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

(B) matrix $A = [a_{ij}]$ is called a skew-symmetric matrix if $A^T = -A$,which implies $a_{ij} = -a_{ji}$ for all $i, j$.
For diagonal elements,we set $i = j$,which gives $a_{ii} = -a_{ii}$.
This implies $2a_{ii} = 0$,so $a_{ii} = 0$ for all $i$.
Therefore,all diagonal elements of a skew-symmetric matrix are zero.

(B) Given the matrix $A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 5 \\ 2 & -5 & 0 \end{bmatrix}$.
To find the transpose $A'$,we interchange the rows and columns:
$A' = \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & -5 \\ -2 & 5 & 0 \end{bmatrix}$.
Now,factor out $-1$ from the matrix $A'$:
$A' = -1 \times \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 5 \\ 2 & -5 & 0 \end{bmatrix}$.
Since the matrix inside is $A$,we have $A' = -A$.
Thus,the matrix $A$ is a skew-symmetric matrix.

(B) The correct property for the transpose of a product of matrices is $(AB)' = B'A'$.
Option $(B)$ states $(AB \dots L)' = A'B' \dots L'$,which is incorrect because the order of matrices is reversed in the transpose of a product.
Option $(D)$ is also technically written incorrectly in the prompt as $(A)' = A$,but assuming it refers to the double transpose property $(A')' = A$,it is a standard identity.
Given the options,$(B)$ is the fundamentally incorrect relation regarding matrix algebra.

(B) Given that $A$ and $B$ are symmetric matrices,we have $A' = A$ and $B' = B$.
Consider the transpose of the matrix $(AB - BA)$:
$(AB - BA)' = (AB)' - (BA)'$
Using the property $(XY)' = Y'X'$,we get:
$(AB - BA)' = B'A' - A'B'$
Since $A' = A$ and $B' = B$,we substitute these into the expression:
$(AB - BA)' = BA - AB$
$(AB - BA)' = -(AB - BA)$
Since the transpose of the matrix $(AB - BA)$ is equal to its negative,$(AB - BA)$ is a skew-symmetric matrix.

(A) To determine the nature of the matrix $M'AM$,we take its transpose:
$(M'AM)' = M'A'(M')'$
Using the property of transpose $(ABC)' = C'B'A'$,we get:
$(M'AM)' = M'A'M$
Since $A$ is a symmetric matrix,we have $A' = A$.
Substituting this into the equation,we get:
$(M'AM)' = M'AM$
Since the transpose of the matrix $M'AM$ is equal to the matrix itself,$M'AM$ is a symmetric matrix.

(B) square matrix $A$ is said to be an orthogonal matrix if $A'A = I = AA'$,where $A'$ is the transpose of $A$ and $I$ is the identity matrix.
Let $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$.
Then,the transpose $A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.
Calculating $AA'$:
$AA' = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} \cos^2 \alpha + \sin^2 \alpha & -\cos \alpha \sin \alpha + \sin \alpha \cos \alpha \\ -\sin \alpha \cos \alpha + \cos \alpha \sin \alpha & \sin^2 \alpha + \cos^2 \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Similarly,$A'A = I$.
Since $AA' = A'A = I$,the matrix in option $(b)$ is an orthogonal matrix.

(D) matrix $M$ is symmetric if $M' = M$.
For option $A$: $(A + A')' = A' + (A')' = A' + A = A + A'$. Thus,$A + A'$ is symmetric.
For option $B$: $(AA')' = (A')'A' = AA'$. Thus,$AA'$ is symmetric.
For option $C$: $(A'A)' = A'(A')' = A'A$. Thus,$A'A$ is symmetric.
For option $D$: $(A - A')' = A' - (A')' = A' - A = -(A - A')$. Since $(A - A')' = -(A - A')$,the matrix $A - A'$ is skew-symmetric,not symmetric.

(B) The transpose of the product of two matrices is equal to the product of their transposes in reverse order.
For any two matrices $A$ and $B$ of the same order,the property is given by $(AB)' = B'A'$.
Therefore,option $(B)$ is the correct property.

(A) Given that $A$ is a symmetric matrix,we have $A^T = A$.
To check if $A^n$ is symmetric,we need to find the transpose of $A^n$,which is $(A^n)^T$.
Using the property of transpose $(A^k)^T = (A^T)^k$,we get:
$(A^n)^T = (A^T)^n$
Since $A^T = A$,we substitute this into the equation:
$(A^n)^T = (A)^n = A^n$
Since $(A^n)^T = A^n$,it follows that $A^n$ is a symmetric matrix.

