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

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

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: 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: Definition of Matrix Addition

According to the fundamental rules of linear algebra, the addition of two matrices is defined only when both matrices possess identical dimensions (same number of rows and columns) [Definition of conformability for addition]. In this instance, both matrices are of the order $2 \times 2$.

The addition is performed element-wise such that if $A = [a_{ij}]$ and $B = [b_{ij}]$, then $A + B = [a_{ij} + b_{ij}]$.

Step 2: Constructing the Addition Operation

We perform the summation of the corresponding entries:

$\begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} a + a & b + b \\ -b + b & a + a \end{bmatrix}$

Step 3: Algebraic Simplification

We simplify each individual entry within the resulting matrix:

• First row, first column ($a_{11}$): $a + a = 2a$
• First row, second column ($a_{12}$): $b + b = 2b$
• Second row, first column ($a_{21}$): $-b + b = 0$ [Additive Inverse Property]
• Second row, second column ($a_{22}$): $a + a = 2a$

Visual Representation of the Resultant Matrix

Final Matrix Assembly

Combining the evaluated terms into the matrix structure:

$\begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}$

Final Solution: $\begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix}$

Step 1: Identifying Matrix Dimensions and Compatibility

Given matrices are:

$B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$ (Order: $2 \times 2$)

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

[By the definition of matrix multiplication, the product $BA$ is defined if the number of columns in $B$ equals the number of rows in $A$. Here, $2 = 2$, thus $BA$ is a valid $2 \times 2$ matrix.]

Step 2: Defining the Multiplication Algorithm

To find the product $BA = C$, where $c_{ij}$ denotes the element in the $i^{th}$ row and $j^{th}$ column, we apply the row-by-column multiplication rule:

$BA = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} = \begin{bmatrix} (b_{11}a_{11} + b_{12}a_{21}) & (b_{11}a_{12} + b_{12}a_{22}) \\ (b_{21}a_{11} + b_{22}a_{21}) & (b_{21}a_{12} + b_{22}a_{22}) \end{bmatrix}$

Step 3: Executing Scalar Operations

Substituting the given values into the algebraic structure:

• $c_{11} = (1 \times 2) + (3 \times 3) = 2 + 9 = 11$
• $c_{12} = (1 \times 4) + (3 \times 2) = 4 + 6 = 10$
• $c_{21} = (-2 \times 2) + (5 \times 3) = -4 + 15 = 11$
• $c_{22} = (-2 \times 4) + (5 \times 2) = -8 + 10 = 2$

Visual Representation of Matrix Transformation

Step 4: Final Matrix Construction

Combining the calculated elements into the resulting $2 \times 2$ matrix:

$BA = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}$

Final Solution: $BA = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}$

Step 1: Defining the Matrix Variables

Let the amount invested in the first bond (5% interest) be $x$ and the amount invested in the second bond (7% interest) be $y$. The total investment is given as:

$x + y = 30,000$

We represent the investment as a row matrix $A$ and the interest rates as a column matrix $B$:

$A = \begin{bmatrix} x & y \end{bmatrix}$

$B = \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix}$

[By the definition of matrix multiplication for total interest calculation: $A \times B = \text{Total Interest}$]

Step 2: Constructing the Matrix Equation

The total interest required is ₹2,000. Applying the multiplication rule for a $1 \times 2$ matrix and a $2 \times 1$ matrix, we obtain a $1 \times 1$ matrix:

$\begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix} = [2000]$

$0.05x + 0.07y = 2000$ --- (Eq. 1)

$x + y = 30,000$ --- (Eq. 2)

Step 3: Solving the System of Equations

From (Eq. 2), we express $x$ in terms of $y$:

$x = 30,000 - y$

Substitute this into (Eq. 1):

$0.05(30,000 - y) + 0.07y = 2000$

$1,500 - 0.05y + 0.07y = 2000$ [Distributive Property]

$0.02y = 2000 - 1500$

$0.02y = 500$

$y = \frac{500}{0.02} = 25,000$

Step 4: Determining the Second Variable

Using the value of $y$ to find $x$:

$x = 30,000 - 25,000$

$x = 5,000$

Visual Representation of Interest Distribution

Step 5: Verification of Results

Final Solution: The trust fund must invest ₹5,000 in the first bond (5% interest) and ₹25,000 in the second bond (7% interest) to obtain an annual total interest of ₹2,000.

Step 1: Scalar Multiplication of Matrices

To solve the equation $2\begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$, we apply the scalar multiplication rule, which states that $k \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} ka & kb \\ kc & kd \end{bmatrix}$ [Per the properties of Scalar Multiplication in Linear Algebra].

Multiplying each matrix by its respective scalar:

$2\begin{bmatrix} x & z \\ y & t \end{bmatrix} = \begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix}$

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

$3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$

Step 2: Matrix Addition and Equation Formulation

The equation now becomes:

$\begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$

Performing element-wise addition on the left-hand side [Per the definition of Matrix Addition]:

$\begin{bmatrix} 2x + 3 & 2z - 3 \\ 2y + 0 & 2t + 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$

Step 3: Solving the System of Linear Equations

Two matrices are equal if and only if their corresponding elements are equal [Per the definition of Matrix Equality]. This yields the following system of equations:

Step 4: Algebraic Verification

Substituting the values back into the original matrix expression: $2\begin{bmatrix} 3 & 9 \\ 6 & 6 \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 6 & 18 \\ 12 & 12 \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix}$. The solution is verified.

