Mix Examples-Determinants and Matrices Questions in English

(B) Given $A=\left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right]$.
The characteristic equation is given by $|A-\lambda I|=0$.
$\left|\begin{array}{ccc}1-\lambda & 1 & 3 \\ 5 & 2-\lambda & 6 \\ -2 & -1 & -3-\lambda\end{array}\right|=0$
Expanding the determinant:
$(1-\lambda)[(2-\lambda)(-3-\lambda) - (-6)] - 1[5(-3-\lambda) - (-12)] + 3[5(-1) - (-2)(2-\lambda)] = 0$
$(1-\lambda)[\lambda^2+\lambda-6+6] - 1[-15-5\lambda+12] + 3[-5+4-2\lambda] = 0$
$(1-\lambda)(\lambda^2+\lambda) - 1(-5\lambda-3) + 3(-2\lambda-1) = 0$
$\lambda^2+\lambda-\lambda^3-\lambda^2 + 5\lambda+3 - 6\lambda-3 = 0$
$-\lambda^3 = 0 \Rightarrow \lambda^3 = 0$.
By the Cayley-Hamilton theorem,every matrix satisfies its characteristic equation,so $A^3 = 0$.
Since $A^3 = 0$,it follows that $A^4 = A^3 \cdot A = 0$ and $A^5 = A^3 \cdot A^2 = 0$.
Therefore,$A+A^3+A^4+A^5+3I = A + 0 + 0 + 0 + 3I = A + 3I$.
$A+3I = \left[\begin{array}{ccc}1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3\end{array}\right] + \left[\begin{array}{ccc}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{array}\right] = \left[\begin{array}{ccc}4 & 1 & 3 \\ 5 & 5 & 6 \\ -2 & -1 & 0\end{array}\right]$.

(D) Let $A = \begin{bmatrix} -x & 14x & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix}$. The inverse $A^{-1}$ is given as $\begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$.
We know that $A \cdot A^{-1} = I$,where $I$ is the identity matrix.
Multiplying the first row of $A$ by the first column of $A^{-1}$:
$(-x)(2) + (14x)(0) + (7x)(1) = 1$ (since the element at $(1,1)$ of $I$ is $1$)
$-2x + 7x = 1 \implies 5x = 1 \implies x = 1/5$.
Now,we need to evaluate the determinant $D = \left|\begin{array}{ccc} x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \end{array}\right|$.
Apply row operations: $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_2$.
$D = \left|\begin{array}{ccc} x & x+1 & x+2 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right|$.
Since two rows are identical,the determinant is $0$.

(A) First,evaluate $A$: $A = \int_{1}^{\sin \theta} \frac{t}{1+t^2} dt = \frac{1}{2} [\ln(1+t^2)]_{1}^{\sin \theta} = \frac{1}{2} \ln(1+\sin^2 \theta) - \frac{1}{2} \ln(2) = \frac{1}{2} \ln(\frac{1+\sin^2 \theta}{2})$.
Next,evaluate $B$: $B = \int_{1}^{\operatorname{cosec} \theta} \frac{1}{t(1+t^2)} dt$. Using partial fractions,$\frac{1}{t(1+t^2)} = \frac{1}{t} - \frac{t}{1+t^2}$.
So,$B = [\ln|t| - \frac{1}{2} \ln(1+t^2)]_{1}^{\operatorname{cosec} \theta} = [\ln(\frac{t}{\sqrt{1+t^2}})]_{1}^{\operatorname{cosec} \theta}$.
Since $\frac{t}{\sqrt{1+t^2}} = \frac{\operatorname{cosec} \theta}{\sqrt{1+\operatorname{cosec}^2 \theta}} = \frac{1}{\sqrt{\sin^2 \theta + 1}}$,we have $B = \ln(\frac{1}{\sqrt{1+\sin^2 \theta}}) - \ln(\frac{1}{\sqrt{2}}) = \ln(\sqrt{\frac{2}{1+\sin^2 \theta}}) = -\frac{1}{2} \ln(\frac{1+\sin^2 \theta}{2}) = -A$.
Thus,$A+B = 0$,which implies $e^{A+B} = e^0 = 1$.
The determinant becomes $\left| \begin{array}{ccc} A & A^2 & -A \\ 1 & (-A)^2 & -1 \\ 1 & A^2+(-A)^2 & -1 \end{array} \right| = \left| \begin{array}{ccc} A & A^2 & -A \\ 1 & A^2 & -1 \\ 1 & 2A^2 & -1 \end{array} \right|$.
Since the first and third columns are proportional (column $3$ is $-1$ times column $1$),the determinant is $0$.

(A)

$A$. Given $A = \begin{bmatrix} \cos^2 37^{\circ} & \cos^2 53^{\circ} & -1 \\ \sin^2 76^{\circ} & -1 & \sin^2 14^{\circ} \\ -1 & \cos^2 28^{\circ} & \cos^2 62^{\circ} \end{bmatrix}$. Using properties of determinants,$|A| = 0$. Thus,$3 - |A| = 3 - 0 = 3$. Matches $(iii)$.

$B$. The period of $\cos(6x-4)$ is $\frac{\pi}{3}$,$\sec(3-4x)$ is $\frac{\pi}{2}$,$\cot(5x+3)$ is $\frac{\pi}{5}$,and $\sin(3x+4)$ is $\frac{2\pi}{3}$. The period of the expression is $LCM(\frac{\pi}{3}, \frac{\pi}{2}, \frac{\pi}{5}, \frac{2\pi}{3}) = 2\pi$. Given $\frac{2k\pi}{5} = 2\pi$,so $k = 5$. Matches $(v)$.

$C$. $y = \frac{1}{2}(\cos x + \sin x)^2 + (\sin x - \cos x)^2 = \frac{3}{2} - \frac{\sin 2x}{2}$. Max value is $\frac{3}{2} + \frac{1}{2} = 2$. Matches $(ii)$.

$D$. If $x+y+z=0$,then $\sin 2x + \sin 2y + \sin 2z = -4\sin x \sin y \sin z$. The expression becomes $\frac{-4\sin x \sin y \sin z}{-\sin x \sin y \sin z} = 4$. Matches $(iv)$.

(A) The matrix $A$ is a diagonal matrix where the diagonal elements are $a_{ii} = k^i$ for $i = 1, 2, \dots, n$.
The trace of $A$,denoted by $m$,is the sum of its diagonal elements:
$m = \sum_{i=1}^{n} k^i = k + k^2 + \dots + k^n = \frac{k(1-k^n)}{1-k}$.
We are given the limit: $\lim_{k \rightarrow 1} \frac{n-m}{1-k} = 171$.
Substituting $m$: $\lim_{k \rightarrow 1} \frac{n - \frac{k(1-k^n)}{1-k}}{1-k} = \lim_{k \rightarrow 1} \frac{n(1-k) - (k - k^{n+1})}{(1-k)^2} = 171$.
Using $L$'Hospital's rule (differentiating numerator and denominator with respect to $k$):
Numerator derivative: $\frac{d}{dk} [n - nk - k + k^{n+1}] = -n - 1 + (n+1)k^n$.
Denominator derivative: $\frac{d}{dk} [(1-k)^2] = 2(1-k)(-1) = -2(1-k)$.
Applying $L$'Hospital's rule again:
$\lim_{k \rightarrow 1} \frac{-n - 1 + (n+1)k^n}{-2(1-k)} = \lim_{k \rightarrow 1} \frac{(n+1)n k^{n-1}}{2} = 171$.
$\frac{n(n+1)}{2} = 171 \Rightarrow n^2 + n - 342 = 0$.
Solving the quadratic equation: $(n+19)(n-18) = 0$.
Since $n > 0$,we have $n = 18$.

