Problem 25
Question
In Problems 25-28, write the given sum as a single-column matrix. $$ 4\left(\begin{array}{r} -1 \\ 2 \end{array}\right)-2\left(\begin{array}{l} 2 \\ 8 \end{array}\right)+3\left(\begin{array}{r} -2 \\ 3 \end{array}\right) $$
Step-by-Step Solution
Verified Answer
\( \begin{array}{r} -14 \\ 1 \end{array} \)
1Step 1: Distribute the Scalars
Begin by multiplying each scalar with its respective matrix. For the first matrix, multiply 4 by each entry:\[4 \left( \begin{array}{r} -1 \ 2 \end{array} \right) = \begin{array}{r} 4(-1) \ 4(2) \end{array} = \begin{array}{r} -4 \ 8 \end{array} \]For the second matrix, multiply -2 by each entry:\[-2 \left( \begin{array}{l} 2 \ 8 \end{array} \right) = \begin{array}{r} -2(2) \ -2(8) \end{array} = \begin{array}{r} -4 \ -16 \end{array} \]For the third matrix, multiply 3 by each entry:\[3 \left( \begin{array}{r} -2 \ 3 \end{array} \right) = \begin{array}{r} 3(-2) \ 3(3) \end{array} = \begin{array}{r} -6 \ 9 \end{array} \]
2Step 2: Addition of the Matrices
Add the resulting matrices together by summing the corresponding entries:\[\left( \begin{array}{r} -4 \ 8 \end{array} \right) + \left( \begin{array}{r} -4 \ -16 \end{array} \right) + \left( \begin{array}{r} -6 \ 9 \end{array} \right) = \begin{array}{r} -4 + (-4) + (-6) \ 8 + (-16) + 9 \end{array}\]Calculate each entry:\[= \begin{array}{r} -4 - 4 - 6 \ 8 - 16 + 9 \end{array} = \begin{array}{r} -14 \ 1 \end{array} \]
3Step 3: Write the Final Column Matrix
The final result is a single column matrix:\[\begin{array}{r} -14 \ 1 \end{array}\] This matrix represents the sum of the original expression after distribution and addition.
Key Concepts
Scalar MultiplicationMatrix AdditionColumn MatrixLinear Algebra
Scalar Multiplication
Scalar multiplication in linear algebra involves multiplying a matrix by a scalar (a single number). Each element of the matrix is scaled by the value of the scalar. In this exercise, we multiply scalars with the matrices:
- The scalar 4 is multiplied with the matrix \(\begin{array}{c} -1 \ 2 \end{array}\) resulting in \(\begin{array}{c} -4 \ 8 \end{array}\).
- The scalar -2 is applied to the matrix \(\begin{array}{c} 2 \ 8 \end{array}\), giving \(\begin{array}{c} -4 \ -16 \end{array}\).
- Lastly, the scalar 3 is distributed to the matrix \(\begin{array}{c} -2 \ 3 \end{array}\), resulting in \(\begin{array}{c} -6 \ 9 \end{array}\).
Matrix Addition
Matrix addition requires adding corresponding elements of two or more matrices. For this, all matrices involved must have the same dimensions (same number of rows and columns).In the given exercise, we added matrices after performing scalar multiplication, shown as:
- For the first element: \(-4 + (-4) + (-6) = -14\).
- For the second element: \(8 + (-16) + 9 = 1\).
Column Matrix
A column matrix (or column vector) is a matrix with only one column. Despite having just one column, a column matrix can have one or more rows, meaning it stacks numerical values vertically.In the exercise, we worked out all calculations to eventually express the result as a column matrix:
- The final column matrix derived was \(\begin{array}{c} -14 \ 1 \end{array}\).
Linear Algebra
Linear algebra is a branch of mathematics that deals with vectors and matrices, and the operations performed on them, such as addition, multiplication, and transformation.
Key components learned through this sort of problem include:
- Understanding how to manipulate matrices with basic arithmetic operations.
- The skill of reading and interpreting matrix notation accurately.
- The application of these skills in computational algorithms and data analysis.
Other exercises in this chapter
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