Problem 23 - Advanced Engineering Mathematics

Question

In Problems 13 and 14, find the entries \(c_{23}\) and \(c_{12}\) for the matrix \(\mathbf{C}=2 \mathbf{A}-3 \mathbf{B}\). If \(\mathbf{A}=\left(\begin{array}{ll}3 & 4 \\ 8 & 1\end{array}\right)\) and \(\mathbf{B}=\left(\begin{array}{rr}5 & 10 \\ -2 & -5\end{array}\right)\), find (a) \((\mathbf{A B})^{T}\) (b) \(\mathbf{B}^{T} \mathbf{A}^{T}\).

Step-by-Step Solution

Answer

(a) \((\mathbf{AB})^T\) is \(\begin{pmatrix}7 & 38 \\ 10 & 75\end{pmatrix}\). (b) \(\mathbf{B}^T\mathbf{A}^T\) is \(\begin{pmatrix}7 & 38 \\ 10 & 75\end{pmatrix}\).

Step 1: Calculate Matrix C

The given matrix \(\mathbf{C}\) is defined as \(2 \mathbf{A} - 3 \mathbf{B}\). First, calculate \(2\mathbf{A}\):\[2\mathbf{A}=2\times \begin{pmatrix}3 & 4 \ 8 & 1\end{pmatrix} = \begin{pmatrix}6 & 8 \ 16 & 2\end{pmatrix}\]Next, calculate \(3\mathbf{B}\):\[3\mathbf{B}=3\times \begin{pmatrix}5 & 10 \ -2 & -5\end{pmatrix} = \begin{pmatrix}15 & 30 \ -6 & -15\end{pmatrix}\]Subtract \(3\mathbf{B}\) from \(2\mathbf{A}\):\[\mathbf{C} = \begin{pmatrix}6 & 8 \ 16 & 2\end{pmatrix} - \begin{pmatrix}15 & 30 \ -6 & -15\end{pmatrix} = \begin{pmatrix}-9 & -22 \ 22 & 17\end{pmatrix}\]

Step 2: Find Entry c_{23} of Matrix C

Since \(c_{23}\) denotes the entry in the second row and third column of matrix \(\mathbf{C}\) and \(\mathbf{C}\) is a 2x2 matrix, it does not contain a third column. Therefore, \(c_{23}\) does not exist in \(\mathbf{C}\).

Step 3: Find Entry c_{12} of Matrix C

Entry \(c_{12}\) for matrix \(\mathbf{C}\) represents the first row, second column. From \(\mathbf{C}=\begin{pmatrix}-9 & -22 \ 22 & 17\end{pmatrix}\), \(c_{12} = -22\).

Step 4: Calculate (AB)^T

Compute the product \(\mathbf{AB}\). Both \(\mathbf{A}\) and \(\mathbf{B}\) are 2x2 matrices:\[\mathbf{AB} = \begin{pmatrix}3 & 4 \ 8 & 1\end{pmatrix}\begin{pmatrix}5 & 10 \ -2 & -5\end{pmatrix} = \begin{pmatrix}3\times5 + 4\times(-2) & 3\times10 + 4\times(-5) \ 8\times5 + 1\times(-2) & 8\times10 + 1\times(-5)\end{pmatrix}\]Calculate each element:\[\mathbf{AB} = \begin{pmatrix}15 - 8 & 30 - 20 \ 40 - 2 & 80 - 5\end{pmatrix} = \begin{pmatrix}7 & 10 \ 38 & 75\end{pmatrix}\]Now find \((\mathbf{AB})^T\) by transposing \(\mathbf{AB}\):\[(\mathbf{AB})^T = \begin{pmatrix}7 & 38 \ 10 & 75\end{pmatrix}\]

