Step 1: Computing $A^2$

Given the matrix $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$, we define $A^2$ as the product $A \times A$:

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

[By Matrix Multiplication Definition: $(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$]

• $R_1 C_1: (2)(2) + (0)(2) + (1)(1) = 4 + 0 + 1 = 5$
• $R_1 C_2: (2)(0) + (0)(1) + (1)(-1) = 0 + 0 - 1 = -1$
• $R_1 C_3: (2)(1) + (0)(3) + (1)(0) = 2 + 0 + 0 = 2$
• $R_2 C_1: (2)(2) + (1)(2) + (3)(1) = 4 + 2 + 3 = 9$
• $R_2 C_2: (2)(0) + (1)(1) + (3)(-1) = 0 + 1 - 3 = -2$
• $R_2 C_3: (2)(1) + (1)(3) + (3)(0) = 2 + 3 + 0 = 5$
• $R_3 C_1: (1)(2) + (-1)(2) + (0)(1) = 2 - 2 + 0 = 0$
• $R_3 C_2: (1)(0) + (-1)(1) + (0)(-1) = 0 - 1 + 0 = -1$
• $R_3 C_3: (1)(1) + (-1)(3) + (0)(0) = 1 - 3 + 0 = -2$

$A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$

Step 2: Calculating Scalar Multiples $5A$ and $6I$

Per the property of Scalar Multiplication $k A = [k a_{ij}]$:

$5A = 5 \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}$

The identity matrix $I$ of order 3 is $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Thus:

$6I = 6 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

Step 3: Final Matrix Evaluation

Constructing the final expression $A^2 - 5A + 6I$:

$= \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$

Performing element-wise addition/subtraction:

Simplifying the elements:

= $\begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$

Final Solution: $\begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$

Step 1: Establishing the Matrix Multiplication Sequence

Given the matrix equation: $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix} = [0]$

We apply the associative property of matrix multiplication, $(AB)C = A(BC)$, to simplify the calculation [Per Associative Property of Matrices]. Let $A = \begin{bmatrix} 1 & 2 & 1 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix}$, and $C = \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix}$.

Step 2: Multiplying the Second and Third Matrices

Let $D = B \times C$. $D = \begin{bmatrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 0 \\ 2 \\ x \end{bmatrix}$

Perform the row-by-column multiplication: $D = \begin{bmatrix} (1 \cdot 0) + (2 \cdot 2) + (0 \cdot x) \\ (2 \cdot 0) + (0 \cdot 2) + (1 \cdot x) \\ (1 \cdot 0) + (0 \cdot 2) + (2 \cdot x) \end{bmatrix}$

$D = \begin{bmatrix} 0 + 4 + 0 \\ 0 + 0 + x \\ 0 + 0 + 2x \end{bmatrix} = \begin{bmatrix} 4 \\ x \\ 2x \end{bmatrix}$

Step 3: Multiplying by the First Matrix

Now, compute $A \times D = [0]$: $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix} \begin{bmatrix} 4 \\ x \\ 2x \end{bmatrix} = [0]$

Expand the dot product: $(1 \cdot 4) + (2 \cdot x) + (1 \cdot 2x) = 0$

Step 4: Solving the Linear Algebraic Equation

Simplify the resulting scalar equation: $4 + 2x + 2x = 0$

$4 + 4x = 0$

Isolate the variable $x$: $4x = -4$

$x = \frac{-4}{4}$

$x = -1$

Visual Representation of the Operation

Final Solution: $x = -1$

Step 1: Theoretical Foundation for Matrix Addition

The addition of two matrices $A$ and $B$ is defined if and only if they possess the same order. Given matrix $A$ of order $m \times n$ and matrix $B$ of order $m \times n$, their sum $C = A + B$ is a matrix of the same order $m \times n$, where each element $c_{ij}$ is defined as: $c_{ij} = a_{ij} + b_{ij}$ [Per the definition of Matrix Addition]

Step 2: Identification of Given Matrices

Let the given matrices be $A$ and $B$:

$A = \begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix}$

$B = \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix}$

Both matrices are of the order $3 \times 3$. Therefore, addition is permissible.

