Problem 22 - Advanced Engineering Mathematics

Question

$$ \begin{aligned} &\text { If } \mathbf{A}=\left(\begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right) \text { and } \mathbf{B}=\left(\begin{array}{rr} -2 & 3 \\ 5 & 7 \end{array}\right), \text { find }\\\ &\text { (a) } \mathbf{A}+\mathbf{B}^{T} \text { , }\\\ &\text { (b) } 2 \mathbf{A}^{T}-\mathbf{B}^{T},(\mathbf{c}) \mathbf{A}^{T}(\mathbf{A}-\mathbf{B}) . \end{aligned} $$

Step-by-Step Solution

Answer

(a) \( \begin{pmatrix} -1 & 7 \\ 5 & 11 \end{pmatrix} \), (b) \( \begin{pmatrix} 4 & -1 \\ 1 & 1 \end{pmatrix} \), (c) \( \begin{pmatrix} -3 & -7 \\ -6 & -14 \end{pmatrix} \)

Step 1: Transpose Matrix B

First, find the transpose of matrix \( \mathbf{B} \). The transpose of a matrix is obtained by swapping its rows and columns. Matrix \( \mathbf{B} = \begin{pmatrix} -2 & 3 \ 5 & 7 \end{pmatrix} \) becomes \( \mathbf{B}^T = \begin{pmatrix} -2 & 5 \ 3 & 7 \end{pmatrix} \).

Step 2: Compute A + B^T

Now add \( \mathbf{A} \) to \( \mathbf{B}^T \). This involves adding corresponding elements of \( \mathbf{A} \) and \( \mathbf{B}^T \).\[ \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} + \begin{pmatrix} -2 & 5 \ 3 & 7 \end{pmatrix} = \begin{pmatrix} 1 + (-2) & 2 + 5 \ 2 + 3 & 4 + 7 \end{pmatrix} = \begin{pmatrix} -1 & 7 \ 5 & 11 \end{pmatrix} \].

Step 3: Transpose Matrix A

Now find the transpose of matrix \( \mathbf{A} \). Swap its rows and columns. Matrix \( \mathbf{A} = \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} \) becomes \( \mathbf{A}^T = \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} \) (note \( \mathbf{A} \) is symmetric so \( \mathbf{A} = \mathbf{A}^T \)).

Step 4: Compute 2A^T

Multiply matrix \( \mathbf{A}^T \) by 2.\[ 2 \cdot \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} = \begin{pmatrix} 2 \cdot 1 & 2 \cdot 2 \ 2 \cdot 2 & 2 \cdot 4 \end{pmatrix} = \begin{pmatrix} 2 & 4 \ 4 & 8 \end{pmatrix} \].

Step 5: Compute 2A^T - B^T

Subtract \( \mathbf{B}^T \) from \( 2\mathbf{A}^T \) by subtracting their corresponding elements.\[ \begin{pmatrix} 2 & 4 \ 4 & 8 \end{pmatrix} - \begin{pmatrix} -2 & 5 \ 3 & 7 \end{pmatrix} = \begin{pmatrix} 2 - (-2) & 4 - 5 \ 4 - 3 & 8 - 7 \end{pmatrix} = \begin{pmatrix} 4 & -1 \ 1 & 1 \end{pmatrix} \].

Step 6: Compute A - B

Subtract \( \mathbf{B} \) from \( \mathbf{A} \) by subtracting their corresponding elements.\[ \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} - \begin{pmatrix} -2 & 3 \ 5 & 7 \end{pmatrix} = \begin{pmatrix} 1 + 2 & 2 - 3 \ 2 - 5 & 4 - 7 \end{pmatrix} = \begin{pmatrix} 3 & -1 \ -3 & -3 \end{pmatrix} \].

Step 7: Compute A^T(A - B)

Multiply \( \mathbf{A}^T \) by \( (\mathbf{A} - \mathbf{B}) \). Perform matrix multiplication by taking the dot product of rows of the first matrix with columns of the second.\[ \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} \begin{pmatrix} 3 & -1 \ -3 & -3 \end{pmatrix} = \begin{pmatrix} (1)(3) + (2)(-3) & (1)(-1) + (2)(-3) \ (2)(3) + (4)(-3) & (2)(-1) + (4)(-3) \end{pmatrix} = \begin{pmatrix} 3 - 6 & -1 - 6 \ 6 - 12 & -2 - 12 \end{pmatrix} = \begin{pmatrix} -3 & -7 \ -6 & -14 \end{pmatrix} \].

