Problem 28
Question
In Problems 25-28, write the given sum as a single-column matrix. $$ \left(\begin{array}{rrr} 1 & -3 & 4 \\ 2 & 5 & -1 \\ 0 & -4 & -2 \end{array}\right)\left(\begin{array}{r} 3 \\ 2 \\ -1 \end{array}\right)+\left(\begin{array}{r} -1 \\ 1 \\ 4 \end{array}\right)-\left(\begin{array}{r} 2 \\ 8 \\ -6 \end{array}\right) $$
Step-by-Step Solution
Verified Answer
The single-column matrix is \(\begin{pmatrix} -10 \\ 10 \\ 4 \end{pmatrix}\).
1Step 1: Matrix Multiplication
First, perform the matrix multiplication of the 3x3 matrix with the 3x1 column vector. Multiply each element of the row of the 3x3 matrix by the corresponding element of the column vector and add them.- First row: \(1 \times 3 + (-3) \times 2 + 4 \times (-1) = 3 - 6 - 4 = -7\)- Second row: \(2 \times 3 + 5 \times 2 + (-1) \times (-1) = 6 + 10 + 1 = 17\)- Third row: \(0 \times 3 + (-4) \times 2 + (-2) \times (-1) = 0 - 8 + 2 = -6\)So, the result of the matrix multiplication is:\[\begin{pmatrix} -7 \ 17 \ -6 \end{pmatrix}\]
2Step 2: Addition of the First Column Matrix
Next, add the result from the matrix multiplication to the given column matrix \(\begin{pmatrix} -1 \ 1 \ 4 \end{pmatrix}\).- Result: \(\begin{pmatrix} -7 \ 17 \ -6 \end{pmatrix} + \begin{pmatrix} -1 \ 1 \ 4 \end{pmatrix} = \begin{pmatrix} -7 + (-1) \ 17 + 1 \ -6 + 4 \end{pmatrix} = \begin{pmatrix} -8 \ 18 \ -2 \end{pmatrix}\)
3Step 3: Subtraction of the Second Column Matrix
Finally, subtract the column matrix \(\begin{pmatrix} 2 \ 8 \ -6 \end{pmatrix}\) from the result of the previous addition.- Result: \(\begin{pmatrix} -8 \ 18 \ -2 \end{pmatrix} - \begin{pmatrix} 2 \ 8 \ -6 \end{pmatrix} = \begin{pmatrix} -8 - 2 \ 18 - 8 \ -2 - (-6) \end{pmatrix} = \begin{pmatrix} -10 \ 10 \ 4 \end{pmatrix}\)
4Step 4: Final Result
The sum of the given expressions as a single column matrix is:\[\begin{pmatrix} -10 \ 10 \ 4 \end{pmatrix}\]
Key Concepts
Matrix MultiplicationColumn MatrixMatrix AdditionMatrix Subtraction
Matrix Multiplication
Matrix multiplication is an essential concept in linear algebra. It involves multiplying rows from the first matrix with columns from the second matrix. The fundamental rule is that the number of columns in the first matrix must match the number of rows in the second matrix. In our example, we are multiplying a 3x3 matrix by a 3x1 matrix (also known as a column vector), which is a valid operation.
To multiply these matrices, follow these steps for each row of the first matrix:
To multiply these matrices, follow these steps for each row of the first matrix:
- Multiply the corresponding elements from the row of the first matrix and the column of the second matrix.
- Add all the products together to get the element in the resulting matrix.
- The first row of the 3x3 matrix is multiplied by the column vector to produce the first element of the result.
- This process is repeated for each row.
Column Matrix
A column matrix is simply a matrix with only one column. It's often called a vector. The primary characteristic is its vertical orientation. In general, you represent a column matrix with size nx1, where n is the number of rows.
Column matrices are particularly useful as they can stand for systems of linear equations. They are also convenient for performing operations like matrix addition or matrix multiplication with other matrices.
In our exercise, both column matrices and a result from multiplication are used in matrix addition and subtraction. This manipulation helps us easily manage operations that require matrices of equal dimensions.
Column matrices are particularly useful as they can stand for systems of linear equations. They are also convenient for performing operations like matrix addition or matrix multiplication with other matrices.
In our exercise, both column matrices and a result from multiplication are used in matrix addition and subtraction. This manipulation helps us easily manage operations that require matrices of equal dimensions.
Matrix Addition
Matrix addition is a straightforward operation involving two matrices of the same dimensions. Each element in one matrix is added to the corresponding element in the other matrix. This component-wise operation results in a new matrix, maintaining the same dimensions as the original matrices.
For matrix addition to be possible, two matrices must have the same dimensions. In our example, after performing matrix multiplication, we added a 3x1 column matrix to another 3x1 column matrix. The steps for addition are:
For matrix addition to be possible, two matrices must have the same dimensions. In our example, after performing matrix multiplication, we added a 3x1 column matrix to another 3x1 column matrix. The steps for addition are:
- Add corresponding elements from each matrix to form a new matrix.
- Ensure the resulting matrix maintains the same dimensions.
Matrix Subtraction
Matrix subtraction works similarly to matrix addition, with the primary difference being that instead of adding, you subtract corresponding elements. As with addition, lawful matrix subtraction requires the matrices involved to have the same dimensions. In our example:
- We subtracted a 3x1 column matrix from another 3x1 column matrix.
- This was done after an addition operation to find the resulting column matrix.
- Subtract each corresponding element of the second matrix from the first.
- Preserve the dimensions of the original matrices in the resultant matrix.
Other exercises in this chapter
Problem 28
In Problems 27 and 28 , use the given matrices to find \((\mathbf{A B})^{-1}\). $$ \mathbf{A}^{-1}=\left(\begin{array}{rrr} 1 & 3 & -15 \\ 0 & -1 & 5 \\ -1 & -2
View solution Problem 28
In Problems 15-28, evaluate the determinant of the given matrix by cofactor expansion. $$ \left(\begin{array}{rrrrr} 2 & 2 & 0 & 0 & -2 \\ 1 & 1 & 6 & 0 & 5 \\
View solution Problem 29
In Problems, the given matrix \(\mathbf{A}\) is symmetric. Find an orthogonal matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\
View solution Problem 29
Determine the size of the matrix \(\mathbf{A}\) such that the given product is defined. $$ \left(\begin{array}{llll} 2 & 1 & 3 & 3 \\ 9 & 6 & 7 & 0 \end{array}\
View solution