Problem 20 - Advanced Engineering Mathematics

Question

$$ \begin{aligned} &\text { If } \mathbf{A}=\left(\begin{array}{lll} 5 & -6 & 7 \end{array}\right), \mathbf{B}=\left(\begin{array}{r} 3 \\ 4 \\ -1 \end{array}\right)\\\ &\text { and } \mathbf{C}=\left(\begin{array}{rrr} 1 & 2 & 4 \\ 0 & 1 & -1 \\ 3 & 2 & 1 \end{array}\right), \text { find }(\mathbf{a}) \mathbf{A B},(\mathbf{b}) \mathbf{B A},(\mathbf{c})(\mathbf{B A}) \mathbf{C},\\\ &\text { (d) }(\mathbf{A B}) \mathbf{C} \text { . } \end{aligned} $$

Step-by-Step Solution

Answer

\( \mathbf{A} \mathbf{B} = -16 \), \( \mathbf{B} \mathbf{A} = \begin{bmatrix} 15 & -18 & 21 \\ 20 & -24 & 28 \\ -5 & 6 & -7 \end{bmatrix} \), \((\mathbf{B} \mathbf{A}) \mathbf{C} = \begin{bmatrix} 78 & 60 &99 \\ 104 & 72 & 124 \\ -26 & -13 & -33 \end{bmatrix} \), \((\mathbf{A} \mathbf{B}) \mathbf{C} = \begin{bmatrix} -16 & -32 & -64 \\ 0 & -16 & 16 \\ -48 & -32 & -16 \end{bmatrix} \).

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

To find the product \( \mathbf{A} \mathbf{B} \), multiply the row vector \( \mathbf{A} \) by the column vector \( \mathbf{B} \):\[ \mathbf{A} \mathbf{B} = \left(\begin{array}{lll} 5 & -6 & 7 \end{array}\right) \left(\begin{array}{c} 3 \ 4 \ -1 \end{array}\right) \].\ Calculating, we get: \[ (5 \times 3) + (-6 \times 4) + (7 \times -1) = 15 - 24 - 7 = -16 \].Thus, \( \mathbf{A} \mathbf{B} = -16 \).

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

To find the product \( \mathbf{B} \mathbf{A} \), multiply the column vector \( \mathbf{B} \) by the row vector \( \mathbf{A} \), resulting in a 3x3 matrix:\[ \mathbf{B} \mathbf{A} = \left(\begin{array}{c} 3 \ 4 \ -1 \end{array}\right) \left(\begin{array}{lll} 5 & -6 & 7 \end{array}\right)\].The resulting matrix is computed as:\[ \begin{bmatrix} 3 \times 5 & 3 \times -6 & 3 \times 7 \ 4 \times 5 & 4 \times -6 & 4 \times 7 \ -1 \times 5 & -1 \times -6 & -1 \times 7 \end{bmatrix} \ = \begin{bmatrix} 15 & -18 & 21 \ 20 & -24 & 28 \ -5 & 6 & -7 \end{bmatrix} \].

Step 3: Compute \((\mathbf{B} \mathbf{A}) \mathbf{C} \)

Using the result from Step 2, multiply matrix \( \mathbf{B} \mathbf{A} \) with matrix \( \mathbf{C} \) (3x3 matrix multiplication):\[ \begin{bmatrix} 15 & -18 & 21 \ 20 & -24 & 28 \ -5 & 6 & -7 \end{bmatrix} \begin{bmatrix} 1 & 2 & 4 \ 0 & 1 & -1 \ 3 & 2 & 1 \end{bmatrix} \].Computing this matrix multiplication:- First Row: \[ (15 \times 1 + (-18) \times 0 + 21 \times 3, 15 \times 2 + (-18) \times 1 + 21 \times 2, 15 \times 4 + (-18) \times -1 + 21 \times 1) = (78, 60, 99) \]- Second Row: \[ (20 \times 1 + (-24) \times 0 + 28 \times 3, 20 \times 2 + (-24) \times 1 + 28 \times 2, 20 \times 4 + (-24) \times -1 + 28 \times 1) = (104, 72, 124) \]- Third Row: \[ (-5 \times 1 + 6 \times 0 + (-7) \times 3, -5 \times 2 + 6 \times 1 + (-7) \times 2, -5 \times 4 + 6 \times -1 + (-7) \times 1) = (-26, -13, -33) \]So, \((\mathbf{B} \mathbf{A}) \mathbf{C} = \begin{bmatrix} 78 & 60 & 99 \ 104 & 72 & 124 \ -26 & -13 & -33 \end{bmatrix} \).

