Special types of matrices, Transpose of matrices and Trace of matrices Questions in English

(N/A) Since $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$.
Part $1$: Assume $AB$ is symmetric.
Then $(AB)^{\prime} = AB$.
Using the property of transpose $(AB)^{\prime} = B^{\prime}A^{\prime}$,we get $B^{\prime}A^{\prime} = AB$.
Since $A^{\prime} = A$ and $B^{\prime} = B$,this implies $BA = AB$.
Part $2$: Conversely,assume $AB = BA$.
We need to show that $AB$ is symmetric,i.e.,$(AB)^{\prime} = AB$.
$(AB)^{\prime} = B^{\prime}A^{\prime}$.
Since $A$ and $B$ are symmetric,$B^{\prime}A^{\prime} = BA$.
Given $BA = AB$,we have $(AB)^{\prime} = AB$.
Thus,$AB$ is symmetric.

(N/A) Given that $A$ and $B$ are symmetric matrices,we have:
$A' = A$ and $B' = B$ ........ $(1)$
We need to check the transpose of $(AB - BA)$:
$(AB - BA)' = (AB)' - (BA)'$
$= B'A' - A'B'$
$= BA - AB$ (Using equation $(1)$)
$= -(AB - BA)$
Since $(AB - BA)' = -(AB - BA)$,it follows that $(AB - BA)$ is a skew-symmetric matrix.

(A) We suppose that $A$ is a symmetric matrix,then $A^{\prime} = A$ ......... $(1)$
Consider
$(B^{\prime}AB)^{\prime} = \{B^{\prime}(AB)\}^{\prime}$
$= (AB)^{\prime}(B^{\prime})^{\prime}$ $[(AB)^{\prime} = B^{\prime}A^{\prime}]$
$= B^{\prime}A^{\prime}(B)$ $[(B^{\prime})^{\prime} = B]$
$= B^{\prime}(A^{\prime}B)$
$= B^{\prime}(AB)$ $[$Using $(1)]$
$\therefore (B^{\prime}AB)^{\prime} = B^{\prime}AB$
Thus,if $A$ is a symmetric matrix,then $B^{\prime}AB$ is a symmetric matrix.
Now,we suppose that $A$ is a skew-symmetric matrix,then $A^{\prime} = -A$ ......... $(2)$
Consider
$(B^{\prime}AB)^{\prime} = \{B^{\prime}(AB)\}^{\prime} = (AB)^{\prime}(B^{\prime})^{\prime}$
$= (B^{\prime}A^{\prime})B$
$= B^{\prime}(-A)B$ $[$Using $(2)]$
$= -B^{\prime}AB$
$\therefore (B^{\prime}AB)^{\prime} = -B^{\prime}AB$
Thus,if $A$ is a skew-symmetric matrix,then $B^{\prime}AB$ is a skew-symmetric matrix.
Hence,the matrix $B^{\prime}AB$ is symmetric or skew-symmetric according as $A$ is symmetric or skew-symmetric.

(C) If $A$ is both a symmetric and a skew-symmetric matrix,then by definition we have $A^{\prime} = A$ and $A^{\prime} = -A$.
Equating the two expressions for $A^{\prime}$,we get $A = -A$.
Adding $A$ to both sides,we get $A + A = 0$,which simplifies to $2A = 0$.
Dividing by $2$,we obtain $A = 0$.
Therefore,$A$ must be a zero matrix.
Hence,the correct answer is $C$.

(B) Given equations are:
$A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix} \quad \dots(1)$
$2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix} \quad \dots(2)$
Multiply equation $(2)$ by $2$:
$4A - 2B = \begin{bmatrix} 4 & -2 & 10 \\ 4 & -2 & 12 \\ 0 & 2 & 4 \end{bmatrix} \quad \dots(3)$
Adding $(1)$ and $(3)$:
$5A = \begin{bmatrix} 1+4 & 2-2 & 0+10 \\ 6+4 & -3-2 & 3+12 \\ -5+0 & 3+2 & 1+4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & -5 & 15 \\ -5 & 5 & 5 \end{bmatrix}$
$A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & -1 & 3 \\ -1 & 1 & 1 \end{bmatrix}$
Now,from $(2)$,$B = 2A - \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$:
$B = \begin{bmatrix} 2 & 0 & 4 \\ 4 & -2 & 6 \\ -2 & 2 & 2 \end{bmatrix} - \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -1 \\ 2 & -1 & 0 \\ -2 & 1 & 0 \end{bmatrix}$
$\operatorname{Tr}(A) = 1 + (-1) + 1 = 1$
$\operatorname{Tr}(B) = 0 + (-1) + 0 = -1$
$\operatorname{Tr}(A) - \operatorname{Tr}(B) = 1 - (-1) = 2$