Step 5: Calculate B^T A^T

First find \(\mathbf{B}^T\):\[\mathbf{B}^T = \begin{pmatrix}5 & -2 \ 10 & -5\end{pmatrix}\]Next, find \(\mathbf{A}^T\):\[\mathbf{A}^T = \begin{pmatrix}3 & 8 \ 4 & 1\end{pmatrix}\]Now compute \(\mathbf{B}^T \mathbf{A}^T\):\[\mathbf{B}^T\mathbf{A}^T = \begin{pmatrix}5 & -2 \ 10 & -5\end{pmatrix}\begin{pmatrix}3 & 8 \ 4 & 1\end{pmatrix} = \begin{pmatrix}5\times3 + (-2)\times4 & 5\times8 + (-2)\times1 \ 10\times3 + (-5)\times4 & 10\times8 + (-5)\times1\end{pmatrix}\]Calculate each element:\[\mathbf{B}^T \mathbf{A}^T = \begin{pmatrix}15 - 8 & 40 - 2 \ 30 - 20 & 80 - 5\end{pmatrix} = \begin{pmatrix}7 & 38 \ 10 & 75\end{pmatrix}\]

Key Concepts

Matrix Transposition

Matrix transposition is a fundamental operation that involves swapping the rows and columns of a matrix. If we have a matrix \( \mathbf{M} \), the transpose of matrix \( \mathbf{M} \), denoted as \( \mathbf{M}^T \), is formed by converting the rows of \( \mathbf{M} \) into columns. Here is a simple way to understand it:

• The first row of the original matrix becomes the first column of the transposed matrix.
• The second row becomes the second column, and so on.

For example, in the given problem, when calculating \( (\mathbf{AB})^T \), the transpose of matrix \( \mathbf{AB} = \begin{pmatrix} 7 & 10 \ 38 & 75 \end{pmatrix} \), we swap rows with columns to get \( (\mathbf{AB})^T = \begin{pmatrix} 7 & 38 \ 10 & 75 \end{pmatrix} \).
Notice how simple it is: just flip the matrix over its diagonal, switching its row and column indices.

Matrix Addition and Subtraction

Matrix addition and subtraction involve combining two matrices by adding or subtracting their corresponding elements. For these operations to be valid, the matrices must have the same dimensions. That means they must have the same number of rows and columns.

• To add matrices \( \mathbf{A} \) and \( \mathbf{B} \), calculate each element in the resulting matrix as follows: \((a_{ij} + b_{ij})\), where \( a_{ij} \) and \( b_{ij} \) are the elements in the \( i \)-th row and \( j \)-th column of matrices \( \mathbf{A} \) and \( \mathbf{B} \), respectively.
• For subtraction, simply use the operation \((a_{ij} - b_{ij})\).

In the given solution, you find \( \mathbf{C} = 2\mathbf{A} - 3\mathbf{B} \) by first scaling matrices \( \mathbf{A} \) and \( \mathbf{B} \) and then subtracting them element-wise, which results in \( \mathbf{C} = \begin{pmatrix} -9 & -22 \ 22 & 17 \end{pmatrix} \).
Each entry in \( \mathbf{C} \) comes from computing \((2a_{ij} - 3b_{ij})\) for corresponding positions in \( \mathbf{A} \) and \( \mathbf{B} \).

Matrix Scalar Multiplication

Matrix scalar multiplication involves multiplying every element of a matrix by a scalar (a single number). This operation is straightforward and results in a new matrix of the same dimensions as the original one.

• To multiply a matrix \( \mathbf{A} \) by a scalar \( k \), simply multiply each element \( a_{ij} \) of \( \mathbf{A} \) by \( k \). The resulting matrix will have elements \( k \cdot a_{ij} \).

For instance, in the problem, we find \( 2\mathbf{A} \) by multiplying each element of matrix \( \mathbf{A} = \begin{pmatrix} 3 & 4 \ 8 & 1 \end{pmatrix} \) by 2, resulting in \( 2\mathbf{A} = \begin{pmatrix} 6 & 8 \ 16 & 2 \end{pmatrix} \). Similarly, \( 3\mathbf{B} \) is obtained by multiplying each element of \( \mathbf{B} = \begin{pmatrix} 5 & 10 \ -2 & -5 \end{pmatrix} \) by 3, giving us \( 3\mathbf{B} = \begin{pmatrix} 15 & 30 \ -6 & -15 \end{pmatrix} \).Scalar multiplication is often the first step in combining matrices to evaluate expressions like \( \mathbf{C} = 2\mathbf{A} - 3\mathbf{B} \), as demonstrated in the exercise.