Mix Examples-Determinants and Matrices Questions in English

(C) Given $f(x) = \left|\begin{array}{ccc} \sin^{2} x & 1+\cos^{2} x & \cos 2x \\ 1+\sin^{2} x & \cos^{2} x & \cos 2x \\ \sin^{2} x & \cos^{2} x & \sin 2x \end{array}\right|$.
Apply the column operation $C_{1} \rightarrow C_{1} + C_{2}$:
$f(x) = \left|\begin{array}{ccc} \sin^{2} x + 1 + \cos^{2} x & 1+\cos^{2} x & \cos 2x \\ 1+\sin^{2} x + \cos^{2} x & \cos^{2} x & \cos 2x \\ \sin^{2} x + \cos^{2} x & \cos^{2} x & \sin 2x \end{array}\right| = \left|\begin{array}{ccc} 2 & 1+\cos^{2} x & \cos 2x \\ 2 & \cos^{2} x & \cos 2x \\ 1 & \cos^{2} x & \sin 2x \end{array}\right|$.
Apply the row operation $R_{1} \rightarrow R_{1} - R_{2}$:
$f(x) = \left|\begin{array}{ccc} 0 & 1 & 0 \\ 2 & \cos^{2} x & \cos 2x \\ 1 & \cos^{2} x & \sin 2x \end{array}\right|$.
Expanding along $R_{1}$:
$f(x) = -1 \times (2 \sin 2x - \cos 2x) = \cos 2x - 2 \sin 2x$.
The maximum value of $a \cos \theta + b \sin \theta$ is $\sqrt{a^{2} + b^{2}}$.
Here $a = 1$ and $b = -2$.
So,$f(x)_{\max} = \sqrt{1^{2} + (-2)^{2}} = \sqrt{1+4} = \sqrt{5}$.

(B) Given $A = XB$,we have $\begin{bmatrix} a_1 \\ a_2 \end{bmatrix} = \frac{1}{\sqrt{3}} \begin{bmatrix} 1 & -1 \\ 1 & k \end{bmatrix} \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$.
Multiplying,we get $\sqrt{3} a_1 = b_1 - b_2$ and $\sqrt{3} a_2 = b_1 + k b_2$.
Squaring and adding these equations:
$3(a_1^2 + a_2^2) = (b_1 - b_2)^2 + (b_1 + k b_2)^2$
$3(a_1^2 + a_2^2) = (b_1^2 + b_2^2 - 2b_1b_2) + (b_1^2 + k^2b_2^2 + 2kb_1b_2)$
$3(a_1^2 + a_2^2) = 2b_1^2 + (1 + k^2)b_2^2 + 2b_1b_2(k - 1)$.
We are given $a_1^2 + a_2^2 = \frac{2}{3}(b_1^2 + b_2^2)$,so $3(a_1^2 + a_2^2) = 2b_1^2 + 2b_2^2$.
Comparing the two expressions for $3(a_1^2 + a_2^2)$:
$2b_1^2 + (1 + k^2)b_2^2 + 2b_1b_2(k - 1) = 2b_1^2 + 2b_2^2$.
This simplifies to $(k^2 + 1)b_2^2 + 2b_1b_2(k - 1) = 2b_2^2$.
$(k^2 - 1)b_2^2 + 2b_1b_2(k - 1) = 0$.
$(k - 1)[(k + 1)b_2^2 + 2b_1b_2] = 0$.
Since $(k^2 + 1)b_2^2 \neq -2b_1b_2$,the term in the bracket cannot be zero for arbitrary $b_1, b_2$ unless $k=1$. Thus,$k = 1$.

(D) Given $P = \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix}$.
First,calculate $5I - 8P$:
$5I - 8P = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 16 & -8 \\ 40 & -24 \end{bmatrix} = \begin{bmatrix} -11 & 8 \\ -40 & 29 \end{bmatrix}$.
Now,calculate powers of $P$:
$P^2 = \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 4-5 & -2+3 \\ 10-15 & -5+9 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix}$.
$P^3 = P^2 \cdot P = \begin{bmatrix} -1 & 1 \\ -5 & 4 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} -2+5 & 1-3 \\ -10+20 & 5-12 \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ 10 & -7 \end{bmatrix}$.
$P^6 = (P^3)^2 = \begin{bmatrix} 3 & -2 \\ 10 & -7 \end{bmatrix} \begin{bmatrix} 3 & -2 \\ 10 & -7 \end{bmatrix} = \begin{bmatrix} 9-20 & -6+14 \\ 30-70 & -20+49 \end{bmatrix} = \begin{bmatrix} -11 & 8 \\ -40 & 29 \end{bmatrix}$.
Comparing $P^n$ with $5I - 8P$,we get $P^6 = 5I - 8P$.
Therefore,$n = 6$.

(D) Let $M = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$. The sum of the diagonal elements of $M^{T}M$ is the trace of $M^{T}M$,which is equal to the sum of the squares of all elements of $M$.
Thus,$a^{2} + b^{2} + c^{2} + d^{2} + e^{2} + f^{2} + g^{2} + h^{2} + i^{2} = 7$.
Since the entries are from $\{0, 1, 2\}$,we consider the possible combinations of squares that sum to $7$:
Case-$I$: Seven $1$'s and two $0$'s.
Number of ways = $\binom{9}{7} = \binom{9}{2} = \frac{9 \times 8}{2} = 36$.
Case-$II$: One $2$ $(2^{2} = 4)$,three $1$'s $(1^{2} = 1)$,and five $0$'s $(0^{2} = 0)$.
Sum = $4 + 1 + 1 + 1 + 0 + 0 + 0 + 0 + 0 = 7$.
Number of ways = $\frac{9!}{1! 3! 5!} = \frac{9 \times 8 \times 7 \times 6}{3 \times 2 \times 1} = 9 \times 4 \times 2 \times 7 = 504$.
Total number of matrices = $36 + 504 = 540$.

(A) Given $A = \begin{bmatrix} x & y & z \\ y & z & x \\ z & x & y \end{bmatrix}$. Since $A$ is symmetric,$A^T = A$.
Given $A^2 = I$,we have $AA^T = I$,which means $A$ is an orthogonal matrix.
For the matrix $A$,the condition $AA^T = I$ implies:
$x^2 + y^2 + z^2 = 1$ and $xy + yz + zx = 0$.
We know that $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$.
Substituting the values,$(x + y + z)^2 = 1 + 2(0) = 1$.
Since $x + y + z > 0$,we have $x + y + z = 1$.
Using the algebraic identity $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - (xy + yz + zx))$,
$x^3 + y^3 + z^3 - 3(2) = (1)(1 - 0)$.
$x^3 + y^3 + z^3 - 6 = 1$.
$x^3 + y^3 + z^3 = 7$.

(C) Given that $A$ is a symmetric matrix,so $A^{T} = A$.
Given that $B$ is a skew-symmetric matrix,so $B^{T} = -B$.
Let $C = A^{2}B^{2} - B^{2}A^{2}$.
Now,consider the transpose of $C$:
$C^{T} = (A^{2}B^{2} - B^{2}A^{2})^{T} = (A^{2}B^{2})^{T} - (B^{2}A^{2})^{T}$.
Using the property $(PQ)^{T} = Q^{T}P^{T}$,we get:
$C^{T} = (B^{2})^{T}(A^{2})^{T} - (A^{2})^{T}(B^{2})^{T}$.
Since $(A^{2})^{T} = (A^{T})^{2} = A^{2}$ and $(B^{2})^{T} = (B^{T})^{2} = (-B)^{2} = B^{2}$,we have:
$C^{T} = B^{2}A^{2} - A^{2}B^{2} = -(A^{2}B^{2} - B^{2}A^{2}) = -C$.
Thus,$C$ is a skew-symmetric matrix.
For any skew-symmetric matrix $C$ of odd order $n$ (here $n = 3$),the determinant is always zero,i.e.,$\det(C) = 0$.
Since the system is $(C)X = O$ and $\det(C) = 0$,the system has infinitely many solutions.

