x + 5y = \beta + 1$.
Applying the condition: $\frac{\alpha}{2} = \frac{5}{-3} = \frac{\beta + 1}{\gamma + 5}$.
From $\frac{\alpha}{2} = \frac{5}{-3}$,we get $\alpha = -\frac{10}{3}$.
Multiplying by $9$,we get $9\alpha = -30$.
From $\frac{5}{-3} = \frac{\beta + 1}{\gamma + 5}$,we get $5(\gamma + 5) = -3(\beta + 1)$.
$5\gamma + 25 = -3\beta - 3$.
$3\beta + 5\gamma = -28$.
Now,we need to find $|9\alpha + 3\beta + 5\gamma|$.
Substituting the values: $|-30 + (-28)| = |-58| = 58$.

(B) For the system of linear equations to have no solution,the determinant of the coefficient matrix must be zero,i.e.,$D = 0$.
$D = \left|\begin{array}{ccc} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & |\lambda| \end{array}\right| = 0$
Expanding along the first row:
$2(1 \cdot |\lambda| - 1(-1)) - 3(1 \cdot |\lambda| - 1(1)) - 1(1(-1) - 1(1)) = 0$
$2(|\lambda| + 1) - 3(|\lambda| - 1) - 1(-2) = 0$
$2|\lambda| + 2 - 3|\lambda| + 3 + 2 = 0$
$-|\lambda| + 7 = 0 \Rightarrow |\lambda| = 7 \Rightarrow \lambda = \pm 7$.
Now,we check the consistency for $\lambda = 7$ and $\lambda = -7$.
For $\lambda = 7$,the third equation is $x - y + 7z = 24$. Adding $(2)$ and $(3)$ gives $3x + 4y = 2$. Solving the system shows it is consistent.
For $\lambda = -7$,the third equation is $x - y + 7z = -32$. Substituting $\lambda = -7$ into the system leads to a contradiction,meaning the system is inconsistent (no solution).
Thus,$\lambda = -7$ is the correct condition.

(B) 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 the consistency condition.
First,we set the determinant of the coefficient matrix to zero:
$\Delta = \begin{vmatrix} 2 & 1 & -1 \\ 1 & -3 & 2 \\ 1 & 4 & \delta \end{vmatrix} = 0$
$2(-3\delta - 8) - 1(\delta - 2) - 1(4 + 3) = 0$
$-6\delta - 16 - \delta + 2 - 7 = 0$
$-7\delta - 21 = 0 \Rightarrow \delta = -3$
Now,for the system to have infinitely many solutions,the augmented matrix must also have a determinant of zero when replacing a column with the constants:
$\Delta_x = \begin{vmatrix} 7 & 1 & -1 \\ 1 & -3 & 2 \\ k & 4 & -3 \end{vmatrix} = 0$
$7(9 - 8) - 1(-3 - 2k) - 1(4 + 3k) = 0$
$7(1) + 3 + 2k - 4 - 3k = 0$
$6 - k = 0 \Rightarrow k = 6$
Thus,$\delta + k = -3 + 6 = 3$.

(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be $0$,and the system must be inconsistent.
The coefficient matrix is $A = \begin{bmatrix} 2 & -3 & 5 \\ 1 & 3 & -1 \\ 3 & -1 & \lambda^2 - |\lambda| \end{bmatrix}$.
Calculating the determinant $\Delta$:
$\Delta = 2(3(\lambda^2 - |\lambda|) - 1) + 3(1(\lambda^2 - |\lambda|) - (-3)) + 5(-1 - 9)$
$= 2(3\lambda^2 - 3|\lambda| - 1) + 3(\lambda^2 - |\lambda| + 3) + 5(-10)$
$= 6\lambda^2 - 6|\lambda| - 2 + 3\lambda^2 - 3|\lambda| + 9 - 50$
$= 9\lambda^2 - 9|\lambda| - 43$.
Setting $\Delta = 0$ gives $9|\lambda|^2 - 9|\lambda| - 43 = 0$.
Let $t = |\lambda|$. Then $9t^2 - 9t - 43 = 0$.
The discriminant $D = (-9)^2 - 4(9)(-43) = 81 + 1548 = 1629 > 0$.
Since $D > 0$,there are two distinct real roots for $t$. The roots are $t = \frac{9 \pm \sqrt{1629}}{18}$.
Since $\sqrt{1629} \approx 40.36$,one root is positive and one is negative.
Since $t = |\lambda| \ge 0$,only the positive root is valid.
Thus,there is only $1$ value for $|\lambda|$,which corresponds to $2$ values of $\lambda$ (i.e.,$\lambda = \pm t$).

(B) For the system of linear equations to have no solution,the determinant $D$ of the coefficient matrix must be zero,and the system must be inconsistent.
First,calculate the determinant $D$:
$D = \begin{vmatrix} 3\sin 3\theta & -1 & 1 \\ 3\cos 2\theta & 4 & 3 \\ 6 & 7 & 7 \end{vmatrix} = 0$
$D = 3\sin 3\theta(28 - 21) + 1(21\cos 2\theta - 18) + 1(21\cos 2\theta - 24) = 0$
$D = 3\sin 3\theta(7) + 21\cos 2\theta - 18 + 21\cos 2\theta - 24 = 0$
$21\sin 3\theta + 42\cos 2\theta - 42 = 0$
Dividing by $21$,we get $\sin 3\theta + 2\cos 2\theta - 2 = 0$.
Using $\cos 2\theta = 1 - 2\sin^2 \theta$ and $\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$:
$(3\sin \theta - 4\sin^3 \theta) + 2(1 - 2\sin^2 \theta) - 2 = 0$
$3\sin \theta - 4\sin^3 \theta + 2 - 4\sin^2 \theta - 2 = 0$
$-4\sin^3 \theta - 4\sin^2 \theta + 3\sin \theta = 0$
$-\sin \theta (4\sin^2 \theta + 4\sin \theta - 3) = 0$
$-\sin \theta (2\sin \theta - 1)(2\sin \theta + 3) = 0$
This gives $\sin \theta = 0$,$\sin \theta = 1/2$,or $\sin \theta = -3/2$ (impossible).
For $\sin \theta = 0$ in $(0, 4\pi)$,$\theta = \pi, 2\pi, 3\pi$.
For $\sin \theta = 1/2$ in $(0, 4\pi)$,$\theta = \pi/6, 5\pi/6, 13\pi/6, 17\pi/6$.
Checking consistency,all these values lead to no solution. Thus,there are $3 + 4 = 7$ values.

(D) For the system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$.
$D = \begin{vmatrix} 8 & 1 & 4 \\ 1 & 1 & 1 \\ \lambda & -3 & 0 \end{vmatrix} = 8(0 - (-3)) - 1(0 - \lambda) + 4(-3 - \lambda) = 0$
$24 + \lambda - 12 - 4\lambda = 0 \Rightarrow 12 - 3\lambda = 0 \Rightarrow \lambda = 4$.
Now,for infinitely many solutions,the augmented matrix must be consistent with rank less than $3$. Using $D_1 = 0$:
$D_1 = \begin{vmatrix} -2 & 1 & 4 \\ 0 & 1 & 1 \\ \mu & -3 & 0 \end{vmatrix} = -2(0 - (-3)) - 1(0 - \mu) + 4(0 - \mu) = 0$
$-6 + \mu - 4\mu = 0 \Rightarrow -3\mu = 6 \Rightarrow \mu = -2$.
The point is $\left(4, -2, -\frac{1}{2}\right)$.
The distance of the point $(x_1, y_1, z_1)$ from the plane $Ax + By + Cz + D = 0$ is given by $\frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$.
Distance $= \frac{|8(4) + 1(-2) + 4(-\frac{1}{2}) + 2|}{\sqrt{8^2 + 1^2 + 4^2}} = \frac{|32 - 2 - 2 + 2|}{\sqrt{64 + 1 + 16}} = \frac{30}{\sqrt{81}} = \frac{30}{9} = \frac{10}{3}$.