(D) Consider $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$.
Each row sum is $6$:
$a_{11} + a_{12} + a_{13} = 6 \dots (i)$
$a_{21} + a_{22} + a_{23} = 6 \dots (ii)$
$a_{31} + a_{32} + a_{33} = 6 \dots (iii)$
From the given condition on diagonal entries:
$a_{11} = a_{12} + a_{21} \dots (iv)$
$a_{22} = a_{23} + a_{32} = 2 \dots (v)$
$a_{33} = a_{13} + a_{31} \dots (vi)$
Since entries are positive integers,from $(v)$,$a_{23} = 1$ and $a_{32} = 1$. Substituting into $(ii)$,$a_{21} + 2 + 1 = 6 \Rightarrow a_{21} = 3$.
From $(iv)$,$a_{11} = a_{12} + 3$. Substituting into $(i)$,$(a_{12} + 3) + a_{12} + a_{13} = 6 \Rightarrow 2a_{12} + a_{13} = 3$. Since $a_{ij} \ge 1$,we must have $a_{12} = 1$,$a_{13} = 1$,and $a_{11} = 4$.
From $(iii)$ and $(vi)$,$a_{31} + a_{32} + a_{33} = 6 \Rightarrow a_{31} + 1 + a_{33} = 6 \Rightarrow a_{31} + a_{33} = 5$. Also $a_{33} = a_{13} + a_{31} = 1 + a_{31}$.
Substituting $a_{33}$ in the sum,$a_{31} + (1 + a_{31}) = 5 \Rightarrow 2a_{31} = 4 \Rightarrow a_{31} = 2$,so $a_{33} = 3$.
Thus,$A = \begin{bmatrix} 4 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{bmatrix}$.
$|A| = 4(6 - 1) - 1(9 - 2) + 1(3 - 4) = 4(5) - 1(7) + 1(-1) = 20 - 7 - 1 = 12$.

(A) Given,$A+B=\left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2\end{array}\right]$ and $AB=\left[\begin{array}{lll}1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1\end{array}\right]$.
We need to find $A^2+B(A+B)$.
Note that $A^2+B(A+B) = A^2+BA+B^2$.
We know that $(A+B)^2 = (A+B)(A+B) = A^2+AB+BA+B^2$.
Thus,$A^2+BA+B^2 = (A+B)^2 - AB$.
First,calculate $(A+B)^2 = \left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2\end{array}\right] \times \left[\begin{array}{lll} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2\end{array}\right] = \left[\begin{array}{lll} 4+1+0 & 2+2+4 & 4+0+4 \\ 2+2+0 & 1+4+0 & 2+0+0 \\ 0+2+0 & 0+4+4 & 0+0+4\end{array}\right] = \left[\begin{array}{lll} 5 & 8 & 8 \\ 4 & 5 & 2 \\ 2 & 8 & 4\end{array}\right]$.
Now,$A^2+B(A+B) = \left[\begin{array}{lll} 5 & 8 & 8 \\ 4 & 5 & 2 \\ 2 & 8 & 4\end{array}\right] - \left[\begin{array}{lll} 1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1\end{array}\right] = \left[\begin{array}{lll} 4 & 6 & 6 \\ 3 & 4 & 2 \\ 1 & 6 & 3\end{array}\right]$.

(C) Let $A = \left[\begin{array}{ccc}0 & 2 & a \\ b & 0 & 4 \\ -3 & c & 0\end{array}\right]$ be a skew-symmetric matrix.
By definition,$A = -A^T$.
$\left[\begin{array}{ccc}0 & 2 & a \\ b & 0 & 4 \\ -3 & c & 0\end{array}\right] = -\left[\begin{array}{ccc}0 & b & -3 \\ 2 & 0 & c \\ a & 4 & 0\end{array}\right] = \left[\begin{array}{ccc}0 & -b & 3 \\ -2 & 0 & -c \\ -a & -4 & 0\end{array}\right]$.
Comparing the corresponding elements,we get:
$-b = 2 \Rightarrow b = -2$
$a = 3$
$c = -4$
Now,substitute these values into the expression:
$\left[\begin{array}{cc}a & b \\ b & a\end{array}\right]\left[\begin{array}{cc}b & c \\ c & b\end{array}\right] = \left[\begin{array}{cc}3 & -2 \\ -2 & 3\end{array}\right]\left[\begin{array}{cc}-2 & -4 \\ -4 & -2\end{array}\right]$
$= \left[\begin{array}{cc}(3)(-2) + (-2)(-4) & (3)(-4) + (-2)(-2) \\ (-2)(-2) + (3)(-4) & (-2)(-4) + (3)(-2)\end{array}\right]$
$= \left[\begin{array}{cc}-6 + 8 & -12 + 4 \\ 4 - 12 & 8 - 6\end{array}\right] = \left[\begin{array}{cc}2 & -8 \\ -8 & 2\end{array}\right]$.

(A) Given,$A B A = B A^2 B$ and $A^3 = I$.
Multiplying both sides of $A B A = B A^2 B$ by $A^2$ on the right,we get:
$A B A \cdot A^2 = B A^2 B \cdot A^2$
$A B A^3 = B A^2 B A^2$
Since $A^3 = I$,we have:
$A B = B A^2 B A^2$
Now,multiply by $B$ on the right:
$A B^2 = B A^2 B A^2 B$
Substitute $A^2 B = A B A$ from the given equation:
$A B^2 = B A^2 B (A B A)$
$A B^2 = B A^2 B A B A$
Using $A^3 = I$,we can simplify further.
Alternatively,note that $A B A = B A^2 B \implies A B = B A^2 B A^{-1}$.
Since $A^3 = I$,$A^{-1} = A^2$.
Thus $A B = B A^2 B A^2$.
By induction or repeated substitution,$A B^4 = B^4 A$.
Therefore,$A B^4 - B^4 A = O_{3 \times 3}$.

(B) Given the matrix $A = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix}$.
Since $A$ satisfies the equation $x^2 + 4x - p = 0$,we have $A^2 + 4A - pI = 0$,where $I$ is the identity matrix $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
First,calculate $A^2$:
$A^2 = \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} (-8)(-8) + (5)(2) & (-8)(5) + (5)(4) \\ (2)(-8) + (4)(2) & (2)(5) + (4)(4) \end{bmatrix} = \begin{bmatrix} 64 + 10 & -40 + 20 \\ -16 + 8 & 10 + 16 \end{bmatrix} = \begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix}$.
Next,calculate $4A$:
$4A = 4 \begin{bmatrix} -8 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix}$.
Now,substitute these into the equation $A^2 + 4A - pI = 0$:
$\begin{bmatrix} 74 & -20 \\ -8 & 26 \end{bmatrix} + \begin{bmatrix} -32 & 20 \\ 8 & 16 \end{bmatrix} - \begin{bmatrix} p & 0 \\ 0 & p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Summing the matrices:
$\begin{bmatrix} 74 - 32 - p & -20 + 20 - 0 \\ -8 + 8 - 0 & 26 + 16 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
$\begin{bmatrix} 42 - p & 0 \\ 0 & 42 - p \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Comparing the elements,we get $42 - p = 0$,which implies $p = 42$.

