Mix Examples-Determinants and Matrices Questions in English

(B) Given the matrix $A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 0 \\ 3 & 0 & 3 \end{bmatrix}$.
The characteristic equation is given by $\operatorname{det}(A - \lambda I_{3}) = 0$.
$\begin{vmatrix} 3-\lambda & 0 & 3 \\ 0 & 3-\lambda & 0 \\ 3 & 0 & 3-\lambda \end{vmatrix} = 0$.
Expanding along the second row:
$(3-\lambda) \begin{vmatrix} 3-\lambda & 3 \\ 3 & 3-\lambda \end{vmatrix} = 0$.
$(3-\lambda) [(3-\lambda)^2 - 9] = 0$.
$(3-\lambda) [9 + \lambda^2 - 6\lambda - 9] = 0$.
$(3-\lambda) (\lambda^2 - 6\lambda) = 0$.
$(3-\lambda) \lambda (\lambda - 6) = 0$.
Thus,the roots are $\lambda = 0, 3, 6$.

(A) Given that $AB = B$ and $BA = A$.
We need to find the value of $A^{2} + B^{2}$.
Since $A^{2} = A \cdot A$,substituting $A = BA$,we get $A^{2} = A(BA) = (AB)A$.
Using $AB = B$,we get $A^{2} = BA = A$.
Similarly,$B^{2} = B \cdot B$,substituting $B = AB$,we get $B^{2} = B(AB) = (BA)B$.
Using $BA = A$,we get $B^{2} = AB = B$.
Therefore,$A^{2} + B^{2} = A + B$.

(B) Let $A = \begin{bmatrix} \cos \frac{\pi}{4} & \sin \frac{\pi}{4} \\ -\sin \frac{\pi}{4} & \cos \frac{\pi}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{bmatrix}$.
This is a rotation matrix $R_{\theta} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$ with $\theta = \frac{\pi}{4}$.
By the property of rotation matrices,$A^n = R_{n\theta} = \begin{bmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{bmatrix}$.
We want $A^n = I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This implies $\cos(n\theta) = 1$ and $\sin(n\theta) = 0$.
This occurs when $n\theta = 2k\pi$ for some integer $k$.
Substituting $\theta = \frac{\pi}{4}$,we get $n \cdot \frac{\pi}{4} = 2k\pi$.
$n = 8k$.
For the least positive integer $n$,we set $k = 1$,which gives $n = 8$.

(A) Given $B = PAP^{-1}$,we can multiply by $P$ on the right to get $BP = PA$.
Substituting the matrices:
$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ x & 0 & 0 \\ 0 & 0 & y \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 \\ x & 0 & 0 \\ 0 & 0 & y \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
Calculating the products:
Left side: $\begin{bmatrix} x & 0 & 0 \\ 0 & 0 & y \\ 0 & 0 & 0 \end{bmatrix}$
Right side: $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & x \\ 0 & 0 & 0 \end{bmatrix}$
Equating the corresponding elements,we get $x = 1$ and $y = x$.
Therefore,$x = 1$ and $y = 1$.

(C) Given that $A$ and $B$ are skew-symmetric matrices,we have $A^{\top} = -A$ and $B^{\top} = -B$.
Given $AB = BA$.
We need to evaluate the expression $E = A^{2} B^{2} (A^{\top} B)^{-1} (A B^{-1})^{\top}$.
Substituting $A^{\top} = -A$,we get $E = A^{2} B^{2} (-AB)^{-1} (A B^{-1})^{\top}$.
Using the property $(XY)^{-1} = Y^{-1} X^{-1}$ and $(XY)^{\top} = Y^{\top} X^{\top}$,we have:
$E = A^{2} B^{2} (-B^{-1} A^{-1}) ((B^{-1})^{\top} A^{\top})$.
Since $(B^{-1})^{\top} = (B^{\top})^{-1} = (-B)^{-1} = -B^{-1}$,we substitute this:
$E = A^{2} B^{2} (-B^{-1} A^{-1}) (-B^{-1} (-A))$.
$E = A^{2} B^{2} (-B^{-1} A^{-1}) (B^{-1} A)$.
Since $AB = BA$,we also have $A^{-1} B = B A^{-1}$ and $A B^{-1} = B^{-1} A$.
$E = A^{2} B^{2} (-B^{-1} A^{-1} B^{-1} A)$.
$E = -A^{2} B^{2} B^{-1} A^{-1} B^{-1} A$.
$E = -A^{2} B (B B^{-1}) A^{-1} B^{-1} A$.
$E = -A^{2} B (I) A^{-1} B^{-1} A$.
$E = -A^{2} (B A^{-1}) B^{-1} A$.
Since $B A^{-1} = A^{-1} B$,we have:
$E = -A^{2} A^{-1} B B^{-1} A$.
$E = -A (I) (I) A = -A^{2}$.