(B) Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Given $\det(A) = ad - bc = -1$.
We know that $\operatorname{Adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Then $A + I = \begin{bmatrix} a+1 & b \\ c & d+1 \end{bmatrix}$ and $\operatorname{Adj}(A) + I = \begin{bmatrix} d+1 & -b \\ -c & a+1 \end{bmatrix}$.
Calculating the product $(A+I)(\operatorname{Adj}(A)+I)$:
$(A+I)(\operatorname{Adj}(A)+I) = \begin{bmatrix} a+1 & b \\ c & d+1 \end{bmatrix} \begin{bmatrix} d+1 & -b \\ -c & a+1 \end{bmatrix} = \begin{bmatrix} (a+1)(d+1)-bc & -(a+1)b + b(a+1) \\ c(d+1) - c(d+1) & -bc + (d+1)(a+1) \end{bmatrix}$.
This simplifies to $\begin{bmatrix} ad+a+d+1-bc & 0 \\ 0 & ad+a+d+1-bc \end{bmatrix}$.
Since $ad-bc = -1$,the diagonal elements are $a+d+1-1 = a+d$.
Thus,the matrix is $\begin{bmatrix} a+d & 0 \\ 0 & a+d \end{bmatrix}$.
The determinant is $(a+d)^2 = 4$.
Therefore,$a+d = 2$ or $a+d = -2$.
Comparing with the options,the sum of the diagonal elements (trace of $A$) can be $2$.

(C) The given system of equations is:
$x+y+z=6$ $(1)$
$2x+5y+\alpha z=\beta$ $(2)$
$x+2y+3z=14$ $(3)$
For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must satisfy the consistency condition.
Let $D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 0$.
$1(15-2\alpha) - 1(6-\alpha) + 1(4-5) = 0$
$15-2\alpha - 6 + \alpha - 1 = 0$
$8 - \alpha = 0 \Rightarrow \alpha = 8$.
Now,substitute $\alpha = 8$ into the system and use the condition for infinite solutions. From $(1)$ and $(3)$:
$x+y = 6-z$
$x+2y = 14-3z$
Subtracting the first from the second: $y = (14-3z) - (6-z) = 8-2z$.
Substituting $y$ into $x+y = 6-z$: $x = 6-z - (8-2z) = z-2$.
Substitute $x, y, z$ into $(2)$:
$2(z-2) + 5(8-2z) + 8z = \beta$
$2z - 4 + 40 - 10z + 8z = \beta$
$36 = \beta$.
Thus,$\alpha + \beta = 8 + 36 = 44$.

(B) Given $AB = 0$,where $A$ and $B$ are $3 \times 3$ non-zero matrices.
Taking the determinant on both sides,$|AB| = |0| = 0$.
Since $|AB| = |A||B| = 0$,it implies that at least one of $|A|$ or $|B|$ must be $0$.
If $|A| \neq 0$,then $A$ is invertible,so $A^{-1}(AB) = A^{-1}(0) \Rightarrow B = 0$,which contradicts the given condition that $B$ is a non-zero matrix.
If $|B| \neq 0$,then $B$ is invertible,so $(AB)B^{-1} = 0(B^{-1}) \Rightarrow A = 0$,which contradicts the given condition that $A$ is a non-zero matrix.
Therefore,$|A| = 0$ and $|B| = 0$.
Since $|A| = 0$,the matrix $A$ is singular,which means the system of linear equations $AX = 0$ has infinitely many solutions.

(C) The given system of linear equations is:
$ax + y = 0$ $(i)$
$x + (a + 10)y = 0$ $(ii)$
For a system of homogeneous linear equations to have at least two distinct solutions (i.e.,non-trivial solutions),the determinant of the coefficient matrix must be zero.
The coefficient matrix is $A = \begin{bmatrix} a & 1 \\ 1 & a + 10 \end{bmatrix}$.
Setting the determinant to zero:
$|A| = a(a + 10) - (1)(1) = 0$
$a^2 + 10a - 1 = 0$
This is a quadratic equation in $a$. The roots are given by the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$a = \frac{-10 \pm \sqrt{10^2 - 4(1)(-1)}}{2(1)}$
$a = \frac{-10 \pm \sqrt{100 + 4}}{2}$
$a = \frac{-10 \pm \sqrt{104}}{2}$
Since the discriminant $D = 104 > 0$,there are $2$ distinct real values of $a$ for which the system has non-trivial (infinitely many) solutions.
Therefore,the correct option is $C$.

(C) The given system of equations is:
$x+2y+3z=3$ ... $(i)$
$4x+3y-4z=4$ ... (ii)
$8x+4y-\lambda z=9+\mu$ ... (iii)
For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must have a rank less than $3$.
First,calculate the determinant of the coefficient matrix $D$:
$D = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 3 & -4 \\ 8 & 4 & -\lambda \end{vmatrix} = 1(-3\lambda + 16) - 2(-4\lambda + 32) + 3(16 - 24) = -3\lambda + 16 + 8\lambda - 64 - 24 = 5\lambda - 72$.
For infinite solutions,$D = 0 \Rightarrow 5\lambda - 72 = 0 \Rightarrow \lambda = \frac{72}{5}$.
Now,consider the augmented matrix $[A|B]$:
$\begin{bmatrix} 1 & 2 & 3 & | & 3 \\ 4 & 3 & -4 & | & 4 \\ 8 & 4 & -\frac{72}{5} & | & 9+\mu \end{bmatrix}$.
Perform row operations: $R_2 \to R_2 - 4R_1$ and $R_3 \to R_3 - 8R_1$:
$\begin{bmatrix} 1 & 2 & 3 & | & 3 \\ 0 & -5 & -16 & | & -8 \\ 0 & -12 & -\frac{72}{5}-24 & | & 9+\mu-24 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 & | & 3 \\ 0 & -5 & -16 & | & -8 \\ 0 & -12 & -\frac{192}{5} & | & \mu-15 \end{bmatrix}$.
For infinite solutions,the third row must be a multiple of the second row. The ratio of coefficients is $\frac{-12}{-5} = 2.4$.
Thus,$\mu - 15 = 2.4 \times (-8) = -19.2 \Rightarrow \mu = 15 - 19.2 = -4.2 = -\frac{21}{5}$.
Therefore,$(\lambda, \mu) = \left(\frac{72}{5}, -\frac{21}{5}\right)$.

(D) The system of equations has a unique solution if the determinant $\Delta \neq 0$.
$\Delta = \begin{vmatrix} a & 2a & -3a \\ 2a+1 & 2a+3 & a+1 \\ 3a+5 & a+5 & a+2 \end{vmatrix}$
Taking $a$ common from the first column:
$\Delta = a \begin{vmatrix} 1 & 2a & -3a \\ 2a+1 & 2a+3 & a+1 \\ 3a+5 & a+5 & a+2 \end{vmatrix}$
Performing row operations $R_2 \to R_2 - (2a+1)R_1$ and $R_3 \to R_3 - (3a+5)R_1$:
$\Delta = a \begin{vmatrix} 1 & 2a & -3a \\ 0 & 2a+3 - 2a(2a+1) & a+1 + 3a(2a+1) \\ 0 & a+5 - 2a(3a+5) & a+2 + 3a(3a+5) \end{vmatrix}$
Simplifying the entries:
$\Delta = a \begin{vmatrix} 1 & 2a & -3a \\ 0 & -4a^2+3 & 6a^2+4a+1 \\ 0 & -6a^2-9a+5 & 9a^2+16a+2 \end{vmatrix}$
Calculating the determinant of the $2 \times 2$ matrix:
$(-4a^2+3)(9a^2+16a+2) - (6a^2+4a+1)(-6a^2-9a+5) = 0$
After expansion,we find $\Delta = 0$ only when $a = 0$. Since $a \in R - \{0\}$,$\Delta$ is never $0$ for any $a$ in the set.
Thus,the system always has a unique solution for all $a \in R - \{0\}$.
Therefore,$S_1 = R - \{0\}$ and $S_2 = \Phi$.

(B) The determinant of the coefficient matrix is $D = \begin{vmatrix} \alpha & 2 & 1 \\ 2\alpha & 3 & 1 \\ 3 & \alpha & 2 \end{vmatrix} = \alpha(6 - \alpha) - 2(4\alpha - 3) + 1(2\alpha^2 - 9) = 6\alpha - \alpha^2 - 8\alpha + 6 + 2\alpha^2 - 9 = \alpha^2 - 2\alpha - 3 = (\alpha - 3)(\alpha + 1)$.
For $D = 0$,we have $\alpha = 3$ or $\alpha = -1$.
If $\alpha = -1$,the system becomes $-x + 2y + z = 1$,$-2x + 3y + z = 1$,$3x - y + 2z = \beta$. Subtracting the first from the second gives $-x + y = 0$,so $x = y$. Substituting into the first gives $x + z = 1$,so $z = 1 - x$. Substituting into the third: $3x - x + 2(1 - x) = \beta \Rightarrow 2 = \beta$. Thus,if $\alpha = -1$ and $\beta \neq 2$,there is no solution. If $\alpha = -1$ and $\beta = 2$,there are infinitely many solutions.
If $\alpha = 3$,the system becomes $3x + 2y + z = 1$,$6x + 3y + z = 1$,$3x + 3y + 2z = \beta$. Subtracting the first from the second gives $3x + y = 0$,so $y = -3x$. Substituting into the first: $3x - 6x + z = 1 \Rightarrow z = 3x + 1$. Substituting into the third: $3x + 3(-3x) + 2(3x + 1) = \beta \Rightarrow 3x - 9x + 6x + 2 = \beta \Rightarrow 2 = \beta$. Thus,if $\alpha = 3$ and $\beta \neq 2$,there is no solution.
Option $B$ states it has no solution for $\alpha = -1$ and for all $\beta \in \mathbb{R}$,which is incorrect because it has infinitely many solutions when $\beta = 2$.

