Step 1: Matrix Subtraction (A - B)

Given matrices $A$ and $B$, the difference $A - B$ is calculated by subtracting corresponding elements: $C_{ij} = A_{ij} - B_{ij}$.

$A - B = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix} - \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$

$A - B = \begin{bmatrix} (-1 - (-4)) & (2 - 1) & (3 - (-5)) \\ (5 - 1) & (7 - 2) & (9 - 0) \\ (-2 - 1) & (1 - 3) & (1 - 1) \end{bmatrix} = \begin{bmatrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \end{bmatrix}$

Step 2: Transposition of (A - B)

The transpose of a matrix, denoted by $M'$, is obtained by swapping rows and columns ($M'_{ij} = M_{ji}$). [Per the definition of Matrix Transpose].

$(A - B)' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$

Step 3: Calculating A' and B'

We transform both initial matrices by interchanging rows and columns:

$A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}$

$B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}$

Step 4: Subtraction of Transposes (A' - B')

Perform element-wise subtraction on the transposed matrices:

$A' - B' = \begin{bmatrix} (-1 - (-4)) & (5 - 1) & (-2 - 1) \\ (2 - 1) & (7 - 2) & (1 - 3) \\ (3 - (-5)) & (9 - 0) & (1 - 1) \end{bmatrix}$

$A' - B' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$

Step 5: Verification and Comparison

Comparing the result from Step 2 and Step 4:

Expression	Resultant Matrix
$(A - B)'$	$\begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$
$A' - B'$	$\begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$

[By the Property of Transpose of Matrices: $(A \pm B)' = A' \pm B'$]. Since both matrices are identical element-wise, the identity is verified.

Final Solution: $(A - B)' = A' - B' = \begin{bmatrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \end{bmatrix}$

Step 1: Theoretical Foundation

Any square matrix $A$ can be uniquely expressed as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$, defined by the following theorem [Per the Theorem on Decomposition of Matrices]:

$A = P + Q$, where:

• $P = \frac{1}{2}(A + A^T)$ (The Symmetric Part, where $P^T = P$)
• $Q = \frac{1}{2}(A - A^T)$ (The Skew-Symmetric Part, where $Q^T = -Q$)

Step 2: Defining the Given Matrix and Transpose

Let $A = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$.

To find the transpose $A^T$, we interchange rows and columns [Definition of Transpose]:

$A^T = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$.

Step 3: Calculating the Symmetric Matrix $P$

$P = \frac{1}{2}(A + A^T)$

$P = \frac{1}{2} \left( \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} + \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} \right)$

$P = \frac{1}{2} \begin{bmatrix} 12 & -4 & 4 \\ -4 & 6 & -2 \\ 4 & -2 & 6 \end{bmatrix} = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}$.

[Justification: $P^T = P$, confirming $P$ is symmetric].

Step 4: Calculating the Skew-Symmetric Matrix $Q$

$Q = \frac{1}{2}(A - A^T)$

$Q = \frac{1}{2} \left( \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} - \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} \right)$

$Q = \frac{1}{2} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.

[Justification: $Q^T = -Q$, confirming $Q$ is skew-symmetric].

Step 5: Synthesizing the Final Expression

The matrix $A$ expressed as the sum of a symmetric and a skew-symmetric matrix is $A = P + Q$:

$\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Final Solution: The matrix $A$ is equal to itself (symmetric) plus a null matrix (skew-symmetric). The result is $\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.

Given Variables & Initial Setup

Let the given matrix be $A$.

$A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$

Any square matrix $A$ can be expressed as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$, where:

$A = P + Q$

$P = \frac{1}{2}(A + A^T)$ [Definition of a symmetric matrix]

$Q = \frac{1}{2}(A - A^T)$ [Definition of a skew-symmetric matrix]

Step 1: Determine the Transpose ($A^T$)

To find the transpose, interchange the rows and columns of matrix $A$:

$A^T = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$

Step 2: Calculate the Symmetric Component ($P$)

First, compute the sum $A + A^T$:

$A + A^T = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} = \begin{bmatrix} 3+3 & 5+1 \\ 1+5 & -1-1 \end{bmatrix} = \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}$

Now, multiply by the scalar $\frac{1}{2}$:

$P = \frac{1}{2} \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$

[Verification: $P^T = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} = P$, which satisfies the condition for symmetry.]

