Problem 32 - Advanced Engineering Mathematics

Question

$$, Suppose \(\mathbf{A}=\left(\begin{array}{rr}2 & 4 \\ -3 & 2\end{array}\right)\) and \(\mathbf{B}=\left(\begin{array}{rr}4 & 10 \\ 2 & 5\end{array}\right)\) Verify the given property by computing the left and right members of the given equality. $$ (\mathbf{A B})^{T}=\mathbf{B}^{T} \mathbf{A}^{T} $$

Step-by-Step Solution

Answer

The property \((\mathbf{A B})^{T} = \mathbf{B}^{T} \mathbf{A}^{T}\) is verified.

Step 1: Compute \(\mathbf{A B}\)

To find the product \(\mathbf{A} \mathbf{B}\), multiply the matrices \(\mathbf{A}\) and \(\mathbf{B}\): \[\mathbf{A} \mathbf{B} = \left(\begin{array}{rr} 2 & 4 \ -3 & 2 \end{array}\right) \left(\begin{array}{rr} 4 & 10 \ 2 & 5 \end{array}\right)\] Calculating each element:- First row, first column: \( 2\times4 + 4\times2 = 8 + 8 = 16 \)- First row, second column: \( 2\times10 + 4\times5 = 20 + 20 = 40 \)- Second row, first column: \(-3\times4 + 2\times2 = -12 + 4 = -8 \)- Second row, second column: \(-3\times10 + 2\times5 = -30 + 10 = -20 \) Thus, \(\mathbf{A B} = \left(\begin{array}{rr} 16 & 40 \ -8 & -20 \end{array}\right)\)

Step 2: Compute \((\mathbf{A B})^{T}\)

Transpose the matrix \(\mathbf{A B}\):\[\mathbf{(A B)}^{T} = \left(\begin{array}{rr} 16 & 40 \ -8 & -20 \end{array}\right)^{T} = \left(\begin{array}{rr} 16 & -8 \ 40 & -20 \end{array}\right)\] Switch rows with columns to get the transpose.

Step 3: Compute \(\mathbf{B}^{T}\) and \(\mathbf{A}^{T}\)

Transpose each matrix individually:\[\mathbf{B}^{T} = \left(\begin{array}{rr} 4 & 10 \ 2 & 5 \end{array}\right)^{T} = \left(\begin{array}{rr} 4 & 2 \ 10 & 5 \end{array}\right)\]\[\mathbf{A}^{T} = \left(\begin{array}{rr} 2 & 4 \ -3 & 2 \end{array}\right)^{T} = \left(\begin{array}{rr} 2 & -3 \ 4 & 2 \end{array}\right)\]Switch rows with columns to obtain each transpose.

Step 4: Compute \(\mathbf{B}^{T} \mathbf{A}^{T}\)

Multiply the matrices \(\mathbf{B}^{T}\) and \(\mathbf{A}^{T}\):\[\mathbf{B}^{T} \mathbf{A}^{T} = \left(\begin{array}{rr} 4 & 2 \ 10 & 5 \end{array}\right) \left(\begin{array}{rr} 2 & -3 \ 4 & 2 \end{array}\right)\]Calculating each element:- First row, first column: \(4\times2 + 2\times4 = 8 + 8 = 16 \)- First row, second column: \(4\times(-3) + 2\times2 = -12 + 4 = -8 \)- Second row, first column: \(10\times2 + 5\times4 = 20 + 20 = 40 \)- Second row, second column: \(10\times(-3) + 5\times2 = -30 + 10 = -20 \) Thus, \(\mathbf{B}^{T} \mathbf{A}^{T} = \left(\begin{array}{rr} 16 & -8 \ 40 & -20 \end{array}\right)\)

Step 5: Compare \((\mathbf{A B})^{T}\) and \(\mathbf{B}^{T} \mathbf{A}^{T}\)

Compare the transposed product \((\mathbf{A B})^{T}\) against the product of transposes \(\mathbf{B}^{T} \mathbf{A}^{T}\). Both are:\[\left(\begin{array}{rr} 16 & -8 \ 40 & -20 \end{array}\right)\]Since both matrices are identical, the property \((\mathbf{A B})^{T} = \mathbf{B}^{T} \mathbf{A}^{T}\) is verified.

Key Concepts

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra. It involves two matrices \(A\) and \(B\), which can be multiplied if the number of columns in \(A\) equals the number of rows in \(B\). The result is a new matrix whose elements are calculated by taking the dot product of corresponding rows of the first matrix and columns of the second matrix. This can be viewed as:

• First row of \(A\) multiplied by the first column of \(B\), producing one element of the product matrix.
• Continue by moving across the row and down through the columns until all elements are calculated.

Let's consider an example. If \(A\) is a 2x2 matrix:\[\mathbf{A} = \begin{pmatrix} 2 & 4 \ -3 & 2 \end{pmatrix}\]And \(B\) is:\[\mathbf{B} = \begin{pmatrix} 4 & 10 \ 2 & 5 \end{pmatrix}\]When multiplying, each element of the resulting matrix is:- The first element is \(2 \times 4 + 4 \times 2 = 16\).- Continue this process for each subsequent element.The resultant matrix \(\mathbf{A} \mathbf{B}\) is \(\begin{pmatrix} 16 & 40 \ -8 & -20 \end{pmatrix}\).This systematic approach is crucial for computations in various applications of linear algebra.

Properties of Matrices

Understanding the properties of matrices helps solve complex problems efficiently in linear algebra. One important property involves the transpose of a product of matrices. The transpose of a single matrix \(C\) is denoted as \(C^{T}\), which involves flipping the matrix over its diagonal, turning its rows into columns. For example:\[(\begin{pmatrix} 16 & 40 \ -8 & -20 \end{pmatrix})^{T} = \begin{pmatrix} 16 & -8 \ 40 & -20 \end{pmatrix}\]A key property is \((AB)^{T} = B^{T} A^{T}\). This means you can transpose a product by switching the order of multiplication and transposing each factor separately.

Compute \(B^{T}\) and \(A^{T}\) individually.

Multiply them in reversed order compared to the original matrices.

By following this order, the resulting matrix proves the equality of \((AB)^{T}\) and \(B^{T}A^{T}\), which is pivotal in matrix operations especially when verifying solutions, as demonstrated in this exercise.

Linear Algebra

Linear Algebra is a key area in mathematics focusing on vector spaces and linear mappings between them. It involves calculations with matrices and vectors, which alongside transposing operations, greatly impacts various fields like engineering, physics, and computer science. Key areas in linear algebra include:

• Matrix operations such as addition, subtraction, multiplication, and transposing.
• Solving systems of linear equations using matrix methods.
• Understanding vector spaces, basis, and dimension.

Transposing matrices and understanding their properties are vital as they aid in transformations and understanding symmetry in problems. Practical applications include: - Graphics transformation in computer graphics where matrices control object movement. - Complex calculations in machine learning, where efficiency in matrix operations is crucial. Transposing errors can lead to incorrect results, underscoring the importance of meticulous operations in mathematical contexts.