(D) For the system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero:
$\Delta = \left|\begin{array}{ccc}1 & 1 & k \\ 2 & 3 & -1 \\ 3 & 4 & 2\end{array}\right| = 0$
Expanding along the first row:
$1(3 \times 2 - (-1) \times 4) - 1(2 \times 2 - (-1) \times 3) + k(2 \times 4 - 3 \times 3) = 0$
$1(6 + 4) - 1(4 + 3) + k(8 - 9) = 0$
$10 - 7 - k = 0$
$3 - k = 0 \Rightarrow k = 3$
Now,substitute $k = 3$ into the second system of equations:
$(3+1)x + (2 \times 3 - 1)y = 7 \Rightarrow 4x + 5y = 7 \dots (1)$
$(2 \times 3 + 1)x + (3+5)y = 10 \Rightarrow 7x + 8y = 10 \dots (2)$
To check for the nature of the solution,find the determinant of the coefficient matrix of this system:
$D = \left|\begin{array}{cc}4 & 5 \\ 7 & 8\end{array}\right| = 32 - 35 = -3 \neq 0$
Since $D \neq 0$,the system has a unique solution.
Solving the equations:
$(2) - (1) \Rightarrow (7x + 8y) - (4x + 5y) = 10 - 7$
$3x + 3y = 3 \Rightarrow x + y = 1$
Thus,the system has a unique solution satisfying $x+y=1$.

(C) For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{vmatrix} = 0$
$1(8 + \alpha) - (-1)(8 - 3\alpha) + 1(-2 - 6) = 0$
$8 + \alpha + 8 - 3\alpha - 8 = 0$
$8 - 2\alpha = 0 \implies \alpha = 4$.
Now,substitute $\alpha = 4$ into the system:
$x - y + z = 5$
$2x + 2y + 4z = 8 \implies x + y + 2z = 4$
$3x - y + 4z = \beta$
Adding the first two equations: $(x - y + z) + (x + y + 2z) = 5 + 4 \implies 2x + 3z = 9$.
For infinitely many solutions,the third equation must be a linear combination of the first two. Let $k_1(x - y + z) + k_2(x + y + 2z) = 3x - y + 4z$.
Comparing coefficients: $k_1 + k_2 = 3$,$-k_1 + k_2 = -1$,$k_1 + 2k_2 = 4$.
Solving $k_1 + k_2 = 3$ and $-k_1 + k_2 = -1$ gives $2k_2 = 2 \implies k_2 = 1$ and $k_1 = 2$.
Thus,$\beta = 2(5) + 1(4) = 14$.
The roots are $\alpha = 4$ and $\beta = 14$.
The quadratic equation is $(x - 4)(x - 14) = x^2 - 18x + 56 = 0$.

(D) The given system of equations is:
$x+y+z=6$ $(1)$
$\alpha x+\beta y+7z=3$ $(2)$
$x+2y+3z=14$ $(3)$
Subtracting $(1)$ from $(3)$,we get $y+2z=8$,so $y=8-2z$. Substituting this into $(1)$,$x+(8-2z)+z=6 \Rightarrow x=z-2$.
Substituting $x=z-2$ and $y=8-2z$ into $(2)$:
$\alpha(z-2)+\beta(8-2z)+7z=3$
$(\alpha-2\beta+7)z = 2\alpha-8\beta+3$.
For a unique solution,the coefficient of $z$ must be non-zero: $\alpha-2\beta+7 \neq 0$.
For infinitely many solutions,both sides must be zero: $\alpha-2\beta+7=0$ and $2\alpha-8\beta+3=0$.
Solving these: $2\alpha-4\beta+14=0$ and $2\alpha-8\beta+3=0$. Subtracting gives $4\beta+11=0 \Rightarrow \beta=-11/4$,then $\alpha=-25/2$. This is a unique point,not on the line $x-2y+7=0$.
Thus,option $D$ is $NOT$ true.

(D) Given the matrix equation $(\alpha \ \beta \ \gamma)\begin{bmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{bmatrix} = (0 \ 0 \ 0)$,we get the following system of linear equations:
$2\alpha + 9\beta + 8\gamma = 0 \quad (1)$
$10\alpha + 3\beta + 4\gamma = 0 \quad (2)$
$8\alpha + 8\beta + 8\gamma = 0 \quad (3)$
From $(3)$,we have $\alpha + \beta + \gamma = 0$,which implies $\gamma = -\alpha - \beta$.
Substitute $\gamma$ into $(1)$:
$2\alpha + 9\beta + 8(-\alpha - \beta) = 0 \implies -6\alpha + \beta = 0 \implies \beta = 6\alpha$.
Now,find $\gamma$ in terms of $\alpha$:
$\gamma = -\alpha - 6\alpha = -7\alpha$.
The point $P(\alpha, 6\alpha, -7\alpha)$ lies on the plane $2x + 4y + 3z = 5$:
$2(\alpha) + 4(6\alpha) + 3(-7\alpha) = 5$
$2\alpha + 24\alpha - 21\alpha = 5$
$5\alpha = 5 \implies \alpha = 1$.
Thus,$\alpha = 1, \beta = 6, \gamma = -7$.
We need to calculate $6\alpha + 9\beta + 7\gamma$:
$6(1) + 9(6) + 7(-7) = 6 + 54 - 49 = 11$.

(D) The system of equations is inconsistent if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.
The coefficient matrix is $A = \begin{bmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{bmatrix}$.
Calculating the determinant $D = \lambda(\lambda^2 - 1) - 1(\lambda - 1) + 1(1 - \lambda) = \lambda(\lambda - 1)(\lambda + 1) - (\lambda - 1) - (\lambda - 1) = (\lambda - 1)(\lambda^2 + \lambda - 2) = (\lambda - 1)(\lambda + 2)(\lambda - 1) = (\lambda - 1)^2(\lambda + 2)$.
Setting $D = 0$,we get $\lambda = 1$ or $\lambda = -2$.
If $\lambda = 1$,the system becomes $x + y + z = 1$,which represents a plane,and all three equations are identical,leading to infinitely many solutions (consistent).
If $\lambda = -2$,the system becomes $-2x + y + z = 1$,$x - 2y + z = 1$,and $x + y - 2z = 1$. Adding these three equations gives $0 = 3$,which is a contradiction,so the system is inconsistent.
Thus,$S = \{-2\}$.
The sum $\sum_{\lambda \in S} (|\lambda|^2 + |\lambda|) = (|-2|^2 + |-2|) = 4 + 2 = 6$.

(A) The determinant of the coefficient matrix is $D = \begin{vmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \end{vmatrix} = a(a^2-1) - 1(a-1) + 1(1-a) = a^3 - 3a + 2 = (a-1)^2(a+2)$.
For $a=1$,the equations become $x+y+z=1$,$x+y+z=1$,$x+y+z=\beta$. If $\beta=1$,there are infinitely many solutions. Thus,option $D$ is correct.
For $a=-2$,$D=0$. The augmented matrix for $\beta=1$ is $\begin{bmatrix} -2 & 1 & 1 & 1 \\ 1 & -2 & 1 & 1 \\ 1 & 1 & -2 & 1 \end{bmatrix}$. Adding the rows gives $0=3$,which is impossible. Thus,there is no solution. Option $B$ is correct.
For $a=2$ and $\beta=1$,$D = (2-1)^2(2+2) = 4 \neq 0$. The system has a unique solution. Using Cramer's rule,$x=y=z = \frac{1}{4}$. Thus $x+y+z = \frac{3}{4}$. Option $C$ is correct.
For $a=2$ and $\beta=-1$,$D=4 \neq 0$,so the system has a unique solution,not infinitely many. Thus,option $A$ is incorrect.