Step 3: Calculate the Skew-Symmetric Component ($Q$)

First, compute the difference $A - A^T$:

$A - A^T = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix} = \begin{bmatrix} 3-3 & 5-1 \\ 1-5 & -1-(-1) \end{bmatrix} = \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}$

Now, multiply by the scalar $\frac{1}{2}$:

$Q = \frac{1}{2} \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

[Verification: $Q^T = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} = -Q$, which satisfies the condition for skew-symmetry.]

Step 4: Expressing the Sum ($P + Q$)

Combining the matrices calculated in Step 2 and Step 3:

$A = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Final Solution: $\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Step 1: Determining Matrix $A$ from Transpose $A'$

Given $A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$. By the property of transpose, $(A')' = A$ [Per the definition of Matrix Transpose: $(A^T)_{ij} = A_{ji}$].

Transposing the rows of $A'$ into columns, we obtain:

$A = \begin{bmatrix} 3 & -1 & 0 \\ 4 & 2 & 1 \end{bmatrix}$

Step 2: Calculating $(A + B)$

Given $B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}$. Since $A$ is of order $2 \times 3$ and $B$ is of order $2 \times 3$, the sum $A+B$ is defined [Per the Condition for Matrix Addition: Matrices must have identical dimensions].

$A + B = \begin{bmatrix} 3 + (-1) & -1 + 2 & 0 + 1 \\ 4 + 1 & 2 + 2 & 1 + 3 \end{bmatrix}$

$A + B = \begin{bmatrix} 2 & 1 & 1 \\ 5 & 4 & 4 \end{bmatrix}$

Step 3: Finding Left-Hand Side (LHS) = $(A + B)'$

Taking the transpose of the sum calculated in Step 2:

$(A + B)' = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$

Step 4: Finding Right-Hand Side (RHS) = $A' + B'$

We are given $A' = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix}$.

First, calculate $B'$ by transposing $B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix}$:

$B' = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix}$

Now, perform the addition $A' + B'$:

$A' + B' = \begin{bmatrix} 3 + (-1) & 4 + 1 \\ -1 + 2 & 2 + 2 \\ 0 + 1 & 1 + 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$

Step 5: Verification of Equality

Expression	Resultant Matrix
LHS: $(A + B)'$	$\begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$
RHS: $A' + B'$	$\begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$

Since the resulting matrices are identical in both entries and dimensions, the property $(A + B)' = A' + B'$ is verified [Per the Theorem of Transpose of Sums].

Final Solution: $(A + B)' = A' + B' = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \end{bmatrix}$

Step 1: Theoretical Definition of Matrix Transpose

For any given matrix $A$ of order $m \times n$, the transpose of the matrix, denoted by $A^T$ or $A'$, is defined as an $n \times m$ matrix obtained by interchanging the rows and columns of $A$. Specifically, if $A = [a_{ij}]$, then $A^T = [a_{ji}]$, where the element at position $(i, j)$ in $A$ moves to position $(j, i)$ in $A^T$ [Per the definition of transpose in linear algebra].

Step 2: Identification of Given Matrix Elements

Let the given matrix $A$ be defined as follows:

$A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$

• Element $a_{11} = 1$
• Element $a_{12} = -1$
• Element $a_{21} = 2$
• Element $a_{22} = 3$

Step 3: Transposition Procedure

To obtain $A^T$, we perform the following mapping:

• Row 1 becomes Column 1: $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$
• Row 2 becomes Column 2: $\begin{bmatrix} 2 \\ 3 \end{bmatrix}$

Consequently, the elements are rearranged such that the element originally in the $(i, j)$ position is placed in the $(j, i)$ position.

Step 4: Construction of the Resultant Matrix

Substituting the values into the new structure:

$A^T = \begin{bmatrix} a_{11} & a_{21} \\ a_{12} & a_{22} \end{bmatrix}$
$A^T = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$

Visual Representation

Final Solution: The transpose of the matrix $\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ is $\begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$.