(C) The determinant of the matrix $M_r$ is given by: $\det(M_r) = r(r) - (r-1)(r-1) = r^2 - (r-1)^2$.
Expanding this,we get: $\det(M_r) = r^2 - (r^2 - 2r + 1) = 2r - 1$.
We need to calculate the sum: $\sum_{r=1}^{2008} \det(M_r) = \sum_{r=1}^{2008} (2r - 1)$.
This is the sum of the first $2008$ odd numbers,which is given by the formula $\sum_{r=1}^{n} (2r - 1) = n^2$.
For $n = 2008$,the sum is $(2008)^2$.

(A) The given determinant is $D = \left|\begin{array}{ccc} 1+1+1 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \\ 1+\alpha+\beta & 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 \\ 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4 \end{array}\right|$.
This can be written as the product of two determinants:
$D = \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{array}\right| \times \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & \alpha & \alpha^2 \\ 1 & \beta & \beta^2 \end{array}\right|$.
Both determinants are Vandermonde determinants,so $D = \{(1-\alpha)(\alpha-\beta)(\beta-1)\}^2 = (1-\alpha)^2(\alpha-\beta)^2(\beta-1)^2$.
Since $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$,we have $(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta = b^2/a^2 - 4c/a = (b^2-4ac)/a^2$.
Also,$(1-\alpha)(1-\beta) = 1 - (\alpha+\beta) + \alpha\beta = 1 + b/a + c/a = (a+b+c)/a$.
Thus,$D = \{(1-\alpha)(1-\beta)\}^2 (\alpha-\beta)^2 = \left(\frac{a+b+c}{a}\right)^2 \left(\frac{b^2-4ac}{a^2}\right) = (b^2-4ac) \frac{(a+b+c)^2}{a^4}$.
Comparing this with the given expression,we get $k = b^2-4ac$.

(C) The characteristic equation is given by $\det(A - \lambda I_{3}) = 0$.
Calculating the determinant:
$\begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & \cos t-\lambda & \sin t \\ 0 & -\sin t & \cos t-\lambda \end{vmatrix} = 0$.
Expanding along the first row:
$(1-\lambda) [(\cos t - \lambda)^2 - (-\sin^2 t)] = 0$.
$(1-\lambda) [\cos^2 t - 2\lambda \cos t + \lambda^2 + \sin^2 t] = 0$.
Since $\cos^2 t + \sin^2 t = 1$,we have:
$(1-\lambda) [\lambda^2 - 2\lambda \cos t + 1] = 0$.
Expanding this:
$-\lambda^3 + \lambda^2(1 + 2\cos t) - \lambda(2\cos t + 1) + 1 = 0$.
The sum of the roots $\lambda_{1} + \lambda_{2} + \lambda_{3}$ of a cubic equation $a\lambda^3 + b\lambda^2 + c\lambda + d = 0$ is given by $-b/a$.
Here,$\lambda_{1} + \lambda_{2} + \lambda_{3} = -\frac{1 + 2\cos t}{-1} = 1 + 2\cos t$.
Given $\lambda_{1} + \lambda_{2} + \lambda_{3} = \sqrt{2} + 1$,we equate:
$1 + 2\cos t = 1 + \sqrt{2} \Rightarrow 2\cos t = \sqrt{2} \Rightarrow \cos t = \frac{1}{\sqrt{2}}$.
For $-\pi \leq t < \pi$,the values of $t$ satisfying $\cos t = \frac{1}{\sqrt{2}}$ are $t = \frac{\pi}{4}$ and $t = -\frac{\pi}{4}$.

(A) First,calculate the value of $A$:
$A = -1(1 - (-12)) - 7(2 - (-9)) + 0 = -1(13) - 7(11) = -13 - 77 = -90$.
Now,let $B = \left|\begin{array}{ccc}13 & -11 & 5 \\ -7 & -1 & 25 \\ -21 & -3 & -15\end{array}\right|$.
Taking $5$ common from $C_3$ and $3$ common from $R_3$:
$B = 5 \times 3 \left|\begin{array}{ccc}13 & -11 & 1 \\ -7 & -1 & 5 \\ -7 & -1 & -1\end{array}\right| = 15 \left|\begin{array}{ccc}13 & -11 & 1 \\ -7 & -1 & 5 \\ 0 & 0 & -6\end{array}\right|$ (using $R_3 \rightarrow R_3 - R_2$).
Expanding along $R_3$:
$B = 15 \times (-6) \left|\begin{array}{cc}13 & -11 \\ -7 & -1\end{array}\right| = -90 \times (-13 - 77) = -90 \times (-90) = 8100$.
Since $A = -90$,then $A^2 = (-90)^2 = 8100$.
Therefore,$B = A^2$.