(A) Given the characteristic equation $A^3-5A^2+7A+I=0$,we have $A^3 = 5A^2-7A-I$.
We need to simplify $P(A) = A^5-6A^4+12A^3-6A^2+2A+2I$.
Divide $P(A)$ by $A^3-5A^2+7A+I$ using polynomial division:
$A^5-6A^4+12A^3-6A^2+2A+2I = (A^2-A)(A^3-5A^2+7A+I) + (0A^2+2A+3I)$.
Since $A^3-5A^2+7A+I=0$,the expression simplifies to $0 + 2A+3I$.
Comparing this with $lA+mI$,we get $l=2$ and $m=3$.
Therefore,$l+m = 2+3 = 5$.

(C) Given $A^2+I=2 A$,we have $A^2=2 A-I$.
Multiplying by $A$,we get $A^3=2 A^2-A=2(2 A-I)-A=4 A-2 I-A=3 A-2 I$.
Now,$A^6 = A^3 \cdot A^3 = (3 A-2 I)(3 A-2 I) = 9 A^2-12 A+4 I$.
Substituting $A^2=2 A-I$,we get $A^6 = 9(2 A-I)-12 A+4 I = 18 A-9 I-12 A+4 I = 6 A-5 I$.
Finally,$A^9 = A^6 \cdot A^3 = (6 A-5 I)(3 A-2 I) = 18 A^2-12 A-15 A+10 I = 18 A^2-27 A+10 I$.
Substituting $A^2=2 A-I$ again,$A^9 = 18(2 A-I)-27 A+10 I = 36 A-18 I-27 A+10 I = 9 A-8 I$.

(A) Given the equation: $I+P+P^2+\ldots+P^{n}=0$ ... $(i)$
We can rewrite this as: $I+P+P^2+\ldots+P^{n-1} = -P^n$ ... $(ii)$
Now,multiply equation $(i)$ by $P^{-1}$ on the left side:
$P^{-1}(I+P+P^2+\ldots+P^{n}) = P^{-1}(0)$
$P^{-1}I + P^{-1}P + P^{-1}P^2 + \ldots + P^{-1}P^n = 0$
$P^{-1} + I + P + \ldots + P^{n-1} = 0$
From equation $(ii)$,we know that $I+P+\ldots+P^{n-1} = -P^n$.
Substituting this into the equation:
$P^{-1} + (-P^n) = 0$
Therefore,$P^{-1} = P^n$.

(B) For matrix $A$ to be symmetric,$A = A^T$,which implies $a=b$,$c=d$,and $3=3$. Thus,$A = \begin{bmatrix} 1 & a & 3 \\ a & 2 & c \\ 3 & c & 4 \end{bmatrix}$.
For matrix $B$ to be skew-symmetric,$B = -B^T$,which implies diagonal elements are $0$. From $B_{13} = -B_{31}$,we get $b = -6$. From $B_{23} = -B_{32}$,we get $-7 = -c$,so $c = 7$.
Substituting these values,we get $a = b = -6$ and $d = c = 7$.
So,$A = \begin{bmatrix} 1 & -6 & 3 \\ -6 & 2 & 7 \\ 3 & 7 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 5 & -6 \\ -5 & 0 & -7 \\ 6 & 7 & 0 \end{bmatrix}$.
Now,calculating the product $AB$:
$AB = \begin{bmatrix} 1 & -6 & 3 \\ -6 & 2 & 7 \\ 3 & 7 & 4 \end{bmatrix} \begin{bmatrix} 0 & 5 & -6 \\ -5 & 0 & -7 \\ 6 & 7 & 0 \end{bmatrix} = \begin{bmatrix} 0+30+18 & 5+0+21 & -6+42+0 \\ 0-10+42 & -30+0+49 & 36-14+0 \\ 0-35+24 & 15+0+28 & -18-49+0 \end{bmatrix} = \begin{bmatrix} 48 & 26 & 36 \\ 32 & 19 & 22 \\ -11 & 43 & -67 \end{bmatrix}$.

(D) Given the matrix $X = \begin{bmatrix} -1 & 2 & b \\ a & 5 & 6 \\ 3 & c & 7 \end{bmatrix}$ is a symmetric matrix,we have $X = X^T$.
Equating the elements,we get:
$\begin{bmatrix} -1 & 2 & b \\ a & 5 & 6 \\ 3 & c & 7 \end{bmatrix} = \begin{bmatrix} -1 & a & 3 \\ 2 & 5 & c \\ b & 6 & 7 \end{bmatrix}$
Comparing corresponding elements,we find $a = 2$,$b = 3$,and $c = 6$.
Now,we need to evaluate the determinant $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = \begin{vmatrix} 2 & 3 & 6 \\ 3 & 6 & 2 \\ 6 & 2 & 3 \end{vmatrix}$.
Expanding along the first row $(R_1)$:
$= 2(6 \times 3 - 2 \times 2) - 3(3 \times 3 - 6 \times 2) + 6(3 \times 2 - 6 \times 6)$
$= 2(18 - 4) - 3(9 - 12) + 6(6 - 36)$
$= 2(14) - 3(-3) + 6(-30)$
$= 28 + 9 - 180$
$= 37 - 180 = -143$.

(C) Given that $A=B+C$,$BC=CB$,and $C^2=0$.
Since $B$ and $C$ commute,we can use the Binomial Theorem for matrices:
$A^k = (B+C)^k = \sum_{r=0}^{k} \binom{k}{r} B^{k-r} C^r$.
Since $C^2=0$,all higher powers $C^r=0$ for $r \ge 2$.
Thus,$A^k = \binom{k}{0} B^k C^0 + \binom{k}{1} B^{k-1} C^1 = B^k + k B^{k-1} C = B^{k-1}(B+kC)$.
Setting $k=2021$,we get:
$A^{2021} = B^{2021-1}(B+2021C) = B^{2020}(B+2021C)$.
Therefore,$B^{2020}[B+(2021)C] = A^{2021}$.

(C) Given $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$.
First,calculate $A^2 = A \cdot A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix}$.
Next,calculate $A^3 = A^2 \cdot A = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix}$.
Now,substitute these into the expression $A^3 - 4A^2 - 6A$:
$= \begin{bmatrix} 41 & 42 & 42 \\ 42 & 41 & 42 \\ 42 & 42 & 41 \end{bmatrix} - 4 \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - 6 \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$
$= \begin{bmatrix} 41-36-6 & 42-32-12 & 42-32-12 \\ 42-32-12 & 41-36-6 & 42-32-12 \\ 42-32-12 & 42-32-12 & 41-36-6 \end{bmatrix}$
$= \begin{bmatrix} -1 & -2 & -2 \\ -2 & -1 & -2 \\ -2 & -2 & -1 \end{bmatrix} = -\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = -A$.