(A) Given the matrix $A = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix}$.
The transpose of matrix $A$,denoted by $A^T$,is obtained by interchanging its rows and columns:
$A^T = \begin{bmatrix} 1 & 5 \\ -2 & 3 \end{bmatrix}$.
Now,we calculate the sum $A + A^T$:
$A + A^T = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix} + \begin{bmatrix} 1 & 5 \\ -2 & 3 \end{bmatrix}$
$A + A^T = \begin{bmatrix} 1+1 & -2+5 \\ 5+(-2) & 3+3 \end{bmatrix}$
$A + A^T = \begin{bmatrix} 2 & 3 \\ 3 & 6 \end{bmatrix}$.
Thus,the correct option is $A$.

(B) Let $A$ be a skew-symmetric matrix of order $n \times n$. By definition,$A^T = -A$.
Taking the determinant on both sides,we have $\det(A^T) = \det(-A)$.
Since $\det(A^T) = \det(A)$ and $\det(-A) = (-1)^n \det(A)$,we get $\det(A) = (-1)^n \det(A)$.
If $n$ is odd,then $(-1)^n = -1$,so $\det(A) = -\det(A)$,which implies $2 \det(A) = 0$,or $\det(A) = 0$.
$A$ matrix with a determinant of $0$ is called a singular matrix.
Therefore,a skew-symmetric matrix of odd order is always singular.

(B) Let $B = AA^T$.
To check if $B$ is symmetric,we find its transpose $B^T$.
$B^T = (AA^T)^T$.
Using the property $(XY)^T = Y^T X^T$,we get:
$B^T = (A^T)^T A^T$.
Since $(A^T)^T = A$,we have:
$B^T = AA^T = B$.
Since $B^T = B$,the matrix $AA^T$ is a symmetric matrix.

(B) Let $A = \begin{bmatrix} 0 & 5 & -7 \\ -5 & 0 & 11 \\ 7 & -11 & 0 \end{bmatrix}$.
To check if the matrix is skew-symmetric,we find its transpose $A'$.
$A' = \begin{bmatrix} 0 & -5 & 7 \\ 5 & 0 & -11 \\ -7 & 11 & 0 \end{bmatrix}$.
Now,factor out $-1$ from the matrix $A'$:
$A' = -1 \begin{bmatrix} 0 & 5 & -7 \\ -5 & 0 & 11 \\ 7 & -11 & 0 \end{bmatrix} = -A$.
Since $A' = -A$,the matrix is a skew-symmetric matrix.

(B) Let $S = A + A^T$.
To check if $S$ is symmetric,we find its transpose:
$S^T = (A + A^T)^T$.
Using the property of transpose $(X + Y)^T = X^T + Y^T$,we get:
$S^T = A^T + (A^T)^T$.
Since $(A^T)^T = A$,we have:
$S^T = A^T + A = A + A^T = S$.
Since $S^T = S$,the matrix $A + A^T$ is a symmetric matrix.

(D) To determine the nature of the matrix $A = \begin{bmatrix} i & 1 - 2i \\ -1 - 2i & 0 \end{bmatrix}$,we calculate the conjugate transpose $(\bar{A})^T$.
First,find the conjugate matrix $\bar{A}$ by changing the sign of the imaginary part of each element:
$\bar{A} = \begin{bmatrix} -i & 1 + 2i \\ -1 + 2i & 0 \end{bmatrix}$.
Next,find the transpose of the conjugate matrix $(\bar{A})^T$:
$(\bar{A})^T = \begin{bmatrix} -i & -1 + 2i \\ 1 + 2i & 0 \end{bmatrix}$.
Now,compare this with $-A$:
$-A = -\begin{bmatrix} i & 1 - 2i \\ -1 - 2i & 0 \end{bmatrix} = \begin{bmatrix} -i & -1 + 2i \\ 1 + 2i & 0 \end{bmatrix}$.
Since $(\bar{A})^T = -A$,the matrix $A$ is skew-hermitian.

(A) Let $A$ be a skew-symmetric matrix of odd order $n$,where $n = 2k + 1$ for some integer $k \ge 0$.
Since $A$ is skew-symmetric,we have $A^T = -A$.
Taking the determinant on both sides,we get $|A^T| = |-A|$.
Using the property $|A^T| = |A|$ and $|cA| = c^n|A|$ for a matrix of order $n$,we have:
$|A| = (-1)^n |A|$.
Since $n$ is odd,$(-1)^n = -1$.
Therefore,$|A| = -|A|$.
$2|A| = 0 \Rightarrow |A| = 0$.