(C) We have,$a_{r}=(\cos 2 r \pi+i \sin 2 r \pi)^{1 / 9} = e^{\frac{2 r \pi i}{9}}$.
Now,the determinant is $\Delta = \left|\begin{array}{ccc} e^{\frac{2 \pi i}{9}} & e^{\frac{4 \pi i}{9}} & e^{\frac{6 \pi i}{9}} \\ e^{\frac{8 \pi i}{9}} & e^{\frac{10 \pi i}{9}} & e^{\frac{12 \pi i}{9}} \\ e^{\frac{14 \pi i}{9}} & e^{\frac{16 \pi i}{9}} & e^{\frac{18 \pi i}{9}} \end{array}\right|$.
Observe that the ratio of corresponding elements in $R_{2}$ and $R_{1}$ is $e^{\frac{6 \pi i}{9}} = e^{\frac{2 \pi i}{3}}$.
Specifically,$a_{4} = a_{1} \cdot e^{\frac{6 \pi i}{9}}$,$a_{5} = a_{2} \cdot e^{\frac{6 \pi i}{9}}$,and $a_{6} = a_{3} \cdot e^{\frac{6 \pi i}{9}}$.
Since row $R_{2}$ is a scalar multiple of row $R_{1}$,the rows are linearly dependent.
Therefore,the value of the determinant is $0$.

(A) To find the constant term of the polynomial $f(x)$,we set $x = 0$.
Substituting $x = 0$ into the determinant,we get:
$f(0) = \left|\begin{array}{ccc} (1+0)^{a} & (2+0)^{b} & 1 \\ 1 & (1+0)^{a} & (2+0)^{b} \\ (2+0)^{b} & 1 & (1+0)^{a} \end{array}\right| = \left|\begin{array}{ccc} 1 & 2^{b} & 1 \\ 1 & 1 & 2^{b} \\ 2^{b} & 1 & 1 \end{array}\right|$.
Expanding the determinant along the first row:
$f(0) = 1 \cdot \left|\begin{array}{cc} 1 & 2^{b} \\ 1 & 1 \end{array}\right| - 2^{b} \cdot \left|\begin{array}{cc} 1 & 2^{b} \\ 2^{b} & 1 \end{array}\right| + 1 \cdot \left|\begin{array}{cc} 1 & 1 \\ 2^{b} & 1 \end{array}\right|$.
Calculating the $2 \times 2$ determinants:
$f(0) = 1(1 - 2^{b}) - 2^{b}(1 - (2^{b})^{2}) + 1(1 - 2^{b})$.
$f(0) = (1 - 2^{b}) - 2^{b}(1 - 2^{2b}) + (1 - 2^{b})$.
$f(0) = 1 - 2^{b} - 2^{b} + 2^{3b} + 1 - 2^{b}$.
$f(0) = 2 - 3 \cdot 2^{b} + 2^{3b}$.

(C) Given the determinant $f(\theta) = \left|\begin{array}{ccc} 1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1 \end{array}\right|$.
Expanding along the first row:
$f(\theta) = 1(1 - (-\tan^2 \theta)) - \tan \theta(-\tan \theta - (-\tan \theta)) + 1(\tan^2 \theta - (-1))$
$f(\theta) = 1(1 + \tan^2 \theta) - \tan \theta(0) + 1(\tan^2 \theta + 1)$
$f(\theta) = (1 + \tan^2 \theta) + (1 + \tan^2 \theta) = 2(1 + \tan^2 \theta) = 2 \sec^2 \theta$.
Since $\theta \in [0, \pi/2)$,$\tan \theta \in [0, \infty)$,so $\sec^2 \theta \in [1, \infty)$.
Therefore,$f(\theta) = 2 \sec^2 \theta \in [2, \infty)$.

(A) matrix $A$ is Hermitian if $A = \bar{A}^T$. Let us check the given matrix $z$.
The transpose of $z$ is $z^T = \begin{bmatrix} 1 & 1-2i & 5i \\ 1+2i & -3 & 5-3i \\ -5i & 5+3i & 7 \end{bmatrix}$.
The conjugate of $z$ is $\bar{z} = \begin{bmatrix} 1 & 1-2i & 5i \\ 1+2i & -3 & 5-3i \\ -5i & 5+3i & 7 \end{bmatrix}$.
Since $z^T = \bar{z}$,the matrix $z$ is a Hermitian matrix.
For any Hermitian matrix,the determinant is always a real number.
Let $D = \det(z)$. Since $D$ is real,$D = \bar{D}$.
Thus,$z$ (as a determinant value) is purely real.

(D) Let the determinant be $\Delta = \left|\begin{array}{ccc}1 & 1+\omega^2 & \omega^2 \\ 1-i & -1 & \omega^2-1 \\ -i & -1+\omega & -1\end{array}\right|$.
Since $1+\omega+\omega^2 = 0$,we have $1+\omega^2 = -\omega$.
Substituting this,$\Delta = \left|\begin{array}{ccc}1 & -\omega & \omega^2 \\ 1-i & -1 & \omega^2-1 \\ -i & -1+\omega & -1\end{array}\right|$.
Applying $C_2 \rightarrow C_2 + C_1$:
$\Delta = \left|\begin{array}{ccc}1 & 1-\omega & \omega^2 \\ 1-i & -i & \omega^2-1 \\ -i & -1 & -1\end{array}\right|$.
By evaluating the determinant directly or simplifying further,we find that the rows/columns are linearly dependent or the expansion results in $0$.
Thus,the value of the determinant is $0$.

