Problem 15 - Advanced Engineering Mathematics

Question

$$ \begin{aligned} &\text { If } \mathbf{A}=\left(\begin{array}{rr} 4 & 5 \\ -6 & 9 \end{array}\right) \text { and } \mathbf{B}=\left(\begin{array}{rr} -2 & 6 \\ 8 & -10 \end{array}\right), \text { find (a) } \mathbf{A}+\mathbf{B}\\\ &\text { (b) } \mathbf{B}-\mathbf{A},(\mathbf{c}) 2 \mathbf{A}+3 \mathbf{B} \end{aligned} $$

Step-by-Step Solution

Answer

(a) \( \mathbf{A} + \mathbf{B} = \left(\begin{array}{rr} 2 & 11 \\ 2 & -1 \end{array}\right) \), (b) \( \mathbf{B} - \mathbf{A} = \left(\begin{array}{rr} -6 & 1 \\ 14 & -19 \end{array}\right) \), (c) \( 2 \mathbf{A} + 3 \mathbf{B} = \left(\begin{array}{rr} 2 & 28 \\ 12 & -12 \end{array}\right) \).

Step 1: Add Matrices A and B

To find \( \mathbf{A} + \mathbf{B} \), we add the corresponding elements of matrices \( \mathbf{A} \) and \( \mathbf{B} \):\[\mathbf{A} + \mathbf{B} = \left(\begin{array}{rr} 4 & 5 \ -6 & 9 \end{array}\right) + \left(\begin{array}{rr} -2 & 6 \ 8 & -10 \end{array}\right)\]This results in:\[= \left(\begin{array}{rr} 4 + (-2) & 5 + 6 \ -6 + 8 & 9 + (-10) \end{array}\right)\]Calculating, we get:\[= \left(\begin{array}{rr} 2 & 11 \ 2 & -1 \end{array}\right)\]

Step 2: Subtract Matrix A from B

For \( \mathbf{B} - \mathbf{A} \), subtract the elements of \( \mathbf{A} \) from the corresponding elements of \( \mathbf{B} \):\[\mathbf{B} - \mathbf{A} = \left(\begin{array}{rr} -2 & 6 \ 8 & -10 \end{array}\right) - \left(\begin{array}{rr} 4 & 5 \ -6 & 9 \end{array}\right)\]This results in:\[= \left(\begin{array}{rr} -2 - 4 & 6 - 5 \ 8 - (-6) & -10 - 9 \end{array}\right)\]Calculating, we get:\[= \left(\begin{array}{rr} -6 & 1 \ 14 & -19 \end{array}\right)\]

Step 3: Calculate 2A

To find \( 2\mathbf{A} \), multiply each element of \( \mathbf{A} \) by 2:\[2 \mathbf{A} = 2 \times \left(\begin{array}{rr} 4 & 5 \ -6 & 9 \end{array}\right) = \left(\begin{array}{rr} 8 & 10 \ -12 & 18 \end{array}\right)\]

Step 4: Calculate 3B

To find \( 3\mathbf{B} \), multiply each element of \( \mathbf{B} \) by 3:\[3 \mathbf{B} = 3 \times \left(\begin{array}{rr} -2 & 6 \ 8 & -10 \end{array}\right) = \left(\begin{array}{rr} -6 & 18 \ 24 & -30 \end{array}\right)\]

Step 5: Add 2A and 3B

To find \( 2 \mathbf{A} + 3 \mathbf{B} \), add the matrices from Step 3 and Step 4:\[2 \mathbf{A} + 3 \mathbf{B} = \left(\begin{array}{rr} 8 & 10 \ -12 & 18 \end{array}\right) + \left(\begin{array}{rr} -6 & 18 \ 24 & -30 \end{array}\right)\]This results in:\[= \left(\begin{array}{rr} 8 + (-6) & 10 + 18 \ -12 + 24 & 18 + (-30) \end{array}\right)\]Calculating, we get:\[= \left(\begin{array}{rr} 2 & 28 \ 12 & -12 \end{array}\right)\]

Key Concepts

Matrix Addition

Matrix addition is a fundamental operation that involves combining two matrices to form a new matrix. This operation is only possible when both matrices have the same dimensions. To perform matrix addition, simply add each corresponding element in the matrices.
For example, given matrices \( \mathbf{A} \) and \( \mathbf{B} \) of dimensions 2x2:

• First element of the resulting matrix will be the sum of the first elements of \( \mathbf{A} \) and \( \mathbf{B} \).
• Continue adding corresponding elements until all are covered.

So, if \( \mathbf{A} = \begin{pmatrix} 4 & 5 \ -6 & 9 \end{pmatrix} \) and \( \mathbf{B} = \begin{pmatrix} -2 & 6 \ 8 & -10 \end{pmatrix} \), the sum \( \mathbf{A} + \mathbf{B} = \begin{pmatrix} 2 & 11 \ 2 & -1 \end{pmatrix} \).
This operation is both commutative and associative, meaning the order of addition doesn’t affect the result.

Matrix Subtraction

Matrix subtraction involves finding the difference between two matrices with identical dimensions. The process is similar to matrix addition, but instead of adding, you subtract each corresponding element.
Take matrices \( \mathbf{A} \) and \( \mathbf{B} \) again:

• Subtract the elements of \( \mathbf{A} \) from \( \mathbf{B} \) one by one.
• Ensure precision to avoid errors in each calculation.

For instance, when \( \mathbf{B} = \begin{pmatrix} -2 & 6 \ 8 & -10 \end{pmatrix} \) and \( \mathbf{A} = \begin{pmatrix} 4 & 5 \ -6 & 9 \end{pmatrix} \), their difference \( \mathbf{B} - \mathbf{A} = \begin{pmatrix} -6 & 1 \ 14 & -19 \end{pmatrix} \).
Unlike addition, subtraction isn’t commutative, so the order matters.

Scalar Multiplication

Scalar multiplication involves multiplying every element of a matrix by a single number, known as the scalar. It’s a straightforward operation and doesn’t require matrices to be the same size, as it only involves one matrix.
Completing a scalar multiplication task:

• Take each element of the matrix \( \mathbf{A} \) and multiply it by the scalar value.
• Repeat the process for every element in the matrix.

For example, with \( 2\mathbf{A} \) where \( \mathbf{A} = \begin{pmatrix} 4 & 5 \ -6 & 9 \end{pmatrix} \), multiply each element by 2 to get \( \begin{pmatrix} 8 & 10 \ -12 & 18 \end{pmatrix} \).
Scalar multiplication can help in changing the scale of a matrix’s elements efficiently.

Element-wise Operations

Element-wise operations extend beyond basic matrix addition and subtraction, allowing for different types of element-by-element operations between matrices of the same size. These can be products, quotients, or various functions applied to corresponding elements.
How element-wise operations work:

• If applying multiplication, take each corresponding element of the matrices and multiply them.
• Consider applying functions such as squaring, doubling, or others to each element respectively.

While the original exercise did not include such operations, element-wise manipulations are common in matrix algebra and useful in various applications. These operations are not always commutative and their results are dependent on the shapes and operations defined.
Mastering these will tremendously help in understanding advanced computational tasks including element-wise transformations in data-processing.