(D) Given $A = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix}$ and $AA^{T} = I_{2}$.
First,find the transpose $A^{T} = \begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix}$.
Now,compute the product $AA^{T}$:
$AA^{T} = \begin{bmatrix} 1 & -\alpha \\ \alpha & \beta \end{bmatrix} \begin{bmatrix} 1 & \alpha \\ -\alpha & \beta \end{bmatrix} = \begin{bmatrix} 1 + \alpha^{2} & \alpha - \alpha\beta \\ \alpha - \alpha\beta & \alpha^{2} + \beta^{2} \end{bmatrix}$.
Since $AA^{T} = I_{2} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,we equate the corresponding elements:
$1 + \alpha^{2} = 1 \Rightarrow \alpha^{2} = 0 \Rightarrow \alpha = 0$.
$\alpha - \alpha\beta = 0 \Rightarrow 0 - 0\beta = 0$ (which is always true).
$\alpha^{2} + \beta^{2} = 1 \Rightarrow 0 + \beta^{2} = 1 \Rightarrow \beta^{2} = 1$.
We need to find the value of $\alpha^{4} + \beta^{4}$:
$\alpha^{4} + \beta^{4} = (0)^{2} + (1)^{2} = 0 + 1 = 1$.

(A) Given,$A^T = A$,$B^T = -B$,$C^T = -C$.
For $(S1)$,let $M = A^{13} B^{26} - B^{26} A^{13}$.
Then,$M^T = (A^{13} B^{26} - B^{26} A^{13})^T = (B^{26})^T (A^{13})^T - (A^{13})^T (B^{26})^T$.
Since $B^T = -B$,$(B^T)^{26} = (-B)^{26} = B^{26}$.
So,$M^T = B^{26} A^{13} - A^{13} B^{26} = -(A^{13} B^{26} - B^{26} A^{13}) = -M$.
Thus,$M$ is skew-symmetric. $(S1)$ is false.
For $(S2)$,let $N = A^{26} C^{13} - C^{13} A^{26}$.
Then,$N^T = (A^{26} C^{13} - C^{13} A^{26})^T = (C^{13})^T (A^{26})^T - (A^{26})^T (C^{13})^T$.
Since $C^T = -C$,$(C^T)^{13} = (-C)^{13} = -C^{13}$.
So,$N^T = (-C^{13}) A^{26} - A^{26} (-C^{13}) = -C^{13} A^{26} + A^{26} C^{13} = N$.
Thus,$N$ is symmetric. $(S2)$ is true.
Therefore,only $S2$ is true.

(D) Given the equation $P^{\top} = 2P + I$.
Taking the transpose on both sides,we get $(P^{\top})^{\top} = (2P + I)^{\top}$.
Since $(P^{\top})^{\top} = P$,we have $P = 2P^{\top} + I$.
Substitute $P^{\top} = 2P + I$ into this equation:
$P = 2(2P + I) + I$.
$P = 4P + 2I + I$.
$P = 4P + 3I$.
Rearranging the terms,we get $3P = -3I$,which implies $P = -I$.
Therefore,for any column matrix $X \neq 0$,we have $PX = (-I)X = -X$.

(D) Given that $X$ and $Y$ are skew-symmetric,$X^T = -X$ and $Y^T = -Y$. Given that $Z$ is symmetric,$Z^T = Z$.
For $(A): (Y^3 Z^4 - Z^4 Y^3)^T = (Z^4)^T (Y^3)^T - (Y^3)^T (Z^4)^T = Z^4 (-Y)^3 - (-Y)^3 Z^4 = -Z^4 Y^3 + Y^3 Z^4 = Y^3 Z^4 - Z^4 Y^3$. This is symmetric.
For $(B): (X^{44} + Y^{44})^T = (X^{44})^T + (Y^{44})^T = (X^T)^{44} + (Y^T)^{44} = (-X)^{44} + (-Y)^{44} = X^{44} + Y^{44}$. This is symmetric.
For $(C): (X^4 Z^3 - Z^3 X^4)^T = (Z^3)^T (X^4)^T - (X^4)^T (Z^3)^T = Z^3 (X^T)^4 - (X^T)^4 Z^3 = Z^3 X^4 - X^4 Z^3 = -(X^4 Z^3 - Z^3 X^4)$. This is skew-symmetric.
For $(D): (X^{23} + Y^{23})^T = (X^{23})^T + (Y^{23})^T = (X^T)^{23} + (Y^T)^{23} = (-X)^{23} + (-Y)^{23} = -(X^{23} + Y^{23})$. This is skew-symmetric.
Thus,$(C)$ and $(D)$ are skew-symmetric.