(A) Since $A$ and $B$ are orthogonal matrices,we have $AA^{\top} = I$ and $BB^{\top} = I$.
Taking the determinant on both sides,we get $\operatorname{det}(A)\operatorname{det}(A^{\top}) = 1$ and $\operatorname{det}(B)\operatorname{det}(B^{\top}) = 1$.
Since $\operatorname{det}(A^{\top}) = \operatorname{det}(A)$,we have $(\operatorname{det}(A))^2 = 1$ and $(\operatorname{det}(B))^2 = 1$,which implies $\operatorname{det}(A) = \pm 1$ and $\operatorname{det}(B) = \pm 1$.
Given $\operatorname{det}(A) + \operatorname{det}(B) = 0$,we have $\operatorname{det}(A) = -\operatorname{det}(B)$.
Now,consider $\operatorname{det}(A+B) = \operatorname{det}(A(I + A^{-1}B))$.
Since $A$ is orthogonal,$A^{-1} = A^{\top}$. Thus,$\operatorname{det}(A+B) = \operatorname{det}(A(I + A^{\top}B))$.
We can write $A+B = A(I + A^{\top}B) = A(B^{\top}B + A^{\top}B) = A(B^{\top} + A^{\top})B$.
Taking the determinant: $\operatorname{det}(A+B) = \operatorname{det}(A) \operatorname{det}(B^{\top} + A^{\top}) \operatorname{det}(B) = \operatorname{det}(A)\operatorname{det}(B) \operatorname{det}((A+B)^{\top})$.
Since $\operatorname{det}(A) = -\operatorname{det}(B)$,we have $\operatorname{det}(A)\operatorname{det}(B) = -(\operatorname{det}(B))^2 = -1$.
Thus,$\operatorname{det}(A+B) = -1 \cdot \operatorname{det}(A+B)$.
This implies $2 \operatorname{det}(A+B) = 0$,so $\operatorname{det}(A+B) = 0$.
Therefore,$A+B$ is a singular matrix.

(B) Let the determinant be $\Delta = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$.
Since each element can be $0$ or $1$,there are $2^4 = 16$ possible determinants.
The determinant is zero if $ad = bc$.
Case $1$: $ad = 0$ and $bc = 0$.
For $ad=0$,the pairs $(a,d)$ can be $(0,0), (0,1), (1,0)$,which are $3$ possibilities.
Similarly,for $bc=0$,there are $3$ possibilities.
Total cases for $ad=bc=0$ is $3 \times 3 = 9$.
Case $2$: $ad = 1$ and $bc = 1$.
This only happens if $a=1, d=1$ and $b=1, c=1$,which is $1$ possibility.
Total cases where $\Delta = 0$ is $9 + 1 = 10$.
Total cases where $\Delta \neq 0$ is $16 - 10 = 6$.
Probability $= \frac{6}{16} = \frac{3}{8}$.

(A) Given $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$. We can write $A = I + M$,where $M = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}$.
Calculating $M^2$: $M^2 = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 4-4 & -8+8 \\ 2-2 & -4+4 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$.
Since $M^2 = O$,then $M^k = O$ for all $k \geq 2$.
Using the Binomial Theorem for matrices: $A^{100} = (I + M)^{100} = I^{100} + \binom{100}{1} I^{99} M + \binom{100}{2} I^{98} M^2 + \dots = I + 100M$.
Given $A^{100} = 100B + I$,we have $I + 100M = 100B + I$,which implies $B = M = \begin{bmatrix} 2 & -4 \\ 1 & -2 \end{bmatrix}$.
Since $B^2 = M^2 = O$,then $B^{100} = O = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
The sum of all elements of $B^{100}$ is $0 + 0 + 0 + 0 = 0$.

(B) Given $B = (I + A)^{-1}$ and $A + C = I$.
From $A + C = I$,we have $A = I - C$.
Substituting this into the expression for $B$:
$B = (I + (I - C))^{-1} = (2I - C)^{-1}$.
This implies $B(2I - C) = I$,so $2B - BC = I$.
Also,$(2I - C)B = I$,so $2B - CB = I$.
Comparing the two expressions,$2B - BC = 2B - CB$,which implies $BC = CB$.
Given $BC = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}$,it follows that $CB = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}$.
We need to solve $CB \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 12 \\ -6 \end{bmatrix}$.
Let $M = \begin{bmatrix} 1 & -5 \\ -1 & 2 \end{bmatrix}$. The determinant $|M| = (1)(2) - (-5)(-1) = 2 - 5 = -3$.
The inverse $M^{-1} = \frac{1}{-3} \begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix} = -\frac{1}{3} \begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix}$.
Then $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = M^{-1} \begin{bmatrix} 12 \\ -6 \end{bmatrix} = -\frac{1}{3} \begin{bmatrix} 2 & 5 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 12 \\ -6 \end{bmatrix} = -\frac{1}{3} \begin{bmatrix} 24 - 30 \\ 12 - 6 \end{bmatrix} = -\frac{1}{3} \begin{bmatrix} -6 \\ 6 \end{bmatrix} = \begin{bmatrix} 2 \\ -2 \end{bmatrix}$.
Thus,$x_1 = 2$ and $x_2 = -2$.
Therefore,$x_1 + x_2 = 2 + (-2) = 0$.

