Problem 23 - College Algebra

Question

Use the matrices below to perform matrix multiplication. \(A=\left[\begin{array}{rr}-1 & 5 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{rrr}3 & 6 & 4 \\ -8 & 0 & 12\end{array}\right], C=\left[\begin{array}{rr}4 & 10 \\ -2 & 6 \\ 5 & 9\end{array}\right], D=\left[\begin{array}{rrr}2 & -3 & 12 \\ 9 & 3 & 1 \\ 0 & 8 & -10\end{array}\right]\) \(\mathrm{CB}\)

Step-by-Step Solution

Answer

CB = \(\begin{bmatrix} 28 & 24 & -8 \\ -18 & 60 & 112 \\ -57 & 30 & 128 \end{bmatrix}\).

Step 1: Verify Validity for Multiplication

Matrix multiplication is possible when the number of columns in the first matrix is equal to the number of rows in the second matrix. Here, for \( C \), which is a \(3 \times 2\) matrix, and \( B \), which is a \(2 \times 3\) matrix, multiplication is valid because the number of columns in \(C\) (2) matches the number of rows in \(B\) (2).

Step 2: Initialize Result Matrix

The resulting matrix from multiplying a \(3 \times 2\) matrix by a \(2 \times 3\) matrix will be a \(3 \times 3\) matrix. Let's denote the resulting matrix as \( CB \) with elements \( r_{ij} \), where \( i \) is the row index and \( j \) is the column index.

Step 3: Calculate the Elements of the First Row of CB

For each element of the first row of \( CB \):- Compute \( r_{11} = 4 \times 3 + (-2) \times (-8) = 12 + 16 = 28 \)- Compute \( r_{12} = 4 \times 6 + (-2) \times 0 = 24 + 0 = 24 \)- Compute \( r_{13} = 4 \times 4 + (-2) \times 12 = 16 + (-24) = -8 \).

Step 4: Calculate the Elements of the Second Row of CB

For each element of the second row of \( CB \):- Compute \( r_{21} = 10 \times 3 + 6 \times (-8) = 30 + (-48) = -18 \)- Compute \( r_{22} = 10 \times 6 + 6 \times 0 = 60 + 0 = 60 \)- Compute \( r_{23} = 10 \times 4 + 6 \times 12 = 40 + 72 = 112 \).

Step 5: Calculate the Elements of the Third Row of CB

For each element of the third row of \( CB \):- Compute \( r_{31} = 5 \times 3 + 9 \times (-8) = 15 + (-72) = -57 \)- Compute \( r_{32} = 5 \times 6 + 9 \times 0 = 30 + 0 = 30 \)- Compute \( r_{33} = 5 \times 4 + 9 \times 12 = 20 + 108 = 128 \).

Step 6: Assemble the Resulting Matrix

Combine the calculated rows to form the resulting matrix:\( CB = \begin{bmatrix} 28 & 24 & -8 \ -18 & 60 & 112 \ -57 & 30 & 128 \end{bmatrix} \).

Key Concepts

Matrix Multiplication Steps

To multiply matrices, we need to follow specific steps to ensure the process is done correctly and efficiently. The first step involves verifying if the multiplication is valid. Matrix multiplication is only possible when the number of columns in the first matrix equals the number of rows in the second matrix.
For example, matrix \( C \) is a \( 3 \times 2 \) matrix and matrix \( B \) is a \( 2 \times 3 \) matrix. Their multiplication is valid because the number of columns in \( C \) matches the number of rows in \( B \). This ensures that each element of the resulting matrix can be calculated.

The second step is to determine the size of the resulting matrix. For \( C \) and \( B \), their resulting matrix \( CB \) will have dimensions \( 3 \times 3 \) as the result of multiplying a \( 3 \times 2 \) matrix by a \( 2 \times 3 \) matrix.
Next, we initialize the result matrix, which will be filled step by step with calculated values based on multiplication and addition rules. This sets the stage for calculating each individual element of the resulting matrix. By following this structured approach, matrix multiplication becomes predictable and methodical.

Matrix Operations

Matrix operations such as multiplication involve combining rows and columns systematically to find the elements of the resulting matrix. Each element in the resulting matrix is found by performing a dot product of the relevant row from the first matrix and the appropriate column from the second matrix.
Let's break down this concept with an easy example step.

Say we want to find the element at the first row and first column (\( r_{11} \)) of \( CB \). We take the first row of \( C \) and the first column of \( B \), then multiply corresponding elements and sum them up:

• Multiply the first element of the first row of \( C \) by the first element of the first column of \( B \).
• Repeat for the second element of the row and column pair, and add the results.

This operation is applied consistently across the elements of both matrices.
This process is repeated for each element in the resulting matrix, allowing to complete the multiplication effectively.

Resulting Matrix Calculation

The resulting matrix, born from our matrix multiplication, is a comprehensive summary of the operations carried out. It models a new matrix where each entry signifies the cumulative impact of the computations. As we've seen, each element of the resulting matrix \( CB \) was calculated by combining corresponding elements from matrices \( C \) and \( B \) with specific operations.
For example, the element \( r_{11} = 28 \) of the resulting matrix is the sum of products \( 4 \times 3 \) and \((-2) \times (-8)\).

This kind of matrix calculation results in:

• \( r_{11} = 28 \)
• \( r_{12} = 24 \)
• \( r_{13} = -8 \)
• \( r_{21} = -18 \)
• \( r_{22} = 60 \)
• \( r_{23} = 112 \)
• \( r_{31} = -57 \)
• \( r_{32} = 30 \)
• \( r_{33} = 128 \)

Each calculation transforms and distills information from the original matrices into the resultant matrix, offering a new perspective or a more compact representation. Understanding this is key to grasping how outcomes of matrix multiplications contribute to broader mathematical problems.