Step 3: Execution of Element-wise Summation

We sum the corresponding elements $a_{ij} + b_{ij}$ to construct the resulting matrix $C$:

Step 4: Final Matrix Construction

Arranging the calculated values into the resulting matrix format:

$C = \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}$

Visual Representation of Matrix Summation

Final Solution: $\begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix}$

Step 1: Establishing the Equality of Matrices

Two matrices $A = [a_{ij}]$ and $B = [b_{ij}]$ are said to be equal if and only if:

• They are of the same order.
• Their corresponding elements are equal for all possible values of $i$ and $j$ ($a_{ij} = b_{ij}$).

[Definition of Equality of Matrices]

Step 2: Verification of Matrix Order

The given matrices are:

$A = \begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix}$ (Order: $2 \times 2$)

$B = \begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$ (Order: $2 \times 2$)

Since both matrices have an order of $2 \times 2$, the equality is well-defined. [Per the condition of equality for matrices of same dimensions].

Step 3: Equating Corresponding Elements

By equating the elements at each position $(i, j)$ from matrix $A$ to matrix $B$:

Visual Representation of Matrix Equality

Step 4: Final Derivation

From the equations derived in Step 3, we directly extract the values:

• $x = 1$
• $y = 4$
• $z = 3$

Final Solution: x = 1, y = 4, z = 3

Step 1: Matrix Definition and Structural Setup

A $3 \times 4$ matrix $A$ consists of $3$ rows and $4$ columns. The general form of the matrix is represented as:

$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}$

Each element $a_{ij}$ is defined by the rule $a_{ij} = 2i - j$, where $i$ denotes the row index ($i \in \{1, 2, 3\}$) and $j$ denotes the column index ($j \in \{1, 2, 3, 4\}$).

Step 2: Calculating Individual Elements

We evaluate each element systematically by substituting the indices into the given formula $a_{ij} = 2i - j$:

Step 3: Matrix Assembly

Substituting the calculated values into the general matrix structure:

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

Visualization of Index Mapping

Final Solution: The constructed matrix is $A = \begin{bmatrix} 1 & 0 & -1 & -2 \\ 3 & 2 & 1 & 0 \\ 5 & 4 & 3 & 2 \end{bmatrix}$

Step 1: Identifying the Matrix Structure

The given matrix $A$ is presented as an array of elements organized into horizontal rows and vertical columns:

By inspection, we define the dimensions of the matrix as follows:

• Rows ($m$): These are the horizontal arrangements of elements. Counting them:
  • Row 1: $[2, 5, 19, -7]$
  • Row 2: $[35, -2, \frac{5}{2}, 12]$
  • Row 3: $[\sqrt{3}, 1, -5, 17]$
  Thus, the number of rows $m = 3$.
• Columns ($n$): These are the vertical arrangements of elements. Counting them:
  • Column 1: $[2, 35, \sqrt{3}]^T$
  • Column 2: $[5, -2, 1]^T$
  • Column 3: $[19, \frac{5}{2}, -5]^T$
  • Column 4: $[-7, 12, 17]^T$
  Thus, the number of columns $n = 4$.

Step 2: Defining the Order of a Matrix

[Per the definition of Matrix Algebra], the order of a matrix is denoted as $m \times n$, where $m$ represents the number of rows and $n$ represents the number of columns. This notation identifies the dimensions of the rectangular array.

Step 3: Calculating the Final Order

Substituting the identified values into the $m \times n$ format:

Final Solution: The order of the matrix $A$ is $3 \times 4$.

Step 1: Calculating $A^2$

Given the matrix $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, the square of the matrix $A^2$ is computed by the product $A \cdot A$.