Step 1: Identifying the Matrix and its Transpose

Given the matrix $A$:

$A = \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$

The transpose of a matrix, denoted as $A'$, is obtained by interchanging the rows and columns of $A$ [Per the Definition of Matrix Transposition].

$A' = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix}$

Step 2: Performing the Matrix Multiplication $A' A$

To verify the identity, we compute the product of $A'$ and $A$:

$A' A = \begin{bmatrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{bmatrix} \begin{bmatrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{bmatrix}$

Following the rule of matrix multiplication (Row $\times$ Column) [Per the Matrix Multiplication Row-Column Rule]:

Element Position	Calculation
(1, 1)	$(\sin \alpha)(\sin \alpha) + (-\cos \alpha)(-\cos \alpha) = \sin^2 \alpha + \cos^2 \alpha$
(1, 2)	$(\sin \alpha)(\cos \alpha) + (-\cos \alpha)(\sin \alpha) = \sin \alpha \cos \alpha - \sin \alpha \cos \alpha$
(2, 1)	$(\cos \alpha)(\sin \alpha) + (\sin \alpha)(-\cos \alpha) = \sin \alpha \cos \alpha - \sin \alpha \cos \alpha$
(2, 2)	$(\cos \alpha)(\cos \alpha) + (\sin \alpha)(\sin \alpha) = \cos^2 \alpha + \sin^2 \alpha$

Step 3: Applying Trigonometric Identities

Applying the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ [Per the Fundamental Trigonometric Identity]:

$A' A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Step 4: Conclusion

Since $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$ (the identity matrix of order 2), the condition $A' A = I$ is satisfied.

Final Solution: $A' A = I$, which confirms that the given matrix $A$ is an orthogonal matrix.

Step 1: Definition of Matrix Transpose

The transpose of a matrix $A$, denoted by $A^T$ or $A'$, is obtained by interchanging the rows and columns of the original matrix. Formally, if $A = [a_{ij}]_{m \times n}$, then $A^T = [a_{ji}]_{n \times m}$. [Per the definition of Transpose of a Matrix in Linear Algebra].

Step 2: Identifying Matrix Elements

Given the matrix $A$ of order $3 \times 3$:

$A = \begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}$

• Row 1: $[-1, 5, 6]$
• Row 2: $[\sqrt{3}, 5, 6]$
• Row 3: $[2, 3, -1]$

Step 3: Transforming Rows into Columns

To compute $A^T$, we assign the elements of the first row to the first column, the second row to the second column, and the third row to the third column.

Original Row Index ($i$)	Original Elements	Transposed Column Index ($j$)
Row 1	$[-1, 5, 6]$	Column 1
Row 2	$[\sqrt{3}, 5, 6]$	Column 2
Row 3	$[2, 3, -1]$	Column 3

Step 4: Final Construction of $A^T$

Placing these into the transposed matrix structure:

$A^T = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}$

Visual Representation of Transposition

Final Solution: The transpose of the matrix is $\begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}$

Given Matrix and Objective

The matrix $A$ is defined as: $A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix}$

To prove that $A$ is a skew-symmetric matrix, we must demonstrate that $A^T = -A$, where $A^T$ is the transpose of matrix $A$ [Per the definition of skew-symmetric matrices].

Step 1: Calculating the Transpose $A^T$

To find the transpose $A^T$, we interchange the rows and columns of matrix $A$ [By the definition of Matrix Transpose, where $(A^T)_{ij} = A_{ji}$].

$A^T = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$

Step 2: Calculating $-A$

To find $-A$, we multiply every element of matrix $A$ by the scalar $-1$ [Per the property of Scalar Multiplication of matrices].

$-A = -1 \cdot \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} -(0) & -(1) & -(-1) \\ -(-1) & -(0) & -(1) \\ -(1) & -(-1) & -(0) \end{bmatrix}$

$-A = \begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$

Step 3: Comparison and Verification

Matrix	Result
$A^T$	$\begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$
$-A$	$\begin{bmatrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \end{bmatrix}$

Since $A^T = -A$, the condition for a skew-symmetric matrix is satisfied [By the formal definition of skew-symmetry: $A^T = -A$]. Furthermore, note that the diagonal elements are all zero, which is a necessary condition for a skew-symmetric matrix.