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

Note that \( \mathbf{A} \mathbf{B} \) is a scalar (-16). To multiply this scalar with matrix \( \mathbf{C} \), each element of \( \mathbf{C} \) should be multiplied by -16:\[ (\mathbf{A} \mathbf{B}) \mathbf{C} = -16 \cdot \begin{bmatrix} 1 & 2 & 4 \ 0 & 1 & -1 \ 3 & 2 & 1 \end{bmatrix} = \begin{bmatrix} -16 & -32 & -64 \ 0 & -16 & 16 \ -48 & -32 & -16 \end{bmatrix} \].

Key Concepts

Vector Multiplication

Vector multiplication, often referred to as the dot product, involves two vectors and results in a single scalar. It fuses both magnitude and direction information. For the vectors \( \mathbf{A} \) and \( \mathbf{B} \) in the exercise, you have a row vector \( \mathbf{A} \) and a column vector \( \mathbf{B} \). This setup allows you to compute a dot product. You multiply corresponding components of the vectors and sum them up, like this:

• Multiply 5 and 3 – thinking of these as results coming from the same dimension on different coordinates.
• Do the same with all subsequent components from each vector: -6 and 4, 7 and -1.
• Add up all these products: producing 15, -24, and -7.

Finally, summing these results gives you \( -16 \) as the complete dot product. This simple method underlines how direction influences the magnitude in vector multiplication.

Scalar Multiplication

Scalar multiplication involves a single number scaling every element of a matrix or vector. In our task, we multiplied the scalar \(-16\) by every component of matrix \( \mathbf{C} \). The scalar operates like a multiplier, expanding or contracting the size of each entry in the matrix:

• For matrix \( \mathbf{C} \), the value 1 becomes -16, since 1 multiplied by -16 is -16.
• Continuing in this manner, 2 turns into -32, 4 turns into -64, and so forth through the entire matrix.
• This simple and essential operation shows how scalar values influence matrix characteristics.

By doing so, multiplication affects the scale but not the structure, maintaining the order of the rows and columns. Scalars can powerfully affect numerical attributes when applied to vectors and matrices.

Linear Algebra

Linear algebra is the backbone of vector and matrix operations. It's foundational for understanding spaces and transformations in mathematics. The operations in this problem exemplify many core properties:

• Commutative Property: Notice here the non-commutative nature of multiplication since \(\mathbf{A} \mathbf{B} eq \mathbf{B} \mathbf{A}\).
• Identity Element: Scalar multiples of an identity matrix don’t change the matrix. In our task, multiplying whole matrices and vectors by scalars alters numerical values but doesn't affect their inherent attributes.
• Dimensional Analysis: Each procedure maintains appropriate dimensional compatibility, a critical aspect of matrix operations: row-by-column setup in multiplication.

Linear algebra enlightens how multiplications align with our understanding of transformations and other complex relations, pivotal in fields like computer science, physics, and economics.

Matrix Operations

Matrix operations form the core structure for manipulating more complex data sets. In mathematics, they extend basic arithmetic to encompass not just numbers, but arrays of numbers. Consider these operations:

• Matrix Multiplication: Multiplying \(\mathbf{B} \mathbf{A}\) with \(\mathbf{C}\), showcasing how complex interactions lead to new matrices. The matrices' dimensions must align correctly: the number of columns in the first must match the number of rows in the second for valid multiplication.
• Additive Identity: When you multiply entire rows by zero, which won’t affect other rows, it illustrates maintaining structure amidst operations.
• Distributive Property: Similar items across matrices keep multiplication additive: adding results from each phase of multiplication to obtain a complete matrix transformation.

Understanding matrix operations allows you to solve high-dimensional problems efficiently, a necessary skill across engineering, computer graphics, simulations, and beyond.