(A) 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 $\Delta$:
$\Delta = \begin{vmatrix} 1 & 1 & a \\ 2 & 5 & 2 \\ 1 & 2 & 3 \end{vmatrix} = 1(15-4) - 1(6-2) + a(4-5) = 11 - 4 - a = 7 - a$.
Setting $\Delta = 0$,we get $7 - a = 0$,so $a = 7$.
Next,for infinitely many solutions,$\Delta_x = 0$ (or $\Delta_y = 0$ or $\Delta_z = 0$):
$\Delta_x = \begin{vmatrix} b & 1 & 7 \\ 6 & 5 & 2 \\ 3 & 2 & 3 \end{vmatrix} = b(15-4) - 1(18-6) + 7(12-15) = 11b - 12 - 21 = 11b - 33$.
Setting $\Delta_x = 0$,we get $11b = 33$,so $b = 3$.
Finally,calculate $2a + 3b$:
$2a + 3b = 2(7) + 3(3) = 14 + 9 = 23$.

(A) The system of equations is given by:
$x+y+z=6$
$x+2y+\alpha z=10$
$x+3y+5z=\beta$
The determinant of the coefficient matrix $D$ is:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{vmatrix} = 1(10-3\alpha) - 1(5-\alpha) + 1(3-2) = 10-3\alpha-5+\alpha+1 = 6-2\alpha$.
For a unique solution,$D \neq 0$,which implies $6-2\alpha \neq 0$,so $\alpha \neq 3$.
If $\alpha=3$,then $D=0$. The system becomes:
$x+y+z=6$
$x+2y+3z=10$
$x+3y+5z=\beta$
Subtracting the first equation from the second: $y+2z=4$.
Subtracting the second equation from the third: $y+2z=\beta-10$.
For the system to have solutions,we must have $4 = \beta-10$,which means $\beta=14$. If $\beta=14$,there are infinitely many solutions. If $\beta \neq 14$,there is no solution.
Option $A$ states the system has a unique solution for $\alpha=3$,which is false because $D=0$ when $\alpha=3$.

(C) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 2 & -1 & 3 \\ 3 & 2 & -1 \\ 4 & 5 & \alpha \end{vmatrix} = 2(2\alpha + 5) + 1(3\alpha + 4) + 3(15 - 8) = 4\alpha + 10 + 3\alpha + 4 + 21 = 7\alpha + 35 = 7(\alpha + 5)$.
For a unique solution,$\Delta \neq 0$,which means $\alpha \neq -5$. Thus,the system has a unique solution for any $\beta$ when $\alpha \neq -5$.
For infinitely many solutions,we require $\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$.
Setting $\Delta = 0$ gives $\alpha = -5$.
Calculating $\Delta_3 = \begin{vmatrix} 2 & -1 & 5 \\ 3 & 2 & 7 \\ 4 & 5 & \beta \end{vmatrix} = 2(2\beta - 35) + 1(3\beta - 28) + 5(15 - 8) = 4\beta - 70 + 3\beta - 28 + 35 = 7\beta - 63 = 7(\beta - 9)$.
Setting $\Delta_3 = 0$ gives $\beta = 9$.
When $\alpha = -5$ and $\beta = 9$,$\Delta = \Delta_1 = \Delta_2 = \Delta_3 = 0$,so the system has infinitely many solutions.
Option $C$ states the system has infinitely many solutions for $\alpha = -6$ and $\beta = 9$,which is incorrect because $\Delta \neq 0$ when $\alpha = -6$.

(C) For the 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 determinants $\Delta_x, \Delta_y, \Delta_z$ must be non-zero.
First,calculate $\Delta = \begin{vmatrix} 6 \lambda & -3 & 3 \\ 2 & 6 \lambda & 4 \\ 3 & 2 & 3 \lambda \end{vmatrix} = 0$.
Expanding along the first row:
$\Delta = 6 \lambda (18 \lambda^2 - 8) + 3 (6 \lambda - 12) + 3 (4 - 18 \lambda) = 0$
$108 \lambda^3 - 48 \lambda + 18 \lambda - 36 + 12 - 54 \lambda = 0$
$108 \lambda^3 - 84 \lambda - 24 = 0$
Dividing by $12$: $9 \lambda^3 - 7 \lambda - 2 = 0$.
By inspection,$\lambda = 1$ is a root. Using synthetic division,$( \lambda - 1 )( 9 \lambda^2 + 9 \lambda + 2 ) = 0$.
$( \lambda - 1 )( 3 \lambda + 1 )( 3 \lambda + 2 ) = 0$.
So,$\lambda \in \{ 1, -1/3, -2/3 \}$.
For these values,we check $\Delta_1 = \begin{vmatrix} 4 \lambda^2 & -3 & 3 \\ 1 & 6 \lambda & 4 \\ \lambda & 2 & 3 \lambda \end{vmatrix} \neq 0$.
For $\lambda = 1$,$\Delta_1 = \begin{vmatrix} 4 & -3 & 3 \\ 1 & 6 & 4 \\ 1 & 2 & 3 \end{vmatrix} = 4(18-8) + 3(3-4) + 3(2-6) = 40 - 3 - 12 = 25 \neq 0$.
For $\lambda = -1/3$,$\Delta_1 \neq 0$. For $\lambda = -2/3$,$\Delta_1 \neq 0$.
Thus,$S = \{ 1, -1/3, -2/3 \}$.
$12 \sum_{\lambda \in S} |\lambda| = 12 ( |1| + |-1/3| + |-2/3| ) = 12 ( 1 + 1/3 + 2/3 ) = 12 ( 2 ) = 24$.

(A) Given the system of equations:
$(i) 7x + 11y + \alpha z = 13$
$(ii) 5x + 4y + 7z = \beta$
$(iii) 175x + 194y + 57z = 361$
For the system to have infinitely many solutions,the third equation must be a linear combination of the first two equations. Let $(iii) = k_1(i) + k_2(ii)$.
Comparing the coefficients of $x$ and $y$:
$7k_1 + 5k_2 = 175$
$11k_1 + 4k_2 = 194$
Solving these equations: Multiply the first by $4$ and the second by $5$:
$28k_1 + 20k_2 = 700$
$55k_1 + 20k_2 = 970$
Subtracting gives $27k_1 = 270$,so $k_1 = 10$.
Substituting $k_1 = 10$ into $7(10) + 5k_2 = 175$,we get $5k_2 = 105$,so $k_2 = 21$.
Now,for $z$ and the constant term:
$10\alpha + 21(7) = 57 \implies 10\alpha + 147 = 57 \implies 10\alpha = -90 \implies \alpha = -9$.
$10(13) + 21\beta = 361 \implies 130 + 21\beta = 361 \implies 21\beta = 231 \implies \beta = 11$.
Therefore,$\alpha + \beta + 2 = -9 + 11 + 2 = 4$.

(A) The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 2 & 4 & 2a \\ 1 & 2 & 3 \\ 2 & -5 & 2 \end{vmatrix}$.
Expanding along the first row: $\Delta = 2(4 + 15) - 4(2 - 6) + 2a(-5 - 4) = 2(19) - 4(-4) + 2a(-9) = 38 + 16 - 18a = 54 - 18a = 18(3 - a)$.
For a unique solution,we require $\Delta \neq 0$,which implies $18(3 - a) \neq 0$,so $a \neq 3$.
If $a \neq 3$,the system has a unique solution for any value of $b$.
Thus,options $B$ and $C$ are correct because $a=6 \neq 3$ and $a=8 \neq 3$.
For infinitely many solutions,we require $\Delta = 0$ and $\Delta_x = \Delta_y = \Delta_z = 0$.
Setting $\Delta = 0$ gives $a = 3$.
Now,calculate $\Delta_x = \begin{vmatrix} b & 4 & 2a \\ 4 & 2 & 3 \\ 8 & -5 & 2 \end{vmatrix}$.
For $a = 3$,$\Delta_x = \begin{vmatrix} b & 4 & 6 \\ 4 & 2 & 3 \\ 8 & -5 & 2 \end{vmatrix} = b(4 + 15) - 4(8 - 24) + 6(-20 - 16) = 19b - 4(-16) + 6(-36) = 19b + 64 - 216 = 19b - 152$.
For $\Delta_x = 0$,$19b = 152$,which gives $b = 8$.
Therefore,the system has infinitely many solutions if $a = 3$ and $b = 8$.
This makes option $D$ correct.
Consequently,option $A$ is $NOT$ correct.

