Problem 29 - Precalculus: Functions and Graphs

Question

Let $$A=\left[\begin{array}{rr}1 & 2 \\\0 & -3\end{array}\right], \quad B=\left[\begin{array}{rr}2 & -1 \\\3 & 1\end{array}\right], \quad C=\left[\begin{array}{rr}3 & 1 \\\\-2 & 0\end{array}\right]$$ Verify the statement. $$A(B+C)=A B+A C$$

Step-by-Step Solution

Answer

The distributive property is verified: \(A(B+C) = AB + AC\).

Step 1: Calculate Matrix B+C

To verify the distributive property, we'll first calculate the sum of matrices B and C. To add matrices, add corresponding elements together.\[B + C = \begin{bmatrix} 2 & -1 \ 3 & 1 \end{bmatrix} + \begin{bmatrix} 3 & 1 \ -2 & 0 \end{bmatrix} = \begin{bmatrix} 2+3 & -1+1 \ 3-2 & 1+0 \end{bmatrix} = \begin{bmatrix} 5 & 0 \ 1 & 1 \end{bmatrix}\]

Step 2: Calculate A(B+C)

Now let's multiply matrix A with the matrix we found, B+C. Use matrix multiplication rules: the element in the ith row and jth column of the product is the dot product of the ith row of the first matrix and the jth column of the second matrix.\[A(B+C) = \begin{bmatrix} 1 & 2 \ 0 & -3 \end{bmatrix} \begin{bmatrix} 5 & 0 \ 1 & 1 \end{bmatrix}\]Calculating the product:- First row, first column: \((1 \times 5) + (2 \times 1) = 5 + 2 = 7\)- First row, second column: \((1 \times 0) + (2 \times 1) = 0 + 2 = 2\)- Second row, first column: \((0 \times 5) + (-3 \times 1) = 0 - 3 = -3\)- Second row, second column: \((0 \times 0) + (-3 \times 1) = 0 - 3 = -3\)So, \[A(B+C) = \begin{bmatrix} 7 & 2 \ -3 & -3 \end{bmatrix}\].

Step 3: Calculate AB

Next, let's calculate the product of matrices A and B.\[AB = \begin{bmatrix} 1 & 2 \ 0 & -3 \end{bmatrix} \begin{bmatrix} 2 & -1 \ 3 & 1 \end{bmatrix}\]Calculating the product:- First row, first column: \((1 \times 2) + (2 \times 3) = 2 + 6 = 8\)- First row, second column: \((1 \times -1) + (2 \times 1) = -1 + 2 = 1\)- Second row, first column: \((0 \times 2) + (-3 \times 3) = 0 - 9 = -9\)- Second row, second column: \((0 \times -1) + (-3 \times 1) = 0 - 3 = -3\)So, \[AB = \begin{bmatrix} 8 & 1 \ -9 & -3 \end{bmatrix}\].

Step 4: Calculate AC

Now let's find the product of matrices A and C.\[AC = \begin{bmatrix} 1 & 2 \ 0 & -3 \end{bmatrix} \begin{bmatrix} 3 & 1 \ -2 & 0 \end{bmatrix}\]Calculating the product:- First row, first column: \((1 \times 3) + (2 \times -2) = 3 - 4 = -1\)- First row, second column: \((1 \times 1) + (2 \times 0) = 1 + 0 = 1\)- Second row, first column: \((0 \times 3) + (-3 \times -2) = 0 + 6 = 6\)- Second row, second column: \((0 \times 1) + (-3 \times 0) = 0 + 0 = 0\)So, \[AC = \begin{bmatrix} -1 & 1 \ 6 & 0 \end{bmatrix}\].

Step 5: Calculate AB+AC

Finally, add the matrices AB and AC that we found in the previous steps.\[AB + AC = \begin{bmatrix} 8 & 1 \ -9 & -3 \end{bmatrix} + \begin{bmatrix} -1 & 1 \ 6 & 0 \end{bmatrix} = \begin{bmatrix} 8 - 1 & 1 + 1 \ -9 + 6 & -3 + 0 \end{bmatrix} = \begin{bmatrix} 7 & 2 \ -3 & -3 \end{bmatrix}\]

Step 6: Verify Equality

We found that \[A(B+C) = \begin{bmatrix} 7 & 2 \ -3 & -3 \end{bmatrix}\] and \[AB + AC = \begin{bmatrix} 7 & 2 \ -3 & -3 \end{bmatrix}\]. Since the two results are identical, the distributive property holds: \(A(B+C) = AB + AC\).

Key Concepts

Distributive Property of Matrices

The distributive property is a fundamental algebraic principle that also applies to matrices. It states that for any matrices \(A\), \(B\), and \(C\), the following equality holds: \[ A(B + C) = AB + AC \] This is similar to how the distributive property works with real numbers. Here's how it breaks down:

• Matrix Addition: First, add matrices \(B\) and \(C\) by adding their respective elements. This is similar to adding numbers.
• Multiplication: Multiply the resulting matrix from the addition by matrix \(A\). Do this element by element, following the specific rules for multiplying matrices.
• Compare Results: Then, separately multiply \(A\) by \(B\) and \(A\) by \(C\), and add the two products. If the first result matches this second one, the distributive property is verified.

The distributive property simplifies complex matrix calculations, making it easier to manipulate and solve matrix problems efficiently. It allows us to distribute multiplication over addition in much the same way as with regular numbers.

Matrix Addition

Matrix addition is a straightforward process where you combine two matrices by adding their corresponding elements. It requires that both matrices have the same dimensions, meaning the same number of rows and columns. Each element from one matrix is added to the corresponding element of the other. For example, adding matrices \(B\) and \(C\) involves:

• Add the first element of the first row of \(B\) with the first element of the first row of \(C\).
• Continue this process for the entire matrix.

The resulting matrix, \(B + C\), is composed of these sums added element-wise. This process is essential when using the distributive property, as it forms the starting point for evaluating expressions like \(A(B + C)\). Matrix addition is not only simple but also essential for many applications in mathematics, physics, and engineering.

Matrix Calculations

Performing matrix calculations involves understanding and applying various operations such as matrix addition, subtraction, multiplication, and more. Each operation has specific rules:

• Matrix Multiplication: Involves taking the dot product of the rows of the first matrix and the columns of the second. It is important to ensure that the number of columns in the first matrix equals the number of rows in the second.
• Element Positioning: The resulting matrix's element is determined by the row number of the first matrix and the column number of the second matrix.
• Matrix Addition and Subtraction: These operations are performed element-wise and require matrices of the same dimension.

Each calculation builds upon these fundamentals. Practicing these operations strengthens your ability to solve complex matrix-related problems. Matrix calculations are integral to various fields that use matrices to model and solve real-world problems, such as graphics, computer science, and systems theory.