Final Solution: Since $A^T = -A$, the matrix $A$ is proven to be a skew-symmetric matrix.

Step 1: Definition of a Symmetric Matrix

A square matrix $A$ is defined as a symmetric matrix if and only if the transpose of the matrix $A$ is equal to the matrix itself. Mathematically, this condition is expressed as:

$A^T = A$

[Definition of Symmetric Matrix: A square matrix $A = [a_{ij}]$ is symmetric if $a_{ij} = a_{ji}$ for all $i, j$.]

Step 2: Representation of Matrix A

Given the matrix:

$A = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$

Step 3: Calculating the Transpose ($A^T$)

To find the transpose $A^T$, we convert the rows of matrix $A$ into the columns of matrix $A^T$.

• Row 1: $(1, -1, 5) \rightarrow$ Column 1
• Row 2: $(-1, 2, 1) \rightarrow$ Column 2
• Row 3: $(5, 1, 3) \rightarrow$ Column 3

$A^T = \begin{bmatrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{bmatrix}$

Step 4: Comparison and Verification

We compare the elements of $A$ and $A^T$ position-wise ($a_{ij}$ with $a_{ji}$):

Element Position	$a_{ij}$ (Original)	$a_{ji}$ (Transpose)
(1,2) and (2,1)	$-1$	$-1$
(1,3) and (3,1)	$5$	$5$
(2,3) and (3,2)	$1$	$1$

Since $a_{ij} = a_{ji}$ for all $i, j \in \{1, 2, 3\}$, it is evident that $A^T = A$.

Step 5: Visual Representation of Symmetry

Final Solution: Since the transpose of the matrix $A$ is identical to the original matrix $A$ ($A^T = A$), the matrix $A$ is confirmed to be a symmetric matrix.

Step 1: Theoretical Foundation of Matrix Transposition

The transpose of a matrix, denoted as $A^T$ or $A'$, is an operation that converts rows into columns and columns into rows. For any matrix $A$ of order $m \times n$, its transpose $A^T$ will have the order $n \times m$. Formally, if $A = [a_{ij}]$, then $A^T = [a_{ji}]$, where the element at the $i$-th row and $j$-th column is shifted to the $j$-th row and $i$-th column [By the definition of matrix transpose].

Step 2: Analysis of the Given Matrix

Let the given matrix be $A$. $A = \begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}$

• Matrix Dimensions: The matrix $A$ consists of 3 rows and 1 column. Therefore, the order is $3 \times 1$.
• Expected Dimensions of $A^T$: Since the order is $m \times n$, the transpose $A^T$ must have an order of $n \times m$, which is $1 \times 3$.

Step 3: Execution of the Transposition

To compute $A^T$, we take the column vector and rewrite it as a row vector:

$A^T = \begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}$

Step 4: Geometric/Structural Representation

Final Solution: The transpose of the matrix $\begin{bmatrix} 5 \\ \frac{1}{2} \\ -1 \end{bmatrix}$ is $\begin{bmatrix} 5 & \frac{1}{2} & -1 \end{bmatrix}$.

Step 1: Determining Matrix $A$

Given the transpose of matrix $A$, denoted as $A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$, we find $A$ by applying the property of transposition: $(A')' = A$ [Property: The transpose of a transpose is the original matrix].