(C) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1 \end{bmatrix}$.
Calculating powers of $A$:
$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}$,$A^3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 8 & 0 \\ 3 & 0 & -1 \end{bmatrix}$,$A^4 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
By induction,for even $n$,$A^n = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and for odd $n$,$A^n = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^n & 0 \\ 3 & 0 & -1 \end{bmatrix}$.
Thus,$A^{20} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $A^{19} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & -1 \end{bmatrix}$.
Substituting into the equation $A^{20} + \alpha A^{19} + \beta A = I_3$ (where $I_3$ is the identity matrix):
$\begin{bmatrix} 1+\alpha+\beta & 0 & 0 \\ 0 & 2^{20}+\alpha 2^{19}+2\beta & 0 \\ 3\alpha+3\beta & 0 & 1-\alpha-\beta \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Comparing elements: $1+\alpha+\beta = 1 \Rightarrow \alpha+\beta = 0 \Rightarrow \beta = -\alpha$.
Also,$3\alpha+3\beta = 0$ (consistent with $\alpha+\beta=0$) and $1-\alpha-\beta = 1 \Rightarrow \alpha+\beta = 0$.
Finally,$2^{20} + \alpha 2^{19} + 2\beta = 4$. Substituting $\beta = -\alpha$:
$2^{20} + \alpha 2^{19} - 2\alpha = 4 \Rightarrow \alpha(2^{19}-2) = 4 - 2^{20}$.
$\alpha = \frac{4 - 2^{20}}{2^{19}-2} = \frac{2(2 - 2^{19})}{2^{19}-2} = -2$.
Since $\beta = -\alpha$,$\beta = 2$.
Therefore,$\beta - \alpha = 2 - (-2) = 4$.

(A) Given $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$.
Calculating powers of $A$:
$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$.
$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$.
$A^4 = A^3 \cdot A = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$.
By induction,$A^n = \begin{bmatrix} 1 & 0 & 0 \\ n-1 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$.
Thus,$A^{2025} - A^{2020} = \begin{bmatrix} 1 & 0 & 0 \\ 2024 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 2019 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
Now check $A^6 - A = \begin{bmatrix} 1 & 0 & 0 \\ 5 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.
Therefore,$A^{2025} - A^{2020} = A^6 - A$.

(A) Given matrix $A = \begin{bmatrix} 0 & 2 \\ K & -1 \end{bmatrix}$.
The given equation is $A(A^{3} + 3I) = 2I$,which implies $A^{4} + 3A = 2I$,or $A^{4} = 2I - 3A$.
The characteristic equation of $A$ is given by $|A - \lambda I| = 0$.
$\begin{vmatrix} -\lambda & 2 \\ K & -1 - \lambda \end{vmatrix} = 0 \Rightarrow \lambda(\lambda + 1) - 2K = 0 \Rightarrow \lambda^{2} + \lambda - 2K = 0$.
By the Cayley-Hamilton theorem,$A^{2} + A - 2KI = 0$,so $A^{2} = 2KI - A$.
Now,$A^{4} = (A^{2})^{2} = (2KI - A)^{2} = 4K^{2}I - 4KA + A^{2}$.
Substituting $A^{2} = 2KI - A$ into the expression for $A^{4}$:
$A^{4} = 4K^{2}I - 4KA + (2KI - A) = (4K^{2} + 2K)I - (4K + 1)A$.
Equating the two expressions for $A^{4}$:
$2I - 3A = (4K^{2} + 2K)I - (4K + 1)A$.
Rearranging terms:
$(4K + 1 - 3)A = (4K^{2} + 2K - 2)I$.
$(4K - 2)A = (4K^{2} + 2K - 2)I$.
$2(2K - 1)A = 2(2K^{2} + K - 1)I$.
$2(2K - 1)A = 2(2K - 1)(K + 1)I$.
If $2K - 1 \neq 0$,then $A = (K + 1)I$,which implies $\begin{bmatrix} 0 & 2 \\ K & -1 \end{bmatrix} = \begin{bmatrix} K+1 & 0 \\ 0 & K+1 \end{bmatrix}$.
This leads to $K+1 = 0$ and $2 = 0$,which is impossible.
Thus,we must have $2K - 1 = 0$,which gives $K = \frac{1}{2}$.

(D) Given $a_{r} = e^{i \frac{2 \pi r}{9}}$.
Note that $a_{r} = (a_{1})^{r}$.
The determinant is $\Delta = \left|\begin{array}{lll}a_{1} & a_{1}^{2} & a_{1}^{3} \\ a_{1}^{4} & a_{1}^{5} & a_{1}^{6} \\ a_{1}^{7} & a_{1}^{8} & a_{1}^{9}\end{array}\right|$.
Taking $a_{1}$ common from $C_{1}$,$a_{1}^{2}$ common from $C_{2}$,and $a_{1}^{3}$ common from $C_{3}$:
$\Delta = a_{1} \cdot a_{1}^{2} \cdot a_{1}^{3} \left|\begin{array}{lll}1 & 1 & 1 \\ a_{1}^{3} & a_{1}^{3} & a_{1}^{3} \\ a_{1}^{6} & a_{1}^{6} & a_{1}^{6}\end{array}\right|$.
Since all columns are identical,the value of the determinant is $0$.

(A) Given the condition $(I-A)^3 = I-A^3$.
Expanding the left side: $I^3 - 3I^2A + 3IA^2 - A^3 = I - A^3$.
Since $I^2 = I$ and $IA = AI = A$,this simplifies to $I - 3A + 3A^2 - A^3 = I - A^3$.
Subtracting $I$ and adding $A^3$ to both sides,we get $3A^2 - 3A = 0$,which implies $A^2 = A$.
Let $A = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix}$. Then $A^2 = \begin{bmatrix} a & b \\ 0 & d \end{bmatrix} \begin{bmatrix} a & b \\ 0 & d \end{bmatrix} = \begin{bmatrix} a^2 & ab+bd \\ 0 & d^2 \end{bmatrix}$.
Equating $A^2 = A$,we get:
$a^2 = a \Rightarrow a \in \{0, 1\}$.
$d^2 = d \Rightarrow d \in \{0, 1\}$.
$ab + bd = b \Rightarrow b(a + d - 1) = 0$.
Case $1$: If $b = 0$,then $a \in \{0, 1\}$ and $d \in \{0, 1\}$. This gives $2 \times 2 = 4$ matrices.
Case $2$: If $b \neq 0$,then $a + d - 1 = 0$,so $a + d = 1$.
Possible pairs $(a, d)$ are $(1, 0)$ and $(0, 1)$.
For each pair,$b$ can be $\{-1, 1\}$ (since $b \neq 0$).
This gives $2 \times 2 = 4$ matrices.
Total number of matrices = $4 + 4 = 8$.

(C) Given $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x$.
For $i \leq j$,$a_{i j}=J_{6+i, 3}-J_{i+3,3} = \int_{0}^{\frac{1}{2}} \frac{x^{6+i}-x^{i+3}}{x^{3}-1} d x = \int_{0}^{\frac{1}{2}} \frac{x^{i+3}(x^{3}-1)}{x^{3}-1} d x = \int_{0}^{\frac{1}{2}} x^{i+3} d x$.
Evaluating the integral: $a_{i j} = \left[ \frac{x^{i+4}}{i+4} \right]_{0}^{\frac{1}{2}} = \frac{(1/2)^{i+4}}{i+4}$.
Thus,$a_{11} = \frac{(1/2)^{5}}{5} = \frac{1}{5 \cdot 2^{5}}$,$a_{12} = \frac{(1/2)^{5}}{5} = \frac{1}{5 \cdot 2^{5}}$,$a_{13} = \frac{(1/2)^{5}}{5} = \frac{1}{5 \cdot 2^{5}}$.
$a_{22} = \frac{(1/2)^{6}}{6} = \frac{1}{6 \cdot 2^{6}}$,$a_{23} = \frac{(1/2)^{6}}{6} = \frac{1}{6 \cdot 2^{6}}$.
$a_{33} = \frac{(1/2)^{7}}{7} = \frac{1}{7 \cdot 2^{7}}$.
Matrix $A = \begin{bmatrix} \frac{1}{5 \cdot 2^{5}} & \frac{1}{5 \cdot 2^{5}} & \frac{1}{5 \cdot 2^{5}} \\ 0 & \frac{1}{6 \cdot 2^{6}} & \frac{1}{6 \cdot 2^{6}} \\ 0 & 0 & \frac{1}{7 \cdot 2^{7}} \end{bmatrix}$.
$|A| = \frac{1}{5 \cdot 2^{5}} \cdot \frac{1}{6 \cdot 2^{6}} \cdot \frac{1}{7 \cdot 2^{7}} = \frac{1}{210 \cdot 2^{18}}$.
We need $|\operatorname{adj} A^{-1}| = |A^{-1}|^{3-1} = |A^{-1}|^{2} = \frac{1}{|A|^{2}} = (210 \cdot 2^{18})^{2} = (2 \cdot 105)^{2} \cdot 2^{36} = 4 \cdot (105)^{2} \cdot 2^{36} = (105)^{2} \cdot 2^{38}$.