(A) Given the equation: $(AB+BA)^{T}+(AB-BA)^{T}=2BA$
Using the property $(X+Y)^T = X^T + Y^T$,we get:
$(AB)^T + (BA)^T + (AB)^T - (BA)^T = 2BA$
$2(AB)^T = 2BA$
$(AB)^T = BA$
$B^T A^T = BA$
If $A$ and $B$ are symmetric,then $A^T = A$ and $B^T = B$. Substituting these,we get $BA = BA$,which is true.
If $A$ and $B$ are skew-symmetric,then $A^T = -A$ and $B^T = -B$. Substituting these,we get $(-B)(-A) = BA$,which is $BA = BA$,which is also true.
Thus,the condition holds if $A$ and $B$ are both symmetric or both skew-symmetric. Since the options provided are limited,the most appropriate conclusion based on the derivation is that the property holds for both cases.

(B) The total number of $3 \times 3$ matrices with entries from $\{-1, 0, 1\}$ is $3^9 = 19683$. Since we are considering only non-zero matrices,the total number of possible matrices is $3^9 - 1 = 19682$.
$A$ skew-symmetric matrix $A$ satisfies $A^T = -A$. For a $3 \times 3$ matrix,this implies the diagonal elements must be $0$. The matrix takes the form:
$\begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$
The entries $a, b, c$ can each be chosen from $\{-1, 0, 1\}$. Thus,there are $3^3 = 27$ such matrices.
Excluding the zero matrix (where $a=b=c=0$),the number of non-zero skew-symmetric matrices is $27 - 1 = 26$.
The probability is $\frac{26}{19682} = \frac{1}{757}$.

(D) For a skew-symmetric matrix $A$,we have $A = -A^T$.
This implies that the diagonal elements must be zero,so $a = e = i = 0$.
Thus,the matrix is $A = \begin{bmatrix} 0 & b & c \\ d & 0 & f \\ g & h & 0 \end{bmatrix}$.
Since $A$ is a skew-symmetric matrix,we have $d = -b$,$g = -c$,and $h = -f$.
Substituting these into the expression $\frac{b}{c}$:
We know that for any skew-symmetric matrix of odd order,the determinant is $0$.
$|A| = 0 \cdot (0 - fh) - b(0 - gf) + c(dh - 0) = 0$.
$bgf + cdh = 0$.
$bgf = -cdh$.
Dividing by $cf$,we get $\frac{b}{c} = \frac{-dh}{fg}$.

(C) For two square matrices $M$ and $N$ of order $3$:
$1$. The matrix $MN - NM$ is skew-symmetric if $M$ and $N$ are symmetric matrices,because $(MN - NM)^T = (MN)^T - (NM)^T = N^T M^T - M^T N^T = NM - MN = -(MN - NM)$.
$2$. The matrix $N^T MN$ is symmetric or skew-symmetric according as $M$ is symmetric or skew-symmetric,because $(N^T MN)^T = N^T M^T (N^T)^T = N^T M^T N$. If $M^T = M$,then $(N^T MN)^T = N^T MN$ (symmetric). If $M^T = -M$,then $(N^T MN)^T = -N^T MN$ (skew-symmetric).
$3$. The product of two symmetric matrices $MN$ is symmetric if and only if $MN = NM$. Since this is not true for all symmetric matrices,statement $(c)$ is false.
$4$. For any two matrices $M$ and $N$,$\text{adj}(MN)$ and $\text{adj}(NM)$ are not necessarily equal.
Thus,the statement that is not true is $(c)$.

(B) Given $a_{i j} = xi + yj$. The matrix $A$ is defined as $A = [a_{i j}]_{n \times n}$.
We can write $A$ as:
$A = \begin{bmatrix} x+y & 2x+y & \dots & nx+y \\ x+2y & 2x+2y & \dots & nx+2y \\ \vdots & \vdots & \ddots & \vdots \\ x+ny & 2x+ny & \dots & nx+ny \end{bmatrix}$
This can be decomposed into:
$A = x \begin{bmatrix} 1 & 2 & \dots & n \\ 1 & 2 & \dots & n \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 2 & \dots & n \end{bmatrix} + y \begin{bmatrix} 1 & 1 & \dots & 1 \\ 2 & 2 & \dots & 2 \\ \vdots & \vdots & \ddots & \vdots \\ n & n & \dots & n \end{bmatrix}$
Let $P = \begin{bmatrix} 1 & 2 & \dots & n \\ 1 & 2 & \dots & n \\ \vdots & \vdots & \ddots & \vdots \\ 1 & 2 & \dots & n \end{bmatrix}$ and $Q = \begin{bmatrix} 1 & 1 & \dots & 1 \\ 2 & 2 & \dots & 2 \\ \vdots & \vdots & \ddots & \vdots \\ n & n & \dots & n \end{bmatrix}$.
Note that $P$ has identical rows, so its rank is $1$. $Q$ has proportional rows, so its rank is $1$.
However, the question asks for properties of $P+Q$ and $P-Q$.
$P+Q = [i+j]_{n \times n}$, which is a symmetric matrix because $(P+Q)^T = [j+i]^T = [i+j] = P+Q$.
$P-Q = [i-j]_{n \times n}$, which is a skew-symmetric matrix because $(P-Q)^T = [j-i]^T = [-(i-j)] = -(P-Q)$.

(A) Given that $B$ is an orthogonal matrix,$B^T = B^{-1}$,which implies $B^T B = I$.
Taking the determinant on both sides,we get $|B^T B| = |I| = 1$.
Since $|B^T| = |B|$,we have $|B|^2 = 1$.
Given $|B| = 1$,we consider the matrix $B-I$.
We know that $|B-I| = |B-I|^T = |B^T - I^T| = |B^T - I|$.
Since $B^T = B^{-1}$,we have $|B^T - I| = |B^{-1} - I| = |B^{-1}(I - B)| = |B^{-1}| |I - B| = \frac{1}{|B|} |-(B-I)| = \frac{1}{1} (-1)^3 |B-I| = -|B-I|$.
Thus,$|B-I| = -|B-I|$,which implies $2|B-I| = 0$,so $|B-I| = 0$.

(B) Given $A = \begin{bmatrix} \frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix}$.
First,we calculate $A^2 = A \times A$.
$A^2 = \begin{bmatrix} \frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix} \begin{bmatrix} \frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix} = \frac{1}{36} \begin{bmatrix} 6 & -12 & -6 \\ -12 & 24 & 12 \\ -6 & 12 & 6 \end{bmatrix} = \begin{bmatrix} \frac{1}{6} & -\frac{1}{3} & -\frac{1}{6} \\ -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix} = A$.
Since $A^2 = A$,it follows that $A^k = A$ for all $k \in N$.
Therefore,$A^{2016l} = A$,$A^{2017m} = A$,and $A^{2018n} = A$.
The given equation becomes $A + A + A = \frac{1}{\alpha} A$,which simplifies to $3A = \frac{1}{\alpha} A$.
Comparing both sides,we get $\frac{1}{\alpha} = 3$,which implies $\alpha = \frac{1}{3}$.