Final Solution: $x = 3, y = 6, z = 9, t = 6$

Step 1: Defining the Variables and Matrix Construction

Let $x$ be the amount invested in the first bond (at 5% interest) and $y$ be the amount invested in the second bond (at 7% interest). Given the total investment constraint:

$x + y = 30,000$ [Linear constraint equation]

We represent the investment amounts as a row matrix $A$ and the interest rates as a column matrix $B$:

$A = \begin{bmatrix} x & y \end{bmatrix}$

$B = \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix}$

Step 2: Setting up the Matrix Equation

The total annual interest is given by the product of the investment matrix and the interest rate matrix. According to the problem, this product must equal $1800$:

$A \times B = [1800]$

$\begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix} = [1800]$

$0.05x + 0.07y = 1800$ [Matrix Multiplication Property: $1 \times 2$ matrix multiplied by $2 \times 1$ matrix yields a $1 \times 1$ scalar]

Step 3: Solving the Simultaneous Equations

We now have a system of two linear equations:

1. $x + y = 30,000 \implies y = 30,000 - x$
2. $0.05x + 0.07y = 1800$

Substitute the value of $y$ from the first equation into the second equation:

$0.05x + 0.07(30,000 - x) = 1800$

$0.05x + 2100 - 0.07x = 1800$ [Distributive Property]

$-0.02x = 1800 - 2100$

$-0.02x = -300$

$x = \frac{-300}{-0.02} = 15,000$

Step 4: Calculating the Second Variable

Using $x = 15,000$ in the first equation:

$15,000 + y = 30,000$

$y = 15,000$

Step 5: Visual Representation of the Investment Split

Final Solution: The trust fund should invest ₹15,000 in the 5% bond and ₹15,000 in the 7% bond to obtain a total annual interest of ₹1800.

Step 1: Scalar Multiplication of Matrices

Given the equation: $x\begin{bmatrix} 2 \\ 3 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$

Perform scalar multiplication by distributing the scalars $x$ and $y$ into their respective matrices [Per the scalar multiplication property of matrices, $k[A] = [kA]$]:

$\begin{bmatrix} 2x \\ 3x \end{bmatrix} + \begin{bmatrix} -y \\ y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$

Step 2: Matrix Addition

Add the two matrices by summing their corresponding elements [Per the definition of matrix addition, where $(A+B)_{ij} = A_{ij} + B_{ij}$]:

$\begin{bmatrix} 2x - y \\ 3x + y \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$

Step 3: Equating Corresponding Elements

By the definition of matrix equality, two matrices are equal if and only if their corresponding elements are equal. This yields the following system of linear equations:

1. $2x - y = 10$ (Equation 1)
2. $3x + y = 5$ (Equation 2)

Step 4: Solving the System of Linear Equations

To eliminate variable $y$, add Equation 1 and Equation 2:

$(2x - y) + (3x + y) = 10 + 5$

$5x = 15$

$x = \frac{15}{5} = 3$

Now, substitute $x = 3$ into Equation 2 to find $y$:

$3(3) + y = 5$

$9 + y = 5$

$y = 5 - 9$

$y = -4$

Step 5: Verification (Geometrical Representation)

The solution can be viewed as finding the linear combination of two vectors $\vec{u} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $\vec{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$ that results in vector $\vec{w} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$.

Final Solution: $x = 3, y = -4$

Step 1: Analyzing Matrix Dimensions

To perform matrix multiplication, the number of columns in the first matrix must equal the number of rows in the second matrix [Definition of Matrix Multiplication].

• Matrix $A = \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}$ is a $2 \times 2$ matrix.
• Matrix $B = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}$ is a $2 \times 3$ matrix.

Since the product $AB$ results in a matrix of dimensions $(2 \times 2) \times (2 \times 3) = 2 \times 3$, the resulting matrix will have 2 rows and 3 columns.

Step 2: Defining the Multiplication Rule

The element at the $i^{th}$ row and $j^{th}$ column of the product matrix $C$ is given by $c_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$ [Matrix Product Summation Notation].

Step 3: Calculating Individual Elements

For a resultant matrix $C = \begin{bmatrix} c_{11} & c_{12} & c_{13} \\ c_{21} & c_{22} & c_{23} \end{bmatrix}$:

Step 4: Constructing the Resultant Matrix

Substituting the computed values into the matrix structure:

$\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$

Visual Representation of Matrix Transformation

Final Solution: $\begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix}$

Step 1: Scalar Multiplication of the First Matrix

Given the equation: $2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$

Apply scalar multiplication [Per the Scalar Multiplication Property of Matrices, where $k[A_{ij}] = [k \cdot A_{ij}]$]:

$2 \times \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} = \begin{bmatrix} 2(1) & 2(3) \\ 2(0) & 2(x) \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix}$