(A) For the system of equations to have infinitely many solutions,the determinant of the coefficient matrix must be zero:
$\Delta = \begin{vmatrix} 2 & 1 & -1 \\ 2 & -5 & \lambda \\ 1 & 2 & -5 \end{vmatrix} = 0$
Expanding along the first row:
$2(25 - 2\lambda) - 1(-10 - \lambda) - 1(4 + 5) = 0$
$50 - 4\lambda + 10 + \lambda - 9 = 0$
$51 - 3\lambda = 0 \Rightarrow 3\lambda = 51 \Rightarrow \lambda = 17$
For infinitely many solutions,the determinant $\Delta_x$ (or $\Delta_y$ or $\Delta_z$) must also be zero:
$\Delta_z = \begin{vmatrix} 2 & 1 & 5 \\ 2 & -5 & \mu \\ 1 & 2 & 7 \end{vmatrix} = 0$
Expanding along the first row:
$2(-35 - 2\mu) - 1(14 - \mu) + 5(4 + 5) = 0$
$-70 - 4\mu - 14 + \mu + 45 = 0$
$-3\mu - 39 = 0 \Rightarrow 3\mu = -39 \Rightarrow \mu = -13$
Now,calculate $(\lambda + \mu)^2 + (\lambda - \mu)^2 = 2(\lambda^2 + \mu^2)$:
$= 2(17^2 + (-13)^2) = 2(289 + 169) = 2(458) = 916$

(D) Given equations are:
$-x+2y-9z=7$ $(1)$
$-x+3y-7z=9$ $(2)$
$-2x+y+5z=8$ $(3)$
$-3x+y+13z=\lambda$ $(4)$
Subtracting $(1)$ from $(2)$:
$(-x+3y-7z) - (-x+2y-9z) = 9-7$
$y+2z=2$ $(5)$
Subtracting $2 \times (1)$ from $(3)$:
$(-2x+y+5z) - 2(-x+2y-9z) = 8-2(7)$
$-3y+23z=-6$ $(6)$
Solving $(5)$ and $(6)$ by multiplying $(5)$ by $3$ and adding to $(6)$:
$3(y+2z) + (-3y+23z) = 3(2) - 6$
$29z = 0 \Rightarrow z=0$
Substituting $z=0$ in $(5)$:
$y=2$
Substituting $y=2, z=0$ in $(1)$:
$-x+2(2)-9(0)=7 \Rightarrow -x+4=7 \Rightarrow x=-3$
So,$(\alpha, \beta, \gamma) = (-3, 2, 0)$.
Substituting these values into $(4)$ to find $\lambda$:
$-3(-3) + 2 + 13(0) = \lambda \Rightarrow 9+2 = \lambda \Rightarrow \lambda = 11$.
The distance of the point $(-3, 2, 0)$ from the plane $2x-2y+z-11=0$ is:
$d = \frac{|2(-3) - 2(2) + 1(0) - 11|}{\sqrt{2^2 + (-2)^2 + 1^2}} = \frac{|-6 - 4 - 11|}{\sqrt{4+4+1}} = \frac{|-21|}{3} = 7$.

(A) Given $A = \begin{bmatrix} 2 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix}$.
Since $AB = [AB_1, AB_2, AB_3]$,we have $AB = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix}$.
We know that $|AB| = |A| |B|$.
First,calculate $|A| = 2(1-0) - 0(1-0) + 1(0-1) = 2 - 1 = 1$.
Calculate $|AB| = 1(3-0) - 2(0-0) + 3(0-0) = 3$.
Since $|A| |B| = |AB|$,we have $1 \times |B| = 3$,so $\alpha = |B| = 3$.
To find $B$,we use $B = A^{-1} (AB)$.
$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1} \begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{bmatrix}$.
$B = \begin{bmatrix} 1 & 0 & -1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 3 & 2 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 2 \\ -1 & 1 & 0 \\ -1 & -2 & -1 \end{bmatrix}$.
The diagonal elements of $B$ are $1, 1, -1$. Thus,$\beta = 1 + 1 - 1 = 1$.
Finally,$\alpha^3 + \beta^3 = 3^3 + 1^3 = 27 + 1 = 28$.

(B) Since $a, b, c$ are in $A.P.$,we have $2b = a + c$,which implies $a - 2b + c = 0$.
This means the line $ax + by + c = 0$ always passes through the fixed point $P(1, -2)$.
For the system of equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be zero.
$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & \alpha \\ 1 & 2 & 3 \end{vmatrix} = 1(15 - 2\alpha) - 1(6 - \alpha) + 1(4 - 5) = 0$.
$15 - 2\alpha - 6 + \alpha - 1 = 0 \Rightarrow 8 - \alpha = 0 \Rightarrow \alpha = 8$.
Now,for infinite solutions,$D_1 = 0$ where $D_1 = \begin{vmatrix} 6 & 1 & 1 \\ \beta & 5 & 8 \\ 4 & 2 & 3 \end{vmatrix} = 0$.
$6(15 - 16) - 1(3\beta - 32) + 1(2\beta - 20) = 0$.
$-6 - 3\beta + 32 + 2\beta - 20 = 0 \Rightarrow -\beta + 6 = 0 \Rightarrow \beta = 6$.
Thus,$Q = (8, 6)$.
The distance $PQ = \sqrt{(8 - 1)^2 + (6 - (-2))^2} = \sqrt{7^2 + 8^2} = \sqrt{49 + 64} = \sqrt{113}$.
Therefore,$(PQ)^2 = 113$.

(A) The system of equations is:
$x+y+z=4\mu$
$x+2y+2\lambda z=10\mu$
$x+3y+4\lambda^2 z=\mu^2+15$
The determinant of the coefficient matrix $\Delta$ is:
$\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 2\lambda \\ 1 & 3 & 4\lambda^2 \end{vmatrix} = 1(8\lambda^2 - 6\lambda) - 1(4\lambda^2 - 2\lambda) + 1(3 - 2) = 4\lambda^2 - 4\lambda + 1 = (2\lambda - 1)^2$.
For a unique solution,$\Delta \neq 0$,which implies $(2\lambda - 1)^2 \neq 0$,so $\lambda \neq \frac{1}{2}$. Thus,option $A$ and $D$ are consistent with this.
When $\lambda = \frac{1}{2}$,$\Delta = 0$. We check for consistency using Cramer's rule or augmented matrix reduction. The augmented matrix is:
$\begin{bmatrix} 1 & 1 & 1 & | & 4\mu \\ 1 & 2 & 1 & | & 10\mu \\ 1 & 3 & 1 & | & \mu^2+15 \end{bmatrix}$
Subtracting $R_1$ from $R_2$ and $R_3$:
$\begin{bmatrix} 1 & 1 & 1 & | & 4\mu \\ 0 & 1 & 0 & | & 6\mu \\ 0 & 2 & 0 & | & \mu^2+15-4\mu \end{bmatrix}$
From $R_2$,$y = 6\mu$. Substituting into $R_3$: $2(6\mu) = \mu^2 - 4\mu + 15 \implies \mu^2 - 16\mu + 15 = 0 \implies (\mu-1)(\mu-15) = 0$.
So,if $\lambda = \frac{1}{2}$ and $\mu \in \{1, 15\}$,the system is consistent (infinite solutions). If $\mu \neq 1, 15$,the system is inconsistent.
Comparing with the options,option $A$ is the incorrect statement because the condition for a unique solution depends only on $\lambda \neq \frac{1}{2}$,regardless of $\mu$.

(D) The coefficient matrix $A$ is given by $\begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{bmatrix}$.
For the system to have a unique solution,$\det(A) \neq 0$.
$\det(A) = 1(2\lambda - 3\lambda^2) - 1(\lambda - \lambda^2) + 1(3 - 2) = 2\lambda - 3\lambda^2 - \lambda + \lambda^2 + 1 = -2\lambda^2 + \lambda + 1 = -(2\lambda+1)(\lambda-1)$.
Thus,$\det(A) = 0$ when $\lambda = 1$ or $\lambda = -1/2$.
If $\lambda \neq 1$ and $\lambda \neq -1/2$,the system has a unique solution for any $\mu$.
If $\lambda = 1$,the equations become $x+y+z=5$,$x+2y+z=9$,and $x+3y+z=\mu$. Subtracting the first from the second gives $y=4$. Subtracting the second from the third gives $y=\mu-9$. Thus,$4 = \mu-9 \Rightarrow \mu=13$. If $\mu=13$,there are infinite solutions. If $\mu \neq 13$,there is no solution.
Option $D$ states the system has a unique solution if $\lambda \neq 1$ and $\mu \neq 13$. This is incorrect because even if $\lambda \neq 1$,if $\lambda = -1/2$,the system may not have a unique solution regardless of $\mu$.