(A) The transpose of a matrix $A$,denoted by $A'$,is obtained by interchanging its rows and columns.
Given $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.
To find $A'$,we swap the elements of the first row with the first column and the second row with the second column.
Thus,$A' = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
Comparing this with the given options,option $A$ is the correct answer.

(D) Given that $A$ and $B$ are skew-symmetric matrices of the same order,we have $A^{\prime} = -A$ and $B^{\prime} = -B$.
Using the property of the transpose of a product,$(AB)^{\prime} = B^{\prime}A^{\prime}$.
Substituting the values of $A^{\prime}$ and $B^{\prime}$,we get $(AB)^{\prime} = (-B)(-A)$.
Therefore,$(AB)^{\prime} = BA$.

(A) Given that $A$ and $B$ are symmetric matrices,we have $A^T = A$ and $B^T = B$.
Consider the matrix $X = AB - BA$.
To check if $X$ is symmetric or skew-symmetric,we find its transpose:
$X^T = (AB - BA)^T = (AB)^T - (BA)^T$.
Using the property $(PQ)^T = Q^T P^T$,we get:
$X^T = B^T A^T - A^T B^T$.
Since $A^T = A$ and $B^T = B$,we substitute these values:
$X^T = BA - AB = -(AB - BA) = -X$.
Since $X^T = -X$,the matrix $AB - BA$ is a skew-symmetric matrix.

(B) Given $A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$.
Then the transpose $A^{\prime} = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$.
Given $A + A^{\prime} = I$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
Adding $A$ and $A^{\prime}$:
$A + A^{\prime} = \begin{bmatrix} \sin \alpha + \sin \alpha & -\cos \alpha + \cos \alpha \\ \cos \alpha - \cos \alpha & \sin \alpha + \sin \alpha \end{bmatrix} = \begin{bmatrix} 2 \sin \alpha & 0 \\ 0 & 2 \sin \alpha \end{bmatrix}$.
Equating to $I$:
$\begin{bmatrix} 2 \sin \alpha & 0 \\ 0 & 2 \sin \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.
This implies $2 \sin \alpha = 1$,so $\sin \alpha = \frac{1}{2}$.
The principal value of $\alpha$ for $\sin \alpha = \frac{1}{2}$ is $\alpha = \frac{\pi}{6}$.

(B) matrix $A$ is skew-symmetric if $A^T = -A$.
For a skew-symmetric matrix,the diagonal elements must be zero,so $a = 0$.
Also,the condition $A_{ij} = -A_{ji}$ must hold for all $i, j$.
Comparing the elements:
$A_{12} = 2$ and $A_{21} = b$,so $b = -2$.
$A_{13} = -3$ and $A_{31} = c$,so $c = -(-3) = 3$.
$A_{23} = 4$ and $A_{32} = -4$,which satisfies $4 = -(-4)$.
Thus,$a = 0$,$b = -2$,and $c = 3$.
Calculating the sum: $a+b+c = 0 + (-2) + 3 = 1$.

(C) matrix $A$ is symmetric if $A = A^T$,which implies $A_{ij} = A_{ji}$ for all $i, j$.
For the given matrix $A = \begin{bmatrix} 5 & 2x+3 \\ x-2 & x+1 \end{bmatrix}$,the condition for symmetry is $A_{12} = A_{21}$.
Equating the elements,we get $2x + 3 = x - 2$.
Subtracting $x$ from both sides,we get $x + 3 = -2$.
Subtracting $3$ from both sides,we get $x = -5$.
Therefore,the correct value of $x$ is $-5$.

(B) For any skew-symmetric matrix $A$ of order $n$,we have $A^T = -A$.
Taking the determinant on both sides,we get $|A^T| = |-A|$.
Since $|A^T| = |A|$ and $|-A| = (-1)^n |A|$,we have $|A| = (-1)^n |A|$.
For a $3 \times 3$ matrix,$n = 3$,so $|A| = (-1)^3 |A| = -|A|$.
This implies $2|A| = 0$,which means $|A| = 0$.

(A) We know that any square matrix $A$ can be expressed as the sum of a symmetric matrix and a skew-symmetric matrix:
$A = \frac{A + A^{\prime}}{2} + \frac{A - A^{\prime}}{2}$.
Here,$\frac{A + A^{\prime}}{2}$ is a symmetric matrix and $\frac{A - A^{\prime}}{2}$ is a skew-symmetric matrix.
Given $A = B + \frac{C}{2}$,where $B$ is skew-symmetric and $C$ is symmetric.
Comparing the two expressions,we identify $B = \frac{A - A^{\prime}}{2}$ and $\frac{C}{2} = \frac{A + A^{\prime}}{2}$.
Multiplying by $2$,we get $C = A + A^{\prime}$.
Therefore,the correct option is $A$.