Thus, transposing the rows and columns of $A'$:

$A = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}$

Step 2: Calculating Matrix $2B$

Given $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$, we perform scalar multiplication by multiplying every element of the matrix by the scalar $2$ [Definition of Scalar Multiplication]:

$2B = 2 \times \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2(-1) & 2(0) \\ 2(1) & 2(2) \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$

Step 3: Calculating $(A + 2B)$

To find the sum of matrices $A$ and $2B$, we add the corresponding elements [Property: Matrix addition is defined for matrices of the same order]:

$A + 2B = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$

$A + 2B = \begin{bmatrix} -2 + (-2) & 1 + 0 \\ 3 + 2 & 2 + 4 \end{bmatrix} = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}$

Step 4: Finding the Transpose $(A + 2B)'$

To find the transpose, we interchange the rows and columns of the resulting matrix from Step 3 [Definition: If $C = [c_{ij}]$, then $C' = [c_{ji}]$]:

$(A + 2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$

Final Solution: $(A + 2B)' = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$

Step 1: Given Matrix and Transpose Definition

Given the matrix $A$ of order $2 \times 2$:

$A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$

The transpose of a matrix, denoted as $A'$, is obtained by interchanging its rows and columns [Per the definition of Matrix Transposition, where $A'_{ij} = A_{ji}$].

$A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$

Step 2: Calculating the Sum $(A + A')$

To compute the sum, we perform element-wise addition [Per the Matrix Addition Property]:

$(A + A') = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix} + \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$

$(A + A') = \begin{bmatrix} (1 + 1) & (5 + 6) \\ (6 + 5) & (7 + 7) \end{bmatrix}$

$(A + A') = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$

Step 3: Verification of Symmetry

A matrix $M$ is defined as symmetric if and only if $M' = M$ [By the definition of a Symmetric Matrix]. Let $S = (A + A')$. We must verify if $S' = S$.

Taking the transpose of $S$:

$S = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$

$S' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$

Comparing $S'$ and $S$, we observe that $S'_{ij} = S_{ij}$ for all $i, j$ [Verification via identity of matrix components].

Visual Representation of Symmetric Matrix $(A + A')$

Final Solution: Since $(A + A')' = (A + A')$, the matrix $(A + A')$ is proven to be a symmetric matrix.

Step 1: Theoretical Foundation

Any square matrix $A$ can be uniquely expressed as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$, defined as follows [Per the Theorem on Decomposition of Matrices]:

$A = P + Q$

Where:

• $P = \frac{1}{2}(A + A^T)$ [Symmetric matrix, as $P^T = P$]
• $Q = \frac{1}{2}(A - A^T)$ [Skew-symmetric matrix, as $Q^T = -Q$]

Step 2: Define Given Matrix and Transpose

Given $A = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}$

The transpose $A^T$ is obtained by interchanging rows and columns:

$A^T = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}$

Step 3: Calculate the Symmetric Matrix ($P$)

$P = \frac{1}{2}(A + A^T) = \frac{1}{2} \left( \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} + \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \right)$

$P = \frac{1}{2} \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0.5 & -2.5 \\ 0.5 & -2 & -2 \\ -2.5 & -2 & 2 \end{bmatrix}$

Step 4: Calculate the Skew-Symmetric Matrix ($Q$)

$Q = \frac{1}{2}(A - A^T) = \frac{1}{2} \left( \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} - \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \right)$

$Q = \frac{1}{2} \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2.5 & 1.5 \\ -2.5 & 0 & 3 \\ -1.5 & -3 & 0 \end{bmatrix}$

Step 5: Verification of Summation

To verify the decomposition, sum $P$ and $Q$:

$P + Q = \begin{bmatrix} 3 & 0.5 & -2.5 \\ 0.5 & -2 & -2 \\ -2.5 & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 2.5 & 1.5 \\ -2.5 & 0 & 3 \\ -1.5 & -3 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} = A$

Final Solution: The matrix is expressed as the sum of the symmetric matrix $\begin{bmatrix} 3 & 0.5 & -2.5 \\ 0.5 & -2 & -2 \\ -2.5 & -2 & 2 \end{bmatrix}$ and the skew-symmetric matrix $\begin{bmatrix} 0 & 2.5 & 1.5 \\ -2.5 & 0 & 3 \\ -1.5 & -3 & 0 \end{bmatrix}$.

Step 1: Given Matrices and Dimensional Analysis

We are given the matrices:

$A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$ (Order: $3 \times 1$)

$B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$ (Order: $1 \times 3$)

[Per the definition of matrix multiplication, the product $AB$ is defined if the number of columns in $A$ equals the number of rows in $B$. Since $A$ is $3 \times 1$ and $B$ is $1 \times 3$, the resulting matrix $AB$ will be of order $3 \times 3$.]