(D) To find $a$,the sum of all coefficients,we set $x=1$ in the expression:
$a = (1-2(1)+2(1)^2)^{2023} \times (3-4(1)^2+2(1)^3)^{2024} = (1)^{2023} \times (1)^{2024} = 1$.
To find $b$,we use $L'\text{H\^opital's Rule}$ as it is a $0/0$ form:
$b = \lim_{x \rightarrow 0} \frac{\frac{d}{dx} \int_0^x \frac{\ln(1+t)}{t^{2024}+1} dt}{\frac{d}{dx} (x^2)} = \lim_{x \rightarrow 0} \frac{\frac{\ln(1+x)}{x^{2024}+1}}{2x} = \frac{1}{2} \lim_{x \rightarrow 0} \left( \frac{\ln(1+x)}{x} \times \frac{1}{x^{2024}+1} \right) = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.
Substituting $a=1$ and $b=1/2$ into the second equation $2bx^2+ax+4=0$:
$2(1/2)x^2 + (1)x + 4 = 0 \implies x^2+x+4=0$.
Since the equations $cx^2+dx+e=0$ and $x^2+x+4=0$ have a common root and the coefficients are real,the ratios of the coefficients must be equal:
$\frac{c}{1} = \frac{d}{1} = \frac{e}{4}$.
Thus,$d:c:e = 1:1:4$.

(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 $0$.
$D = \begin{vmatrix} 1 & -2 & 1 \\ 2 & \alpha & 3 \\ 3 & -1 & \beta \end{vmatrix} = 1(\alpha\beta + 3) + 2(2\beta - 9) + 1(-2 - 3\alpha) = \alpha\beta - 3\alpha + 4\beta - 17 = 0 \implies \alpha\beta - 3\alpha + 4\beta = 17 \dots (1)$
$D_2 = \begin{vmatrix} 1 & -4 & 1 \\ 2 & 5 & 3 \\ 3 & 3 & \beta \end{vmatrix} = 1(5\beta - 9) + 4(2\beta - 9) + 1(6 - 15) = 5\beta - 9 + 8\beta - 36 - 9 = 13\beta - 54 = 0 \implies \beta = \frac{54}{13}$.
Substituting $\beta = \frac{54}{13}$ into equation $(1)$:
$\alpha(\frac{54}{13}) - 3\alpha + 4(\frac{54}{13}) = 17$
$\frac{54\alpha - 39\alpha + 216}{13} = 17$
$15\alpha + 216 = 221 \implies 15\alpha = 5 \implies \alpha = \frac{1}{3}$.
Now,calculate $12\alpha + 13\beta = 12(\frac{1}{3}) + 13(\frac{54}{13}) = 4 + 54 = 58$.

(A) The given equations show that the vectors $v_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$,$v_2 = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$,and $v_3 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ are eigenvectors of $A$ with corresponding eigenvalues $\lambda_1 = 2$,$\lambda_2 = 4$,and $\lambda_3 = 2$.
Since these three vectors are linearly independent,they form a basis for $\mathbb{R}^3$.
The matrix $A$ can be diagonalized as $A = PDP^{-1}$,where $D = \text{diag}(2, 4, 2)$ and $P$ is the matrix with columns $v_1, v_2, v_3$.
The system is $(A-3I)X = B$. The matrix $(A-3I)$ has eigenvalues $\lambda_i - 3$,which are $2-3 = -1$,$4-3 = 1$,and $2-3 = -1$.
Since none of the eigenvalues of $(A-3I)$ are $0$,the determinant $|A-3I| = (-1)(1)(-1) = 1 \neq 0$.
Because the determinant is non-zero,the matrix $(A-3I)$ is invertible.
Therefore,the system $(A-3I)X = B$ has a unique solution.

(B) 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 have a rank less than $3$.
The coefficient matrix is $A = \begin{bmatrix} 2 & 3 & -1 \\ 1 & \alpha & 3 \\ 3 & -1 & \beta \end{bmatrix}$.
Setting $|A| = 0$:
$2(\alpha\beta + 3) - 3(\beta - 9) - 1(-1 - 3\alpha) = 0$
$2\alpha\beta + 6 - 3\beta + 27 + 1 + 3\alpha = 0$
$2\alpha\beta + 3\alpha - 3\beta + 34 = 0$ (Equation $1$)
Since the system has infinitely many solutions,the planes must be linearly dependent. We can express the third equation as a linear combination of the first two: $L_3 = c_1 L_1 + c_2 L_2$.
Comparing coefficients:
$2c_1 + c_2 = 3$
$3c_1 + c_2\alpha = -1$
$-c_1 + 3c_2 = \beta$
$5c_1 - 4c_2 = 7$
Solving the system of equations for $c_1$ and $c_2$ using the first and last equations:
$c_2 = 3 - 2c_1$
$5c_1 - 4(3 - 2c_1) = 7 \implies 5c_1 - 12 + 8c_1 = 7 \implies 13c_1 = 19 \implies c_1 = \frac{19}{13}$
$c_2 = 3 - 2(\frac{19}{13}) = \frac{39 - 38}{13} = \frac{1}{13}$
Substituting $c_1$ and $c_2$ into the other equations:
$3(\frac{19}{13}) + \alpha(\frac{1}{13}) = -1 \implies 57 + \alpha = -13 \implies \alpha = -70$
$-(\frac{19}{13}) + 3(\frac{1}{13}) = \beta \implies \beta = \frac{-16}{13}$
Finally,$13\alpha\beta = 13(-70)(\frac{-16}{13}) = 70 \times 16 = 1120$.

(B) Given the system of equations:
$x+2y+3z=5$
$2x+3y+z=9$
$4x+3y+\lambda z=\mu$
For the system to have an infinite number of solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and $\Delta_1 = \Delta_2 = \Delta_3 = 0$.
First,calculate $\Delta$:
$\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 4 & 3 & \lambda \end{vmatrix} = 1(3\lambda - 3) - 2(2\lambda - 4) + 3(6 - 12) = 0$
$3\lambda - 3 - 4\lambda + 8 - 18 = 0$
$-\lambda - 13 = 0 \Rightarrow \lambda = -13$
Now,calculate $\Delta_1$ using $\lambda = -13$:
$\Delta_1 = \begin{vmatrix} 5 & 2 & 3 \\ 9 & 3 & 1 \\ \mu & 3 & -13 \end{vmatrix} = 5(-39 - 3) - 2(-117 - \mu) + 3(27 - 3\mu) = 0$
$5(-42) + 234 + 2\mu + 81 - 9\mu = 0$
$-210 + 315 - 7\mu = 0$
$105 - 7\mu = 0 \Rightarrow \mu = 15$
Finally,calculate $\lambda + 2\mu = -13 + 2(15) = -13 + 30 = 17$.

(C) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $ D $ must be $ 0 $,and the augmented determinants $ D_1, D_2, D_3 $ must also be $ 0 $.
$ D = \begin{vmatrix} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{vmatrix} = 0 $
$ 11(-117 + 95) - 1(-78 - 40) + \lambda(-38 - 24) = 0 $
$ 11(-22) + 118 - 62\lambda = 0 $
$ -242 + 118 = 62\lambda $
$ 62\lambda = -124 \Rightarrow \lambda = -2 $
Now,for $ D_1 = 0 $:
$ D_1 = \begin{vmatrix} -5 & 1 & -2 \\ 3 & 3 & 5 \\ \mu & -19 & -39 \end{vmatrix} = 0 $
$ -5(-117 + 95) - 1(-117 - 5\mu) - 2(-57 - 3\mu) = 0 $
$ -5(-22) + 117 + 5\mu + 114 + 6\mu = 0 $
$ 110 + 231 + 11\mu = 0 $
$ 11\mu = -341 \Rightarrow \mu = -31 $
Finally,calculate $ \lambda^4 - \mu $:
$ \lambda^4 - \mu = (-2)^4 - (-31) = 16 + 31 = 47 $

(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,calculate $D$:
$D = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 5 \\ 1 & 2 & m \end{vmatrix} = 1(5m - 10) - 1(2m - 5) + 1(4 - 5) = 5m - 10 - 2m + 5 - 1 = 3m - 6$.
Setting $D = 0$,we get $3m - 6 = 0 \Rightarrow m = 2$.
Next,calculate $D_3$ with $m=2$:
$D_3 = \begin{vmatrix} 1 & 1 & 4 \\ 2 & 5 & 17 \\ 1 & 2 & n \end{vmatrix} = 1(5n - 34) - 1(2n - 17) + 4(4 - 5) = 5n - 34 - 2n + 17 - 4 = 3n - 21$.
Setting $D_3 = 0$,we get $3n - 21 = 0 \Rightarrow n = 7$.
Now,check the options with $m=2$ and $n=7$:
$m^2 + n^2 - mn = 2^2 + 7^2 - (2)(7) = 4 + 49 - 14 = 39$.
Thus,the values satisfy $m^2 + n^2 - mn = 39$.