(C) Given $A = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$.
The transpose of matrix $A$,denoted by $A'$,is obtained by interchanging its rows and columns.
Thus,$A' = \begin{bmatrix} 1 & 2 & 3 \end{bmatrix}$.
Now,we calculate the product $AA'$:
$AA' = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 1 \times 1 & 1 \times 2 & 1 \times 3 \\ 2 \times 1 & 2 \times 2 & 2 \times 3 \\ 3 \times 1 & 3 \times 2 & 3 \times 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix}$.

(B) The transpose of the product of two matrices is equal to the product of their transposes in reverse order.
Mathematically,for any two matrices $A$ and $B$ such that the product $AB$ is defined,the property of the transpose is given by:
$(AB)' = B'A'$.
Therefore,the correct option is $B$.

(C) Let $B = A - A^T$. To check if $B$ is symmetric or skew-symmetric,we find its transpose $B^T$.
$B^T = (A - A^T)^T$
Using the property $(X - Y)^T = X^T - Y^T$,we get:
$B^T = A^T - (A^T)^T$
Since $(A^T)^T = A$,we have:
$B^T = A^T - A$
$B^T = -(A - A^T)$
$B^T = -B$
Since $B^T = -B$,the matrix $A - A^T$ is a skew-symmetric matrix.

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

(A) Given that $A$ and $B$ are symmetric matrices,we have $A' = A$ and $B' = B$.
To check if $ABA$ is symmetric,we find its transpose:
$(ABA)' = A'B'A'$
Using the property $(XYZ)' = Z'Y'X'$,we get:
$(ABA)' = A'B'A'$
Since $A' = A$ and $B' = B$,we substitute these into the expression:
$(ABA)' = ABA$
Since the transpose of $ABA$ is equal to $ABA$ itself,$ABA$ is a symmetric matrix.

(D) The trace of a square matrix $A$,denoted by $t_r(A)$,is defined as the sum of its diagonal elements.
Properties of the trace include:
$1$. $t_r(A + B) = t_r(A) + t_r(B)$,which is true.
$2$. $t_r(\alpha A) = \alpha t_r(A)$,which is true for any scalar $\alpha \in R$.
$3$. $t_r(A^T) = t_r(A)$,which is true because the diagonal elements remain unchanged under transposition.
$4$. $t_r(AB) = t_r(BA)$ is a fundamental property of the trace of products of matrices. Therefore,the statement $t_r(AB) \ne t_r(BA)$ is incorrect.

(A) square matrix $A$ is orthogonal if $AA^T = A^TA = I$,where $I$ is the identity matrix. This implies that the sum of squares of elements in each row (or column) is $1$,and the sum of products of corresponding elements of any two distinct rows (or columns) is $0$.
Let us check option $A$:
Row $1$: $(6/7)^2 + (2/7)^2 + (-3/7)^2 = (36+4+9)/49 = 49/49 = 1$.
Row $2$: $(2/7)^2 + (3/7)^2 + (6/7)^2 = (4+9+36)/49 = 49/49 = 1$.
Row $3$: $(3/7)^2 + (-6/7)^2 + (2/7)^2 = (9+36+4)/49 = 49/49 = 1$.
Dot product of Row $1$ and Row $2$: $(6/7)(2/7) + (2/7)(3/7) + (-3/7)(6/7) = (12+6-18)/49 = 0/49 = 0$.
Dot product of Row $2$ and Row $3$: $(2/7)(3/7) + (3/7)(-6/7) + (6/7)(2/7) = (6-18+12)/49 = 0/49 = 0$.
Dot product of Row $1$ and Row $3$: $(6/7)(3/7) + (2/7)(-6/7) + (-3/7)(2/7) = (18-12-6)/49 = 0/49 = 0$.
Since all conditions are satisfied,the matrix in option $A$ is orthogonal.

(C) matrix $A$ is called nilpotent if there exists a positive integer $m$ such that $A^m = 0$,where $0$ is the zero matrix.
For option $C$,let $A = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}$.
Then $A^2 = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 \times 0 + 0 \times 1 & 0 \times 0 + 0 \times 0 \\ 1 \times 0 + 0 \times 1 & 1 \times 0 + 0 \times 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Since $A^2 = 0$,the matrix $A$ is a nilpotent matrix of index $2$.