(B) The matrix $A$ is given by:
$A = \begin{bmatrix} 1 & -x & 2x+1 \\ -x & 1 & -x \\ 2x+1 & -x & 1 \end{bmatrix}$
Calculating the determinant $|A|$:
$|A| = 1(1 - x^2) + x(-x + x(2x+1)) + (2x+1)(x^2 - (2x+1))$
$|A| = 1 - x^2 - x^2 + 2x^3 + x^2 + (2x+1)(x^2 - 2x - 1)$
$|A| = 2x^3 - x^2 + 1 + (2x^3 - 4x^2 - 2x + x^2 - 2x - 1)$
$|A| = 4x^3 - 4x^2 - 4x = f(x)$
To find the critical points,set $f'(x) = 0$:
$f'(x) = 12x^2 - 8x - 4 = 4(3x^2 - 2x - 1) = 4(3x+1)(x-1) = 0$
Thus,$x = 1$ and $x = -\frac{1}{3}$.
Evaluating $f(x)$ at these points:
$f(1) = 4(1)^3 - 4(1)^2 - 4(1) = 4 - 4 - 4 = -4$ (Minimum value)
$f(-\frac{1}{3}) = 4(-\frac{1}{27}) - 4(\frac{1}{9}) - 4(-\frac{1}{3}) = -\frac{4}{27} - \frac{12}{27} + \frac{36}{27} = \frac{20}{27}$ (Maximum value)
The sum of the maximum and minimum values is:
$-\frac{88}{27} = -4 + \frac{20}{27} = \frac{-108 + 20}{27} = -\frac{88}{27}$

(D) Given $A = \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix}$.
Any square matrix $A$ can be written as $A = P + Q$,where $P = \frac{A + A^T}{2}$ is symmetric and $Q = \frac{A - A^T}{2}$ is skew-symmetric.
$P = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix} + \begin{bmatrix} 2 & a \\ 3 & 0 \end{bmatrix} \right) = \begin{bmatrix} 2 & \frac{3+a}{2} \\ \frac{3+a}{2} & 0 \end{bmatrix}$.
$Q = \frac{1}{2} \left( \begin{bmatrix} 2 & 3 \\ a & 0 \end{bmatrix} - \begin{bmatrix} 2 & a \\ 3 & 0 \end{bmatrix} \right) = \begin{bmatrix} 0 & \frac{3-a}{2} \\ \frac{a-3}{2} & 0 \end{bmatrix}$.
Given $\operatorname{det}(Q) = 9$,we have $0 - \left( \frac{3-a}{2} \right) \left( \frac{a-3}{2} \right) = 9$.
$\Rightarrow \frac{(a-3)^2}{4} = 9 \Rightarrow (a-3)^2 = 36 \Rightarrow a-3 = \pm 6$.
Thus,$a = 9$ or $a = -3$.
Now,$\operatorname{det}(P) = 0 - \left( \frac{3+a}{2} \right)^2 = -\frac{(a+3)^2}{4}$.
For $a = 9$,$\operatorname{det}(P) = -\frac{(9+3)^2}{4} = -\frac{144}{4} = -36$.
For $a = -3$,$\operatorname{det}(P) = -\frac{(-3+3)^2}{4} = 0$.
The sum of all possible values of $\operatorname{det}(P)$ is $-36 + 0 = -36$.
The modulus of the sum is $|-36| = 36$.

(D) Let the matrix be $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. The determinant is $|A| = ad - bc$.
The total number of ways to choose $4$ numbers from $6$ faces of a die is $6^4 = 1296$.
We require all entries $a, b, c, d$ to be distinct. The number of ways to choose $4$ distinct numbers from ${1, 2, 3, 4, 5, 6}$ and arrange them in the matrix is $^6P_4 = 6 \times 5 \times 4 \times 3 = 360$.
For the matrix to be nonsingular,$|A| \neq 0$,which means $ad \neq bc$.
We count the cases where $ad = bc$ with distinct $a, b, c, d$:
$1$. $6 \times 1 = 2 \times 3$: The sets are ${1, 2, 3, 6}$. Possible arrangements $(a, b, c, d)$ such that $ad=bc$ are $(6, 2, 3, 1), (6, 3, 2, 1), (1, 2, 3, 6), (1, 3, 2, 6), (2, 6, 1, 3), (3, 6, 1, 2), (2, 1, 6, 3), (3, 1, 6, 2)$. Total $8$ cases.
$2$. $6 \times 2 = 3 \times 4$: The sets are ${2, 3, 4, 6}$. Possible arrangements $(a, b, c, d)$ such that $ad=bc$ are $(6, 3, 4, 2), (6, 4, 3, 2), (2, 3, 4, 6), (2, 4, 3, 6), (3, 6, 2, 4), (4, 6, 2, 3), (3, 2, 6, 4), (4, 2, 6, 3)$. Total $8$ cases.
Total cases where $ad = bc$ with distinct entries is $8 + 8 = 16$.
Number of favorable cases = (Total ways with distinct entries) - (Cases where $ad = bc$) = $360 - 16 = 344$.
Required probability = $\frac{344}{1296} = \frac{43}{162}$.

(A) Let $A = \begin{bmatrix} 0 & i \\ 1 & 0 \end{bmatrix}$ and $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
The given condition is $A^n X = X$ for all $a, b, c, d \in R$.
Since $X$ can be any $2 \times 2$ matrix,we can choose $X = I$ (the identity matrix),which implies $A^n = I$.
Now,calculate powers of $A$:
$A^2 = \begin{bmatrix} 0 & i \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} i & 0 \\ 0 & i \end{bmatrix} = iI$.
$A^4 = (A^2)^2 = (iI)^2 = i^2 I = -I$.
$A^8 = (A^4)^2 = (-I)^2 = I$.
Thus,$A^n = I$ if and only if $n$ is a multiple of $8$.
We need to find the number of $2$-digit numbers $n$ such that $n$ is a multiple of $8$.
The $2$-digit multiples of $8$ are $16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96$.
Counting these,we have $11$ such numbers.

(A) Given that $A^{5}=B^{5}$ and $A^{3} B^{2}=A^{2} B^{3}$.
Subtracting the second equation from the first,we get:
$A^{5}-A^{3} B^{2} = B^{5}-A^{2} B^{3}$
$A^{3}(A^{2}-B^{2}) = B^{5}-A^{2} B^{3}$
Rearranging the terms:
$A^{3}(A^{2}-B^{2}) + B^{3}(A^{2}-B^{2}) = 0$
$(A^{3}+B^{3})(A^{2}-B^{2}) = 0$
Since $(A^{2}-B^{2})$ is an invertible matrix,we can post-multiply by its inverse $(A^{2}-B^{2})^{-1}$:
$(A^{3}+B^{3})(A^{2}-B^{2})(A^{2}-B^{2})^{-1} = 0 \cdot (A^{2}-B^{2})^{-1}$
$A^{3}+B^{3} = 0$
Therefore,the determinant of the zero matrix is $|A^{3}+B^{3}| = |0| = 0$.

(B) Given $A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$.
We can observe that $A = I + N$,where $I$ is the identity matrix and $N = \begin{bmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$.
Note that $N^2 = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $N^3 = 0$.
Using the binomial expansion,$A^n = (I+N)^n = I + nN + \frac{n(n-1)}{2}N^2$.
Thus,$A^n = \begin{bmatrix} 1 & n & n + \frac{n(n-1)}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & n & \frac{n^2+n}{2} \\ 0 & 1 & n \\ 0 & 0 & 1 \end{bmatrix}$.
The sum of elements of $A^n$ is $S_n = 1 + n + \frac{n^2+n}{2} + 0 + 1 + n + 0 + 0 + 1 = 3 + 2n + \frac{n^2+n}{2} = 3 + \frac{5n+n^2}{2}$.
Sum of elements of $M = \sum_{n=1}^{20} S_n = \sum_{n=1}^{20} (3 + \frac{5}{2}n + \frac{1}{2}n^2) = 3(20) + \frac{5}{2} \frac{20(21)}{2} + \frac{1}{2} \frac{20(21)(41)}{6}$.
$= 60 + 525 + 1435 = 2020$.