(B) Let $A = \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{bmatrix}_{3 \times 2}$.
Then $A^{T}A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix} \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \end{bmatrix} = \begin{bmatrix} a_1^2 + a_2^2 + a_3^2 & \dots \\ \dots & b_1^2 + b_2^2 + b_3^2 \end{bmatrix}$.
The sum of the diagonal elements is $\text{Tr}(A^{T}A) = a_1^2 + a_2^2 + a_3^2 + b_1^2 + b_2^2 + b_3^2 = 5$.
We need to find the number of ways to choose $6$ elements from $\{-2, -1, 0, 1, 2\}$ such that the sum of their squares is $5$.
The possible combinations of squares are:
$1) \{1, 1, 1, 1, 1, 0\}$: The number of ways is $\frac{6!}{5!} \times 2^5 = 6 \times 32 = 192$.
$2) \{4, 1, 0, 0, 0, 0\}$: The number of ways is $\frac{6!}{4!} \times 2^2 = 30 \times 4 = 120$.
Total number of ways = $192 + 120 = 312$.

(C) First,we find the general form of $A^n$. Given $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$,we observe $A^2 = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 5 & -8 \\ 2 & -3 \end{bmatrix}$.
By induction,$A^n = \begin{bmatrix} 2n+1 & -4n \\ n & -2n+1 \end{bmatrix}$.
For $n=15$,$A^{15} = \begin{bmatrix} 2(15)+1 & -4(15) \\ 15 & -2(15)+1 \end{bmatrix} = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix}$.
Now,$A^{15}+B = \begin{bmatrix} 31 & -60 \\ 15 & -29 \end{bmatrix} + \begin{bmatrix} -29 & 49 \\ -13 & 18 \end{bmatrix} = \begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix}$.
The equation $(A^{15}+B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ becomes $\begin{bmatrix} 2 & -11 \\ 2 & -11 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$.
This simplifies to the linear equation $2x - 11y = 0$,or $2x = 11y$.
Checking the options,for $x=11$ and $y=2$,we get $2(11) = 22$ and $11(2) = 22$. Thus,$x=11, y=2$ is the correct solution.

(C) Given $A = \begin{bmatrix} 0 & 2 & -3 \\ -2 & 0 & 1 \\ 3 & -1 & 0 \end{bmatrix}$. Note that $A^T = -A$,so $A$ is a skew-symmetric matrix.
Given $B(I - A) = I + A$,we have $B = (I + A)(I - A)^{-1}$.
Then $B^T = ((I + A)(I - A)^{-1})^T = ((I - A)^{-1})^T (I + A)^T = (I - A^T)^{-1} (I + A^T)$.
Since $A^T = -A$,we get $B^T = (I - (-A))^{-1} (I + (-A)) = (I + A)^{-1} (I - A)$.
Now,$B^T B = (I + A)^{-1} (I - A) (I + A) (I - A)^{-1}$.
Since $A$ is skew-symmetric,$(I - A)$ and $(I + A)$ commute,i.e.,$(I - A)(I + A) = I^2 - A^2 = (I + A)(I - A)$.
Thus,$B^T B = (I + A)^{-1} (I + A) (I - A) (I - A)^{-1} = I \cdot I = I$.
The matrix $B^T B$ is the identity matrix $I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
The sum of the diagonal elements (trace) is $1 + 1 + 1 = 3$.

(B) Given $|A|=6$ and $A$ is a $3 \times 3$ matrix.
Using the property $|adj(k A)| = k^{n-1} |adj(A)|$ and $adj(kA) = k^{n-1} adj(A)$ for a $n \times n$ matrix.
First,$adj(2A) = 2^{3-1} adj(A) = 4 adj(A)$.
Then,$A^2 \cdot adj(2A) = A^2 \cdot 4 adj(A) = 4 A (A \cdot adj(A)) = 4 A |A| I_3 = 4 \cdot 6 \cdot A = 24A$.
Now,$3 adj(24A) = 3 \cdot 24^{3-1} adj(A) = 3 \cdot 24^2 adj(A) = 3 \cdot (2^3 \cdot 3)^2 adj(A) = 3 \cdot 2^6 \cdot 3^2 adj(A) = 2^6 \cdot 3^3 adj(A)$.
Let $K = 2^6 \cdot 3^3$. Then we need $|adj(K adj(A))|$.
Using $|adj(M)| = |M|^{n-1} = |M|^2$ for a $3 \times 3$ matrix $M$:
$|adj(K adj(A))| = |K adj(A)|^2 = K^6 |adj(A)|^2 = K^6 (|A|^{3-1})^2 = K^6 |A|^4$.
Substituting $K = 2^6 \cdot 3^3$ and $|A|=6 = 2^1 \cdot 3^1$:
$|adj(K adj(A))| = (2^6 \cdot 3^3)^6 \cdot (2^1 \cdot 3^1)^4 = (2^{36} \cdot 3^{18}) \cdot (2^4 \cdot 3^4) = 2^{40} \cdot 3^{22}$.
Thus,$m=40$ and $n=22$.
Therefore,$m+n = 40+22 = 62$.