(B) Given,$\operatorname{adj} A = |A| B$.
Since $A$ is non-singular,$\operatorname{adj} A = |A| A^{-1}$.
Thus,$|A| B = |A| A^{-1}$,which implies $B = A^{-1}$.
Therefore,$A B = A A^{-1} = I$,where $I$ is the identity matrix.
Now,consider the sum $S = \sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k}(A B)^k C\right)$.
Since $(A B)^k = I^k = I$,the expression becomes $S = \sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k} I C\right) = \sum_{k=1}^{\infty} \frac{1}{3^k} \operatorname{tr}(C)$.
The trace of $C$ is $\operatorname{tr}(C) = 4 + (-2) + 6 = 8$.
So,$S = 8 \sum_{k=1}^{\infty} \left(\frac{1}{3}\right)^k$.
This is an infinite geometric series with first term $a = \frac{1}{3}$ and common ratio $r = \frac{1}{3}$.
The sum is $S = 8 \left( \frac{1/3}{1 - 1/3} \right) = 8 \left( \frac{1/3}{2/3} \right) = 8 \times \frac{1}{2} = 4$.

(B) Let the eigenvalues of the $2 \times 2$ matrix $A$ be $\lambda_1$ and $\lambda_2$.
Given $\operatorname{det} A = \lambda_1 \lambda_2 = -21$.
The eigenvalues of $A^3$ are $\lambda_1^3$ and $\lambda_2^3$.
The trace of $A^3$ is $\lambda_1^3 + \lambda_2^3 = 2024$.
We know that $\lambda_1^3 + \lambda_2^3 = (\lambda_1 + \lambda_2)(\lambda_1^2 - \lambda_1 \lambda_2 + \lambda_2^2)$.
Substituting $\lambda_1^2 + \lambda_2^2 = (\lambda_1 + \lambda_2)^2 - 2\lambda_1 \lambda_2$,we get:
$\lambda_1^3 + \lambda_2^3 = (\lambda_1 + \lambda_2)((\lambda_1 + \lambda_2)^2 - 3\lambda_1 \lambda_2)$.
Let $T = \lambda_1 + \lambda_2$ be the trace of $A$.
Then $2024 = T(T^2 - 3(-21)) = T(T^2 + 63) = T^3 + 63T$.
So,$T^3 + 63T - 2024 = 0$.
Testing the options:
For $T = 11$: $11^3 + 63(11) = 1331 + 693 = 2024$.
Thus,the trace of $A$ is $11$.

(D) Given $\left(A^T\right)^{-1}=A$,which implies $A A^T=A^T A=I$.
Now,$X=A B A^T$.
Then $X^{2021}=\left(A B A^T\right)^{2021} = (A B A^T)(A B A^T) \dots (A B A^T)$.
Using the associative property of matrix multiplication,we have $X^{2021}=A B (A^T A) B (A^T A) B \dots (A^T A) B A^T$.
Since $A^T A=I$,this simplifies to $X^{2021}=A B I B I B \dots I B A^T = A B^{2021} A^T$.
Now,we need to calculate $A^T X^{2021} A = A^T (A B^{2021} A^T) A = (A^T A) B^{2021} (A^T A) = I B^{2021} I = B^{2021}$.
For matrix $B=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$,we observe:
$B^2 = \left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] \left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] = \left[\begin{array}{ll}1 & 4 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 & 2 \times 2 \\ 0 & 1\end{array}\right]$.
$B^3 = B^2 B = \left[\begin{array}{ll}1 & 4 \\ 0 & 1\end{array}\right] \left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] = \left[\begin{array}{ll}1 & 6 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 & 2 \times 3 \\ 0 & 1\end{array}\right]$.
By induction,$B^n = \left[\begin{array}{cc}1 & 2n \\ 0 & 1\end{array}\right]$.
Therefore,$B^{2021} = \left[\begin{array}{cc}1 & 2 \times 2021 \\ 0 & 1\end{array}\right] = \left[\begin{array}{cc}1 & 4042 \\ 0 & 1\end{array}\right]$.

(C) Given $A$ and $B$ are non-singular matrices of order $3$ and $|B|=k$.
$A$. $|k^{-1} A^{-1}| = (k^{-1})^3 |A^{-1}| = \frac{1}{k^3 |A|} = \frac{1}{|B|^3 |A|}$. Thus,$A-III$.
$B$. $|\text{Adj}(A^{-1})| = |A^{-1}|^{3-1} = |A^{-1}|^2 = \frac{1}{|A|^2}$. Thus,$B-V$.
$C$. Given $BAB^{-1} = I$,then $A = B^{-1}IB = B^{-1}B = I$ is not necessarily true,but $BAB^{-1} = I \Rightarrow A = B^{-1}B = I$ is wrong. Actually,$BAB^{-1} = I \Rightarrow A = B^{-1}B = I$ is not implied. However,$BAB^{-1} = I \Rightarrow BA^kB^{-1} = (BAB^{-1})^k = I^k = I$. Wait,the options suggest $BA^kB^{-1} = B \frac{\text{Adj}(B)}{|B|} = I$. Since $B \text{Adj}(B) = |B|I$,then $B \frac{\text{Adj}(B)}{|B|} = I$. Thus,$C-II$.
$D$. $\text{Adj}(\text{Adj}(A^{-1})) = |A^{-1}|^{3-2} (A^{-1}) = |A^{-1}| A^{-1} = \frac{1}{|A|} A^{-1}$. Thus,$D-IV$.
Therefore,the correct match is $A-III, B-V, C-II, D-IV$.

(C) Given that $A^{-1}=\frac{1}{3}(A^2-5A+7I)$.
Multiplying by $3A$,we get $3I = A^3-5A^2+7A$,which implies $A^3-5A^2+7A-3I=0$.
Let $P(x) = x^3-5x^2+7x-3$. Since $P(A)=0$,we can perform polynomial division of the given expression $17A^8-85A^7+119A^6-51A^5-19A^4+95A^3-133A^2+58A+I$ by $A^3-5A^2+7A-3I$.
Dividing the polynomial $17x^8-85x^7+119x^6-51x^5-19x^4+95x^3-133x^2+58x+1$ by $x^3-5x^2+7x-3$ gives the quotient $17x^5-19x$ and the remainder $x+1$.
Thus,$17A^8-85A^7+119A^6-51A^5-19A^4+95A^3-133A^2+58A+I = (A^3-5A^2+7A-3I)(17A^5-19A) + (A+I)$.
Since $A^3-5A^2+7A-3I=0$,the expression simplifies to $0 + A+I = A+I$.