(C) Let $A = \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix} \begin{bmatrix} -1 & a \\ 0 & b \end{bmatrix} = \begin{bmatrix} 1 & -a + ab \\ 0 & b^2 \end{bmatrix}$.
For $A^{n(n+1)} = I$ to hold for all $n \in \{1, 2, \ldots, 100\}$,we examine the condition $A^{n(n+1)} = I$.
If $b = 1$,then $A^2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Since $n(n+1)$ is always even for any integer $n \ge 1$,$A^{n(n+1)} = (A^2)^{\frac{n(n+1)}{2}} = I^{\frac{n(n+1)}{2}} = I$.
Thus,if $b = 1$,$A \in T_n$ for all $n$.
If $b \neq 1$,then $A^2 = \begin{bmatrix} 1 & a(b-1) \\ 0 & b^2 \end{bmatrix}$.
For $A^{n(n+1)} = I$,we require $b^{n(n+1)} = 1$ and the top-right entry to be $0$.
Since $b \in \{1, 2, \ldots, 100\}$,$b^{n(n+1)} = 1$ implies $b = 1$ (as $b > 0$).
Therefore,only matrices with $b = 1$ satisfy the condition for all $n$.
With $b = 1$,$a$ can be any value in $\{1, 2, \ldots, 100\}$.
There are $100$ such elements.

(B) The set $S$ consists of square roots of odd integers from $1$ to $50$. Thus,$S = \{\sqrt{1}, \sqrt{3}, \dots, \sqrt{49}\}$. The number of terms in $S$ is $25$.
For the matrix $A = \begin{bmatrix} 1 & 0 & a \\ -1 & 1 & 0 \\ -a & 0 & 1 \end{bmatrix}$,the determinant is $|A| = 1(1 - 0) - 0 + a(0 - (-a)) = 1 + a^2$.
We know that $\operatorname{det}(\operatorname{adj} A) = |A|^{n-1}$,where $n$ is the order of the matrix. Here $n=3$,so $\operatorname{det}(\operatorname{adj} A) = |A|^2 = (1 + a^2)^2$.
We need to calculate $\sum_{a \in S} (1 + a^2)^2$. Since $a = \sqrt{n}$,$a^2 = n$. The sum becomes $\sum_{n \in \{1, 3, \dots, 49\}} (1 + n)^2$.
Let $n = 2k - 1$ for $k = 1, 2, \dots, 25$. Then $1 + n = 1 + 2k - 1 = 2k$.
The sum is $\sum_{k=1}^{25} (2k)^2 = 4 \sum_{k=1}^{25} k^2$.
Using the formula $\sum_{k=1}^{N} k^2 = \frac{N(N+1)(2N+1)}{6}$,we get $4 \times \frac{25(26)(51)}{6} = 4 \times 25 \times 13 \times 17 = 22100$.
Given $\sum \operatorname{det}(\operatorname{adj} A) = 100 \lambda$,we have $22100 = 100 \lambda$,which implies $\lambda = 221$.

(A) Given $A = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}$.
Calculating powers of $A$:
$A^2 = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} -4 & 0 \\ 0 & -4 \end{bmatrix} = -4I$.
$A^3 = A^2 \cdot A = -4A$.
$A^4 = (A^2)^2 = (-4I)^2 = 16I$.
In general,$A^{2k} = (-4)^k I$ and $A^{2k-1} = (-4)^{k-1} A$.
$M = \sum_{k=1}^{10} A^{2k} = \sum_{k=1}^{10} (-4)^k I = I \sum_{k=1}^{10} (-4)^k$. Since $M$ is a scalar multiple of the identity matrix $I$,$M$ is symmetric.
$N = \sum_{k=1}^{10} A^{2k-1} = \sum_{k=1}^{10} (-4)^{k-1} A = A \sum_{k=1}^{10} (-4)^{k-1}$. Since $A$ is skew-symmetric,$N$ is a scalar multiple of $A$,so $N$ is skew-symmetric.
$N^2 = (\text{scalar} \cdot A)^2 = \text{scalar}^2 \cdot A^2 = \text{scalar}^2 \cdot (-4I)$,which is a scalar multiple of $I$,hence $N^2$ is symmetric.
Since $M$ and $N^2$ are both symmetric and commute (as both are scalar multiples of $I$),their product $MN^2$ is also symmetric.
Since $M$ and $N^2$ are scalar matrices,$MN^2$ is a scalar matrix,which is symmetric but not necessarily the identity matrix.

(C) Given that $A$ is a $3 \times 3$ matrix,so $|A| = \Delta$.
Using the property $|\operatorname{adj}(M)| = |M|^{n-1}$ where $n=3$,we have $|\operatorname{adj}(M)| = |M|^2$.
Given $|\operatorname{adj}(24A)| = |\operatorname{adj}(3 \operatorname{adj}(2A))|$.
$|24A|^2 = |3 \operatorname{adj}(2A)|^2$.
Since $|kA| = k^n|A|$,$|24A| = 24^3|A|$.
So,$(24^3|A|)^2 = (3^3 |\operatorname{adj}(2A)|)^2$.
$(24^3|A|)^2 = (27 |2A|^2)^2$.
Since $|2A| = 2^3|A| = 8|A|$,we have $|\operatorname{adj}(2A)| = (8|A|)^2 = 64|A|^2$.
Substituting this back: $(24^3|A|)^2 = (27 \times 64|A|^2)^2$.
$(24^3|A|)^2 = (1728|A|^2)^2$.
Since $24^3 = 13824$,we have $(13824|A|)^2 = (1728|A|^2)^2$.
$13824|A| = 1728|A|^2$ (assuming $|A| \neq 0$).
$|A| = \frac{13824}{1728} = 8$.
We need to find $|A^2| = |A|^2 = 8^2 = 64 = 2^6$.

(B) Given $a = \alpha - i \beta$,where $\alpha, \beta \in \mathbb{R}$.
The system of equations is:
$4ix + (1 + i)y = 0$
$8(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3})x + \bar{a}y = 0$
Since the system has more than one solution,the determinant of the coefficient matrix must be zero:
$\begin{vmatrix} 4i & 1 + i \\ 8e^{i2\pi/3} & \bar{a} \end{vmatrix} = 0$
$4i\bar{a} - (1 + i)8e^{i2\pi/3} = 0$
$4i(\alpha + i\beta) - 8(1 + i)(\cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3}) = 0$
$4i\alpha - 4\beta - 8(1 + i)(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 0$
$4i\alpha - 4\beta - 4(1 + i)(-1 + i\sqrt{3}) = 0$
$4i\alpha - 4\beta - 4(-1 + i\sqrt{3} - i - \sqrt{3}) = 0$
$i\alpha - \beta - (-1 - \sqrt{3} + i(\sqrt{3} - 1)) = 0$
$i\alpha - \beta + 1 + \sqrt{3} - i(\sqrt{3} - 1) = 0$
Equating real and imaginary parts to zero:
Real part: $-\beta + 1 + \sqrt{3} = 0 \Rightarrow \beta = \sqrt{3} + 1$
Imaginary part: $\alpha - (\sqrt{3} - 1) = 0 \Rightarrow \alpha = \sqrt{3} - 1$
Therefore,$\frac{\alpha}{\beta} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 + 1 - 2\sqrt{3}}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$.

(C) Given $AB = I$,taking the determinant on both sides,we get $|A||B| = |I| = 1$.
Since $|A| = \frac{1}{8}$,we have $\frac{1}{8}|B| = 1$,which implies $|B| = 8$.
We need to find $|\operatorname{adj}(B \operatorname{adj}(2A))|$.
Using the property $|\operatorname{adj}(M)| = |M|^{n-1}$ for an $n \times n$ matrix,here $n=3$,so $|\operatorname{adj}(M)| = |M|^2$.
Thus,$|\operatorname{adj}(B \operatorname{adj}(2A))| = |B \operatorname{adj}(2A)|^2 = |B|^2 |\operatorname{adj}(2A)|^2$.
Since $|\operatorname{adj}(2A)| = |2A|^{3-1} = |2A|^2 = (2^3 |A|)^2 = (8 \times \frac{1}{8})^2 = 1^2 = 1$.
Substituting the values,we get $|B|^2 \times (1)^2 = 8^2 \times 1 = 64$.