(A) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ and the determinants $D_1, D_2, D_3$ must all be zero.
First,we calculate $D_3 = 0$:
$D_3 = \begin{vmatrix} 2 & 7 & 3 \\ 3 & 2 & 4 \\ 1 & \mu & -1 \end{vmatrix} = 2(-2 - 4\mu) - 7(-3 - 4) + 3(3\mu - 2) = 0$
$-4 - 8\mu + 49 + 9\mu - 6 = 0$
$\mu + 39 = 0 \Rightarrow \mu = -39$
Next,we calculate $D = 0$ with $\mu = -39$:
$D = \begin{vmatrix} 2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & -39 & 32 \end{vmatrix} = 2(64 + 195) - 7(96 - 5) + \lambda(-117 - 2) = 0$
$2(259) - 7(91) - 119\lambda = 0$
$518 - 637 - 119\lambda = 0$
$-119 - 119\lambda = 0 \Rightarrow \lambda = -1$
Finally,we find $(\lambda - \mu)$:
$\lambda - \mu = -1 - (-39) = -1 + 39 = 38$.

(B) Given the characteristic equation $A^3 - 4A^2 + A + 21I = 0$.
The characteristic polynomial is $P(\lambda) = \lambda^3 - 4\lambda^2 + \lambda + 21 = 0$.
The sum of the eigenvalues (trace of $A$) is equal to the coefficient of $\lambda^2$ with a sign change,so $\text{tr}(A) = 2 + 3 + b = 4$,which gives $b = -1$.
The determinant of $A$ is equal to the constant term of the characteristic polynomial with a sign change (for $3 \times 3$ matrix),so $|A| = -21$.
Calculating the determinant: $|A| = 2(3b - 5) - a(b - 0) + 0 = 6b - 10 - ab = -21$.
Substituting $b = -1$: $6(-1) - 10 - a(-1) = -21 \Rightarrow -6 - 10 + a = -21 \Rightarrow -16 + a = -21 \Rightarrow a = -5$.
Finally,$2a + 3b = 2(-5) + 3(-1) = -10 - 3 = -13$.

(B) For the system of equations to have infinitely many solutions,the determinant of the coefficient matrix $\Delta$ must be $0$,and $\Delta_x = \Delta_y = \Delta_z = 0$.
First,calculate $\Delta = \begin{vmatrix} 1 & 4 & -1 \\ 7 & 9 & \mu \\ 5 & 1 & 2 \end{vmatrix} = 0$.
Expanding along the first row: $1(18-\mu) - 4(14-5\mu) - 1(7-45) = 0$.
$18 - \mu - 56 + 20\mu + 38 = 0$.
$19\mu = 0 \Rightarrow \mu = 0$.
Now,for $\Delta_x = 0$:
$\Delta_x = \begin{vmatrix} \lambda & 4 & -1 \\ -3 & 9 & 0 \\ -1 & 1 & 2 \end{vmatrix} = 0$.
Expanding along the first row: $\lambda(18-0) - 4(-6-0) - 1(-3+9) = 0$.
$18\lambda + 24 - 6 = 0$.
$18\lambda = -18 \Rightarrow \lambda = -1$.
Finally,calculate $(2\mu + 3\lambda) = 2(0) + 3(-1) = -3$.

(A) The given system of equations is:
$(1)$ $3x + 5y + \lambda z = 3$
$(2)$ $7x + 11y - 9z = 2$
$(3)$ $97x + 155y - 189z = \mu$
For the system to have infinitely many solutions,the determinant of the coefficient matrix must be zero,and the augmented matrix must be consistent.
Let $D = \begin{vmatrix} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{vmatrix} = 0$.
Expanding along the first row:
$3(11 \times -189 - (-9 \times 155)) - 5(7 \times -189 - (-9 \times 97)) + \lambda(7 \times 155 - 11 \times 97) = 0$
$3(-2079 + 1395) - 5(-1323 + 873) + \lambda(1085 - 1067) = 0$
$3(-684) - 5(-450) + 18\lambda = 0$
$-2052 + 2250 + 18\lambda = 0$
$198 + 18\lambda = 0 \implies \lambda = -11$.
Now,substitute $\lambda = -11$ into the system:
$(1)$ $3x + 5y - 11z = 3$
$(2)$ $7x + 11y - 9z = 2$
$(3)$ $97x + 155y - 189z = \mu$
For infinite solutions,the third equation must be a linear combination of the first two. Let $(3) = a(1) + b(2)$:
$3a + 7b = 97$
$5a + 11b = 155$
Solving this: $15a + 35b = 485$ and $15a + 33b = 465$. Subtracting gives $2b = 20 \implies b = 10$. Then $3a + 70 = 97 \implies 3a = 27 \implies a = 9$.
Thus,$\mu = 9(3) + 10(2) = 27 + 20 = 47$.
Wait,checking the consistency: $9(-11) + 10(-9) = -99 - 90 = -189$. This matches the coefficient of $z$ in $(3)$.
So,$\mu = 47$ and $\lambda = -11$.
$\mu + 2\lambda = 47 + 2(-11) = 47 - 22 = 25$.

(A) The system of equations is given by $AX = B$,where $A = \begin{bmatrix} 1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4 \end{bmatrix}$.
First,calculate the determinant of the coefficient matrix $D = |A| = 1(4-6) + 2(-4+2) + 3(3-1) = 1(-2) + 2(-2) + 3(2) = -2 - 4 + 6 = 0$.
Since $D = 0$,the system either has no solution or infinitely many solutions.
Now,calculate $D_1 = \left|\begin{array}{ccc}-1 & -2 & 3 \\ k & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = -1(4-6) + 2(4k+2) + 3(3k-1) = 2 + 8k + 4 + 9k - 3 = 17k + 3$.
Wait,let's re-evaluate $D_1$: $D_1 = -1(4-6) + 2(4k+2) + 3(3k-1) = 2 + 8k + 4 + 9k - 3 = 17k + 3$. Actually,let's use row operations: $R_3 \to R_3 - R_1$ and $R_2 \to R_2 + R_1$: $\left|\begin{array}{ccc}1 & -2 & 3 \\ 0 & -1 & 1 \\ 0 & -1 & 1\end{array}\right| = 0$.
For the system to have no solution,at least one of $D_1, D_2, D_3$ must be non-zero.
$D_1 = \left|\begin{array}{ccc}-1 & -2 & 3 \\ k & 1 & -2 \\ 1 & -3 & 4\end{array}\right| = -1(4-6) + 2(4k+2) + 3(-3k-1) = 2 + 8k + 4 - 9k - 3 = 3 - k$.
$D_1 = 0$ only if $k = 3$. If $k \neq 3$,$D_1 \neq 0$,so the system has no solution.
Thus,Statement-$1$ is True. Statement-$2$ is True as it correctly identifies $D=0$ which is the condition for the system to have no solution or infinite solutions.

(C) The given system of equations is:
$ax + 2y = \lambda$
$3x - 2y = \mu$
The determinant of the coefficient matrix is:
$\Delta = \begin{vmatrix} a & 2 \\ 3 & -2 \end{vmatrix} = -2a - 6 = -2(a + 3)$.
For a unique solution,$\Delta \neq 0$,which implies $a \neq -3$. Thus,statement $(B)$ is correct.
If $a = -3$,then $\Delta = 0$. The system will have either no solution or infinitely many solutions.
We calculate $\Delta_1 = \begin{vmatrix} \lambda & 2 \\ \mu & -2 \end{vmatrix} = -2\lambda - 2\mu = -2(\lambda + \mu)$.
If $\lambda + \mu = 0$,then $\Delta_1 = 0$. Since $\Delta = 0$ and $\Delta_1 = 0$,the system has infinitely many solutions. Thus,statement $(C)$ is correct.
If $\lambda + \mu \neq 0$,then $\Delta_1 \neq 0$. Since $\Delta = 0$ and $\Delta_1 \neq 0$,the system has no solution. Thus,statement $(D)$ is correct.
Therefore,statements $(B)$,$(C)$,and $(D)$ are correct.