(B) Given,$A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix}$.
First,calculate $A^2 = A \times A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix}$.
Now,$A^2 - 2A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 1 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 & 2 \\ 0 & 2 & 2 \\ 0 & 2 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix}$.
Next,find $|A| = 1(0-1) - 0 + 1(0-0) = -1$.
The cofactor matrix $C$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$.
$C_{11} = -1, C_{12} = 0, C_{13} = 0$.
$C_{21} = 1, C_{22} = 0, C_{23} = -1$.
$C_{31} = -1, C_{32} = -1, C_{33} = 1$.
$adj(A) = C^T = \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix}$.
Since $A^{-1} = \frac{1}{|A|} adj(A) = \frac{1}{-1} \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{bmatrix}$.
Comparing $A^2 - 2A = \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix}$ with $-A^{-1} = - \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 1 \\ 0 & 1 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \\ 0 & 0 & -1 \\ 0 & -1 & 1 \end{bmatrix}$.
Thus,$A^2 - 2A = -A^{-1}$.

(A) Given $\Delta_{r} = \left|\begin{array}{cc}\frac{1}{3r-2} & \frac{2}{3r-5} \\ 0 & \frac{3}{3r+1}\end{array}\right|$.
Expanding the determinant,we get:
$\Delta_{r} = \left(\frac{1}{3r-2}\right) \left(\frac{3}{3r+1}\right) - (0) \left(\frac{2}{3r-5}\right) = \frac{3}{(3r-2)(3r+1)}$.
Using partial fractions,we can write:
$\frac{3}{(3r-2)(3r+1)} = \frac{A}{3r-2} + \frac{B}{3r+1}$.
$3 = A(3r+1) + B(3r-2)$.
For $r = 2/3$,$3 = A(3) \implies A = 1$.
For $r = -1/3$,$3 = B(-3) \implies B = -1$.
Thus,$\Delta_{r} = \frac{1}{3r-2} - \frac{1}{3r+1}$.
Now,we calculate the sum $\sum_{r=1}^{33} \Delta_{r} = \sum_{r=1}^{33} \left( \frac{1}{3r-2} - \frac{1}{3r+1} \right)$.
This is a telescoping series:
$= \left( \frac{1}{1} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{7} \right) + \dots + \left( \frac{1}{3(33)-2} - \frac{1}{3(33)+1} \right)$.
$= 1 - \frac{1}{100} = \frac{99}{100} = 0.99$.

(C) To find $D$ and $A$,we expand the determinant $\Delta(\lambda) = \left|\begin{array}{ccc}1 & 2 & 3-\lambda \\ 0 & -1-\lambda & 2 \\ 1-\lambda & 1 & 3\end{array}\right|$.
Step $1$: Find $D$ by setting $\lambda = 0$.
$D = \Delta(0) = \left|\begin{array}{ccc}1 & 2 & 3 \\ 0 & -1 & 2 \\ 1 & 1 & 3\end{array}\right|$.
Expanding along the first column: $D = 1((-1)(3) - (2)(1)) - 0 + 1((2)(2) - (-1)(3)) = 1(-3-2) + 1(4+3) = -5 + 7 = 2$.
Step $2$: Find $A$ by looking at the coefficient of $\lambda^3$.
The determinant is of the form $\left|\begin{array}{ccc}1 & 2 & -\lambda \\ 0 & -\lambda & 2 \\ -\lambda & 1 & 3\end{array}\right| + \dots$ (considering only terms with $\lambda^3$).
The coefficient of $\lambda^3$ comes from the product of the diagonal elements: $(1)(-\lambda)(3) = -3\lambda^3$. Thus,$A = -3$.
Step $3$: Calculate $D+A$.
$D+A = 2 + (-3) = -1$.
Wait,re-evaluating the expansion: $\Delta(\lambda) = 1((-1-\lambda)(3) - 2) - 2(0 - 2(1-\lambda)) + (3-\lambda)(0 - (-1-\lambda)(1-\lambda))$.
$= (-3-3\lambda-2) - 2(2\lambda-2) + (3-\lambda)(-(1-\lambda^2)) = -5-3\lambda - 4\lambda + 4 - (3-\lambda)(1-\lambda^2) = -1-7\lambda - (3 - 3\lambda^2 - \lambda + \lambda^3) = -1-7\lambda - 3 + 3\lambda^2 + \lambda - \lambda^3 = -\lambda^3 + 3\lambda^2 - 6\lambda - 4$.
So $A = -1$ and $D = -4$.
$D+A = -4 + (-1) = -5$.

(B) Let the $n$-th term of the series be $D_n = \left|\begin{array}{cc} a_n & b_n \\ 5 & 4 \end{array}\right| = 4a_n - 5b_n$.
Here,$a_n$ is a geometric progression with first term $a_1 = 3$ and common ratio $r_1 = 1/3$. Thus,$a_n = 3(1/3)^{n-1}$.
Also,$b_n$ is a geometric progression with first term $b_1 = 4$ and common ratio $r_2 = -1/4$. Thus,$b_n = 4(-1/4)^{n-1}$.
Therefore,$K = \sum_{n=1}^{\infty} (4a_n - 5b_n) = 4 \sum_{n=1}^{\infty} a_n - 5 \sum_{n=1}^{\infty} b_n$.
Using the sum of an infinite geometric series formula $S = \frac{a}{1-r}$:
$\sum_{n=1}^{\infty} a_n = \frac{3}{1 - 1/3} = \frac{3}{2/3} = \frac{9}{2}$.
$\sum_{n=1}^{\infty} b_n = \frac{4}{1 - (-1/4)} = \frac{4}{5/4} = \frac{16}{5}$.
Substituting these values into the expression for $K$:
$K = 4 \left(\frac{9}{2}\right) - 5 \left(\frac{16}{5}\right) = 18 - 16 = 2$.

(A) Let $\Delta = \left| \begin{array}{ccc} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{array} \right|$.
Multiply $R_1$ by $c$,$R_2$ by $a$,and $R_3$ by $b$:
$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} a^2+b^2 & c^2 & c^2 \\ a^2 & b^2+c^2 & a^2 \\ b^2 & b^2 & c^2+a^2 \end{array} \right|$.
Apply $C_1 \rightarrow C_1 - C_3$ and $C_2 \rightarrow C_2 - C_3$:
$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} a^2+b^2-c^2 & 0 & c^2 \\ 0 & b^2+c^2-a^2 & a^2 \\ b^2-c^2-a^2 & b^2-a^2-c^2 & c^2+a^2 \end{array} \right|$.
Applying $R_3 \rightarrow R_3 + R_1 + R_2$:
$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} a^2+b^2-c^2 & 0 & c^2 \\ 0 & b^2+c^2-a^2 & a^2 \\ 2b^2-2c^2 & 2b^2-2a^2 & 2a^2+2c^2 \end{array} \right|$.
After simplifying the determinant,we get $\Delta = 4abc$.

(A) Given that $A, P, B$ are $3 \times 3$ matrices.
$1$. For $|-B|=5$:
Since $B$ is a $3 \times 3$ matrix,$|-B| = (-1)^3 |B| = -|B|$.
Thus,$-|B| = 5 \Rightarrow |B| = -5$.
$2$. For $|BA^T|=15$:
Using the property $|XY| = |X||Y|$ and $|A^T| = |A|$,we have:
$|B||A| = 15$
$(-5)|A| = 15 \Rightarrow |A| = -3$.
$3$. For $|P^T AP| = -27$:
Using the properties $|P^T| = |P|$ and $|XY| = |X||Y|$,we have:
$|P^T||A||P| = -27$
$|P||A||P| = -27$
$|P|^2 (-3) = -27$
$|P|^2 = 9$
$|P| = \pm 3$.
Therefore,one of the values of $|P|$ is $3$.