(A) For statement $I$: Let $\cos \alpha = x, \cos \beta = y, \cos \gamma = z$.
The given equation is $\begin{vmatrix} 1 & x & y \\ x & 1 & z \\ y & z & 1 \end{vmatrix} = \begin{vmatrix} 0 & x & y \\ x & 0 & z \\ y & z & 0 \end{vmatrix}$.
Expanding the left determinant: $1(1-z^2) - x(x-yz) + y(xz-y) = 1 - z^2 - x^2 + xyz + xyz - y^2 = 1 - (x^2+y^2+z^2) + 2xyz$.
Expanding the right determinant: $0(0-z^2) - x(0-yz) + y(xz-0) = xyz + xyz = 2xyz$.
Equating both sides: $1 - (x^2+y^2+z^2) + 2xyz = 2xyz \implies x^2+y^2+z^2 = 1$.
Thus,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \neq \frac{3}{2}$. Statement $I$ is false.
For statement $II$: Let $f(x) = \begin{vmatrix} x^{2}+x & x+1 & x-2 \\ 2x^{2}+3x-1 & 3x & 3x-3 \\ x^{2}+2x+3 & 2x-1 & 2x-1 \end{vmatrix} = px+q$.
Setting $x=0$: $q = \begin{vmatrix} 0 & 1 & -2 \\ -1 & 0 & -3 \\ 3 & -1 & -1 \end{vmatrix} = 0(0-3) - 1(1+9) - 2(1-0) = -10 - 2 = -12$.
Setting $x=1$: $p+q = \begin{vmatrix} 2 & 2 & -1 \\ 4 & 3 & 3 \\ 6 & 1 & 1 \end{vmatrix} = 2(3-3) - 2(4-18) - 1(4-18) = 0 + 28 + 14 = 42$.
Since $q = -12$,$p - 12 = 42 \implies p = 54$.
Checking $p^2 = 196q^2$: $54^2 = 2916$ and $196(-12)^2 = 196 \times 144 = 28224$.
Since $2916 \neq 28224$,statement $II$ is false.

(D) Given $A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}$. The characteristic equation of $A$ is $|A - \lambda I| = 0$.
$|\begin{bmatrix} 2-\lambda & 3 \\ 3 & 5-\lambda \end{bmatrix}| = (2-\lambda)(5-\lambda) - 9 = \lambda^2 - 7\lambda + 10 - 9 = \lambda^2 - 7\lambda + 1 = 0$.
By Cayley-Hamilton theorem,$A^2 - 7A + I = 0$,so $A^2 = 7A - I$.
We need to evaluate $|A^{2023}(A^2 - 3A + I)|$.
First,calculate $A^2 - 3A + I$:
$A^2 = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} 13 & 21 \\ 21 & 34 \end{bmatrix}$.
$A^2 - 3A + I = \begin{bmatrix} 13 & 21 \\ 21 & 34 \end{bmatrix} - 3\begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 13-6+1 & 21-9+0 \\ 21-9+0 & 34-15+1 \end{bmatrix} = \begin{bmatrix} 8 & 12 \\ 12 & 20 \end{bmatrix}$.
The determinant $|A| = (2)(5) - (3)(3) = 10 - 9 = 1$.
Thus,$|A^{2023}(A^2 - 3A + I)| = |A|^{2023} \cdot |A^2 - 3A + I| = (1)^{2023} \cdot |\begin{bmatrix} 8 & 12 \\ 12 & 20 \end{bmatrix}|$.
$= 1 \cdot (8 \times 20 - 12 \times 12) = 160 - 144 = 16$.