(B) Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$,where $a, b, c, d \in \{0, 1, 2, 3, 4, 5\}$. The sum $S = a + b + c + d = p$,where $p \in \{3, 5, 7\}$.
Case $(i): S = 3$. The number of non-negative integer solutions to $a + b + c + d = 3$ is given by $\binom{3+4-1}{4-1} = \binom{6}{3} = 20$.
Case $(ii): S = 5$. The number of non-negative integer solutions to $a + b + c + d = 5$ is given by $\binom{5+4-1}{4-1} = \binom{8}{3} = 56$.
Case $(iii): S = 7$. The number of non-negative integer solutions to $a + b + c + d = 7$ with $a, b, c, d \le 5$ is calculated using inclusion-exclusion. Total solutions without restriction is $\binom{7+4-1}{4-1} = \binom{10}{3} = 120$. Subtract cases where at least one variable $\ge 6$. If $a \ge 6$,let $a = a' + 6$,then $a' + b + c + d = 1$,which has $\binom{1+4-1}{4-1} = \binom{4}{3} = 4$ solutions. Since there are $4$ variables,total invalid cases = $4 \times 4 = 16$. Thus,$120 - 16 = 104$.
Total number of matrices = $20 + 56 + 104 = 180$.

(A) Given $A = \begin{bmatrix} 1+i & 1 \\ -i & 0 \end{bmatrix}$.
Calculate $A^2$:
$A^2 = \begin{bmatrix} 1+i & 1 \\ -i & 0 \end{bmatrix} \begin{bmatrix} 1+i & 1 \\ -i & 0 \end{bmatrix} = \begin{bmatrix} (1+i)^2 - i & 1+i \\ -i(1+i) & -i \end{bmatrix} = \begin{bmatrix} 2i - i & 1+i \\ -i+1 & -i \end{bmatrix} = \begin{bmatrix} i & 1+i \\ 1-i & -i \end{bmatrix}$.
Calculate $A^4 = (A^2)^2$:
$A^4 = \begin{bmatrix} i & 1+i \\ 1-i & -i \end{bmatrix} \begin{bmatrix} i & 1+i \\ 1-i & -i \end{bmatrix} = \begin{bmatrix} i^2 + (1+i)(1-i) & i(1+i) - i(1+i) \\ i(1-i) - i(1-i) & (1-i)(1+i) + i^2 \end{bmatrix} = \begin{bmatrix} -1 + 2 & 0 \\ 0 & 2 - 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
We want $A^n = A$. Since $A^4 = I$,we have $A^{4k+1} = (A^4)^k \cdot A = I^k \cdot A = A$.
Thus,$n$ must be of the form $4k+1$ where $k \ge 0$.
Given $1 \le n \le 100$,we have $1 \le 4k+1 \le 100 \Rightarrow 0 \le 4k \le 99 \Rightarrow 0 \le k \le 24.75$.
Since $k$ is an integer,$k \in \{0, 1, 2, \ldots, 24\}$.
There are $25$ such values of $k$,so there are $25$ elements in the set.

(C) Given $A = \begin{bmatrix} 2 & -1 \\ 0 & 2 \end{bmatrix}$.
First,we find $\operatorname{adj} A$. For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$,the adjoint is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.
Thus,$\operatorname{adj} A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$.
The expression for $B$ is given by the binomial expansion: $B = (I - \operatorname{adj} A)^{5}$.
Calculate $I - \operatorname{adj} A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix}$.
Let $M = \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix}$. We need to find $M^{5}$.
$M^{2} = \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$.
$M^{3} = M^{2} \cdot M = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & -3 \\ 0 & -1 \end{bmatrix}$.
$M^{4} = M^{2} \cdot M^{2} = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix}$.
$M^{5} = M^{4} \cdot M = \begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} -1 & -1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & -5 \\ 0 & -1 \end{bmatrix}$.
The sum of all elements of matrix $B$ is $(-1) + (-5) + 0 + (-1) = -7$.

(D) Given $M = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix}$.
Calculating $M^2$: $M^2 = \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix} \begin{bmatrix} 0 & -\alpha \\ \alpha & 0 \end{bmatrix} = \begin{bmatrix} -\alpha^2 & 0 \\ 0 & -\alpha^2 \end{bmatrix} = -\alpha^2 I$.
Now,$N = \sum_{k=1}^{49} M^{2k} = M^2 + M^4 + \dots + M^{98}$.
Since $M^2 = -\alpha^2 I$,we have $M^{2k} = (M^2)^k = (-\alpha^2)^k I$.
Thus,$N = \sum_{k=1}^{49} (-\alpha^2)^k I = I \sum_{k=1}^{49} (-\alpha^2)^k$.
This is a geometric progression with first term $a = -\alpha^2$ and common ratio $r = -\alpha^2$ for $49$ terms.
$N = I \left( \frac{-\alpha^2(1 - (-\alpha^2)^{49})}{1 - (-\alpha^2)} \right) = I \left( \frac{-\alpha^2(1 + \alpha^{98})}{1 + \alpha^2} \right)$.
Given $(I - M^2)N = -2I$.
Since $M^2 = -\alpha^2 I$,$I - M^2 = I - (-\alpha^2 I) = (1 + \alpha^2)I$.
Substituting this into the equation: $(1 + \alpha^2)I \cdot \left( \frac{-\alpha^2(1 + \alpha^{98})}{1 + \alpha^2} \right) I = -2I$.
$(1 + \alpha^2) \cdot \frac{-\alpha^2(1 + \alpha^{98})}{1 + \alpha^2} = -2$.
$-\alpha^2(1 + \alpha^{98}) = -2$.
$\alpha^2(1 + \alpha^{98}) = 2$.
If $\alpha = 1$,then $1^2(1 + 1^{98}) = 1(1 + 1) = 2$. This satisfies the equation.
Thus,the positive integral value of $\alpha$ is $1$.

(A) Let $A = I + B$,where $I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & a & a \\ 0 & 0 & b \\ 0 & 0 & 0 \end{bmatrix}$.
Calculating $B^2 = \begin{bmatrix} 0 & 0 & ab \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $B^3 = 0$.
Using the Binomial Theorem,$A^n = (I + B)^n = I + nB + \frac{n(n-1)}{2} B^2$.
$A^n = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & na & na \\ 0 & 0 & nb \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & \frac{n(n-1)ab}{2} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & na & na + \frac{n(n-1)ab}{2} \\ 0 & 1 & nb \\ 0 & 0 & 1 \end{bmatrix}$.
Comparing with the given matrix $\begin{bmatrix} 1 & 48 & 2160 \\ 0 & 1 & 96 \\ 0 & 0 & 1 \end{bmatrix}$:
$na = 48$,$nb = 96$,and $na + \frac{n(n-1)ab}{2} = 2160$.
From $na = 48$ and $nb = 96$,we get $b = 2a$.
Substituting into the third equation: $48 + \frac{n(n-1)a(2a)}{2} = 2160 \Rightarrow 48 + n(n-1)a^2 = 2160 \Rightarrow n(n-1)a^2 = 2112$.
Since $na = 48$,$a = \frac{48}{n}$. Substituting this: $n(n-1)(\frac{48}{n})^2 = 2112 \Rightarrow (n-1) \frac{2304}{n} = 2112 \Rightarrow 2304n - 2304 = 2112n \Rightarrow 192n = 2304 \Rightarrow n = 12$.
Then $a = \frac{48}{12} = 4$ and $b = \frac{96}{12} = 8$.
Thus,$n + a + b = 12 + 4 + 8 = 24$.

(A) Given $A = \begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix}$. Calculating $A^2$:
$A^2 = \begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & -1 & -1 \\ 1 & 0 & -1 \\ 1 & -1 & 0 \end{bmatrix} = A$.
Thus, $A^n = A$ for all $n \geq 1$.
Now, $B = A - I = \begin{bmatrix} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{bmatrix}$.
Calculating $B^2$:
$B^2 = \begin{bmatrix} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 & -1 \\ 1 & -1 & -1 \\ 1 & -1 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 1 & 1 \\ -1 & 1 & 1 \\ -1 & 1 & 1 \end{bmatrix} = -B$.
Then $B^3 = -B^2 = B$, $B^4 = -B$, $B^5 = B$, and in general $B^n = B$ for odd $n$ and $B^n = -B$ for even $n$.
The equation is $A + \omega^n B^n = A + B$.
This implies $\omega^n B^n = B$.
Case $1$: $n$ is odd. Then $B^n = B$, so $\omega^n B = B \Rightarrow \omega^n = 1$.
Since $\omega = e^{i2\pi/3}$, $\omega^n = 1$ implies $n$ is a multiple of $3$.
So $n \in \{3, 9, 15, \ldots, 99\}$. This is an arithmetic progression with $a=3, d=6, l=99$.
$99 = 3 + (k-1)6 \Rightarrow 96 = (k-1)6 \Rightarrow 16 = k-1 \Rightarrow k = 17$.
Case $2$: $n$ is even. Then $B^n = -B$, so $\omega^n (-B) = B \Rightarrow \omega^n = -1$.
$\omega^n = -1$ implies $n$ is an odd multiple of $3$ (e.g., $3, 9, \ldots$), but $n$ must be even, so no solution here.
Thus, there are $17$ values.