(C) matrix $A$ is orthogonal if $AA^T = I$.
Calculating $A^T = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$.
Then $AA^T = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} = \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & 0 \\ 0 & \sin^2 \theta + \cos^2 \theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$.
Since $AA^T = I$,the matrix $A$ is an orthogonal matrix for all $\theta \in \mathbb{R}$.
Thus,option $C$ is correct.

(C) The trace of a matrix,$Tr(M)$,is the sum of its diagonal elements.
For $A + 2B$,the diagonal elements are $1, -3, 1$. Thus,$Tr(A + 2B) = 1 - 3 + 1 = -1$.
Using the property $Tr(A + 2B) = Tr(A) + 2Tr(B)$,we have $Tr(A) + 2Tr(B) = -1$.
For $2A - B$,the diagonal elements are $2, -1, 2$. Thus,$Tr(2A - B) = 2 - 1 + 2 = 3$.
Using the property $Tr(2A - B) = 2Tr(A) - Tr(B)$,we have $2Tr(A) - Tr(B) = 3$.
Let $Tr(A) = x$ and $Tr(B) = y$.
We have the system of equations:
$x + 2y = -1$
$2x - y = 3$
Multiplying the second equation by $2$,we get $4x - 2y = 6$.
Adding this to the first equation: $(x + 2y) + (4x - 2y) = -1 + 6 \implies 5x = 5 \implies x = 1$.
Substituting $x = 1$ into $2x - y = 3$: $2(1) - y = 3 \implies 2 - y = 3 \implies y = -1$.
Therefore,$Tr(A) - Tr(B) = x - y = 1 - (-1) = 2$.

(A) Given $Q = PAP^T$ and $P$ is an orthogonal matrix,so $P^TP = PP^T = I$.
We need to find $X = P^TQ^{2005}P$.
First,calculate $Q^2 = (PAP^T)(PAP^T) = PA(P^TP)AP^T = PA(I)AP^T = PA^2P^T$.
By induction,$Q^n = PA^nP^T$.
Thus,$Q^{2005} = PA^{2005}P^T$.
Substituting this into the expression for $X$:
$X = P^T(PA^{2005}P^T)P = (P^TP)A^{2005}(P^TP) = I \cdot A^{2005} \cdot I = A^{2005}$.
Since $A$ is a periodic matrix with period $4$,$A^{k+1} = A$ where $k=4$. Thus $A^5 = A$.
We can write $2005 = 4 \times 501 + 1$.
Therefore,$A^{2005} = A^{4 \times 501 + 1} = (A^4)^{501} \cdot A = I^{501} \cdot A = A$.
Hence,$X = A$.

(B) The trace of a matrix,$Tr(M)$,is the sum of its diagonal elements.
Given $2A + B = \begin{bmatrix} 1 & 0 & 3 \\ -1 & 4 & 6 \\ 2 & 5 & 2 \end{bmatrix}$,the trace is $Tr(2A + B) = 1 + 4 + 2 = 7$.
Since $Tr(2A + B) = 2Tr(A) + Tr(B)$,we have $2Tr(A) + Tr(B) = 7$ (Equation $1$).
Given $A - 2B = \begin{bmatrix} 2 & -1 & 5 \\ 0 & 3 & 6 \\ 1 & 2 & 1 \end{bmatrix}$,the trace is $Tr(A - 2B) = 2 + 3 + 1 = 6$.
Since $Tr(A - 2B) = Tr(A) - 2Tr(B)$,we have $Tr(A) - 2Tr(B) = 6$ (Equation $2$).
Multiplying Equation $1$ by $2$: $4Tr(A) + 2Tr(B) = 14$.
Adding this to Equation $2$: $(4Tr(A) + 2Tr(B)) + (Tr(A) - 2Tr(B)) = 14 + 6 \Rightarrow 5Tr(A) = 20 \Rightarrow Tr(A) = 4$.
Substituting $Tr(A) = 4$ into Equation $1$: $2(4) + Tr(B) = 7 \Rightarrow 8 + Tr(B) = 7 \Rightarrow Tr(B) = -1$.
Finally,$Tr(A) - Tr(B) = 4 - (-1) = 4 + 1 = 5$.