(A) Given that $ A = \begin{bmatrix} \cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta \end{bmatrix} $.
The transpose of $ A $ is $ A^{T} = \begin{bmatrix} \cos 2 \theta & \sin 2 \theta \\ -\sin 2 \theta & \cos 2 \theta \end{bmatrix} $.
Adding $ A $ and $ A^{T} $,we get:
$ A+A^{T} = \begin{bmatrix} \cos 2 \theta + \cos 2 \theta & -\sin 2 \theta + \sin 2 \theta \\ \sin 2 \theta - \sin 2 \theta & \cos 2 \theta + \cos 2 \theta \end{bmatrix} = \begin{bmatrix} 2 \cos 2 \theta & 0 \\ 0 & 2 \cos 2 \theta \end{bmatrix} $.
This can be written as $ (2 \cos 2 \theta) I $,where $ I $ is the identity matrix.
Given $ A+A^{T} = I $,we have $ (2 \cos 2 \theta) I = I $.
Comparing the elements,we get $ 2 \cos 2 \theta = 1 $,which implies $ \cos 2 \theta = \frac{1}{2} $.
Since $ \cos \frac{\pi}{3} = \frac{1}{2} $,we have $ 2 \theta = \frac{\pi}{3} $,which gives $ \theta = \frac{\pi}{6} $.

(A) Given that $ A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} $.
Then the transpose of $ A $ is $ A^{\prime} = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} $.
Now,calculating the product $ A A^{\prime} $:
$ A A^{\prime} = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} $
$ = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & -\cos \alpha \sin \alpha + \sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha + \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} $
Using the trigonometric identity $ \cos^2 \alpha + \sin^2 \alpha = 1 $:
$ = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I $.

(A) Given,$A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.
The transpose of matrix $A$ is $A^{\prime} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$.
Now,calculate the product $A \cdot A^{\prime}$:
$A \cdot A^{\prime} = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$
Performing matrix multiplication:
$A \cdot A^{\prime} = \begin{bmatrix} (\cos \theta)(\cos \theta) + (\sin \theta)(\sin \theta) & (\cos \theta)(-\sin \theta) + (\sin \theta)(\cos \theta) \\ (-\sin \theta)(\cos \theta) + (\cos \theta)(\sin \theta) & (-\sin \theta)(-\sin \theta) + (\cos \theta)(\cos \theta) \end{bmatrix}$
Using the trigonometric identity $\cos^{2} \theta + \sin^{2} \theta = 1$:
$A \cdot A^{\prime} = \begin{bmatrix} \cos^{2} \theta + \sin^{2} \theta & -\sin \theta \cos \theta + \sin \theta \cos \theta \\ -\sin \theta \cos \theta + \sin \theta \cos \theta & \sin^{2} \theta + \cos^{2} \theta \end{bmatrix}$
$A \cdot A^{\prime} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$
Thus,$A \cdot A^{\prime} = I$,which is the identity matrix.

(D) Given,$A^T = -A$.
Let $P = A^{2021}$.
Then,$P^T = (A^{2021})^T = (A^T)^{2021}$.
Substituting $A^T = -A$,we get $P^T = (-A)^{2021} = (-1)^{2021} A^{2021}$.
Since $2021$ is an odd number,$(-1)^{2021} = -1$.
Therefore,$P^T = -A^{2021} = -P$.
Since $P^T = -P$,$A^{2021}$ is a skew symmetric matrix.

(B) Given that $P$ and $Q$ are symmetric matrices,we have $P^{\prime} = P$ and $Q^{\prime} = Q$.
To determine the nature of $PQ - QP$,we take its transpose:
$(PQ - QP)^{\prime} = (PQ)^{\prime} - (QP)^{\prime}$
Using the property $(AB)^{\prime} = B^{\prime}A^{\prime}$,we get:
$(PQ - QP)^{\prime} = Q^{\prime}P^{\prime} - P^{\prime}Q^{\prime}$
Since $P^{\prime} = P$ and $Q^{\prime} = Q$,this becomes:
$(PQ - QP)^{\prime} = QP - PQ$
$(PQ - QP)^{\prime} = -(PQ - QP)$
Since the transpose of the matrix is equal to its negative,$PQ - QP$ is a skew-symmetric matrix.

(D) Given that $B$ is a skew-symmetric matrix,we have $B^{\prime} = -B$.
To check if $A^{\prime} B A$ is symmetric or skew-symmetric,we take its transpose:
$(A^{\prime} B A)^{\prime} = A^{\prime} B^{\prime} (A^{\prime})^{\prime}$
Using the property $(XYZ)^{\prime} = Z^{\prime} Y^{\prime} X^{\prime}$,we get:
$(A^{\prime} B A)^{\prime} = A^{\prime} B^{\prime} A$
Since $B^{\prime} = -B$,we substitute this into the expression:
$(A^{\prime} B A)^{\prime} = A^{\prime} (-B) A = -(A^{\prime} B A)$
Since the transpose of the matrix $A^{\prime} B A$ is equal to its negative,$A^{\prime} B A$ is a skew-symmetric matrix.