(A) Given the system of equations:
$x+2y+z=7$
$x+\alpha z=11$
$2x-3y+\beta z=\gamma$
The determinant of the coefficient matrix is $\Delta = \begin{vmatrix} 1 & 2 & 1 \\ 1 & 0 & \alpha \\ 2 & -3 & \beta \end{vmatrix} = 1(0 - (-3\alpha)) - 2(\beta - 2\alpha) + 1(-3 - 0) = 3\alpha - 2\beta + 4\alpha - 3 = 7\alpha - 2\beta - 3$.
If $\beta = \frac{1}{2}(7\alpha - 3)$,then $\Delta = 0$.
For $(P)$: $\beta = \frac{1}{2}(7\alpha - 3)$ and $\gamma = 28$. Calculating $\Delta_x, \Delta_y, \Delta_z$,we find they are all $0$. Thus,the system has infinitely many solutions. $(P \rightarrow 3)$.
For $(Q)$: $\beta = \frac{1}{2}(7\alpha - 3)$ and $\gamma \neq 28$. Since $\Delta = 0$ and at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero,the system has no solution. $(Q \rightarrow 2)$.
For $(R)$: $\beta \neq \frac{1}{2}(7\alpha - 3)$,so $\Delta \neq 0$. The system has a unique solution. $(R \rightarrow 1)$.
For $(S)$: $\beta \neq \frac{1}{2}(7\alpha - 3)$ and $\alpha = 1, \gamma = 28$. Since $\Delta \neq 0$,there is a unique solution. Substituting $x=11, y=-2, z=0$ into the equations: $11+2(-2)+0 = 7$ (True),$11+1(0) = 11$ (True),$2(11)-3(-2)+\beta(0) = 22+6 = 28 = \gamma$ (True). Thus,$(x=11, y=-2, z=0)$ is the unique solution. $(S \rightarrow 4)$.

(A) First,we construct the matrix $M$ based on the condition that $a_{ij} = 1$ if $i$ divides $(j+1)$,and $0$ otherwise.
For $i=1$: $j+1$ is divisible by $1$ for $j=1, 2, 3$. Thus,$a_{11}=1, a_{12}=1, a_{13}=1$.
For $i=2$: $j+1$ is divisible by $2$ for $j=1, 3$. Thus,$a_{21}=1, a_{22}=0, a_{23}=1$.
For $i=3$: $j+1$ is divisible by $3$ for $j=2$. Thus,$a_{31}=0, a_{32}=1, a_{33}=0$.
So,$M = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$.
$(A)$ Calculate the determinant: $|M| = 1(0-1) - 1(0-0) + 1(1-0) = -1 + 1 = 0$. Since $|M| = 0$,$M$ is singular and not invertible. ($A$ is false).
$(B)$ We solve $M X = -X$,which is $(M + I)X = 0$.
$M+I = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{bmatrix}$. The determinant $|M+I| = 2(1-1) - 1(1-0) + 1(1-0) = 0$. Since the determinant is $0$,there exists a non-zero solution $X$. ($B$ is true).
$(C)$ The set $\{X \in \mathbb{R}^3 : MX = 0, X \neq 0\}$ represents the non-trivial null space of $M$. Since $|M| = 0$,the null space is non-trivial. ($C$ is true).
$(D)$ $M - 2I = \begin{bmatrix} 1-2 & 1 & 1 \\ 1 & 0-2 & 1 \\ 0 & 1 & 0-2 \end{bmatrix} = \begin{bmatrix} -1 & 1 & 1 \\ 1 & -2 & 1 \\ 0 & 1 & -2 \end{bmatrix}$.
$|M-2I| = -1(4-1) - 1(-2-0) + 1(1-0) = -3 + 2 + 1 = 0$. Thus,$(M-2I)$ is not invertible. ($D$ is false).
Therefore,the correct statements are $B$ and $C$.

(D) The system has infinitely many solutions if the determinant of the coefficient matrix $D = 0$ and the augmented matrix satisfies the consistency condition.
First,calculate the determinant $D$:
$D = \begin{vmatrix} 1 & \alpha & \alpha^2 \\ \alpha & 1 & \alpha \\ \alpha^2 & \alpha & 1 \end{vmatrix} = 1(1 - \alpha^2) - \alpha(\alpha - \alpha^3) + \alpha^2(\alpha^2 - \alpha^2) = 1 - \alpha^2 - \alpha^2 + \alpha^4 = \alpha^4 - 2\alpha^2 + 1 = (\alpha^2 - 1)^2$.
Setting $D = 0$ gives $(\alpha^2 - 1)^2 = 0$,so $\alpha^2 = 1$,which means $\alpha = 1$ or $\alpha = -1$.
Case $1$: If $\alpha = 1$,the system becomes:
$x + y + z = 1$
$x + y + z = -1$
$x + y + z = 1$
This is inconsistent because $1 \neq -1$,so there are no solutions.
Case $2$: If $\alpha = -1$,the system becomes:
$x - y + z = 1$
$-x + y - z = -1$
$x - y + z = 1$
All three equations are equivalent to $x - y + z = 1$,which represents a plane,thus there are infinitely many solutions.
Therefore,$\alpha = -1$.
Then,$1 + \alpha + \alpha^2 = 1 + (-1) + (-1)^2 = 1 - 1 + 1 = 1$.

(D) Given the system of equations:
$1) 3x - y - z = 0$
$2) -3x + z = 0$
$3) -3x + 2y + z = 0$
From equation $(2)$,we get $z = 3x$.
Substituting $z = 3x$ into equation $(1)$:
$3x - y - 3x = 0 \Rightarrow y = 0$.
Checking with equation $(3)$:
$-3x + 2(0) + 3x = 0$,which is $0 = 0$. This is consistent.
Thus,any point $(x, y, z)$ satisfying the system is of the form $(a, 0, 3a)$ where $a$ is an integer.
We are given the condition $x^2 + y^2 + z^2 \leq 100$.
Substituting the point $(a, 0, 3a)$:
$a^2 + 0^2 + (3a)^2 \leq 100$
$a^2 + 9a^2 \leq 100$
$10a^2 \leq 100$
$a^2 \leq 10$
Since $a$ is an integer,the possible values for $a$ are $a \in \{-3, -2, -1, 0, 1, 2, 3\}$.
Counting these values,we get $7$ possible points.

(A) system of linear equations $AX = B$ can have either a unique solution,no solution,or infinitely many solutions.
It is mathematically impossible for a system of linear equations to have exactly two distinct solutions.
If a system has more than one solution,it must have infinitely many solutions.
Therefore,the number of such matrices $A$ is $0$.

(A) For the system to have at least one solution,the determinant $\Delta$ must be non-zero,or if $\Delta = 0$,then $\Delta_1 = \Delta_2 = \Delta_3 = 0$.
For the given system:
$\Delta = \begin{vmatrix} -1 & 2 & 5 \\ 2 & -4 & 3 \\ 1 & -2 & 2 \end{vmatrix} = -1(-8+6) - 2(4-3) + 5(-4+4) = 2 - 2 + 0 = 0$.
For the system to have a solution,we must have $\Delta_1 = \Delta_2 = \Delta_3 = 0$.
Calculating $\Delta_1 = \begin{vmatrix} b_1 & 2 & 5 \\ b_2 & -4 & 3 \\ b_3 & -2 & 2 \end{vmatrix} = b_1(-8+6) - 2(2b_2-3b_3) + 5(-2b_2+4b_3) = -2b_1 - 4b_2 + 6b_3 - 10b_2 + 20b_3 = -2b_1 - 14b_2 + 26b_3 = 0$.
This simplifies to $b_1 + 7b_2 - 13b_3 = 0$. Thus,$S = \{ [b_1, b_2, b_3]^T : b_1 + 7b_2 - 13b_3 = 0 \}$.
For each option,we check if the system has a solution for all $[b_1, b_2, b_3]^T \in S$. If $\Delta \neq 0$,it always has a unique solution.
$(A)$ $\Delta = \begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 2 & 6 \end{vmatrix} = 1(24-10) - 2(0-5) + 3(0-4) = 14 + 10 - 12 = 12 \neq 0$. Solution exists.
$(B)$ $\Delta = \begin{vmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{vmatrix} = 0$. Checking $\Delta_1$: $\begin{vmatrix} b_1 & 1 & 3 \\ b_2 & 2 & 6 \\ b_3 & -1 & -3 \end{vmatrix} = b_1(0) - 1(-3b_2-6b_3) + 3(-b_2-2b_3) = 3b_2+6b_3-3b_2-6b_3 = 0$. Since $\Delta_1 = \Delta_2 = \Delta_3 = 0$,it has solutions,but we must check if it holds for all $S$. Actually,for $(B)$,the equations are dependent,and it has solutions for all $b_i$.
$(C)$ $\Delta = 0$ and $\Delta_1 = \Delta_2 = \Delta_3 = 0$. Solution exists.
$(D)$ $\Delta = \begin{vmatrix} 1 & 2 & 5 \\ 2 & 0 & 3 \\ 1 & 4 & -5 \end{vmatrix} = 1(0-12) - 2(-10-3) + 5(8-0) = -12 + 26 + 40 = 54 \neq 0$. Solution exists.
Thus,$(A), (B), (C), (D)$ all have solutions.