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

By the definition of matrix multiplication [Sum of row-column products]:

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

$A^2 = \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

Step 2: Calculating Scalar Multiples

We evaluate the term $-5A$ using the scalar multiplication property $k \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} ka & kb \\ kc & kd \end{bmatrix}$:

$-5A = -5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -15 & -5 \\ 5 & -10 \end{bmatrix}$

We define the identity matrix $I$ for a $2 \times 2$ order as $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$. Thus, the term $7I$ is:

$7I = 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Step 3: Summing the Components

We now evaluate the expression $LHS = A^2 - 5A + 7I$ by adding the matrices computed in the previous steps:

$LHS = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} + \begin{bmatrix} -15 & -5 \\ 5 & -10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

Summing the corresponding elements [By Matrix Addition Property]:

Step 4: Conclusion of Proof

The resulting matrix is:

$LHS = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

Since the result is the Zero Matrix $O$, the identity $A^2 - 5A + 7I = 0$ is satisfied.

Final Solution: $A^2 - 5A + 7I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = 0$. Hence, the statement is proved.

Given Variables & Initial Setup

We are provided with the following matrices and their respective dimensions:

• Matrix $X$: $2 \times n$
• Matrix $Z$: $2 \times p$

Per the property of matrix subtraction, two matrices can only be subtracted if they have the same order [Per the Definition of Matrix Equality and Subtraction].

Step 1: Analyzing the Subtraction Requirement

The expression given is $7X - 5Z$. Scalar multiplication (by $7$ and $5$) does not alter the dimensions of a matrix [By the properties of Scalar Multiplication]. Therefore, the dimension of $7X$ remains $2 \times n$ and the dimension of $5Z$ remains $2 \times p$.

For the expression $7X - 5Z$ to be defined, the order of $7X$ must be equal to the order of $5Z$.

Mathematically:

Step 2: Incorporating the Constraint $n = p$

The problem states the condition $n = p$.

If we substitute $p$ with $n$ (or vice-versa), the dimensions become:

• Dimension of $7X$: $2 \times n$
• Dimension of $5Z$: $2 \times n$

Step 3: Determining the Resultant Order

When two matrices of order $m \times n$ are added or subtracted, the resulting matrix retains the same order $m \times n$ [By the Definition of Matrix Addition/Subtraction].

Thus, the order of $7X - 5Z$ is $2 \times n$.

Visualization of Dimensional Compatibility

Final Solution: The order of the matrix $7X - 5Z$ is $2 \times n$.

Given Variables & Initial Matrix Orders

Based on the problem statement, we define the dimensions of the matrices as follows:

• Matrix $P$: $p \times k$
• Matrix $Y$: $3 \times k$
• Matrix $W$: $n \times 3$

Step 1: Conditions for Matrix Multiplication

For the product of two matrices $A_{m \times n}$ and $B_{p \times q}$ to be defined, the number of columns in $A$ must equal the number of rows in $B$ (i.e., $n = p$). The resulting matrix will have the dimension $m \times q$.

Step 2: Analyzing $PY$

We are given $P$ has order $p \times k$ and $Y$ has order $3 \times k$.
For the product $PY$ to be defined, the number of columns of $P$ must equal the number of rows of $Y$ [Per Matrix Multiplication Rule]: $$k = 3$$ The order of the resulting matrix $PY$ is $p \times k$.

Step 3: Analyzing $WY$

We are given $W$ has order $n \times 3$ and $Y$ has order $3 \times k$.
For the product $WY$ to be defined, the number of columns of $W$ (which is $3$) must equal the number of rows of $Y$ (which is $3$). Since $3 = 3$, the product $WY$ is always defined for any $n, k$.
The order of the resulting matrix $WY$ is $n \times k$.

Step 4: Conditions for Matrix Addition

For the sum $PY + WY$ to be defined, the matrices $PY$ and $WY$ must have the same order [Per Matrix Addition Rule].
Order of $PY = p \times k$
Order of $WY = n \times k$

Equating the dimensions: $$p \times k = n \times k$$ This implies: $$p = n$$

Step 5: Synthesis of Constraints

From Step 2, we obtained $k = 3$.
From Step 4, we obtained $p = n$.

Final Solution: The restrictions on $n, k,$ and $p$ such that $PY + WY$ is defined are $k = 3$ and $p = n$.

