Problem 17 - Advanced Engineering Mathematics

Question

If \(\mathbf{A}=\left(\begin{array}{rr}2 & -3 \\ -5 & 4\end{array}\right)\) and \(\mathbf{B}=\left(\begin{array}{rr}-1 & 6 \\ 3 & 2\end{array}\right)\), find \((\mathbf{a}) \mathbf{A B}\) (b) \(\mathbf{B A},(\mathbf{c}) \mathbf{A}^{2}=\mathbf{A} \mathbf{A}\) (d) \(\mathbf{B}^{2}=\mathbf{B B}\)

Step-by-Step Solution

Answer

\( \mathbf{A} \mathbf{B} = \left(\begin{array}{rr} -11 & 6 \\ 17 & -22 \end{array}\right) \), \( \mathbf{B} \mathbf{A} = \left(\begin{array}{rr} -32 & 27 \\ -4 & -1 \end{array}\right) \), \( \mathbf{A}^2 = \left(\begin{array}{rr} 19 & -18 \\ -30 & 31 \end{array}\right) \), \( \mathbf{B}^2 = \left(\begin{array}{rr} 19 & 6 \\ 3 & 22 \end{array}\right) \)."}

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

To find \( \mathbf{A} \mathbf{B} \), multiply each element in the rows of matrix \( \mathbf{A} \) by the corresponding elements in the columns of matrix \( \mathbf{B} \) and sum the products:\[ \mathbf{A} \mathbf{B} = \left(\begin{array}{rr} 2 & -3 \ -5 & 4 \end{array}\right) \left(\begin{array}{rr} -1 & 6 \ 3 & 2 \end{array}\right) \]The resulting matrix is:- For the first row, first column: \( 2(-1) + (-3)(3) = -2 - 9 = -11 \)- For the first row, second column: \( 2(6) + (-3)(2) = 12 - 6 = 6 \)- For the second row, first column: \( -5(-1) + 4(3) = 5 + 12 = 17 \)- For the second row, second column: \( -5(6) + 4(2) = -30 + 8 = -22 \)Therefore, \( \mathbf{A} \mathbf{B} = \left(\begin{array}{rr} -11 & 6 \ 17 & -22 \end{array}\right) \).

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

To compute \( \mathbf{B} \mathbf{A} \), multiply matrix \( \mathbf{B} \) by matrix \( \mathbf{A} \):\[ \mathbf{B} \mathbf{A} = \left(\begin{array}{rr} -1 & 6 \ 3 & 2 \end{array}\right) \left(\begin{array}{rr} 2 & -3 \ -5 & 4 \end{array}\right) \]The resulting matrix is:- For the first row, first column: \( -1(2) + 6(-5) = -2 - 30 = -32 \)- For the first row, second column: \( -1(-3) + 6(4) = 3 + 24 = 27 \)- For the second row, first column: \( 3(2) + 2(-5) = 6 - 10 = -4 \)- For the second row, second column: \( 3(-3) + 2(4) = -9 + 8 = -1 \)Thus, \( \mathbf{B} \mathbf{A} = \left(\begin{array}{rr} -32 & 27 \ -4 & -1 \end{array}\right) \).

Step 3: Compute \( \mathbf{A}^2 \)

To find \( \mathbf{A}^2 \), multiply matrix \( \mathbf{A} \) by itself:\[ \mathbf{A}^2 = \left(\begin{array}{rr} 2 & -3 \ -5 & 4 \end{array}\right) \left(\begin{array}{rr} 2 & -3 \ -5 & 4 \end{array}\right) \]The resulting matrix is:- For the first row, first column: \( 2(2) + (-3)(-5) = 4 + 15 = 19 \)- For the first row, second column: \( 2(-3) + (-3)(4) = -6 - 12 = -18 \)- For the second row, first column: \( -5(2) + 4(-5) = -10 - 20 = -30 \)- For the second row, second column: \( -5(-3) + 4(4) = 15 + 16 = 31 \)Therefore, \( \mathbf{A}^2 = \left(\begin{array}{rr} 19 & -18 \ -30 & 31 \end{array}\right) \).