Step 2: Matrix Addition

Add the two matrices on the left-hand side [Per the definition of Matrix Addition, where $[A_{ij}] + [B_{ij}] = [A_{ij} + B_{ij}]$]:

$\begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 + y & 6 + 0 \\ 0 + 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix}$

Step 3: Equating Corresponding Elements

Set the resulting matrix equal to the matrix on the right-hand side:

$\begin{bmatrix} 2 + y & 6 \\ 1 & 2x + 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$

By the Principle of Equality of Matrices, we equate corresponding elements:

• For position $(1,1)$: $2 + y = 5$
• For position $(2,2)$: $2x + 2 = 8$

Step 4: Solving for Variables

Solving for $y$:

$2 + y = 5$

$y = 5 - 2$ [By Subtraction Property of Equality]

$y = 3$

Solving for $x$:

$2x + 2 = 8$

$2x = 8 - 2$ [By Subtraction Property of Equality]

$2x = 6$

$x = \frac{6}{2}$ [By Division Property of Equality]

$x = 3$

Final Solution: $x = 3, y = 3$

Step 1: Establishing the Matrix Representation

To determine the total revenue using matrix multiplication, we must define the inventory as a row matrix $A$ and the prices as a column matrix $P$.

• Quantity ($Q$): Since the quantities are given in dozens, we convert them to absolute numbers by multiplying by 12.
• Chemistry: $10 \times 12 = 120$
• Physics: $8 \times 12 = 96$
• Economics: $10 \times 12 = 120$

Representing these as a $1 \times 3$ row matrix $A$:

$A = \begin{bmatrix} 120 & 96 & 120 \end{bmatrix}$

Step 2: Defining the Price Matrix

The selling prices per book are given as ₹80, ₹60, and ₹40. Representing these as a $3 \times 1$ column matrix $P$:

$P = \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}$

Step 3: Calculating Total Revenue via Matrix Multiplication

The total amount ($T$) received is given by the matrix product $T = A \times P$ [Per the definition of Matrix Multiplication for total sum calculation].

$T = \begin{bmatrix} 120 & 96 & 120 \end{bmatrix} \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix}$

Step 4: Executing the Dot Product

Perform the multiplication: $T = (120 \times 80) + (96 \times 60) + (120 \times 40)$

• $120 \times 80 = 9600$
• $96 \times 60 = 5760$
• $120 \times 40 = 4800$

$T = 9600 + 5760 + 4800$

$T = 20160$

Visual Representation of Matrix Product

Final Solution: The total amount the bookshop will receive is ₹ 20,160.

Step 1: Dimensional Analysis of the Matrices

To compute the product of two matrices, we must first determine the order of the matrices involved. Let the product be defined as $C = A \times B$.

• Matrix $A$ is a column vector with $3$ rows and $1$ column. Order: $3 \times 1$.
• Matrix $B$ is a row vector with $1$ row and $3$ columns. Order: $1 \times 3$.

According to the rule of matrix multiplication, the product $AB$ is defined if and only if the number of columns in $A$ equals the number of rows in $B$ [Per Matrix Multiplication Conformability Criterion]. Here, $1 = 1$, so the product is defined and the resulting matrix $C$ will have the order $3 \times 3$.

Step 2: Constructing the Multiplication Mapping

The element $c_{ij}$ in the resulting matrix $C$ is obtained by the dot product of the $i^{th}$ row of $A$ and the $j^{th}$ column of $B$.

$C = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \times \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} = \begin{bmatrix} (1)(2) & (1)(3) & (1)(4) \\ (2)(2) & (2)(3) & (2)(4) \\ (3)(2) & (3)(3) & (3)(4) \end{bmatrix}$

Step 3: Calculating Scalar Elements

Performing the arithmetic operations for each entry:

Step 4: Geometric Visualization of the Outer Product

Final Solution: $\begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix}$

Step 1: Define the Matrices

Let $A$ and $B$ be two $3 \times 3$ matrices defined as follows:

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

Step 2: Calculate the Left Hand Side (LHS) Product: $A \times B$

The product of two matrices $C = AB$ is obtained by taking the dot product of rows of $A$ and columns of $B$.

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

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

$AB = \begin{bmatrix} -1+0+6 & 1-2+9 & 0+2+12 \\ 0+0+0 & 0-1+0 & 0+1+0 \\ -1+0+0 & 1-1+0 & 0+1+0 \end{bmatrix} = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$

Step 3: Calculate the Right Hand Side (RHS) Product: $B \times A$

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

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

$BA = \begin{bmatrix} -1+0+0 & -2+1+0 & -3+0+0 \\ 0+0+1 & 0-1+1 & 0+0+0 \\ 2+0+4 & 4+3+4 & 6+0+0 \end{bmatrix} = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$

Step 4: Comparison of Products

We compare the resulting matrices $AB$ and $BA$:

Since the corresponding elements of the resulting matrices are not equal, $AB \neq BA$.

Step 5: Visual Representation (Conceptual Verification)

Final Solution: Through the systematic evaluation of products $AB = \begin{bmatrix} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{bmatrix}$ and $BA = \begin{bmatrix} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{bmatrix}$, it is proven that $AB \neq BA$.