Problem 13 - Advanced Engineering Mathematics

Question

$$ \begin{aligned} &\text { In Problems } \underline{\phantom{xxx}} , \text { find the entries } c_{23} \text { and } c_{12} \text { for the matrix }\\\ &\mathbf{C}=2 \mathbf{A}-3 \mathbf{B} \end{aligned} $$ $$ \mathbf{A}=\left(\begin{array}{rrr} 2 & 3 & -1 \\ -1 & 6 & 0 \end{array}\right), \mathbf{B}=\left(\begin{array}{rrr} 4 & -2 & 6 \\ 1 & 3 & -3 \end{array}\right) $$

Step-by-Step Solution

Answer

\(c_{23} = 9\), \(c_{12} = 12\).

Step 1: Multiply Matrix A by Scalar

To find \(\mathbf{C} = 2\mathbf{A} - 3\mathbf{B}\), start by calculating \(2\mathbf{A}\). Multiply each element of matrix \(\mathbf{A}\) by 2.\[ 2\mathbf{A} = 2 \cdot \begin{pmatrix} 2 & 3 & -1 \ -1 & 6 & 0 \end{pmatrix} = \begin{pmatrix} 4 & 6 & -2 \ -2 & 12 & 0 \end{pmatrix}\]

Step 2: Multiply Matrix B by Scalar

Next, calculate \(3\mathbf{B}\) by multiplying each element of matrix \(\mathbf{B}\) by 3.\[ 3\mathbf{B} = 3 \cdot \begin{pmatrix} 4 & -2 & 6 \ 1 & 3 & -3 \end{pmatrix} = \begin{pmatrix} 12 & -6 & 18 \ 3 & 9 & -9 \end{pmatrix}\]

Step 3: Subtract Matrices

Subtract \(3\mathbf{B}\) from \(2\mathbf{A}\) to find \(\mathbf{C}\).\[ \mathbf{C} = 2\mathbf{A} - 3\mathbf{B} = \begin{pmatrix} 4 & 6 & -2 \ -2 & 12 & 0 \end{pmatrix} - \begin{pmatrix} 12 & -6 & 18 \ 3 & 9 & -9 \end{pmatrix} = \begin{pmatrix} 4-12 & 6+6 & -2-18 \ -2-3 & 12-9 & 0+9 \end{pmatrix}\]

Step 4: Simplify the Resultant Matrix

Perform the arithmetic operation for each entry of the resultant matrix \(\mathbf{C}\) from Step 3.\[ \mathbf{C} = \begin{pmatrix} -8 & 12 & -20 \ -5 & 3 & 9 \end{pmatrix}\]

Step 5: Identify Matrix Entries c23 and c12

In the matrix \(\mathbf{C}\), locate the entry \(c_{23}\) (second row, third column) and entry \(c_{12}\) (first row, second column).\(c_{23} = 9\) and \(c_{12} = 12\).

Key Concepts

Scalar Multiplication

Scalar multiplication is a fundamental operation in matrix algebra. It involves multiplying every element of a matrix by a given scalar (which is a single number). This process scales the matrix by changing each of its elements by the same factor.
For example, consider a matrix \( extbf{A}\) given by

• \(\begin{pmatrix} 2 & 3 & -1 \ -1 & 6 & 0 \end{pmatrix}\)

To multiply this matrix by scalar 2, multiply each element by 2:

• Row 1: \(2 \times 2 = 4\), \(2 \times 3 = 6\), \(2 \times -1 = -2\)
• Row 2: \(2 \times -1 = -2\), \(2 \times 6 = 12\), \(2 \times 0 = 0\)

The resulting matrix after performing scalar multiplication is:

• \(\begin{pmatrix} 4 & 6 & -2 \ -2 & 12 & 0 \end{pmatrix}\)

This scaled matrix is essential for further operations like matrix addition and subtraction.

Matrix Addition

Matrix addition is the process of adding two matrices by adding their corresponding elements. It is crucial to ensure that both matrices are of the same size, otherwise, addition cannot be performed.
For instance, if we have two matrices \( extbf{X}\) and \( extbf{Y}\) both of dimensions 2x3:

• \(\textbf{X} = \begin{pmatrix} x_{11} & x_{12} & x_{13} \ x_{21} & x_{22} & x_{23} \end{pmatrix}\)
• \(\textbf{Y} = \begin{pmatrix} y_{11} & y_{12} & y_{13} \ y_{21} & y_{22} & y_{23} \end{pmatrix}\)

The sum \(\textbf{X} + \textbf{Y}\) will be:

• \(\begin{pmatrix} x_{11} + y_{11} & x_{12} + y_{12} & x_{13} + y_{13} \ x_{21} + y_{21} & x_{22} + y_{22} & x_{23} + y_{23} \end{pmatrix}\)

Each element is simply the sum of the corresponding elements from \(\textbf{X}\) and \(\textbf{Y}\). This operation can be extended to any number of matrices of the same dimensions.

Matrix Subtraction

Matrix subtraction is similar to matrix addition but involves subtracting corresponding elements from two matrices instead. This means we can only subtract two matrices if they have exactly the same dimensions, much like matrix addition.
To perform matrix subtraction, take each element from the first matrix and subtract the corresponding element from the second matrix. Let's consider two matrices \(\textbf{M}\) and \(\textbf{N}\):

• \(\textbf{M} = \begin{pmatrix} m_{11} & m_{12} & m_{13} \ m_{21} & m_{22} & m_{23} \end{pmatrix}\)
• \(\textbf{N} = \begin{pmatrix} n_{11} & n_{12} & n_{13} \ n_{21} & n_{22} & n_{23} \end{pmatrix}\)

The difference \(\textbf{M} - \textbf{N}\) will be:

• \(\begin{pmatrix} m_{11} - n_{11} & m_{12} - n_{12} & m_{13} - n_{13} \ m_{21} - n_{21} & m_{22} - n_{22} & m_{23} - n_{23} \end{pmatrix}\)

Each element in the resulting matrix is computed by subtracting the element in \(\textbf{N}\) from the corresponding element in \(\textbf{M}\). This operation can be used in various applications such as solving systems of linear equations.

Entry Identification

In matrix operations, entry identification refers to locating a specific element within a matrix. Each element in a matrix is defined uniquely by its row and column position. This position is referred to using a subscript notation. For example, the element located in the first row and second column is denoted as \(c_{12}\).
Consider the matrix \(\textbf{C}\):

• \(\begin{pmatrix} -8 & 12 & -20 \ -5 & 3 & 9 \end{pmatrix}\)

To find \(c_{23}\), look at the second row and third column which contains the value 9. Similarly, \(c_{12}\) is found in the first row and second column, containing 12.
Remembering how to locate entries accurately is essential when performing calculations or solving matrix-related problems. These indices allow us to refer to and manipulate specific elements directly.