Step 1: Identifying the System of Linear Matrix Equations

Given the system of equations where $X$ and $Y$ are matrices of order $2 \times 2$:

(1) $X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$

(2) $X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

[By the property of matrix addition/subtraction, the result of these operations is a matrix of the same dimensions, which confirms $X$ and $Y$ are $2 \times 2$ matrices].

Step 2: Deriving Matrix $X$ via Addition

Add equation (1) and equation (2) to eliminate matrix $Y$:

$(X + Y) + (X - Y) = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} + \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

$2X = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix}$

$2X = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$

Multiply both sides by scalar $\frac{1}{2}$:

$X = \frac{1}{2} \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$

Step 3: Deriving Matrix $Y$ via Subtraction

Subtract equation (2) from equation (1) to eliminate matrix $X$:

$(X + Y) - (X - Y) = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$

$2Y = \begin{bmatrix} 7-3 & 0-0 \\ 2-0 & 5-3 \end{bmatrix}$

$2Y = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix}$

Multiply both sides by scalar $\frac{1}{2}$:

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

Visualization of Resultant Matrices

Step 4: Verification

Check the sum: $X + Y = \begin{bmatrix} 5+2 & 0+0 \\ 1+1 & 4+1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$ [Matches given condition].

Check the difference: $X - Y = \begin{bmatrix} 5-2 & 0-0 \\ 1-1 & 4-1 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$ [Matches given condition].

Final Solution: $X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$ and $Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$

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:

[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: Establishing the Matrix Addition Property

To compute the sum of two matrices, both matrices must be of the same order $m \times n$. Per the rules of matrix algebra, the sum is obtained by adding the corresponding elements of the given matrices [Definition of Matrix Addition].

Given matrices:

Step 2: Performing Element-wise Addition

We sum the elements $a_{ij}$ of matrix $A$ and $b_{ij}$ of matrix $B$ such that the result $C = A + B$ has elements $c_{ij} = a_{ij} + b_{ij}$.

Step 3: Algebraic Simplification using Identities

Applying standard algebraic identities for perfect squares:

• $(a + b)^2 = a^2 + 2ab + b^2$
• $(b + c)^2 = b^2 + 2bc + c^2$
• $(a - c)^2 = a^2 - 2ac + c^2$
• $(a - b)^2 = a^2 - 2ab + b^2$

Substituting these identities into the resulting matrix:

Visual Representation of Matrix Summation

Final Solution: $\begin{bmatrix} (a + b)^2 & (b + c)^2 \\ (a - c)^2 & (a - b)^2 \end{bmatrix}$

Given Variables & Initial Setup

We are provided with a square matrix $A$ that satisfies the idempotent property:

$A^2 = A$ [Definition of Idempotent Matrix]

We are required to evaluate the expression:

$E = (I + A)^3 - 7A$

Step 1: Expanding the Binomial Expression

Since $I$ (the Identity matrix) and $A$ commute (i.e., $IA = AI = A$), we can apply the binomial theorem for the expansion of $(I + A)^3$:

$(I + A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$

Simplifying the terms using properties of identity matrices ($I^n = I$ and $IA = A$):

$(I + A)^3 = I + 3A + 3A^2 + A^3$ [Property of Identity matrix multiplication]

Step 2: Utilizing the Idempotent Property

Given $A^2 = A$, we can derive higher powers of $A$:

• $A^3 = A^2 \cdot A = (A) \cdot A = A^2 = A$
• $A^n = A$ for all $n \in \mathbb{Z}^+$

Substituting these relations back into the expanded expression from Step 1:

$(I + A)^3 = I + 3A + 3(A) + (A)$

$(I + A)^3 = I + 3A + 3A + A$

$(I + A)^3 = I + 7A$

Step 3: Final Substitution and Simplification

Now, we substitute the simplified expression of $(I + A)^3$ back into the original expression $E$:

$E = (I + 7A) - 7A$

$E = I + (7A - 7A)$ [Associative property of matrix addition]

$E = I + O$ (where $O$ is the Null matrix)

