Problem 17
Question
For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{r} 2 \\ -7 \end{array}\right], \quad B=\left[\begin{array}{rr} 3 & -2 \\ 1 & 0 \\ -1 & 4 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
\(AB\) does not exist; \(BA = \begin{bmatrix} 20 \\ 2 \\ -30 \end{bmatrix}\).
1Step 1: Determine if Product AB Exists
To multiply two matrices, the number of columns in the first matrix must equal the number of rows in the second matrix. Matrix \(A\) is \(2 \times 1\) and matrix \(B\) is \(3 \times 2\). Since the number of columns in \(A\) (1) does not equal the number of rows in \(B\) (3), the product \(AB\) does not exist.
2Step 2: Determine if Product BA Exists
Similarly, for the product \(BA\) to exist, the number of columns in \(B\) (2) must equal the number of rows in \(A\) (2). Since these numbers are equal, the product \(BA\) does exist, and will result in a \(3 \times 1\) matrix.
3Step 3: Calculate the Product BA
Matrix \(B\) is \(3 \times 2\), and matrix \(A\) is \(2 \times 1\). The product \(BA\) is calculated as follows:\[BA = \begin{bmatrix} 3 & -2 \ 1 & 0 \ -1 & 4 \end{bmatrix} \begin{bmatrix} 2 \ -7 \end{bmatrix} = \begin{bmatrix} (3)(2) + (-2)(-7) \ (1)(2) + (0)(-7) \ (-1)(2) + (4)(-7) \end{bmatrix} \]Simplifying each row, this becomes:\[\begin{bmatrix} 6 + 14 \ 2 + 0 \ -2 - 28 \end{bmatrix} = \begin{bmatrix} 20 \ 2 \ -30 \end{bmatrix}\]
4Step 4: Present the Solution
The product \(AB\) does not exist due to incompatible dimensions, but the product \(BA\) does exist and is equal to the matrix:\[\begin{bmatrix} 20 \ 2 \ -30 \end{bmatrix}\]
Key Concepts
Matrix DimensionsExistence of Matrix ProductMatrix Calculation Steps
Matrix Dimensions
Understanding matrix dimensions is crucial when working with matrix multiplication. A matrix's dimensions are denoted as "rows by columns". For example, a matrix with 2 rows and 1 column is called a 2 \( \times \) 1 matrix. Similarly, a 3 \( \times \) 2 matrix has 3 rows and 2 columns.
Before performing any multiplication, it's important to check the dimensions of both matrices involved. Only some matrices can be multiplied together - they must be compatible in a specific way:
Before performing any multiplication, it's important to check the dimensions of both matrices involved. Only some matrices can be multiplied together - they must be compatible in a specific way:
- The number of columns in the first matrix must match the number of rows in the second matrix.
Existence of Matrix Product
Not all matrix products are possible, and the key factor is compatibility between their dimensions. Let's break this down with the provided exercise. Matrix \( A \) has dimensions of 2 \( \times \) 1, and matrix \( B \) has dimensions of 3 \( \times \) 2.
For the product \( AB \) to exist, the number of columns in \( A \) (which is 1) must equal the number of rows in \( B \) (which is 3).
Since these dimensions do not match, \( AB \) cannot be calculated.
Conversely, for the product \( BA \), the number of columns in \( B \) (2) must match the number of rows in \( A \) (2). These dimensions are equal, so the product \( BA \) is possible and results in a matrix of dimensions 3 \( \times \) 1.
For the product \( AB \) to exist, the number of columns in \( A \) (which is 1) must equal the number of rows in \( B \) (which is 3).
Since these dimensions do not match, \( AB \) cannot be calculated.
Conversely, for the product \( BA \), the number of columns in \( B \) (2) must match the number of rows in \( A \) (2). These dimensions are equal, so the product \( BA \) is possible and results in a matrix of dimensions 3 \( \times \) 1.
Matrix Calculation Steps
To multiply matrices and obtain the product, follow a step-by-step approach. We will calculate the product \( BA \) based on the given matrices in the exercise:
1. **Identify the Compatible Pairs**: Matrix \( B \) is 3 \( \times \) 2 and matrix \( A \) is 2 \( \times \) 1. Since their respective dimensions align for multiplication, \( BA \) can be calculated and will yield a 3 \( \times \) 1 matrix.
2. **Multiply Each Element Appropriately**: Each row element of \( B \) is multiplied by each column element of \( A \) and summed up:
\[\begin{bmatrix} 20 \ 2 \ -30 \end{bmatrix}\]
Understanding these steps is key to mastering matrix multiplication and confidently determining the existence and result of a matrix product.
1. **Identify the Compatible Pairs**: Matrix \( B \) is 3 \( \times \) 2 and matrix \( A \) is 2 \( \times \) 1. Since their respective dimensions align for multiplication, \( BA \) can be calculated and will yield a 3 \( \times \) 1 matrix.
2. **Multiply Each Element Appropriately**: Each row element of \( B \) is multiplied by each column element of \( A \) and summed up:
- First Row: \((3)(2) + (-2)(-7) = 6 + 14 = 20\)
- Second Row: \((1)(2) + (0)(-7) = 2 + 0 = 2\)
- Third Row: \((-1)(2) + (4)(-7) = -2 - 28 = -30\)
\[\begin{bmatrix} 20 \ 2 \ -30 \end{bmatrix}\]
Understanding these steps is key to mastering matrix multiplication and confidently determining the existence and result of a matrix product.
Other exercises in this chapter
Problem 16
For Problems \(13-26\), compute \(A B\) and \(B A\). $$ A=\left[\begin{array}{rr} 5 & 0 \\ -2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rr} -3 & 6 \\
View solution Problem 16
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 17
For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} 1 & 1 \\ 1 & -1 \end{array}\right] $$
View solution Problem 17
For Problems \(13-26\), compute \(A B\) and \(B A\). $$ A=\left[\begin{array}{ll} 2 & -4 \\ 1 & -2 \end{array}\right], \quad B=\left[\begin{array}{rr} 1 & -2 \\
View solution