(C) To determine the type of matrix $A$,we calculate $A^2$:
$A^2 = A \times A = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \end{bmatrix}$
$A^2 = \begin{bmatrix} (\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) & (\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) \\ (-\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) & (-\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}}) \end{bmatrix}$
$A^2 = \begin{bmatrix} \frac{1}{2} - \frac{1}{2} & \frac{1}{2} - \frac{1}{2} \\ -\frac{1}{2} + \frac{1}{2} & -\frac{1}{2} + \frac{1}{2} \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
Since $A^2 = O$ (the null matrix),the matrix $A$ is a nilpotent matrix.

(B) Given $A = \begin{bmatrix} 1 & -1 & 2 \\ -2 & 3 & -3 \\ 4 & -4 & 5 \end{bmatrix}$.
Then $A^T = \begin{bmatrix} 1 & -2 & 4 \\ -1 & 3 & -4 \\ 2 & -3 & 5 \end{bmatrix}$.
First,calculate $AA^T$:
$AA^T = \begin{bmatrix} 1 & -1 & 2 \\ -2 & 3 & -3 \\ 4 & -4 & 5 \end{bmatrix} \begin{bmatrix} 1 & -2 & 4 \\ -1 & 3 & -4 \\ 2 & -3 & 5 \end{bmatrix} = \begin{bmatrix} 6 & -11 & 18 \\ -11 & 22 & -35 \\ 18 & -35 & 57 \end{bmatrix}$.
Next,calculate $A + A^T$:
$A + A^T = \begin{bmatrix} 1+1 & -1-2 & 2+4 \\ -2-1 & 3+3 & -3-4 \\ 4+2 & -4-3 & 5+5 \end{bmatrix} = \begin{bmatrix} 2 & -3 & 6 \\ -3 & 6 & -7 \\ 6 & -7 & 10 \end{bmatrix}$.
Finally,calculate $AA^T - (A + A^T)$:
$\begin{bmatrix} 6 & -11 & 18 \\ -11 & 22 & -35 \\ 18 & -35 & 57 \end{bmatrix} - \begin{bmatrix} 2 & -3 & 6 \\ -3 & 6 & -7 \\ 6 & -7 & 10 \end{bmatrix} = \begin{bmatrix} 4 & -8 & 12 \\ -8 & 16 & -28 \\ 12 & -28 & 47 \end{bmatrix}$.
Thus,the correct option is $B$.

(C) Given $A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7 \end{bmatrix}$.
We need to find $\operatorname{Tr}(A^2-A)$.
First,calculate $A-I$:
$A-I = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 3 \\ 1 & 6 & 9 \\ 2 & 3 & 6 \end{bmatrix}$.
Now,calculate $A(A-I)$:
$A^2-A = \begin{bmatrix} 1 & 1 & 3 \\ 1 & 7 & 9 \\ 2 & 3 & 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 3 \\ 1 & 6 & 9 \\ 2 & 3 & 6 \end{bmatrix}$.
The diagonal elements of the resulting matrix are:
$d_{11} = (1)(0) + (1)(1) + (3)(2) = 0 + 1 + 6 = 7$.
$d_{22} = (1)(1) + (7)(6) + (9)(3) = 1 + 42 + 27 = 70$.
$d_{33} = (2)(3) + (3)(9) + (7)(6) = 6 + 27 + 42 = 75$.
The trace is the sum of the diagonal elements:
$\operatorname{Tr}(A^2-A) = 7 + 70 + 75 = 152$.

(A) The trace of a square matrix is defined as the sum of its principal diagonal elements.
Given the matrix $A = \begin{bmatrix} 1 & -5 & 7 \\ 0 & 7 & 9 \\ 11 & 8 & 9 \end{bmatrix}$.
The diagonal elements are $a_{11} = 1$,$a_{22} = 7$,and $a_{33} = 9$.
The trace $tr(A) = a_{11} + a_{22} + a_{33}$.
$tr(A) = 1 + 7 + 9 = 17$.
Therefore,the trace of the matrix is $17$.