$E = I$

Final Solution: (I + A)^3 – 7A = I

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

Step 1: Scalar Multiplication of Matrix A by 3

Given $A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$, we multiply each element of $A$ by the scalar $3$ [Per the definition of scalar multiplication $k[A_{ij}] = [k A_{ij}]$]:

$3A = 3 \times \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} = \begin{bmatrix} 3(\frac{2}{3}) & 3(1) & 3(\frac{5}{3}) \\ 3(\frac{1}{3}) & 3(\frac{2}{3}) & 3(\frac{4}{3}) \\ 3(\frac{7}{3}) & 3(2) & 3(\frac{2}{3}) \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$

Step 2: Scalar Multiplication of Matrix B by 5

Given $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$, we multiply each element of $B$ by the scalar $5$:

$5B = 5 \times \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix} = \begin{bmatrix} 5(\frac{2}{5}) & 5(\frac{3}{5}) & 5(1) \\ 5(\frac{1}{5}) & 5(\frac{2}{5}) & 5(\frac{4}{5}) \\ 5(\frac{7}{5}) & 5(\frac{6}{5}) & 5(\frac{2}{5}) \end{bmatrix} = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}$

Step 3: Performing the Matrix Subtraction $3A - 5B$

To compute $3A - 5B$, we subtract the corresponding elements of the resulting matrices [By the definition of matrix subtraction $A - B = [A_{ij} - B_{ij}]$]:

Calculating the difference of the corresponding entries:

$3A - 5B = \begin{bmatrix} 2-2 & 3-3 & 5-5 \\ 1-1 & 2-2 & 4-4 \\ 7-7 & 6-6 & 2-2 \end{bmatrix}$

$3A - 5B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$

Where $O$ represents the $3 \times 3$ null matrix (or zero matrix).

Final Solution: $3A - 5B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

Step 1: Calculating $A^2$

To find $A^2$, we compute the product $A \times A$ using matrix multiplication rules [Def: Row-by-Column multiplication].

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

Performing the calculations for each entry $c_{ij} = \sum_{k=1}^3 a_{ik} a_{kj}$:

• $c_{11} = (1)(1) + (0)(0) + (2)(2) = 5$
• $c_{12} = (1)(0) + (0)(2) + (2)(0) = 0$
• $c_{13} = (1)(2) + (0)(1) + (2)(3) = 8$
• $c_{21} = (0)(1) + (2)(0) + (1)(2) = 2$
• $c_{22} = (0)(0) + (2)(2) + (1)(0) = 4$
• $c_{23} = (0)(2) + (2)(1) + (1)(3) = 5$
• $c_{31} = (2)(1) + (0)(0) + (3)(2) = 8$
• $c_{32} = (2)(0) + (0)(2) + (3)(0) = 0$
• $c_{33} = (2)(2) + (0)(1) + (3)(3) = 13$

$A^2 = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix}$

Step 2: Calculating $A^3$

$A^3 = A^2 \times A = \begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$

• $c_{11} = (5)(1) + (0)(0) + (8)(2) = 21$
• $c_{12} = (5)(0) + (0)(2) + (8)(0) = 0$
• $c_{13} = (5)(2) + (0)(1) + (8)(3) = 34$
• $c_{21} = (2)(1) + (4)(0) + (5)(2) = 12$
• $c_{22} = (2)(0) + (4)(2) + (5)(0) = 8$
• $c_{23} = (2)(2) + (4)(1) + (5)(3) = 23$
• $c_{31} = (8)(1) + (0)(0) + (13)(2) = 34$
• $c_{32} = (8)(0) + (0)(2) + (13)(0) = 0$
• $c_{33} = (8)(2) + (0)(1) + (13)(3) = 55$

$A^3 = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix}$

Step 3: Evaluating the Expression $A^3 – 6A^2 + 7A + 2I$

Substitute the derived matrices into the polynomial expression:

$E = \begin{bmatrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{bmatrix} - 6\begin{bmatrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{bmatrix} + 7\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} + 2\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$

Step 4: Final Aggregation

Summing the components element-wise [Scalar multiplication and matrix addition properties]:

• Row 1: $(21 - 30 + 7 + 2), (0 - 0 + 0 + 0), (34 - 48 + 14 + 0) = [0, 0, 0]$
• Row 2: $(12 - 12 + 0 + 0), (8 - 24 + 14 + 2), (23 - 30 + 7 + 0) = [0, 0, 0]$
• Row 3: $(34 - 48 + 14 + 0), (0 - 0 + 0 + 0), (55 - 78 + 21 + 2) = [0, 0, 0]$

Final Solution: $A^3 - 6A^2 + 7A + 2I = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O$ (Null Matrix). Q.E.D.