(D) Given $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $A = A^{-1}$.
This implies $A^2 = I$,so $\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This gives the system of equations:
$1) a^2 + bc = 1$
$2) b(a + d) = 0$
$3) c(a + d) = 0$
$4) bc + d^2 = 1$
From $(1)$ and $(4)$,$a^2 = d^2$,so $a = d$ or $a = -d$.
Case $I$: $a = -d$. Then $b(a - a) = 0$ is always true. We need $a^2 + bc = 1$.
If $a = 0$,then $d = 0$ and $bc = 1$. Since $b, c \in \{-1, 0, 1, \ldots, 10\}$,$bc = 1$ implies $(b, c) = (1, 1)$ or $(-1, -1)$. ($2$ pairs).
If $a = 1$,then $d = -1$ and $1 + bc = 1 \Rightarrow bc = 0$. This means $b=0$ ($12$ values for $c$) or $c=0$ ($12$ values for $b$). Excluding $(0,0)$ which is counted twice,we have $12 + 12 - 1 = 23$ pairs.
If $a = -1$,then $d = 1$ and $1 + bc = 1 \Rightarrow bc = 0$. Similarly,$23$ pairs.
Case $II$: $a = d$. Then $b(2a) = 0$ and $c(2a) = 0$. If $a \neq 0$,then $b = c = 0$. Since $a^2 = 1$,$a = 1$ or $a = -1$. This gives $(1, 1)$ and $(-1, -1)$ for $(a, d)$. ($2$ pairs).
If $a = 0$,then $d = 0$,which leads to $bc = 1$,already covered in Case $I$.
Total = $2 + 23 + 23 + 2 = 50$.

(A) The characteristic equation of a $2 \times 2$ matrix $A$ is given by $\lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = 0$.
By the Cayley-Hamilton theorem,every square matrix satisfies its own characteristic equation,so $A^2 - \operatorname{tr}(A)A + \det(A)I = O$.
Given the equation $A^2 + \gamma A + 18I = O$,we compare this with the characteristic equation $A^2 - \operatorname{tr}(A)A + \det(A)I = O$.
Comparing the constant terms,we get $\det(A) = 18$.
Thus,the determinant of matrix $A$ is $18$.

(A) Given $A = \begin{bmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{bmatrix}$.
Applying $R_{3} \rightarrow R_{3} + R_{1}$,we get $|A| = \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \alpha+\beta+\gamma & \alpha+\beta+\gamma & \alpha+\beta+\gamma \end{vmatrix}$.
Taking $(\alpha+\beta+\gamma)$ common from $R_{3}$,we have $|A| = (\alpha+\beta+\gamma) \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ 1 & 1 & 1 \end{vmatrix}$.
Using the determinant of a Vandermonde matrix,$|A| = -(\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$.
Since $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))))| = |A|^{(n-1)^4} = |A|^{2^4} = |A|^{16}$,where $n=3$.
Substituting this into the given equation: $\frac{|A|^{16}}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}} = 2^{32} \times 3^{16}$.
This simplifies to $(\alpha+\beta+\gamma)^{16} = 2^{32} \times 3^{16} = (2^2 \times 3)^{16} = 12^{16}$.
Thus,$\alpha+\beta+\gamma = 12$.
Since $\alpha, \beta, \gamma \in \mathbb{N}$,the number of positive integer solutions is $\binom{12-1}{3-1} = \binom{11}{2} = 55$.
We must exclude cases where $\alpha, \beta, \gamma$ are not distinct. If $\alpha=\beta=\gamma$,then $3\alpha=12 \Rightarrow \alpha=4$,which is $1$ case $(4,4,4)$.
If two are equal,say $\alpha=\beta$,then $2\alpha+\gamma=12$. Possible values for $\alpha$ are $1, 2, 3, 5$ (since $\alpha=4$ gives $\gamma=4$). There are $4$ such pairs for each permutation of $(\alpha, \beta, \gamma)$,totaling $4 \times 3 = 12$ cases.
Total distinct tuples = $55 - 1 - 12 = 42$.

(D) The characteristic equation of matrix $A$ is given by $|A - \lambda I| = 0$.
$\begin{vmatrix} 1 - \lambda & 2 \\ -2 & -5 - \lambda \end{vmatrix} = 0$
$(1 - \lambda)(-5 - \lambda) - (2)(-2) = 0$
$-5 - \lambda + 5\lambda + \lambda^{2} + 4 = 0$
$\lambda^{2} + 4\lambda - 1 = 0$
By the Cayley-Hamilton theorem,every square matrix satisfies its characteristic equation,so $A^{2} + 4A - I = 0$,which implies $A^{2} + 4A = I$.
Multiplying by $2$,we get $2A^{2} + 8A = 2I$.
We are given $\alpha A^{2} + \beta A = 2I$.
Comparing the coefficients,we get $\alpha = 2$ and $\beta = 8$.
Therefore,$\alpha + \beta = 2 + 8 = 10$.

(A) Let $A = [a_{ij}]$ be a $3 \times 3$ matrix where $a_{ij} \in \{-1, 0, 1\}$.
The sum of the diagonal elements of $A^{T}A$ is given by the trace of $A^{T}A$,denoted as $\operatorname{Tr}(A^{T}A)$.
We know that $\operatorname{Tr}(A^{T}A) = \sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij}^{2}$.
Given that $\operatorname{Tr}(A^{T}A) = 6$,we have $\sum_{i=1}^{3} \sum_{j=1}^{3} a_{ij}^{2} = 6$.
Since $a_{ij} \in \{-1, 0, 1\}$,$a_{ij}^{2}$ can only be $0$ or $1$.
For the sum of nine such squares to be $6$,exactly $6$ entries must be $\pm 1$ and $3$ entries must be $0$.
First,we choose $3$ positions out of $9$ to be $0$,which can be done in $\binom{9}{3}$ ways.
For the remaining $6$ positions,each can be either $1$ or $-1$,which gives $2^{6}$ possibilities.
Therefore,the total number of such matrices is $\binom{9}{3} \times 2^{6} = 84 \times 64 = 5376$.

(A) Given $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$.
$A + B = \begin{bmatrix} \beta + 1 & 0 \\ 3 & \alpha \end{bmatrix}$.
$(A + B)^{2} = \begin{bmatrix} \beta + 1 & 0 \\ 3 & \alpha \end{bmatrix} \begin{bmatrix} \beta + 1 & 0 \\ 3 & \alpha \end{bmatrix} = \begin{bmatrix} (\beta + 1)^{2} & 0 \\ 3(\beta + 1) + 3\alpha & \alpha^{2} \end{bmatrix}$.
$A^{2} = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix} = \begin{bmatrix} -1 & -1 - \alpha \\ 2 + 2\alpha & \alpha^{2} - 2 \end{bmatrix}$.
For $\alpha_{1}$,$(A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 1 - \alpha \\ 4 + 2\alpha & \alpha^{2} \end{bmatrix}$.
Comparing elements,$(\beta + 1)^{2} = 1 \implies \beta + 1 = \pm 1$. Also,$1 - \alpha = 0 \implies \alpha_{1} = 1$.
For $\alpha_{2}$,$(A + B)^{2} = B^{2} = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} \beta^{2} + 1 & \beta \\ \beta & 1 \end{bmatrix}$.
Comparing elements,the $(1,2)$ position gives $0 = \beta$,and the $(2,2)$ position gives $\alpha_{2}^{2} = 1$. From the $(2,1)$ position,$3(\beta + 1) + 3\alpha = \beta$. Substituting $\beta = 0$,we get $3(1) + 3\alpha = 0 \implies 3\alpha = -3 \implies \alpha_{2} = -1$.
Thus,$|\alpha_{1} - \alpha_{2}| = |1 - (-1)| = |2| = 2$.