Step 2: Calculating the Product $AB$

Applying the row-by-column multiplication rule:

$AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \begin{bmatrix} -1 & 2 & 1 \end{bmatrix} = \begin{bmatrix} (1)(-1) & (1)(2) & (1)(1) \\ (-4)(-1) & (-4)(2) & (-4)(1) \\ (3)(-1) & (3)(2) & (3)(1) \end{bmatrix}$

$AB = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}$

Step 3: Finding the Transpose of the Product $(AB)'$

To find the transpose, we interchange rows and columns [Per the definition of Transpose Matrix $(A^T)_{ij} = A_{ji}$]:

$(AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Step 4: Finding Transposes of $A$ and $B$ individually

For $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$, the transpose $A' = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$.

For $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$, the transpose $B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$.

Step 5: Calculating the Product $B'A'$

We now compute the product of $B' (3 \times 1)$ and $A' (1 \times 3)$:

$B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \begin{bmatrix} 1 & -4 & 3 \end{bmatrix} = \begin{bmatrix} (-1)(1) & (-1)(-4) & (-1)(3) \\ (2)(1) & (2)(-4) & (2)(3) \\ (1)(1) & (1)(-4) & (1)(3) \end{bmatrix}$

$B'A' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Step 6: Verification

Expression	Result
$(AB)'$	$\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$
$B'A'$	$\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Comparing the matrices from Step 3 and Step 5, we observe that $(AB)' = B'A'$.

Final Solution: The property $(AB)' = B'A'$ is verified as both sides yield the matrix $\begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$.

Step 1: Identifying the Transpose of Matrix $A$

Given the matrix $A = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$, the transpose matrix $A'$ is obtained by interchanging the rows and columns [Definition of Transpose].

$A' = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$

Step 2: Setting up the Product $A'A$

We perform the matrix multiplication $A'A$ using the row-by-column method [Matrix Multiplication Rule: $(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$].

$A'A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$

Step 3: Executing Matrix Multiplication

We compute each element of the resulting $2 \times 2$ matrix:

• Element $(1,1) = (\cos \alpha)(\cos \alpha) + (-\sin \alpha)(-\sin \alpha) = \cos^2 \alpha + \sin^2 \alpha$
• Element $(1,2) = (\cos \alpha)(\sin \alpha) + (-\sin \alpha)(\cos \alpha) = \cos \alpha \sin \alpha - \sin \alpha \cos \alpha$
• Element $(2,1) = (\sin \alpha)(\cos \alpha) + (\cos \alpha)(-\sin \alpha) = \sin \alpha \cos \alpha - \cos \alpha \sin \alpha$
• Element $(2,2) = (\sin \alpha)(\sin \alpha) + (\cos \alpha)(\cos \alpha) = \sin^2 \alpha + \cos^2 \alpha$

Step 4: Applying Trigonometric Identities

Utilizing the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ and the property of additive inverses:

$\cos^2 \alpha + \sin^2 \alpha = 1$	$\cos \alpha \sin \alpha - \sin \alpha \cos \alpha = 0$
$\sin \alpha \cos \alpha - \cos \alpha \sin \alpha = 0$	$\sin^2 \alpha + \cos^2 \alpha = 1$

Step 5: Final Matrix Verification

Substituting the computed values back into the matrix structure:

$A'A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Since $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is the Identity matrix of order 2, denoted as $I$ [Definition of Identity Matrix].

Final Solution: $A'A = I$ is verified.

Theoretical Foundation

Any square matrix $A$ can be uniquely expressed as the sum of a symmetric matrix ($P$) and a skew-symmetric matrix ($Q$) using the following theorem:

$A = P + Q = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$

Where:

• $P = \frac{1}{2}(A + A^T)$ is symmetric, satisfying $P^T = P$ [Per Theorem of Symmetric Matrices].
• $Q = \frac{1}{2}(A - A^T)$ is skew-symmetric, satisfying $Q^T = -Q$ [Per Theorem of Skew-Symmetric Matrices].