Step 4: Compute \( \mathbf{B}^2 \)

To compute \( \mathbf{B}^2 \), multiply matrix \( \mathbf{B} \) by itself:\[ \mathbf{B}^2 = \left(\begin{array}{rr} -1 & 6 \ 3 & 2 \end{array}\right) \left(\begin{array}{rr} -1 & 6 \ 3 & 2 \end{array}\right) \]The resulting matrix is:- For the first row, first column: \( -1(-1) + 6(3) = 1 + 18 = 19 \)- For the first row, second column: \( -1(6) + 6(2) = -6 + 12 = 6 \)- For the second row, first column: \( 3(-1) + 2(3) = -3 + 6 = 3 \)- For the second row, second column: \( 3(6) + 2(2) = 18 + 4 = 22 \)Thus, \( \mathbf{B}^2 = \left(\begin{array}{rr} 19 & 6 \ 3 & 22 \end{array}\right) \).

Key Concepts

Matrix Algebra

Matrix algebra is a cornerstone of linear algebra and involves operations like addition, subtraction, and, notably, multiplication of matrices. When we talk about multiplying matrices, it's not as straightforward as multiplying individual numbers. Instead, it's a systematic combination of rows and columns.
For two matrices, say \( \mathbf{A} \) and \( \mathbf{B} \), to be multiplied, the number of columns in \( \mathbf{A} \) must equal the number of rows in \( \mathbf{B} \). This results in a new matrix where each element is the sum of the products of corresponding elements from the rows of the first matrix and the columns of the second matrix.
For example, matrix \( \mathbf{A} \) of size \( 2 \times 2 \) can be multiplied with another \( 2 \times 2 \) matrix such as \( \mathbf{B} \). The element in the "top-left" corner of the resulting matrix is computed by multiplying the first row of \( \mathbf{A} \) by the first column of \( \mathbf{B} \). This highlights the fundamental aspect of matrix multiplication: each entry in the resulting matrix is product of a row with a column.

• The resulting matrix from \( \mathbf{A} \mathbf{B} \) is formed by systematically computing similar pairs of row and column products.
• Matrix multiplication is associative, meaning \( (\mathbf{A}\mathbf{B})\mathbf{C} = \mathbf{A}(\mathbf{B}\mathbf{C}) \).
• It is not, however, commutative. In other words, usually \( \mathbf{A}\mathbf{B} eq \mathbf{B}\mathbf{A} \).

Linear Transformations

Linear transformations are one of the foundational concepts behind matrix operations. Essentially, a linear transformation is a function between two vector spaces that preserves vector addition and scalar multiplication. In simpler terms, it's a tool that can stretch, shrink, rotate, or reflect vectors without tearing them apart or distorting their relative positions.
In mathematics, matrices often represent linear transformations. When you multiply a vector by a matrix, you're applying a linear transformation defined by that matrix. This is particularly useful in various applications, like computer graphics where transforming shapes in space is required.
Given a matrix \( \mathbf{A} \) and a vector \( \mathbf{v} \), the product \( \mathbf{A} \mathbf{v} \) gives a new vector that results from the linear transformation of \( \mathbf{v} \) by \( \mathbf{A} \). It is powerful because it can be represented in matrix form where operations on vectors are equivalent to matrix operations.

• Linear transformations maintain the "straightness" and "flatness" of any geometric shape or space they're applied to.
• They are utilized heavily in computer sciences for transforming images, sounds, and animations.
• Understanding linear transformations helps in grasping advanced mathematical concepts like eigenvalues and eigenvectors.

Elementary Matrices

Elementary matrices are the building blocks of more complex matrix operations. Each represents a simple transformation equivalent to a single elementary row operation. Elementary matrices are derived by performing a basic row operation on the identity matrix.
There are three main types of elementary operations and their corresponding matrices:

• Row switching: Swaps two rows in a matrix. The corresponding elementary matrix is produced by swapping the same rows in an identity matrix.
• Row multiplication: Multiplies all entries of a row by a nonzero constant. The result is an elementary matrix with that row and all others unchanged.
• Row addition: Adds a multiple of one row to another row. The elementary matrix remains mostly an identity matrix with a multiple added in one position.

Elementary matrices are essential because they simplify complex operations like Gaussian elimination or finding the inverse of a matrix. By understanding how these matrices function, one can perform systematic changes to matrices without altering their fundamental properties, much like editing points without moving the entire grid.