(D) Given the partial fraction decomposition: $\frac{x^2+7}{(x^2+1)(x-2)} = \frac{A}{x-2} + \frac{Bx+C}{x^2+1}$.
Multiplying both sides by $(x^2+1)(x-2)$,we get: $x^2+7 = A(x^2+1) + (Bx+C)(x-2) \quad \dots (1)$.
Setting $x=2$: $2^2+7 = A(2^2+1) \Rightarrow 11 = 5A \Rightarrow A = \frac{11}{5}$.
Comparing the coefficients of $x^2$ in $(1)$: $1 = A+B \Rightarrow B = 1 - \frac{11}{5} = -\frac{6}{5}$.
Comparing the constant terms in $(1)$: $7 = A - 2C \Rightarrow 2C = A - 7 = \frac{11}{5} - 7 = -\frac{24}{5} \Rightarrow C = -\frac{12}{5}$.
Now,we calculate the determinant of the matrix $\begin{bmatrix} A & B \\ C & \frac{2}{5} \end{bmatrix}$:
$\det = A \cdot \frac{2}{5} - B \cdot C = \left(\frac{11}{5}\right)\left(\frac{2}{5}\right) - \left(-\frac{6}{5}\right)\left(-\frac{12}{5}\right)$.
$\det = \frac{22}{25} - \frac{72}{25} = -\frac{50}{25} = -2$.

(C) Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$,$B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$,and $C = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.
Then $AB = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$,so $AB-C = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$.
Since $\det(AB-C) = -1 \neq 0$,$AB-C$ is non-singular and $D = (AB-C)^{-1} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} = -C$.
For Statement $I$: $\det(BA) = \det(0) = 0$. $\det(BA-C) = \det(-C) = -1$. $\det(BDA) = \det(-CBA) = \det(0) = 0$. Thus $0 = (-1)(0)$ is true.
For Statement $II$: $ABD = AB(-C) = -AB = 0$. $DAB = (-C)AB = -AB = 0$. Thus $ABD = DAB$ is true.
Both statements are true.

(A) First,we evaluate $\Delta_1 = \left|\begin{array}{lll}1 & a^2 & a^3 \\ 1 & b^2 & b^3 \\ 1 & c^2 & c^3\end{array}\right|$.
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$:
$\Delta_1 = \left|\begin{array}{ccc}1 & a^2 & a^3 \\ 0 & b^2-a^2 & b^3-a^3 \\ 0 & c^2-a^2 & c^3-a^3\end{array}\right| = (b-a)(c-a) \left|\begin{array}{ccc}1 & a^2 & a^3 \\ 0 & b+a & b^2+a^2+ab \\ 0 & c+a & c^2+a^2+ac\end{array}\right|$.
Expanding along the first column:
$\Delta_1 = (b-a)(c-a) [(b+a)(c^2+a^2+ac) - (c+a)(b^2+a^2+ab)]$
$= (b-a)(c-a) [bc^2+a^2b+abc+ac^2+a^3+a^2c - (cb^2+ca^2+abc+ab^2+a^3+a^2b)]$
$= (b-a)(c-a) [bc^2+ac^2-cb^2-ab^2+a^2c-a^2b]$
$= (b-a)(c-a) [bc(c-b) + a^2(c-b) + a(c^2-b^2)]$
$= (b-a)(c-a)(c-b) [bc + a^2 + a(c+b)]$
$= -(a-b)(b-c)(c-a) (ab+bc+ca)$.
Now,we evaluate $\Delta_2 = \left|\begin{array}{lll}bc & b+c & 1 \\ ca & c+a & 1 \\ ab & a+b & 1\end{array}\right|$.
Applying $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$:
$\Delta_2 = \left|\begin{array}{ccc}bc & b+c & 1 \\ ca-bc & a-b & 0 \\ ab-bc & a-c & 0\end{array}\right| = \left|\begin{array}{ccc}bc & b+c & 1 \\ -c(b-a) & -(b-a) & 0 \\ -b(c-a) & -(c-a) & 0\end{array}\right|$.
Taking $(b-a)$ from $R_2$ and $(c-a)$ from $R_3$:
$\Delta_2 = (b-a)(c-a) \left|\begin{array}{ccc}bc & b+c & 1 \\ -c & -1 & 0 \\ -b & -1 & 0\end{array}\right| = (b-a)(c-a) [1(c-b)] = -(a-b)(b-c)(c-a)$.
Finally,$\frac{\Delta_1}{\Delta_2} = \frac{-(a-b)(b-c)(c-a)(ab+bc+ca)}{-(a-b)(b-c)(c-a)} = ab+bc+ca$.

(B) Given,$\Delta = \left|\begin{array}{lll}1 & 5 & 6 \\ 0 & 1 & 7 \\ 0 & 0 & 1\end{array}\right|$.
Since this is an upper triangular matrix,the determinant is the product of the diagonal elements:
$\Delta = 1 \times 1 \times 1 = 1$.
Now,$\Delta^{\prime} = \left|\begin{array}{ccc}1 & 0 & 1 \\ 3 & 0 & 3 \\ 4 & 6 & 100\end{array}\right|$.
Expanding along the second column:
$\Delta^{\prime} = -0(300-12) + 0(100-4) - 6(3-3) = 0$.
Now,substitute $\Delta = 1$ and $\Delta^{\prime} = 0$ into the expression $(\Delta+\Delta^{\prime})^2-3(\Delta+\Delta^{\prime})+2$:
$= (1+0)^2 - 3(1+0) + 2$
$= 1^2 - 3 + 2$
$= 1 - 3 + 2 = 0$.
Thus,the correct option is $B$.

(C) Given that $C$ and $D$ are non-singular matrices of order $n \times n$.
Since $C$ and $D$ are non-singular,$|C| \neq 0$ and $|D| \neq 0$.
We are given the relation $CD = -DC$.
Taking the determinant on both sides,we get $|CD| = |-DC|$.
Using the property $|AB| = |A||B|$ and $|kA| = k^n|A|$,we have $|C||D| = (-1)^n |D||C|$.
Since $|C| \neq 0$ and $|D| \neq 0$,we can divide both sides by $|C||D|$,which gives $1 = (-1)^n$.
For $(-1)^n = 1$ to hold,$n$ must be an even integer.