(B) Given that $A$ is a symmetric matrix,so $A^T = A$.
Given that $B$ is a skew-symmetric matrix,so $B^T = -B$.
We need to check the nature of the matrix $M = AB - BA$.
Taking the transpose of $M$:
$M^T = (AB - BA)^T = (AB)^T - (BA)^T$
Using the property $(XY)^T = Y^T X^T$:
$M^T = B^T A^T - A^T B^T$
Substituting $A^T = A$ and $B^T = -B$:
$M^T = (-B)(A) - (A)(-B)$
$M^T = -BA + AB = AB - BA = M$
Since $M^T = M$,the matrix $AB - BA$ is a symmetric matrix.

(B) Given that $A$ is a symmetric matrix,$A^T = A$.
Given that $B$ is a skew-symmetric matrix,$B^T = -B$.
We are given $A + B = \begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix} \quad (1)$.
Taking the transpose of both sides:
$(A + B)^T = \begin{bmatrix} 2 & 3 \\ 5 & -1 \end{bmatrix}^T \implies A^T + B^T = \begin{bmatrix} 2 & 5 \\ 3 & -1 \end{bmatrix}$.
Substituting $A^T = A$ and $B^T = -B$,we get $A - B = \begin{bmatrix} 2 & 5 \\ 3 & -1 \end{bmatrix} \quad (2)$.
Adding equations $(1)$ and $(2)$:
$2A = \begin{bmatrix} 2+2 & 3+5 \\ 5+3 & -1-1 \end{bmatrix} = \begin{bmatrix} 4 & 8 \\ 8 & -2 \end{bmatrix} \implies A = \begin{bmatrix} 2 & 4 \\ 4 & -1 \end{bmatrix}$.
Subtracting equation $(2)$ from $(1)$:
$2B = \begin{bmatrix} 2-2 & 3-5 \\ 5-3 & -1-(-1) \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} \implies B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
Now,calculating $AB$:
$AB = \begin{bmatrix} 2 & 4 \\ 4 & -1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} (2)(0) + (4)(1) & (2)(-1) + (4)(0) \\ (4)(0) + (-1)(1) & (4)(-1) + (-1)(0) \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ -1 & -4 \end{bmatrix}$.

(N/A) Given the matrix $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$.
To find the transpose $A'$,we interchange the rows and columns of $A$:
$A' = \begin{bmatrix} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{bmatrix}$.
Now,to find the transpose of $A'$,denoted as $(A')'$,we interchange the rows and columns of $A'$:
$(A')' = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$.
Comparing this result with the original matrix $A$,we see that $(A')' = A$.
Hence,the property is verified.

(N/A) Given $A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & -1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$.
First,calculate $A+B$:
$A+B = \begin{bmatrix} 3+2 & \sqrt{3}-1 & 2+2 \\ 4+1 & 2+2 & 0+4 \end{bmatrix} = \begin{bmatrix} 5 & \sqrt{3}-1 & 4 \\ 5 & 4 & 4 \end{bmatrix}$.
Now,find the transpose $(A+B)^{\prime}$:
$(A+B)^{\prime} = \begin{bmatrix} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{bmatrix}$.
Next,find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \begin{bmatrix} 3 & 4 \\ \sqrt{3} & 2 \\ 2 & 0 \end{bmatrix}$,$B^{\prime} = \begin{bmatrix} 2 & 1 \\ -1 & 2 \\ 2 & 4 \end{bmatrix}$.
Finally,calculate $A^{\prime} + B^{\prime}$:
$A^{\prime} + B^{\prime} = \begin{bmatrix} 3+2 & 4+1 \\ \sqrt{3}-1 & 2+2 \\ 2+2 & 0+4 \end{bmatrix} = \begin{bmatrix} 5 & 5 \\ \sqrt{3}-1 & 4 \\ 4 & 4 \end{bmatrix}$.
Since $(A+B)^{\prime} = A^{\prime} + B^{\prime}$,the property is verified.

(A) Given $A = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 & -6 \end{bmatrix}$.
First,calculate the product $AB$:
$AB = \begin{bmatrix} -2 \\ 4 \\ 5 \end{bmatrix} \begin{bmatrix} 1 & 3 & -6 \end{bmatrix} = \begin{bmatrix} -2 & -6 & 12 \\ 4 & 12 & -24 \\ 5 & 15 & -30 \end{bmatrix}$.
Now,find the transpose of the product $(AB)^{\prime}$:
$(AB)^{\prime} = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$.
Next,find the transposes of $A$ and $B$:
$A^{\prime} = \begin{bmatrix} -2 & 4 & 5 \end{bmatrix}$ and $B^{\prime} = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix}$.
Now,calculate the product $B^{\prime} A^{\prime}$:
$B^{\prime} A^{\prime} = \begin{bmatrix} 1 \\ 3 \\ -6 \end{bmatrix} \begin{bmatrix} -2 & 4 & 5 \end{bmatrix} = \begin{bmatrix} -2 & 4 & 5 \\ -6 & 12 & 15 \\ 12 & -24 & -30 \end{bmatrix}$.
Comparing the results,we see that $(AB)^{\prime} = B^{\prime} A^{\prime}$. Hence,the property is verified.