(A) Given,$A = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$.
The transpose of $A$ is $A^T = \begin{bmatrix} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{bmatrix}$.
Now,calculate the product $A A^T$:
$A A^T = \begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & 2 & 0 \\ -3 & -3 & -1 \\ 4 & 4 & 1 \end{bmatrix}$
$A A^T = \begin{bmatrix} (9+9+16) & (6+9+16) & (0+3+4) \\ (6+9+16) & (4+9+16) & (0+3+4) \\ (0+3+4) & (0+3+4) & (0+1+1) \end{bmatrix}$
$A A^T = \begin{bmatrix} 34 & 31 & 7 \\ 31 & 29 & 7 \\ 7 & 7 & 2 \end{bmatrix}$.
Since $(A A^T)^T = A A^T$,the matrix $A A^T$ is a symmetric matrix.

(C) Given that $A$ and $B$ are symmetric matrices,we have $A^{T} = A$ and $B^{T} = B$.
We are given $X = AB + BA$ and $Y = AB - BA$.
We need to find $(XY)^{T}$.
Using the property of transpose $(XY)^{T} = Y^{T} X^{T}$.
First,let us find $X^{T}$ and $Y^{T}$:
$X^{T} = (AB + BA)^{T} = (AB)^{T} + (BA)^{T} = B^{T}A^{T} + A^{T}B^{T} = BA + AB = X$.
$Y^{T} = (AB - BA)^{T} = (AB)^{T} - (BA)^{T} = B^{T}A^{T} - A^{T}B^{T} = BA - AB = -(AB - BA) = -Y$.
Now,$(XY)^{T} = Y^{T} X^{T}$.
Substituting the values of $Y^{T}$ and $X^{T}$,we get:
$(XY)^{T} = (-Y)(X) = -YX$.

(C) Given,$\left(A^T-\frac{1}{2} I\right)\left(A-\frac{1}{2} I\right) = \left(A^T+\frac{1}{2} I\right)\left(A+\frac{1}{2} I\right) = I$.
Expanding both sides:
$A^T A - \frac{1}{2} A^T - \frac{1}{2} A + \frac{1}{4} I = A^T A + \frac{1}{2} A^T + \frac{1}{2} A + \frac{1}{4} I$.
Subtracting $A^T A + \frac{1}{4} I$ from both sides,we get:
$-\frac{1}{2} A^T - \frac{1}{2} A = \frac{1}{2} A^T + \frac{1}{2} A$.
Rearranging the terms:
$0 = A^T + A$.
Therefore,$A^T = -A$.
This is the condition for a skew-symmetric matrix. Hence,$A$ is a skew-symmetric matrix.

(D) Given that $A$ is a skew-symmetric matrix,we have $A^T = -A$.
For $A^{2n}$:
$(A^{2n})^T = (A^T)^{2n} = (-A)^{2n} = (-1)^{2n} A^{2n} = A^{2n}$.
Since $(A^{2n})^T = A^{2n}$,$A^{2n}$ is a symmetric matrix. Thus,statement $1$ is false.
For $A^{2n+1}$:
$(A^{2n+1})^T = (A^T)^{2n+1} = (-A)^{2n+1} = (-1)^{2n+1} A^{2n+1} = -A^{2n+1}$.
Since $(A^{2n+1})^T = -A^{2n+1}$,$A^{2n+1}$ is a skew-symmetric matrix. Thus,statement $2$ is true.
Therefore,option $D$ is correct.

(B) Any square matrix $A$ can be expressed as the sum of a symmetric matrix $B$ and a skew-symmetric matrix $C$,where $B = \frac{1}{2}(A + A^T)$ and $C = \frac{1}{2}(A - A^T)$.
Given $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 4 \end{bmatrix}$.
Then $A^T = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & -1 \\ 1 & 0 & 4 \end{bmatrix}$.
Now,$C = \frac{1}{2}(A - A^T) = \frac{1}{2} \left( \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & -1 & 4 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & -1 \\ 1 & 0 & 4 \end{bmatrix} \right)$.
$C = \frac{1}{2} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0.5 \\ 0 & -0.5 & 0 \end{bmatrix}$.
Thus,the correct option is $B$.

(A) Given $(A+B)(A-B) = A^2 - B^2$.
Expanding the left side,we get $A^2 - AB + BA - B^2 = A^2 - B^2$.
This implies $-AB + BA = 0$,or $AB = BA$.
Calculating $AB$: $\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x & y \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} x+2 & y+4 \\ 2x+1 & 2y+2 \end{bmatrix}$.
Calculating $BA$: $\begin{bmatrix} x & y \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} x+2y & 2x+y \\ 5 & 4 \end{bmatrix}$.
Equating the two matrices:
$2x+1 = 5 \Rightarrow 2x = 4 \Rightarrow x = 2$.
$2y+2 = 4 \Rightarrow 2y = 2 \Rightarrow y = 1$.
Given $C = \begin{bmatrix} x & 2 \\ 1 & y \end{bmatrix}$,the trace of $C$ is the sum of its diagonal elements: $\operatorname{Trace}(C) = x + y = 2 + 1 = 3$.