Step 1: Identifying the Matrix and its Transpose

Given the matrix $A$:

$A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$

The transpose $A^T$ is obtained by interchanging rows and columns:

$A^T = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}$

Step 2: Calculating the Symmetric Matrix ($P$)

$P = \frac{1}{2}(A + A^T) = \frac{1}{2} \left( \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} + \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix} \right)$

$P = \frac{1}{2} \begin{bmatrix} 1+1 & 5-1 \\ -1+5 & 2+2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix}$

$P = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}$

Step 3: Calculating the Skew-Symmetric Matrix ($Q$)

$Q = \frac{1}{2}(A - A^T) = \frac{1}{2} \left( \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} - \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix} \right)$

$Q = \frac{1}{2} \begin{bmatrix} 1-1 & 5-(-1) \\ -1-5 & 2-2 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix}$

$Q = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$

Step 4: Verification and Representation

To verify, we perform the addition $P + Q$:

$\begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix} = \begin{bmatrix} 1+0 & 2+3 \\ 2-3 & 2+0 \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} = A$

Final Solution: The matrix $\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$ is expressed as the sum of a symmetric and skew-symmetric matrix as $\begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$.

Step 1: Establishing Definitions and Properties

Given that $A$ and $B$ are symmetric matrices of the same order $n \times n$, by definition, their transposes are equal to the matrices themselves:

• $A^T = A$ [Definition of Symmetric Matrix]
• $B^T = B$ [Definition of Symmetric Matrix]

We are required to determine the nature of the matrix $C$, defined as:

$C = AB - BA$

Step 2: Testing for Symmetry or Skew-Symmetry

To classify the matrix $C$, we evaluate its transpose $C^T$ using the properties of matrix transpose operations:

$C^T = (AB - BA)^T$

Applying the linearity property of transposes, $(X - Y)^T = X^T - Y^T$ [Property of Transpose]:

$C^T = (AB)^T - (BA)^T$

Applying the reversal law of transposes, $(AB)^T = B^T A^T$ [Property of Transpose]:

$C^T = B^T A^T - A^T B^T$

Step 3: Algebraic Substitution and Simplification

Substitute the symmetry conditions ($A^T = A$ and $B^T = B$) into the expression:

$C^T = (B)(A) - (A)(B)$

$C^T = BA - AB$

Factor out the negative sign to relate it back to the original matrix $C$:

$C^T = -(AB - BA)$

$C^T = -C$

Step 4: Theoretical Conclusion

By definition, a matrix $M$ is skew-symmetric if $M^T = -M$ [Definition of Skew-Symmetric Matrix]. Since we have demonstrated that $(AB - BA)^T = -(AB - BA)$, the matrix $AB - BA$ satisfies the criteria for a skew-symmetric matrix.

Final Solution: The matrix $AB – BA$ is a skew-symmetric matrix.

Step 1: Identifying Matrix Dimensions and Initial Setup

Given matrices:

$A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$ (Order: $3 \times 1$)

$B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$ (Order: $1 \times 3$)

To verify the property $(AB)' = B'A'$, we must perform matrix multiplication and transposition [Per the definition of Matrix Transposition and Matrix Product].

Step 2: Calculating the Product Matrix $AB$

The product $AB$ of a $(3 \times 1)$ matrix and a $(1 \times 3)$ matrix results in a $(3 \times 3)$ matrix.

$AB = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} \times \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$

$AB = \begin{bmatrix} (0)(1) & (0)(5) & (0)(7) \\ (1)(1) & (1)(5) & (1)(7) \\ (2)(1) & (2)(5) & (2)(7) \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \end{bmatrix}$

Step 3: Calculating $(AB)'$

To find the transpose $(AB)'$, swap rows and columns [Per the property $A'_{ij} = A_{ji}$].

$(AB)' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

Step 4: Calculating $B'$ and $A'$

$B' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix}$ (Order: $3 \times 1$)

$A' = \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}$ (Order: $1 \times 3$)

Step 5: Calculating $B'A'$

The product $B'A'$ is a $(3 \times 1)$ matrix multiplied by a $(1 \times 3)$ matrix, resulting in a $(3 \times 3)$ matrix.