Key Concepts

Matrix Addition

Adding matrices is like combining similar terms in a simple equation. To add two matrices, each matrix must be the same size, which means having the same number of rows and columns. For each corresponding element from both matrices, you simply add them together. Say we have two matrices, \( \mathbf{A} \) and \( \mathbf{B} \), with dimensions 2x2:

• \( \mathbf{A} = \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} \)
• \( \mathbf{B} = \begin{pmatrix} -2 & 5 \ 3 & 7 \end{pmatrix} \)

To find \( \mathbf{A} + \mathbf{B} \), we add each element from \( \mathbf{A} \) with the corresponding element in \( \mathbf{B} \):

• The top-left element becomes \( 1 + (-2) = -1 \)
• The top-right element becomes \( 2 + 5 = 7 \)
• The bottom-left element becomes \( 2 + 3 = 5 \)
• The bottom-right element becomes \( 4 + 7 = 11 \)

Putting them together, we get the matrix \( \begin{pmatrix} -1 & 7 \ 5 & 11 \end{pmatrix} \). It's that straightforward!

Matrix Transposition

Matrix transposition involves flipping a matrix over its diagonal. This means swapping the row and column indices of the matrix.Consider matrix \( \mathbf{B} \):

• \( \mathbf{B} = \begin{pmatrix} -2 & 3 \ 5 & 7 \end{pmatrix} \)

To get the transpose, denoted as \( \mathbf{B}^T \), convert the first row of \( \mathbf{B} \) into the first column, and the second row into the second column. Thus, \( \mathbf{B}^T \) becomes:

• \( \mathbf{B}^T = \begin{pmatrix} -2 & 5 \ 3 & 7 \end{pmatrix} \)

This operation is especially useful in linear algebra, where transposing a matrix can reveal symmetry and other properties.

Matrix Multiplication

Multiplying matrices involves a slightly more complex operation than addition or subtraction. Let's explore it step by step. For two matrices that can be multiplied, the number of columns in the first matrix must match the number of rows in the second. With our matrices \( \mathbf{A}^T \) and \( \mathbf{A} - \mathbf{B} \), this condition is satisfied:

• \( \mathbf{A}^T = \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} \)
• \( \mathbf{A} - \mathbf{B} = \begin{pmatrix} 3 & -1 \ -3 & -3 \end{pmatrix} \)

Each element of the resulting matrix is the dot product of corresponding rows and columns. For example, the top-left element is obtained by calculating \((1 \times 3) + (2 \times -3) = 3 - 6 = -3\). Other elements are calculated similarly, resulting in:

• Final matrix: \( \begin{pmatrix} -3 & -7 \ -6 & -14 \end{pmatrix} \)

This operation finds widespread use in computer graphics, solving systems of equations, and more.

Matrix Subtraction

Matrix subtraction is quite analogous to matrix addition, with the main difference being that you subtract correlation elements instead of adding them. Let's consider matrices \( \mathbf{A} \) and \( \mathbf{B} \):

• \( \mathbf{A} = \begin{pmatrix} 1 & 2 \ 2 & 4 \end{pmatrix} \)
• \( \mathbf{B} = \begin{pmatrix} -2 & 3 \ 5 & 7 \end{pmatrix} \)

To compute \( \mathbf{A} - \mathbf{B} \), subtract each element of \( \mathbf{B} \) from \( \mathbf{A} \):

• Top-left: \( 1 - (-2) = 3 \)
• Top-right: \( 2 - 3 = -1 \)
• Bottom-left: \( 2 - 5 = -3 \)
• Bottom-right: \( 4 - 7 = -3 \)

This gives us the resulting matrix: \( \begin{pmatrix} 3 & -1 \ -3 & -3 \end{pmatrix} \). This straightforward operation is essential when solving linear equations and systems.