(A) Given $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 4 & 3 & 2 \\ 3 & 4 & 5\end{array}\right]$.
First,find the transpose $A^T$:
$A^T=\left[\begin{array}{lll}1 & 4 & 3 \\ 2 & 3 & 4 \\ 3 & 2 & 5\end{array}\right]$.
Now,calculate $(A+A^T)$:
$A+A^T=\left[\begin{array}{lll}1+1 & 2+4 & 3+3 \\ 4+2 & 3+3 & 2+4 \\ 3+3 & 4+2 & 5+5\end{array}\right]=\left[\begin{array}{lll}2 & 6 & 6 \\ 6 & 6 & 6 \\ 6 & 6 & 10\end{array}\right]$.
Next,calculate $(A-A^T)$:
$A-A^T=\left[\begin{array}{lll}1-1 & 2-4 & 3-3 \\ 4-2 & 3-3 & 2-4 \\ 3-3 & 4-2 & 5-5\end{array}\right]=\left[\begin{array}{lll}0 & -2 & 0 \\ 2 & 0 & -2 \\ 0 & 2 & 0\end{array}\right]$.
Finally,multiply the two matrices:
$(A+A^T)(A-A^T)=\left[\begin{array}{lll}2 & 6 & 6 \\ 6 & 6 & 6 \\ 6 & 6 & 10\end{array}\right]\left[\begin{array}{lll}0 & -2 & 0 \\ 2 & 0 & -2 \\ 0 & 2 & 0\end{array}\right]$
$= \left[\begin{array}{lll}0+12+0 & -4+0+12 & 0-12+0 \\ 0+12+0 & -12+0+12 & 0-12+0 \\ 0+12+0 & -12+0+20 & 0-12+0\end{array}\right]$
$= \left[\begin{array}{lll}12 & 8 & -12 \\ 12 & 0 & -12 \\ 12 & 8 & -12\end{array}\right] = 4\left[\begin{array}{lll}3 & 2 & -3 \\ 3 & 0 & -3 \\ 3 & 2 & -3\end{array}\right]$.
Thus,the correct option is $A$.

(B) We know that for any square matrix $P$,the trace of its transpose is equal to the trace of the matrix itself,i.e.,$\operatorname{Tr}(P) = \operatorname{Tr}(P^T)$.
Also,$\operatorname{Tr}(P - P^T) = \operatorname{Tr}(P) - \operatorname{Tr}(P^T) = 0$ and $\operatorname{Tr}(P + P^T) = \operatorname{Tr}(P) + \operatorname{Tr}(P^T) = 2\operatorname{Tr}(P)$.
Substituting these into the given equation:
$0 + 2\operatorname{Tr}(P) + \frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P)} + \operatorname{Tr}(P) \times \operatorname{Tr}(P) = 0$
Since $\operatorname{Tr}(P) \neq 0$,we have $\frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P)} = 1$.
Thus,the equation becomes: $2\operatorname{Tr}(P) + 1 + (\operatorname{Tr}(P))^2 = 0$.
This is a quadratic equation in terms of $\operatorname{Tr}(P)$:
$(\operatorname{Tr}(P))^2 + 2\operatorname{Tr}(P) + 1 = 0$
$(\operatorname{Tr}(P) + 1)^2 = 0$
$\operatorname{Tr}(P) = -1$.

(B) Given that $A+B$ is symmetric,$(A+B)^T = A+B \Rightarrow A^T+B^T = A+B$.
Given that $A-B$ is skew-symmetric,$(A-B)^T = -(A-B) \Rightarrow A^T-B^T = -A+B$.
Adding these two equations: $2A^T = 2B \Rightarrow B = A^T$.
Subtracting these two equations: $2B^T = 2A \Rightarrow B^T = A$.
Since $A = \left[\begin{array}{ccc}-1 & 2 & 3 \\ 4 & 3 & -2 \\ 3 & -4 & 5\end{array}\right]$,then $B = A^T = \left[\begin{array}{ccc}-1 & 4 & 3 \\ 2 & 3 & -4 \\ 3 & -2 & 5\end{array}\right]$.
Given $D = C^T$,and $C = \left[\begin{array}{ccc}0 & 1 & -2 \\ 2 & -1 & 0 \\ 0 & 2 & 1\end{array}\right]$,then $D = \left[\begin{array}{ccc}0 & 2 & 0 \\ 1 & -1 & 2 \\ -2 & 0 & 1\end{array}\right]$.
Now,$B+D = \left[\begin{array}{ccc}-1 & 4 & 3 \\ 2 & 3 & -4 \\ 3 & -2 & 5\end{array}\right] + \left[\begin{array}{ccc}0 & 2 & 0 \\ 1 & -1 & 2 \\ -2 & 0 & 1\end{array}\right] = \left[\begin{array}{ccc}-1 & 6 & 3 \\ 3 & 2 & -2 \\ 1 & -2 & 6\end{array}\right]$.

