Problem 6
Question
For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ 2 B+3 D $$
Step-by-Step Solution
Verified Answer
\(2B + 3D = \begin{bmatrix} -2 & 3 \\ 25 & -14 \end{bmatrix}\)
1Step 1: Scalar Multiplication of Matrix B
First, we need to multiply matrix \(B\) by the scalar \(2\). Perform this operation element-wise across all elements of matrix \(B\): \[B = \begin{bmatrix} 2 & -3 \ 5 & -1 \end{bmatrix}\]Multiply each element by \(2\): \[2B = \begin{bmatrix} 2 \times 2 & -3 \times 2 \ 5 \times 2 & -1 \times 2 \end{bmatrix} = \begin{bmatrix} 4 & -6 \ 10 & -2 \end{bmatrix}\]
2Step 2: Scalar Multiplication of Matrix D
Next, we multiply matrix \(D\) by the scalar \(3\). Perform this operation element-wise across all elements of matrix \(D\): \[D = \begin{bmatrix} -2 & 3 \ 5 & -4 \end{bmatrix}\]Multiply each element by \(3\): \[3D = \begin{bmatrix} 3 \times (-2) & 3 \times 3 \ 3 \times 5 & 3 \times (-4) \end{bmatrix} = \begin{bmatrix} -6 & 9 \ 15 & -12 \end{bmatrix}\]
3Step 3: Addition of the Two Resultant Matrices
Finally, add the resultant matrices from Step 1 (\(2B\)) and Step 2 (\(3D\)) together element-wise: \[2B = \begin{bmatrix} 4 & -6 \ 10 & -2 \end{bmatrix}\] \[3D = \begin{bmatrix} -6 & 9 \ 15 & -12 \end{bmatrix}\]Add them element-wise:\[2B + 3D = \begin{bmatrix} 4 + (-6) & -6 + 9 \ 10 + 15 & -2 + (-12) \end{bmatrix} = \begin{bmatrix} -2 & 3 \ 25 & -14 \end{bmatrix}\]
Key Concepts
Scalar MultiplicationMatrix AdditionElement-wise Operations
Scalar Multiplication
Scalar multiplication is one of the fundamental operations performed on matrices. It involves multiplying every element of a matrix by a constant value, called a scalar. This operation modifies the scale of the matrix while maintaining its original structure.
When multiplying a matrix by a scalar, each element in the matrix is scaled uniformly. So a matrix with dimensions of 2x2, like matrix \(B\) in our exercise, is modified by multiplying each of its elements by the scalar, which in this case is 2:
\[ \begin{bmatrix} 2 & -3 \ 5 & -1 \end{bmatrix} \times 2 = \begin{bmatrix} 4 & -6 \ 10 & -2 \end{bmatrix} \]
This operation retains the original matrix size, but alters each individual value by the scalar multiple.
When multiplying a matrix by a scalar, each element in the matrix is scaled uniformly. So a matrix with dimensions of 2x2, like matrix \(B\) in our exercise, is modified by multiplying each of its elements by the scalar, which in this case is 2:
\[ \begin{bmatrix} 2 & -3 \ 5 & -1 \end{bmatrix} \times 2 = \begin{bmatrix} 4 & -6 \ 10 & -2 \end{bmatrix} \]
This operation retains the original matrix size, but alters each individual value by the scalar multiple.
- Scalars can be positive, negative, or even zero, affecting the resulting matrix similarly.
- Is scalar multiplication commutative? Yes, but only because matrices and scalars follow standard constants' rules.
- It's essentially multiplying the matrix by a number and it scales the entire matrix while keeping the relative proportions intact.
Matrix Addition
Matrix addition is another crucial operation that involves adding two matrices together. However, note that this is only possible when both matrices have the same dimensions. Each element of the first matrix is added to the corresponding element of the second matrix.
In the original problem, after performing scalar multiplication on matrices \(B\) and \(D\), their resulting matrices \(2B\) and \(3D\) are added together, element-wise.
Given:
\[2B + 3D = \begin{bmatrix} 4 + (-6) & -6 + 9 \ 10 + 15 & -2 + (-12) \end{bmatrix} = \begin{bmatrix} -2 & 3 \ 25 & -14 \end{bmatrix}\]
Points to remember:
In the original problem, after performing scalar multiplication on matrices \(B\) and \(D\), their resulting matrices \(2B\) and \(3D\) are added together, element-wise.
Given:
- \(2B = \begin{bmatrix} 4 & -6 \ 10 & -2 \end{bmatrix}\)
- \(3D = \begin{bmatrix} -6 & 9 \ 15 & -12 \end{bmatrix}\)
\[2B + 3D = \begin{bmatrix} 4 + (-6) & -6 + 9 \ 10 + 15 & -2 + (-12) \end{bmatrix} = \begin{bmatrix} -2 & 3 \ 25 & -14 \end{bmatrix}\]
Points to remember:
- Ensure the matrices are of the same size for this operation.
- In matrix addition, the order in which you add doesn’t matter; it’s commutative.
- This operation relies on corresponding positions in each matrix.
Element-wise Operations
Element-wise operations refer to operations that apply to each corresponding element of two matrices. This concept is pivotal when dealing with operations like addition or multiplication that apply the operation directly to individual pairs of numbers from two or more given matrices.
Both scalar multiplication and matrix addition are performed as element-wise operations in the given exercise. For example, with the matrix addition of \(2B\) and \(3D\), each number from one matrix is directly added to the number directly across from it in another matrix.
Element-wise operations are essential:
Both scalar multiplication and matrix addition are performed as element-wise operations in the given exercise. For example, with the matrix addition of \(2B\) and \(3D\), each number from one matrix is directly added to the number directly across from it in another matrix.
Element-wise operations are essential:
- They allow operations like addition and multiplication while retaining each element’s relative position.
- These operations are directly aligned with simple numerical operations but are adapted for the matrix format.
- You must ensure the matrices involved are the same shape, otherwise these operations can't be performed.
Other exercises in this chapter
Problem 6
For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rr} 7 & -4 \\ -5 & 9 \\ -1 & 2 \end{array}\right], \quad B=\left[\be
View solution Problem 6
For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} 1 & -2 \\ 4 & -3 \end{array}\right] $$
View solution Problem 6
For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{arra
View solution Problem 7
For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rr} -1 & 0 \\ 2 & 3 \\ -5 & -4 \\ -7 & 11 \end{array}\right], \quad
View solution