$B'A' = \begin{bmatrix} 1 \\ 5 \\ 7 \end{bmatrix} \times \begin{bmatrix} 0 & 1 & 2 \end{bmatrix}$

$B'A' = \begin{bmatrix} (1)(0) & (1)(1) & (1)(2) \\ (5)(0) & (5)(1) & (5)(2) \\ (7)(0) & (7)(1) & (7)(2) \end{bmatrix} = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

Step 6: Comparative Analysis

Comparing the results from Step 3 and Step 5:

Expression	Resultant Matrix
$(AB)'$	$\begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$
$B'A'$	$\begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$

Final Solution: Since $(AB)' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$ and $B'A' = \begin{bmatrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \end{bmatrix}$, it is verified that $(AB)' = B'A'$.

Step 1: Identifying the Transpose Matrix

Given the matrix $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$, the transpose $A'$ is obtained by interchanging the rows and columns [Definition of Transpose Matrix].

$A' = \begin{bmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{bmatrix}$

Step 2: Performing Matrix Addition

Calculate the sum $A + A'$ by adding corresponding elements [Addition of Matrices Rule]:

$A + A' = \begin{bmatrix} \cos \alpha + \cos \alpha & -\sin \alpha + \sin \alpha \\ \sin \alpha - \sin \alpha & \cos \alpha + \cos \alpha \end{bmatrix}$

$A + A' = \begin{bmatrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{bmatrix}$

Step 3: Equating to the Identity Matrix

The problem states $A + A' = I$, where $I$ is the identity matrix of order 2, $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

$\begin{bmatrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Step 4: Solving for the Variable $\alpha$

By comparing corresponding elements [Equality of Matrices Principle]:

$2\cos \alpha = 1$

$\cos \alpha = \frac{1}{2}$

Since $\cos \alpha = \cos(\frac{\pi}{3})$, the general solution is $\alpha = 2n\pi \pm \frac{\pi}{3}$ for $n \in \mathbb{Z}$. Within the principal domain, $\alpha = \frac{\pi}{3}$.

Final Solution: The value of $\alpha$ is $\frac{\pi}{3}$ (or generally $2n\pi \pm \frac{\pi}{3}$).

Step 1: Calculating the Sum $A + B$

Given matrices:

$A = \begin{bmatrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{bmatrix}$, $B = \begin{bmatrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{bmatrix}$

To compute the sum $A + B$, we add corresponding elements [Per the definition of Matrix Addition]:

$A + B = \begin{bmatrix} (-1 - 4) & (2 + 1) & (3 - 5) \\ (5 + 1) & (7 + 2) & (9 + 0) \\ (-2 + 1) & (1 + 3) & (1 + 1) \end{bmatrix} = \begin{bmatrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \end{bmatrix}$

Step 2: Finding the Transpose of $(A + B)$

The transpose of a matrix is obtained by interchanging its rows and columns [Per the property $(A^T)_{ij} = A_{ji}$].

$(A + B)' = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$

Step 3: Calculating Transposes $A'$ and $B'$

Transposing matrix $A$:

$A' = \begin{bmatrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \end{bmatrix}$

Transposing matrix $B$:

$B' = \begin{bmatrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \end{bmatrix}$

Step 4: Adding $A'$ and $B'$

$A' + B' = \begin{bmatrix} (-1 - 4) & (5 + 1) & (-2 + 1) \\ (2 + 1) & (7 + 2) & (1 + 3) \\ (3 - 5) & (9 + 0) & (1 + 1) \end{bmatrix} = \begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$

Step 5: Comparative Analysis and Verification

Left Hand Side $(A+B)'$	Right Hand Side $A' + B'$
$\begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$	$\begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$

Since the resulting matrix from the Left Hand Side is identical to the matrix from the Right Hand Side, the property $(A + B)' = A' + B'$ is verified.

Final Solution: The identity $(A + B)' = A' + B'$ is verified as both sides yield $\begin{bmatrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \end{bmatrix}$.