Problem 12 - Discrete Mathematics with Applications

Question

Let \(A=\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 2 & 3\end{array}\right], B=\left[\begin{array}{ccc}0 & -2 & 5 \\ 0 & 0 & 1\end{array}\right],\) and \(C=\left[\begin{array}{ccc}-3 & 0 & 0 \\ 0 & 1 & 2\end{array}\right] .\) Find each. $$3 B-2 C$$

Step-by-Step Solution

Answer

\(3B - 2C = \left[\begin{array}{ccc}6 & -6 & 15 \\\ 0 & -2 & -1\end{array}\right]\)

Step 1: Scalar Multiplication of B and C

Perform scalar multiplication for both B and C with the given scalars 3 and -2, respectively. We do this simply by multiplying each element by the scalar. For B, we get: \[ 3B = \left[\begin{array}{ccc}3(0) & 3(-2) & 3(5) \\\ 3(0) & 3(0) & 3(1)\end{array}\right] = \left[\begin{array}{ccc}0 & -6 & 15 \\\ 0 & 0 & 3\end{array}\right] \] For C, we get: \[ -2C = \left[\begin{array}{ccc}-2(-3) & -2(0) & -2(0) \\\ -2(0) & -2(1) & -2(2)\end{array}\right] = \left[\begin{array}{ccc}6 & 0 & 0 \\\ 0 & -2 & -4\end{array}\right] \]

Step 2: Matrix Addition

Now, we need to add the two resulting matrices from Step 1: (3B) and (-2C). To do this, add the corresponding elements in the matrices together. This will give us: \[ 3B - 2C = \left[\begin{array}{ccc}(0+6) & (-6+0) & (15+0) \\\ (0+0) & (0-2) & (3-4)\end{array}\right] = \left[\begin{array}{ccc}6 & -6 & 15 \\\ 0 & -2 & -1\end{array}\right] \] So, \(3B - 2C = \left[\begin{array}{ccc}6 & -6 & 15 \\\ 0 & -2 & -1\end{array}\right]\).

Key Concepts

Scalar Multiplication

Scalar multiplication is an essential part of matrix algebra and it's very straightforward. You multiply every element of a matrix by a given number, which is called a scalar. For instance, if you have a matrix like

• \[A = \begin{bmatrix}1 & 0 & -1 \0 & 2 & 3\end{bmatrix}\]

And you want to multiply by 3, you simply take each number in the matrix and multiply it by 3:

• \[3A = \begin{bmatrix}3(1) & 3(0) & 3(-1) \3(0) & 3(2) & 3(3)\end{bmatrix} = \begin{bmatrix}3 & 0 & -3 \0 & 6 & 9\end{bmatrix}\]

This operation doesn't change the size of the matrix; it only scales the values by the scalar. It's quite helpful for adjusting the magnitude of a matrix's values without changing its structure.

Matrix Addition

Matrix addition is one of the simplest matrix operations. When adding matrices, it's crucial that they are the same dimension, meaning they have the same number of rows and columns. For example:

• \[A = \begin{bmatrix}1 & 2 \3 & 4\end{bmatrix}\]
• \[B = \begin{bmatrix}5 & 6 \7 & 8\end{bmatrix}\]

To add matrix \( A \) and \( B \), you merely add the corresponding elements from both matrices as follows:

• \[A + B = \begin{bmatrix}1+5 & 2+6 \3+7 & 4+8\end{bmatrix} = \begin{bmatrix}6 & 8 \10 & 12\end{bmatrix}\]

This operation results in another matrix of the same dimension where each element is the sum of the corresponding elements of the original matrices. Thus, addition is straightforward and keeps matrices within their dimensional space.

Matrix Operations

Matrix operations encompass a variety of actions you can perform on matrices, perhaps some of the most common being addition, subtraction, and scalar multiplication. Let's focus on a combined operation: scalar multiplication followed by addition. This is crucial in many practical scenarios, especially in computer graphics and solving system equations.Consider matrices \( B \) and \( C \):

• \[3B - 2C\]
• \[3B = \begin{bmatrix}0 & -6 & 15 \0 & 0 & 3\end{bmatrix}\]
• \[-2C = \begin{bmatrix}6 & 0 & 0 \0 & -2 & -4\end{bmatrix}\]

When you perform the addition (or subtraction) of these resulting matrices, you execute it by adding corresponding elements:

• \[3B - 2C = \begin{bmatrix}0+6 & -6+0 & 15+0 \0+0 & 0-2 & 3-4\end{bmatrix} = \begin{bmatrix}6 & -6 & 15 \0 & -2 & -1\end{bmatrix}\]

Combining scalar multiplication with addition/subtraction in steps like these is foundational in solving equations involving matrices. This example highlights how consecutive operations retain the matrices' dimensions allowing for controlled modifications of their data.