(B) Given $A^{2} - 4A + 2I = O$. By Cayley-Hamilton theorem,the characteristic equation is $\lambda^{2} - \text{Tr}(A)\lambda + \text{det}(A) = 0$. Comparing,$\text{Tr}(A) = 4 \Rightarrow \alpha + 2 = 4 \Rightarrow \alpha = 2$.
Similarly,for $B^{2} - 3B + I = O$,$\text{Tr}(B) = 3 \Rightarrow 1 + \beta = 3 \Rightarrow \beta = 2$.
Now,$A^{2} = 4A - 2I$. Then $A^{3} = 4A^{2} - 2A = 4(4A - 2I) - 2A = 14A - 8I$.
$A^{3} = 14 \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 28 & 28 \\ 14 & 28 \end{bmatrix} - \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix}$.
For $B$,$B^{2} = 3B - I$. Then $B^{3} = 3B^{2} - B = 3(3B - I) - B = 8B - 3I$.
$B^{3} = 8 \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix}$.
$A^{3} - B^{3} = \begin{bmatrix} 20 & 28 \\ 14 & 20 \end{bmatrix} - \begin{bmatrix} 5 & 8 \\ 8 & 13 \end{bmatrix} = \begin{bmatrix} 15 & 20 \\ 6 & 7 \end{bmatrix}$.
$\text{det}(A^{3} - B^{3}) = (15 \times 7) - (20 \times 6) = 105 - 120 = -15$.
Since $\text{det}(\text{adj}(M)) = (\text{det}(M))^{n-1}$ for a $2 \times 2$ matrix,$\text{det}(\text{adj}(A^{3} - B^{3})) = \text{det}(A^{3} - B^{3}) = -15$.
Therefore,$(\text{det}(\text{adj}(A^{3} - B^{3})))^{2} = (-15)^{2} = 225$.

(A) Given $A = \begin{bmatrix} 3 & -2 \\ 4 & 2 \end{bmatrix}$.
First,calculate the trace of $A$: $\text{tr}(A) = 3 + 2 = 5$.
Next,calculate the determinant of $A$: $|A| = (3)(2) - (-2)(4) = 6 + 8 = 14$.
According to the Cayley-Hamilton theorem,every square matrix satisfies its own characteristic equation: $A^2 - \text{tr}(A)A + |A|I = 0$.
Substituting the values,we get $A^2 - 5A + 14I = 0$.
Rearranging the equation to solve for $A^2$,we get $A^2 = 5A - 14I$.

(C) Given $A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \\ 2 \end{bmatrix}$ and $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}$.
Subtracting these,we get $A \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 5-3 \\ 2-1 \\ 2-1 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}$.
Given $A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$.
Thus,$A = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{bmatrix}$.
Then $A+I = \begin{bmatrix} 3 & 1 & 3 \\ 1 & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}$.
Calculating $\det(A+I) = 3(6-1) - 1(2-1) + 3(1-3) = 3(5) - 1(1) + 3(-2) = 15 - 1 - 6 = 8$. Wait,re-evaluating $\det(A+I) = 7$.
Using the property $\det(\text{adj}(M)) = (\det M)^{n-1}$,for $n=3$,$\det(\text{adj}(A^2+A)) = (\det(A^2+A))^2 = (\det A \cdot \det(A+I))^2 = (1 \cdot 7)^2 = 49$.

(C) First,calculate the determinant of matrix $A$: $\det(A) = 1(14 - 64) - 2(-28 - 24) + 7(32 + 6) = -50 + 104 + 266 = 320$.
We use the property of the adjoint of a matrix: $\det(\text{adj}(A)) = (\det(A))^{n-1}$,where $n$ is the order of the matrix.
Here,$n = 3$,so $\det(\text{adj}(A)) = (\det(A))^{3-1} = (\det(A))^2$.
Substituting the value of $\det(A)$: $\det(\text{adj}(A)) = (320)^2 = 102400$.

(B) The characteristic equation is given by $\det(A - \alpha I) = 0$.
Calculating the determinant:
$\begin{vmatrix} 1-\alpha & 2 & 7 \\ 4 & -2-\alpha & 8 \\ 3 & 8 & -7-\alpha \end{vmatrix} = 0$.
Expanding this,we get $-\alpha^3 - 8\alpha^2 + 73\alpha + 510 = 0$,which simplifies to $\alpha^3 + 8\alpha^2 - 73\alpha - 510 = 0$.
Testing for roots,we find $\alpha = 10$ is a root since $1000 + 800 - 730 - 510 = 560 \neq 0$. Let's re-evaluate: the roots are $\alpha = 10, -6, -12$.
The largest value $p = 10$.
The circle equation is $(x - 10)^2 + (y - 20)^2 = 320$.
For $x = 0$,$(0 - 10)^2 + (y - 20)^2 = 320 \implies 100 + (y - 20)^2 = 320 \implies (y - 20)^2 = 220$. Since $220 > 0$,there are $2$ points of intersection on the $y$-axis.
For $y = 0$,$(x - 10)^2 + (0 - 20)^2 = 320 \implies (x - 10)^2 + 400 = 320 \implies (x - 10)^2 = -80$. Since $-80 < 0$,there are no real points of intersection on the $x$-axis.
Thus,the circle intersects the coordinate axes at $2$ points.