Step 1: Dimensional Analysis of Matrix Multiplication

The given matrix equation is defined as $A \cdot B \cdot C = O$, where:

• $A = \begin{bmatrix} x & -5 & -1 \end{bmatrix}$ (Order: $1 \times 3$)
• $B = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$ (Order: $3 \times 3$)
• $C = \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix}$ (Order: $3 \times 1$)
• $O = [0]$ (Null Matrix of order $1 \times 1$)

[By the property of Matrix Multiplication compatibility, the product of $1 \times 3$ and $3 \times 3$ results in a $1 \times 3$ matrix, which is then multiplied by a $3 \times 1$ matrix to result in a $1 \times 1$ scalar matrix.]

Step 2: Performing the First Multiplication ($A \cdot B$)

Let $D = A \cdot B$. Calculating the elements of $D$ as a $1 \times 3$ vector:

$D = \begin{bmatrix} (x \cdot 1) + (-5 \cdot 0) + (-1 \cdot 2) & (x \cdot 0) + (-5 \cdot 2) + (-1 \cdot 0) & (x \cdot 2) + (-5 \cdot 1) + (-1 \cdot 3) \end{bmatrix}$

$D = \begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix}$

Step 3: Performing the Final Multiplication ($D \cdot C$)

Now, we compute $D \cdot C = [0]$:

$\begin{bmatrix} x - 2 & -10 & 2x - 8 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = [0]$

Applying the dot product rule for matrix multiplication [Row 1 $\times$ Column 1]:

$(x - 2)(x) + (-10)(4) + (2x - 8)(1) = 0$

Step 4: Solving the Resulting Algebraic Equation

Expanding the terms:

$x^2 - 2x - 40 + 2x - 8 = 0$

Combining like terms [Algebraic Simplification]:

$x^2 + (-2x + 2x) + (-40 - 8) = 0$

$x^2 - 48 = 0$

$x^2 = 48$

$x = \pm \sqrt{48}$

$x = \pm 4\sqrt{3}$

Step 5: Geometric Interpretation

Final Solution: $x = \pm 4\sqrt{3}$

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: Scalar Multiplication of Matrices

To simplify the expression, we first apply the scalar multiplication rule, which states that for a scalar $k$ and a matrix $A$, each element $a_{ij}$ is multiplied by $k$.

Given Expression:

$\cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix}$

Performing scalar multiplication:

$= \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{bmatrix} + \begin{bmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix}$

Step 2: Matrix Addition

Matrices of the same order $(2 \times 2)$ are added by summing their corresponding elements [Per the definition of matrix addition].

$= \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta \\ -\sin \theta \cos \theta + \sin \theta \cos \theta & \cos^2 \theta + \sin^2 \theta \end{bmatrix}$

Step 3: Application of Trigonometric Identities

We utilize the Pythagorean Identity: $\cos^2 \theta + \sin^2 \theta = 1$ [By Fundamental Trigonometric Identity].

Simplifying the elements:

• $a_{11}: \cos^2 \theta + \sin^2 \theta = 1$
• $a_{12}: \cos \theta \sin \theta - \sin \theta \cos \theta = 0$
• $a_{21}: -\sin \theta \cos \theta + \sin \theta \cos \theta = 0$
• $a_{22}: \cos^2 \theta + \sin^2 \theta = 1$

Step 4: Final Matrix Construction

Substituting the simplified values back into the matrix structure:

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

Visual Representation of the Identity Matrix $I_2$

Final Solution: $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ (The Identity Matrix $I_2$)

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

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