(D) Given $X = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $A = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix}$.
First,calculate $A^{2}$:
$A^{2} = \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 2 & 3 \\ 0 & 1 & 6 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Now,calculate $A^{4} = A^{2} \cdot A^{2} = \begin{bmatrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 12 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
By induction,for any even $k$,$A^{k} = \begin{bmatrix} 1 & 0 & 3k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
We are given $X^{T} A^{k} X = 33$. Substituting the expression for $A^{k}$:
$\begin{bmatrix} 1 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 3k \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 33$
$\begin{bmatrix} 1 & 1 & 3k+1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 33$
$1 + 1 + 3k + 1 = 33$
$3k + 3 = 33$
$3k = 30 \implies k = 10$.
Since $10$ is an even number,this solution is valid.

(A) Given $\Delta = \left|\begin{array}{ccc} p! & (p+1)! & (p+2)! \\ (p+1)! & (p+2)! & (p+3)! \\ (p+2)! & (p+3)! & (p+4)!\end{array}\right|$.
Taking $p!$,$(p+1)!$,and $(p+2)!$ common from rows $1, 2, 3$ respectively:
$\Delta = p!(p+1)!(p+2)! \left|\begin{array}{ccc} 1 & p+1 & (p+2)(p+1) \\ 1 & p+2 & (p+3)(p+2) \\ 1 & p+3 & (p+4)(p+3)\end{array}\right|$.
Applying $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$:
$\Delta = p!(p+1)!(p+2)! \left|\begin{array}{ccc} 1 & p+1 & (p+1)(p+2) \\ 0 & 1 & 2(p+2) \\ 0 & 1 & 2(p+3)\end{array}\right|$.
Expanding along the first column:
$\Delta = p!(p+1)!(p+2)! [1 \cdot (2(p+3) - 2(p+2))] = p!(p+1)!(p+2)! [2] = 2 \cdot p!(p+1)!(p+2)!$.
Since $p$ and $p+2$ are primes,$p!$ contains $p$ exactly once (for $p \ge 2$).
$(p+1)! = (p+1)p!$,so $p$ appears in $p!$ and $(p+1)!$,thus $p^2$ divides $p!(p+1)!$. Also $(p+2)!$ contains $p$ once. So $p^3$ divides $\Delta$,hence $\alpha = 3$.
$(p+2)!$ contains $(p+2)$ once. Thus $(p+2)^1$ divides $\Delta$,hence $\beta = 1$.
The sum $\alpha + \beta = 3 + 1 = 4$.

(B) Given the equation: $-3 x^4 + \operatorname{det}\begin{bmatrix} 1 & x & x^2 \\ 1 & x^2 & x^4 \\ 1 & x^3 & x^6 \end{bmatrix} = 0$.
First,we evaluate the determinant $D = \operatorname{det}\begin{bmatrix} 1 & x & x^2 \\ 1 & x^2 & x^4 \\ 1 & x^3 & x^6 \end{bmatrix}$.
Using the property of Vandermonde-like determinants or direct expansion:
$D = 1(x^2 \cdot x^6 - x^4 \cdot x^3) - x(1 \cdot x^6 - x^4 \cdot 1) + x^2(1 \cdot x^3 - x^2 \cdot 1)$
$D = (x^8 - x^7) - x(x^6 - x^4) + x^2(x^3 - x^2)$
$D = x^8 - x^7 - x^7 + x^5 + x^5 - x^4 = x^8 - 2x^7 + 2x^5 - x^4$.
Substituting this into the original equation:
$-3x^4 + x^8 - 2x^7 + 2x^5 - x^4 = 0$
$x^8 - 2x^7 + 2x^5 - 4x^4 = 0$.
Factoring the expression:
$x^4(x^4 - 2x^3 + 2x - 4) = 0$
$x^4(x^3(x - 2) + 2(x - 2)) = 0$
$x^4(x^3 + 2)(x - 2) = 0$.
Setting each factor to zero:
$x^4 = 0 \implies x = 0$
$x^3 + 2 = 0 \implies x^3 = -2 \implies x = \sqrt[3]{-2}$ (not an integer)
$x - 2 = 0 \implies x = 2$.
Since $x$ must be an integer,the possible values are $x = 0$ and $x = 2$.
Thus,there are $2$ such integers.

(A) The matrix $A$ is a $3 \times 3$ matrix with entries from the set $S = \{-1000, -999, \ldots, 1000\}$. The total number of elements in $S$ is $2001$.
The total number of possible matrices $A$ is $(2001)^9$.
Case $1$: $A^2 = -I$. If $A$ is a $3 \times 3$ matrix,the characteristic equation is given by $\det(A - \lambda I) = 0$. By the Cayley-Hamilton theorem,$A$ satisfies its characteristic equation. If $A^2 = -I$,then the minimal polynomial divides $x^2 + 1$. Since the degree of the minimal polynomial must divide the dimension $3$,and $x^2+1$ has no real roots,this is impossible for a $3 \times 3$ matrix with real (integer) entries. Thus,the number of such matrices is $0$.
Case $2$: $A$ is a diagonal matrix. Let $A = \text{diag}(a, b, c)$. The number of such matrices is $(2001)^3$.
Therefore,the total number of favorable outcomes is $(2001)^3$.
The probability $P$ is given by $P = \frac{(2001)^3}{(2001)^9} = \frac{1}{(2001)^6}$.
We have $P = \frac{1}{(2001)^6} = \frac{1}{(2000 + 1)^6} = \frac{1}{2000^6 (1 + \frac{1}{2000})^6} = \frac{1}{64 \times 10^{18} (1 + \frac{1}{2000})^6}$.
Since $(1 + \frac{1}{2000})^6 > 1$,it follows that $P < \frac{1}{64 \times 10^{18}} < \frac{1}{10^{18}}$.
Thus,$P < \frac{1}{10^{18}}$.

(C) Given $A^{-1} = \begin{bmatrix} 1 & 2017 & 2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8 \end{bmatrix}$.
First,calculate the determinant $|A^{-1}|$.
Applying $R_1 \rightarrow R_1 - R_2$:
$|A^{-1}| = \begin{vmatrix} 0 & 0 & -2 \\ 1 & 2017 & 4 \\ 1 & 2018 & 8 \end{vmatrix}$.
Expanding along $R_1$:
$|A^{-1}| = -2(2018 - 2017) = -2(1) = -2$.
We know that $|A| = \frac{1}{|A^{-1}|} = \frac{1}{-2} = -0.5$.
For a matrix $A$ of order $n=3$,$|kA| = k^n |A|$.
Thus,$|2A| = 2^3 |A| = 8|A| = 8 \times (-0.5) = -4$.
Similarly,$|2A^{-1}| = 2^3 |A^{-1}| = 8|A^{-1}| = 8 \times (-2) = -16$.
Therefore,$|2A| - |2A^{-1}| = -4 - (-16) = -4 + 16 = 12$.

(D) Given that $P^2 = P$. This is an idempotent matrix.
We use the binomial expansion for $(I+P)^n$:
$(I+P)^n = \sum_{k=0}^{n} \binom{n}{k} I^{n-k} P^k$
Since $I^m = I$ for any $m \ge 1$ and $P^k = P$ for all $k \ge 1$ (because $P^2=P, P^3=P^2P=PP=P$,etc.),we have:
$(I+P)^n = I + \sum_{k=1}^{n} \binom{n}{k} P$
$(I+P)^n = I + P \left( \sum_{k=1}^{n} \binom{n}{k} \right)$
We know that $\sum_{k=0}^{n} \binom{n}{k} = 2^n$,so $\sum_{k=1}^{n} \binom{n}{k} = 2^n - \binom{n}{0} = 2^n - 1$.
Therefore,$(I+P)^n = I + (2^n - 1)P$.

(B) Given that $A B = B A$,$A^k = I$,and $B^l = 0$ for some positive integers $k$ and $l$.
Since $B^l = 0$,taking the determinant on both sides,we get $\operatorname{det}(B^l) = \operatorname{det}(0) = 0$.
Using the property $\operatorname{det}(B^l) = (\operatorname{det}(B))^l$,we have $(\operatorname{det}(B))^l = 0$,which implies $\operatorname{det}(B) = 0$.
Now,consider the determinant of the product $A B$:
$\operatorname{det}(A B) = \operatorname{det}(A) \times \operatorname{det}(B)$.
Since $\operatorname{det}(B) = 0$,it follows that $\operatorname{det}(A B) = \operatorname{det}(A) \times 0 = 0$.
Thus,the correct option is $(b)$.