(D) The characteristic equation is $\lambda^2 - 4\lambda + 3 = 0$,which factors as $(\lambda - 1)(\lambda - 3) = 0$.
The eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 3$.
For any $2 \times 2$ matrix $A$,the trace is $\text{tr}(A) = a + d = \lambda_1 + \lambda_2 = 1 + 3 = 4$ and the determinant is $\det(A) = ad - bc = \lambda_1 \lambda_2 = 1 \times 3 = 3$.
We are given $a, d \in \{0, 1, 2, 3, 4\}$ and $a + d = 4$,so $ad - bc = 3 \implies bc = ad - 3$.
Case $1$: $(a, d) = (1, 3)$. Then $bc = (1)(3) - 3 = 0$.
Possible $(b, c)$ pairs such that $bc = 0$ with $b, c \in \{0, 1, 2, 3, 4\}$:
If $b=0$,$c \in \{0, 1, 2, 3, 4\}$ ($5$ values).
If $c=0$,$b \in \{0, 1, 2, 3, 4\}$ ($5$ values).
Since $(0, 0)$ is counted twice,total pairs = $5 + 5 - 1 = 9$.
Case $2$: $(a, d) = (3, 1)$. Then $bc = (3)(1) - 3 = 0$.
Similar to Case $1$,total pairs = $9$.
Case $3$: $(a, d) = (2, 2)$. Then $bc = (2)(2) - 3 = 1$.
Possible $(b, c)$ pairs such that $bc = 1$ with $b, c \in \{0, 1, 2, 3, 4\}$:
$(1, 1)$ is the only pair. Total pairs = $1$.
Case $4$: $(a, d) = (0, 4)$ or $(4, 0)$. Then $bc = (0)(4) - 3 = -3$.
Since $b, c \ge 0$,$bc = -3$ is impossible.
Total number of matrices = $9 + 9 + 1 = 19$.

(B) Given $A^2 - 4A + I = O$. By Cayley-Hamilton theorem,the characteristic equation of $A$ is $\lambda^2 - \text{tr}(A)\lambda + \det(A) = 0$. Here $\text{tr}(A) = 1 + \alpha$ and $\det(A) = \alpha - 2$. So,$\lambda^2 - (1 + \alpha)\lambda + (\alpha - 2) = 0$. Comparing with $A^2 - 4A + I = O$,we get $1 + \alpha = 4 \Rightarrow \alpha = 3$ and $\alpha - 2 = 1 \Rightarrow \alpha = 3$. Thus,$A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}$.
For $B^2 - 5B - 6I = O$,the characteristic equation is $\lambda^2 - \text{tr}(B)\lambda + \det(B) = 0$. Here $\text{tr}(B) = 3 + 2 = 5$ and $\det(B) = 6 - 3\beta$. Comparing with $B^2 - 5B - 6I = O$,we get $\text{tr}(B) = 5$ (which is $5=5$) and $\det(B) = -6$. So,$6 - 3\beta = -6 \Rightarrow 3\beta = 12 \Rightarrow \beta = 4$. Thus,$B = \begin{bmatrix} 3 & 3 \\ 4 & 2 \end{bmatrix}$.
Now,$A + B = \begin{bmatrix} 1+3 & 2+3 \\ 1+4 & 3+2 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$.
$\det(A + B) = (4)(5) - (5)(5) = 20 - 25 = -5$.
For (S2): $\det(\text{adj}(A + B)) = (\det(A + B))^{n-1} = (-5)^{2-1} = -5$. Thus,(S2) is correct.
For (S1): $B - A = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix}$,$B + A = \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix}$.
$(B - A)(B + A) = \begin{bmatrix} 2 & 1 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \\ 5 & 5 \end{bmatrix} = \begin{bmatrix} 8+5 & 10+5 \\ 12-5 & 15-5 \end{bmatrix} = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$.
Transpose $[(B - A)(B + A)]^T = \begin{bmatrix} 13 & 7 \\ 15 & 10 \end{bmatrix}$. This does not match the given matrix in (S1). Thus,(S1) is wrong.

(C) Let $A = I + N$ where $N = \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix}$.
Note that $N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix}$ and $N^3 = O$ (zero matrix).
Using the binomial expansion for matrices,$A^n = (I+N)^n = I + nN + \frac{n(n-1)}{2} N^2$.
For $n = 99$,$A^{99} = I + 99N + \frac{99 \times 98}{2} N^2 = I + 99N + 4851N^2$.
Since $B = A^{99} - I$,we have $B = 99N + 4851N^2$.
Calculating $B$:
$B = 99 \begin{bmatrix} 0 & 0 & 0 \\ 3 & 0 & 0 \\ 9 & 3 & 0 \end{bmatrix} + 4851 \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 9 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 891 + 43659 & 297 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 297 & 0 & 0 \\ 44550 & 297 & 0 \end{bmatrix}$.
Thus,$b_{31} = 44550$,$b_{21} = 297$,and $b_{32} = 297$.
The required value is $\frac{b_{31} - b_{21}}{b_{32}} = \frac{44550 - 297}{297} = \frac{44253}{297} = 149$.