Problem 14
Question
For the following exercises, use the matrices below to perform scalar multiplication. $$ A=\left[\begin{array}{cc}{4} & {6} \\ {13} & {12}\end{array}\right], B=\left[\begin{array}{cc}{3} & {9} \\ {21} & {12} \\ {0} & {64}\end{array}\right], C=\left[\begin{array}{cccc}{16} & {3} & {7} & {18} \\\ {90} & {5} & {3} & {29}\end{array}\right], D=\left[\begin{array}{ccc}{18} & {12} & {13} \\ {8} & {14} & {6} \\ {7} & {4} & {21}\end{array}\right] $$ $$ -2 B $$
Step-by-Step Solution
Verified Answer
The resulting matrix after multiplying matrix B by -2 is \(\begin{bmatrix} -6 & -18 \\ -42 & -24 \\ 0 & -128 \end{bmatrix}\).
1Step 1: Understand Scalar Multiplication
Scalar multiplication involves multiplying every entry in the matrix by a scalar (constant). For this exercise, the scalar is \(-2\) and the matrix to be multiplied is \(B\).
2Step 2: Write Original Matrix B
Matrix \(B\) is given as: \[B = \begin{bmatrix} 3 & 9 \ 21 & 12 \ 0 & 64 \end{bmatrix}\]
3Step 3: Multiply Each Element by Scalar -2
Multiply each element of matrix \(B\) by \(-2\):- First row, first column: \(3 \times (-2) = -6\)- First row, second column: \(9 \times (-2) = -18\)- Second row, first column: \(21 \times (-2) = -42\)- Second row, second column: \(12 \times (-2) = -24\)- Third row, first column: \(0 \times (-2) = 0\)- Third row, second column: \(64 \times (-2) = -128\)
4Step 4: Write the Resulting Matrix
The matrix after scalar multiplication by \(-2\) is: \[-2B = \begin{bmatrix} -6 & -18 \ -42 & -24 \ 0 & -128 \end{bmatrix}\]
Key Concepts
Matrices and Their StructureScalar Multiplication of a MatrixBasic Matrix Operations
Matrices and Their Structure
A matrix is a rectangular grid of numbers arranged in rows and columns. Each number in a matrix is called an element or entry. Matrices are a foundational aspect of linear algebra and are used to solve systems of linear equations, among many other applications in various fields such as physics, computer science, and economics.
Matrices can be any size, from 1x1 to any larger dimensions, such as 2x2, 3x3, or even larger. The size of a matrix is defined by the number of rows and columns it has. For example, a matrix with two rows and three columns is a 2x3 matrix. Knowing this structure helps in carrying out different operations, such as addition, subtraction, and multiplication, involving matrices.
Matrices can be any size, from 1x1 to any larger dimensions, such as 2x2, 3x3, or even larger. The size of a matrix is defined by the number of rows and columns it has. For example, a matrix with two rows and three columns is a 2x3 matrix. Knowing this structure helps in carrying out different operations, such as addition, subtraction, and multiplication, involving matrices.
- The horizontal arrays of elements are called rows.
- The vertical arrays of elements are called columns.
- An individual element is usually defined by its row and column position, using an index.
Scalar Multiplication of a Matrix
Scalar multiplication involves multiplying every element of a matrix by a single number, known as a scalar. This operation is fundamental and helps in modifying the scaling and dimensions within matrix operations. Let's delve into this basic yet powerful tool in linear algebra.
In scalar multiplication, you take each element of the matrix and multiply it by the scalar. This is simpler than it might sound! For example, if our scalar is \(-2\), each element of the matrix \(B\) would be multiplied by \(-2\). This means that every number within the matrix becomes \(-2\) times its original value.
In scalar multiplication, you take each element of the matrix and multiply it by the scalar. This is simpler than it might sound! For example, if our scalar is \(-2\), each element of the matrix \(B\) would be multiplied by \(-2\). This means that every number within the matrix becomes \(-2\) times its original value.
- Start by identifying the matrix you wish to multiply.
- Multiply each entry individually by the scalar value.
- The resulting matrix will have the same dimensions as the original matrix.
Basic Matrix Operations
Matrix operations are fundamental tools in algebra that include addition, subtraction, and multiplication (both scalar and matrix multiplication). They allow for the transformation and manipulation of data represented in matrix form. These operations are essential for both theoretical mathematical developments and practical applications, such as in computer graphics, engineering, and more.
Here, we focus on the contexts required to understand scalar multiplication, yet similar insights will apply when dealing with other matrix operations. Matrix operations require matrices to have certain dimensions so that they can be combined effectively:
Here, we focus on the contexts required to understand scalar multiplication, yet similar insights will apply when dealing with other matrix operations. Matrix operations require matrices to have certain dimensions so that they can be combined effectively:
- Addition/Subtraction: Operate element-wise, meaning each element in one matrix is added or subtracted from the respective element in another matrix. For this to happen, the matrices must have the same dimensions.
- Scalar Multiplication: As covered, involves changing the scale of each element by multiplying them all by a constant number.
- Matrix Multiplication: Requires more complex rules, such as the number of columns in the first matrix needing to equal the number of rows in the second matrix. This operation results in a new matrix with different dimensions.
Other exercises in this chapter
Problem 14
Solve the system of nonlinear equations using elimination. $$ \begin{array}{r} y^{2}-x^{2}=9 \\ 3 x^{2}+2 y^{2}=8 \end{array} $$
View solution Problem 14
Use the matrices below to perform scalar multiplication. \(A=\left[\begin{array}{rr}4 & 6 \\ 13 & 12\end{array}\right], B=\left[\begin{array}{rr}3 & 9 \\ 21 & 1
View solution Problem 14
Solve each system by substitution. $$ \begin{aligned} 4 x-3 y+5 z &=31 \\ -x+2 y+4 z &=20 \\ x+5 y-2 z &=-29 \end{aligned} $$
View solution Problem 14
Find the decomposition of the partial fraction for the nonrepeating linear factors. \(\frac{10 x}{x^{2}-25}\)
View solution