Any square matrix $B$ can be expressed as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$,where $P = \frac{1}{2}(B + B')$ and $Q = \frac{1}{2}(B - B')$.
Given $B = \left[\begin{array}{rrr}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right]$,its transpose is $B' = \left[\begin{array}{rrr}2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3\end{array}\right]$.
Calculating $P = \frac{1}{2}(B + B') = \frac{1}{2} \left( \left[\begin{array}{rrr}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right] + \left[\begin{array}{rrr}2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3\end{array}\right] \right) = \frac{1}{2} \left[\begin{array}{rrr}4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6\end{array}\right] = \left[\begin{array}{rrr}2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3\end{array}\right]$.
Since $P' = P$,$P$ is a symmetric matrix.
Calculating $Q = \frac{1}{2}(B - B') = \frac{1}{2} \left( \left[\begin{array}{rrr}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right] - \left[\begin{array}{rrr}2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3\end{array}\right] \right) = \frac{1}{2} \left[\begin{array}{rrr}0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0\end{array}\right] = \left[\begin{array}{rrr}0 & -\frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0\end{array}\right]$.
Since $Q' = -Q$,$Q$ is a skew-symmetric matrix.
Thus,$B = P + Q = \left[\begin{array}{rrr}2 & -\frac{3}{2} & -\frac{3}{2} \\ -\frac{3}{2} & 3 & 1 \\ -\frac{3}{2} & 1 & -3\end{array}\right] + \left[\begin{array}{rrr}0 & -\frac{1}{2} & -\frac{5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0\end{array}\right]$.

(A) Let $A = \left[\begin{array}{c}5 \\ \frac{1}{2} \\ -1\end{array}\right]$.
To find the transpose of a matrix,we interchange its rows and columns.
The given matrix $A$ is a column matrix of order $3 \times 1$.
Its transpose $A^T$ will be a row matrix of order $1 \times 3$.
Thus,$A^T = \left[\begin{array}{lll}5 & \frac{1}{2} & -1\end{array}\right]$.

(A) Let $A = \left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$.
To find the transpose of a matrix,we interchange its rows and columns.
The first row of $A$ is $(1, -1)$,which becomes the first column of $A^T$.
The second row of $A$ is $(2, 3)$,which becomes the second column of $A^T$.
Therefore,$A^T = \left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]$.

(A) To find the transpose of a matrix,we interchange its rows and columns.
Let $A = \left[\begin{array}{ccc}-1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1\end{array}\right]$.
The transpose of $A$,denoted by $A^T$,is obtained by writing the first row of $A$ as the first column of $A^T$,the second row of $A$ as the second column of $A^T$,and the third row of $A$ as the third column of $A^T$.
Thus,$A^T = \left[\begin{array}{ccc}-1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1\end{array}\right]$.

(A) We have:
$A^{\prime}=\begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}, B^{\prime}=\begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix}$
$A+B=\begin{bmatrix}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{bmatrix}=\begin{bmatrix}-5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2\end{bmatrix}$
$\therefore (A+B)^{\prime}=\begin{bmatrix}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{bmatrix}$
$A^{\prime}+B^{\prime}=\begin{bmatrix}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{bmatrix}+\begin{bmatrix}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{bmatrix}=\begin{bmatrix}-5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2\end{bmatrix}$
Since $(A+B)^{\prime} = A^{\prime}+B^{\prime}$,the property is verified.

(A) First,calculate $A-B$:
$A-B = \left[\begin{array}{rrr}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right] - \left[\begin{array}{rrr}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right] = \left[\begin{array}{rrr}3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0\end{array}\right]$
Now,find the transpose $(A-B)^{\prime}$:
$(A-B)^{\prime} = \left[\begin{array}{rrr}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]$
Next,find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \left[\begin{array}{rrr}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right]$,$B^{\prime} = \left[\begin{array}{rrr}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right]$
Now,calculate $A^{\prime}-B^{\prime}$:
$A^{\prime}-B^{\prime} = \left[\begin{array}{rrr}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right] - \left[\begin{array}{rrr}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right] = \left[\begin{array}{rrr}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]$
Since $(A-B)^{\prime} = A^{\prime}-B^{\prime}$,the property is verified.