(D) Given $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$.
First,find the transpose $A'$ by interchanging rows and columns:
$A' = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix}$.
Now,calculate the product $AA'$:
$AA' = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} = \begin{bmatrix} (1+4+9) & (4+10+18) & (7+16+27) \\ (4+10+18) & (16+25+36) & (28+40+54) \\ (7+16+27) & (28+40+54) & (49+64+81) \end{bmatrix} = \begin{bmatrix} 14 & 32 & 50 \\ 32 & 77 & 122 \\ 50 & 122 & 194 \end{bmatrix}$.
We know that for any matrix $B$,$(B')' = B$. Since $AA'$ is a symmetric matrix (as $(AA')' = (A')'A' = AA'$),we have $(AA')' = AA'$.
Therefore,$(AA')' = \begin{bmatrix} 14 & 32 & 50 \\ 32 & 77 & 122 \\ 50 & 122 & 194 \end{bmatrix}$.

(B) Since $A$ and $B$ are symmetric matrices,we have $A^{\prime} = A$ and $B^{\prime} = B$.
Let $P = AB - BA$.
Taking the transpose of $P$:
$P^{\prime} = (AB - BA)^{\prime} = (AB)^{\prime} - (BA)^{\prime}$
Using the property $(XY)^{\prime} = Y^{\prime}X^{\prime}$,we get:
$P^{\prime} = B^{\prime}A^{\prime} - A^{\prime}B^{\prime}$
Substituting $A^{\prime} = A$ and $B^{\prime} = B$:
$P^{\prime} = BA - AB = -(AB - BA) = -P$.
Since $P^{\prime} = -P$,the matrix $AB - BA$ is a skew-symmetric matrix.

(C) matrix $A$ is symmetric if $A = A^T$,where $A^T$ is the transpose of matrix $A$.
Given $A = \begin{bmatrix} 3 & x-1 \\ 2x+3 & x+2 \end{bmatrix}$.
The transpose $A^T$ is obtained by interchanging rows and columns: $A^T = \begin{bmatrix} 3 & 2x+3 \\ x-1 & x+2 \end{bmatrix}$.
Since $A = A^T$,we equate the corresponding elements:
$x-1 = 2x+3$.
Subtracting $x$ from both sides: $-1 = x+3$.
Subtracting $3$ from both sides: $x = -4$.
Therefore,the value of $x$ is $-4$.

(A) matrix $M$ is symmetric if $M^{T} = M$.
Consider the matrix $M = A + A^{T}$.
Taking the transpose of $M$,we get $M^{T} = (A + A^{T})^{T}$.
Using the property $(X + Y)^{T} = X^{T} + Y^{T}$,we have $M^{T} = A^{T} + (A^{T})^{T}$.
Since $(A^{T})^{T} = A$,we get $M^{T} = A^{T} + A = A + A^{T} = M$.
Since $M^{T} = M$,the matrix $A + A^{T}$ is symmetric.

(B) For a matrix $A$ to be orthogonal,the rows must be mutually orthogonal unit vectors. Let the rows be $\vec{r}_1, \vec{r}_2, \vec{r}_3$.
$1$. For $\vec{r}_1 = (0, a, a)$,we have $|\vec{r}_1|^2 = 0^2 + a^2 + a^2 = 2a^2 = 1 \Rightarrow a = \pm \frac{1}{\sqrt{2}}$.
$2$. For $\vec{r}_2 = (2b, b, -b)$,we have $|\vec{r}_2|^2 = (2b)^2 + b^2 + (-b)^2 = 4b^2 + b^2 + b^2 = 6b^2 = 1 \Rightarrow b = \pm \frac{1}{\sqrt{6}}$.
$3$. For $\vec{r}_3 = (c, -c, c)$,we have $|\vec{r}_3|^2 = c^2 + (-c)^2 + c^2 = 3c^2 = 1 \Rightarrow c = \pm \frac{1}{\sqrt{3}}$.
Checking orthogonality: $\vec{r}_1 \cdot \vec{r}_2 = 0(2b) + a(b) + a(-b) = ab - ab = 0$.
$\vec{r}_1 \cdot \vec{r}_3 = 0(c) + a(-c) + a(c) = -ac + ac = 0$.
$\vec{r}_2 \cdot \vec{r}_3 = 2b(c) + b(-c) + (-b)(c) = 2bc - bc - bc = 0$.
Thus,the values are $a = \pm \frac{1}{\sqrt{2}}, b = \pm \frac{1}{\sqrt{6}}, c = \pm \frac{1}{\sqrt{3}}$.