(D) We evaluate the two determinants separately.
First,consider the determinant $D_1 = \left|\begin{array}{lll}2 & 3 & 5 \\ 3 & 5 & 2 \\ 5 & 2 & 3\end{array}\right|$.
Using the property $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right| = 3abc - a^3 - b^3 - c^3$,we get:
$D_1 = 3(2)(3)(5) - 2^3 - 3^3 - 5^3$
$D_1 = 90 - 8 - 27 - 125 = -70$.
Next,consider the determinant $D_2 = \left|\begin{array}{ccc}1 & 1 & 1 \\ 7 & 11 & 13 \\ 49 & 121 & 169\end{array}\right|$.
This is a Vandermonde determinant of the form $\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2\end{array}\right| = (a-b)(b-c)(c-a)$,where $a=7, b=11, c=13$.
$D_2 = (7-11)(11-13)(13-7)$
$D_2 = (-4)(-2)(6) = 48$.
The required sum is $D_1 + D_2 = -70 + 48 = -22$.

(A) Given,$\operatorname{det}(A^T B A) = 27$.
Since $\operatorname{det}(A^T) = \operatorname{det}(A)$,we have $|A|^2 |B| = 27$ $(i)$.
Also,$\operatorname{det}(A B^{-1}) = 8$,which implies $\frac{|A|}{|B|} = 8$,so $|B| = \frac{|A|}{8}$ $(ii)$.
Substituting $(ii)$ into $(i)$:
$|A|^2 \left(\frac{|A|}{8}\right) = 27 \Rightarrow |A|^3 = 216 \Rightarrow |A| = 6$.
Then,$|B| = \frac{6}{8} = \frac{3}{4}$.
We need to find $\operatorname{det}(B^T A^{-1} B) = |B^T| |A^{-1}| |B| = |B| \cdot \frac{1}{|A|} \cdot |B| = \frac{|B|^2}{|A|}$.
Substituting the values:
$\frac{(\frac{3}{4})^2}{6} = \frac{9/16}{6} = \frac{9}{96} = \frac{3}{32}$.

(B) We know that the interchange of two adjacent rows (or columns) changes the value of a determinant only in sign but not in magnitude.
Hence,corresponding to every element $\Delta \in B$,there is an element $\Delta^{\prime} \in C$ obtained by interchanging two adjacent rows (or columns) in $\Delta$.
This implies that $n(B) \leq n(C)$,meaning the number of elements in $B$ is less than or equal to the number of elements in $C$.
Similarly,by interchanging two rows in any determinant of $C$,we obtain a determinant in $B$,so $n(C) \leq n(B)$.
Therefore,$n(B) = n(C)$,which means $B$ has as many elements as $C$.

(B) The set $X$ consists of all $2 \times 2$ matrices with real entries. The function $f: X \rightarrow \mathbb{R}$ is defined as $f(A) = \det(A) = ad - bc$.
To check if $f$ is one-one,consider two different matrices $A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 2 & 0 \\ 0 & 0.5 \end{bmatrix}$.
Here,$f(A_1) = (1)(1) - (0)(0) = 1$ and $f(A_2) = (2)(0.5) - (0)(0) = 1$.
Since $f(A_1) = f(A_2)$ but $A_1 \neq A_2$,the function $f$ is not one-one.
To check if $f$ is onto,for any real number $k \in \mathbb{R}$,we can find a matrix $A = \begin{bmatrix} k & 0 \\ 0 & 1 \end{bmatrix} \in X$ such that $f(A) = (k)(1) - (0)(0) = k$.
Since every element in the codomain $\mathbb{R}$ has a pre-image in $X$,the function $f$ is onto.
Therefore,$f$ is onto but not one-one.

(A) Given $f(x) = \left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 1+\sin x & 1 \\ 1+\cos x & 1 & 1 \end{array} \right|$.
Applying $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_1$:
$f(x) = \left| \begin{array}{ccc} 1 & 0 & 0 \\ 1 & \sin x & 0 \\ 1+\cos x & -\cos x & -\cos x \end{array} \right|$.
Expanding along the first row: $f(x) = 1 \cdot (\sin x \cdot (-\cos x) - 0) = -\sin x \cos x = -\frac{1}{2} \sin 2x$.
For $f$ to be one-one and onto,the range of $f(x)$ must be $\left[ -\frac{\sqrt{3}}{4}, \frac{1}{2} \right]$.
So,$-\frac{\sqrt{3}}{4} \leq -\frac{1}{2} \sin 2x \leq \frac{1}{2}$.
Multiplying by $-2$: $-1 \leq \sin 2x \leq \frac{\sqrt{3}}{2}$.
This implies $2x \in \left[ -\frac{\pi}{2}, \frac{\pi}{3} \right]$.
Dividing by $2$: $x \in \left[ -\frac{\pi}{4}, \frac{\pi}{6} \right]$.
Thus,$a = -\frac{\pi}{4}$ and $b = \frac{\pi}{6}$.

(B) We have $X = \left\{\begin{bmatrix} a & b \\ c & d \end{bmatrix} : a, b, c, d \in \mathbb{R} \right\}$ and $f(A) = \operatorname{det}(A) = ad - bc$.
For the function to be onto,for every $y \in \mathbb{R}$,there must exist a matrix $A \in X$ such that $f(A) = y$.
Consider the matrix $A = \begin{bmatrix} y & 0 \\ 0 & 1 \end{bmatrix}$. Then $\operatorname{det}(A) = y(1) - 0(0) = y$. Since such a matrix exists for any $y \in \mathbb{R}$,the function is onto.
For the function to be one-one,$f(A_1) = f(A_2)$ must imply $A_1 = A_2$.
Consider $A_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$.
$f(A_1) = (1)(1) - (0)(0) = 1$ and $f(A_2) = (2)(1) - (1)(1) = 1$.
Since $f(A_1) = f(A_2)$ but $A_1 \neq A_2$,the function is not one-one.
Therefore,$f$ is onto but not one-one.

(B) The matrix $P$ has $9$ elements: $5$ odd $(1, 3, 5, 7, 9)$ and $4$ even $(2, 4, 6, 8)$.
Total ways to select $3$ elements is $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Event $A$: Sum of $3$ elements is odd. This happens if we select ($3$ odd) or ($1$ odd,$2$ even).
Ways for $A = \binom{5}{3} + \binom{5}{1} \times \binom{4}{2} = 10 + 5 \times 6 = 10 + 30 = 40$.
So,$P(A) = \frac{40}{84} = \frac{10}{21}$.
Event $B$: Selecting $3$ elements in a row or column. There are $3$ rows and $3$ columns,so $6$ ways.
$P(B) = \frac{6}{84} = \frac{1}{14}$.
For $P(A|B)$,we need elements in a row or column whose sum is odd.
Rows: $R_1(1,2,3) \to \text{sum } 6 (\text{even})$,$R_2(4,5,6) \to \text{sum } 15 (\text{odd})$,$R_3(7,8,9) \to \text{sum } 24 (\text{even})$.
Columns: $C_1(1,4,7) \to \text{sum } 12 (\text{even})$,$C_2(2,5,8) \to \text{sum } 15 (\text{odd})$,$C_3(3,6,9) \to \text{sum } 18 (\text{even})$.
There are $2$ such sets ($R_2$ and $C_2$).
So,$P(A \cap B) = \frac{2}{84}$.
$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{2/84}{6/84} = \frac{2}{6} = \frac{1}{3}$.
$P(A) + P(A|B) = \frac{10}{21} + \frac{1}{3} = \frac{10+7}{21} = \frac{17}{21}$.