(A) We know that $A = (A^{\prime})^{\prime}$.
Therefore,$A = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}$.
Given $B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}$,we find $B^{\prime} = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}$.
Now,$A+B = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}$.
Taking the transpose,$(A+B)^{\prime} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$.
Next,$A^{\prime} + B^{\prime} = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$.
Since $(A+B)^{\prime} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$ and $A^{\prime} + B^{\prime} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$,it is verified that $(A+B)^{\prime} = A^{\prime} + B^{\prime}$.

(A) Given $A^{\prime} = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$,we find $A = (A^{\prime})^{\prime} = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}$.
Now,$A-B = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} = \begin{bmatrix} 4 & -3 & -1 \\ 3 & 0 & -2 \end{bmatrix}$.
Therefore,$(A-B)^{\prime} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$.
Next,$B^{\prime} = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}$.
Then,$A^{\prime}-B^{\prime} = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$.
Since $(A-B)^{\prime} = A^{\prime}-B^{\prime} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix}$,the property is verified.

(A) First,we calculate the product $AB$:
$AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \begin{bmatrix} 1 & 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 \times 1 & 0 \times 5 & 0 \times 7 \\ 1 \times 1 & 1 \times 5 & 1 \times 7 \\ 2 \times 1 & 2 \times 5 & 2 \times 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}$
Now,we find the transpose $(AB)^{\prime}$ by interchanging rows and columns:
$(AB)^{\prime} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$
Next,we find $A^{\prime}$ and $B^{\prime}$:
$A^{\prime} = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}$
$B^{\prime} = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}$
Now,we calculate the product $B^{\prime} A^{\prime}$:
$B^{\prime} A^{\prime} = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \begin{bmatrix} 0 & 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 \times 0 & 1 \times 1 & 1 \times 2 \\ 5 \times 0 & 5 \times 1 & 5 \times 2 \\ 7 \times 0 & 7 \times 1 & 7 \times 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$
Since $(AB)^{\prime} = B^{\prime} A^{\prime}$,the property is verified.

(N/A) matrix $A$ is symmetric if $A^{\prime} = A$,where $A^{\prime}$ is the transpose of matrix $A$.
Given $A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$.
To find the transpose $A^{\prime}$,we interchange the rows and columns of $A$:
$A^{\prime} = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$.
Since $A^{\prime} = A$,the matrix $A$ is a symmetric matrix.

(N/A) matrix $A$ is said to be skew-symmetric if $A^{\prime} = -A$.
Given $A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}$.
The transpose of matrix $A$,denoted by $A^{\prime}$,is obtained by interchanging its rows and columns:
$A^{\prime} = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$.
Now,factor out $-1$ from the matrix $A^{\prime}$:
$A^{\prime} = -1 \times \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} = -A$.
Since $A^{\prime} = -A$,the matrix $A$ is a skew-symmetric matrix.

(N/A) Given: $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$.
First,find the transpose of matrix $A$,denoted as $A^{\prime}$:
$A^{\prime} = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$.
Now,calculate the sum $(A + A^{\prime})$:
$A + A^{\prime} = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} 1+1 & 5+6 \\ 6+5 & 7+7 \end{bmatrix} = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$.
To verify if $(A + A^{\prime})$ is symmetric,we check if $(A + A^{\prime})^{\prime} = (A + A^{\prime})$:
$(A + A^{\prime})^{\prime} = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}^{\prime} = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$.
Since $(A + A^{\prime})^{\prime} = (A + A^{\prime})$,it is proved that $(A + A^{\prime})$ is a symmetric matrix.

(N/A) Given $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$.
The transpose of $A$ is $A^{\prime} = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$.
Now,calculate $A - A^{\prime}$:
$A - A^{\prime} = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} - \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
Let $B = A - A^{\prime} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$.
Now,find the transpose of $B$:
$B^{\prime} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$.
Observe that $B^{\prime} = -\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = -B$.
Since $(A - A^{\prime})^{\prime} = -(A - A^{\prime})$,it is verified that $(A - A^{\prime})$ is a skew-symmetric matrix.