(B) Given $A = \left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$.
Calculating powers of $A$:
$A^2 = \left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right] \left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right] = \left[\begin{array}{cc}i^2 & 0 \\ 0 & i^2\end{array}\right] = \left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] = -I$.
$A^3 = A^2 \cdot A = (-I) \cdot A = -A$.
$A^4 = A^2 \cdot A^2 = (-I) \cdot (-I) = I$.
Since $A^4 = I$,the powers of $A$ repeat in a cycle of $4$: $I, A, -I, -A, I, \dots$.
The sum of any four consecutive terms is $I + A + A^2 + A^3 = I + A - I - A = 0$.
The series is $S = I + A + A^2 + \dots + A^{2010}$.
There are $2011$ terms in total. Since $2011 = 4 \times 502 + 3$,the sum consists of $502$ groups of $4$ terms (each summing to $0$) plus the remaining $3$ terms:
$S = 502(0) + (I + A + A^2) = I + A - I = A$.
Therefore,$S = \left[\begin{array}{ll}0 & i \\ i & 0\end{array}\right]$.

(A) Let $P(x)$ be a cubic polynomial. We are given $P(x) = 2x$ for $x = 1, 2, 3, 4$.
Define a new polynomial $Q(x) = P(x) - 2x$.
Since $P(x)$ is a cubic polynomial,$Q(x)$ is also a polynomial of degree at most $3$.
From the given conditions,$Q(1) = P(1) - 2(1) = 2 - 2 = 0$,$Q(2) = P(2) - 2(2) = 4 - 4 = 0$,$Q(3) = P(3) - 2(3) = 6 - 6 = 0$,and $Q(4) = P(4) - 2(4) = 8 - 8 = 0$.
Thus,$1, 2, 3, 4$ are the roots of $Q(x)$.
Since $Q(x)$ is a polynomial of degree at most $3$ and it has $4$ distinct roots,$Q(x)$ must be the zero polynomial.
Therefore,$P(x) - 2x = 0$,which implies $P(x) = 2x$.
However,$P(x) = 2x$ is a polynomial of degree $1$,not $3$.
Thus,there is no cubic polynomial $P(x)$ that satisfies the given conditions.
Hence,the number of such cubic polynomials is $0$.

(D) Given the equation: $A^2 + B = A^2 B$
Rearrange the terms to group $B$ on one side:
$A^2 = A^2 B - B$
$A^2 = (A^2 - I)B$
Alternatively,rearrange to factorize:
$A^2 B - B = A^2$
$B(A^2 - I) = A^2$
Consider the expression $(A^2 - I)(B - I) = A^2 B - A^2 - B + I$
Substitute $A^2 B = A^2 + B$ into the expression:
$(A^2 - I)(B - I) = (A^2 + B) - A^2 - B + I = I$
Since $(A^2 - I)(B - I) = I$,it implies that the matrices $(A^2 - I)$ and $(B - I)$ commute.
Therefore,$(A^2 - I)(B - I) = (B - I)(A^2 - I) = I$
Expanding $(B - I)(A^2 - I) = I$:
$B A^2 - B - A^2 + I = I$
$B A^2 = A^2 + B$
Since $A^2 + B = A^2 B$,we conclude that:
$A^2 B = B A^2$

(B) Given the matrix is singular,its determinant is $0$:
$\Delta = \begin{vmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{vmatrix} = 0$
Expanding along the first row:
$\alpha^2(c-b) - \alpha(c-a) + (b-a) = 0$
This is the same as the given equation $(a-c)x^2 + (b-a)x + (c-b) = 0$ if we note that $\alpha=1$ is a root because $(a-c) + (b-a) + (c-b) = 0$.
Let $X = a-c$,$Y = b-a$,and $Z = c-b$. Note that $X+Y+Z = 0$.
The expression is $\frac{X^2}{YZ} + \frac{Y^2}{XZ} + \frac{Z^2}{XY} = \frac{X^3 + Y^3 + Z^3}{XYZ}$.
Since $X+Y+Z = 0$,we have the identity $X^3 + Y^3 + Z^3 = 3XYZ$.
Therefore,the expression becomes $\frac{3XYZ}{XYZ} = 3$.

(B) Given $A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix}$.
Using the property $\log_a b = \frac{\ln b}{\ln a}$,we can write $A = \begin{bmatrix} 1 & \frac{\ln y}{\ln x} & \frac{\ln z}{\ln x} \\ \frac{\ln x}{\ln y} & 2 & \frac{\ln z}{\ln y} \\ \frac{\ln x}{\ln z} & \frac{\ln y}{\ln z} & 3 \end{bmatrix}$.
Multiply $R_1$ by $\ln x$,$R_2$ by $\ln y$,and $R_3$ by $\ln z$:
$|A| = \frac{1}{\ln x \ln y \ln z} \begin{vmatrix} \ln x & \ln y & \ln z \\ \ln x & 2 \ln y & \ln z \\ \ln x & \ln y & 3 \ln z \end{vmatrix}$.
Factoring out $\ln x, \ln y, \ln z$ from the columns:
$|A| = \frac{\ln x \ln y \ln z}{\ln x \ln y \ln z} \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 3 \end{vmatrix} = 1(6-1) - 1(3-1) + 1(1-2) = 5 - 2 - 1 = 2$.
We know that $|\operatorname{adj}(\operatorname{adj} M)| = |M|^{(n-1)^2}$,where $n$ is the order of the matrix.
Here $n=3$,so $|\operatorname{adj}(\operatorname{adj} A^2)| = |A^2|^{(3-1)^2} = |A^2|^4 = (|A|^2)^4 = |A|^8$.
Since $|A| = 2$,$|A|^8 = 2^8$.

(B) The $n^{\text{th}}$ term of an $A$.$P$. is given by $T_n = a_1 + (n-1)d$.
Given $A_1, A_2, A_3$ have first terms $A, A+1, A+2$ and common difference $d$:
$a = A + (7-1)d = A + 6d$
$b = (A+1) + (9-1)d = A + 1 + 8d$
$c = (A+2) + (17-1)d = A + 2 + 16d$
Given $a = 29$,so $A + 6d = 29$.
The determinant equation is:
$\left|\begin{array}{lll} A+6d & 7 & 1 \\ 2(A+1+8d) & 17 & 1 \\ A+2+16d & 17 & 1\end{array}\right| + 70 = 0$
Subtracting row $2$ from row $3$:
$\left|\begin{array}{lll} A+6d & 7 & 1 \\ 2A+2+16d & 17 & 1 \\ -A & 0 & 0\end{array}\right| + 70 = 0$
Expanding along the third row:
$-(-A) \times (7 - 17) + 70 = 0 \Rightarrow 10A + 70 = 0 \Rightarrow A = -7$.
Since $A + 6d = 29$,we have $-7 + 6d = 29 \Rightarrow 6d = 36 \Rightarrow d = 6$.
Now,$a = 29$,$b = -7 + 1 + 8(6) = 42$,$c = -7 + 2 + 16(6) = 91$.
First term of new $A$.$P$. is $c - a - b = 91 - 29 - 42 = 20$.
Common difference is $\frac{d}{12} = \frac{6}{12} = 0.5$.
Sum of first $20$ terms $S_{20} = \frac{20}{2} [2(20) + (20-1)(0.5)] = 10 [40 + 9.5] = 10 [49.5] = 495$.

(D) Given $A^2 = 3A + \alpha I$.
Multiplying by $A$,we get $A^3 = 3A^2 + \alpha A$.
Substituting $A^2 = 3A + \alpha I$ into the expression for $A^3$:
$A^3 = 3(3A + \alpha I) + \alpha A = 9A + 3\alpha I + \alpha A = (9 + \alpha)A + 3\alpha I$.
Now,multiplying by $A$ again to find $A^4$:
$A^4 = (9 + \alpha)A^2 + 3\alpha A$.
Substituting $A^2 = 3A + \alpha I$ again:
$A^4 = (9 + \alpha)(3A + \alpha I) + 3\alpha A$.
$A^4 = (27 + 3\alpha)A + (9\alpha + \alpha^2)I + 3\alpha A$.
$A^4 = (27 + 6\alpha)A + (9\alpha + \alpha^2)I$.
Comparing this with $A^4 = 21A + \beta I$,we get:
$27 + 6\alpha = 21 \Rightarrow 6\alpha = -6 \Rightarrow \alpha = -1$.
And $\beta = 9\alpha + \alpha^2 = 9(-1) + (-1)^2 = -9 + 1 = -8$.
Thus,$\alpha = -1$ and $\beta = -8$.