Given the matrix $A = \left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right].$
First,find the transpose $A^{\prime}$ by interchanging rows and columns:
$A^{\prime} = \left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right].$
Now,calculate $A+A^{\prime}$:
$A+A^{\prime} = \left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right] + \left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right].$
Therefore,$\frac{1}{2}(A+A^{\prime}) = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right].$
Next,calculate $A-A^{\prime}$:
$A-A^{\prime} = \left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right] - \left[\begin{array}{ccc}0 & -a & -b \\ a & 0 & -c \\ b & c & 0\end{array}\right] = \left[\begin{array}{ccc}0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0\end{array}\right].$
Therefore,$\frac{1}{2}(A-A^{\prime}) = \left[\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right].$

(A) Let $A = \left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$.
Then,the transpose $A^{\prime} = \left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$.
Any square matrix $A$ can be written as $A = P + Q$,where $P = \frac{1}{2}(A + A^{\prime})$ is a symmetric matrix and $Q = \frac{1}{2}(A - A^{\prime})$ is a skew-symmetric matrix.
First,calculate $A + A^{\prime}$:
$A + A^{\prime} = \left[\begin{array}{ccc}6+6 & -2-2 & 2+2 \\ -2-2 & 3+3 & -1-1 \\ 2+2 & -1-1 & 3+3\end{array}\right] = \left[\begin{array}{ccc}12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6\end{array}\right]$.
Thus,$P = \frac{1}{2}(A + A^{\prime}) = \left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right]$.
Since $P^{\prime} = P$,$P$ is a symmetric matrix.
Next,calculate $A - A^{\prime}$:
$A - A^{\prime} = \left[\begin{array}{ccc}6-6 & -2-(-2) & 2-2 \\ -2-(-2) & 3-3 & -1-(-1) \\ 2-2 & -1-(-1) & 3-3\end{array}\right] = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$.
Thus,$Q = \frac{1}{2}(A - A^{\prime}) = \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$.
Since $Q^{\prime} = -Q$,$Q$ is a skew-symmetric matrix.
Finally,$A = P + Q = \left[\begin{array}{ccc}6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3\end{array}\right] + \left[\begin{array}{ccc}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right]$.

(N/A) Let $A = \left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right]$. Then the transpose of $A$ is $A^{\prime} = \left[\begin{array}{cc}1 & -1 \\ 5 & 2\end{array}\right]$.
Any square matrix $A$ can be written as $A = P + Q$,where $P = \frac{1}{2}(A + A^{\prime})$ is a symmetric matrix and $Q = \frac{1}{2}(A - A^{\prime})$ is a skew-symmetric matrix.
First,calculate $A + A^{\prime} = \left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right] + \left[\begin{array}{cc}1 & -1 \\ 5 & 2\end{array}\right] = \left[\begin{array}{cc}2 & 4 \\ 4 & 4\end{array}\right]$.
Thus,$P = \frac{1}{2}(A + A^{\prime}) = \left[\begin{array}{cc}1 & 2 \\ 2 & 2\end{array}\right]$. Since $P^{\prime} = P$,$P$ is symmetric.
Next,calculate $A - A^{\prime} = \left[\begin{array}{cc}1 & 5 \\ -1 & 2\end{array}\right] - \left[\begin{array}{cc}1 & -1 \\ 5 & 2\end{array}\right] = \left[\begin{array}{cc}0 & 6 \\ -6 & 0\end{array}\right]$.
Thus,$Q = \frac{1}{2}(A - A^{\prime}) = \left[\begin{array}{cc}0 & 3 \\ -3 & 0\end{array}\right]$. Since $Q^{\prime} = -Q$,$Q$ is skew-symmetric.
Therefore,$A = P + Q = \left[\begin{array}{cc}1 & 2 \\ 2 & 2\end{array}\right] + \left[\begin{array}{cc}0 & 3 \\ -3 & 0\end{array}\right]$.

(D) Given $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.
The transpose of matrix $A$ is $A^{\prime} = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$.
Given the condition $A + A^{\prime} = I$,where $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the identity matrix.
Adding $A$ and $A^{\prime}$:
$\begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} + \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 2 \cos \alpha & 0 \\ 0 & 2 \cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Comparing the corresponding elements,we get:
$2 \cos \alpha = 1$
$\cos \alpha = \frac{1}{2}$
Since $\cos \alpha = \frac{1}{2}$,the value of $\alpha$ is $\frac{\pi}{3}$.