Problem 23 - Precalculus: Mathematical for Calculus

Question

The matrices \(A, B, C, D, E, F,\) and \(G\) are defined as $$\begin{array}{l} A=\left[\begin{array}{rr} 2 & -5 \\ 0 & 7 \end{array}\right] \quad B=\left[\begin{array}{rrrr} 3 & \frac{1}{2} & 5 \\ 1 & -1 & 3 \end{array}\right] \quad C=\left[\begin{array}{rrr} 2 & -\frac{5}{2} & 0 \\ 0 & 2 & -3 \end{array}\right] \\ D=\left[\begin{array}{rrr} 7 & 3 \end{array}\right] & E=\left[\begin{array}{l} 1 \\ 2 \\ 0 \end{array}\right] \\ F=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \quad G=\left[\begin{array}{rrr} 5 & -3 & 10 \\ 6 & 1 & 0 \\ -5 & 2 & 2 \end{array}\right] \end{array}$$ Carry out the indicated algebraic operation, or explain why it cannot be performed. $$2 C-6 B$$

Step-by-Step Solution

Answer

The result of the operation \( 2C - 6B \) is \( \begin{bmatrix} -14 & -8 & -30 \\ -6 & 10 & -24 \end{bmatrix} \).

Step 1: Matrix Multiplication by a Scalar for Matrix C

First, we multiply each entry of matrix \( C \) by 2 to perform the operation \( 2C \). The matrix \( C \) is given by: \[ C = \begin{bmatrix} 2 & -\frac{5}{2} & 0 \ 0 & 2 & -3 \end{bmatrix} \]. The multiplication gives:\[ 2C = \begin{bmatrix} 2 \times 2 & 2 \times -\frac{5}{2} & 2 \times 0 \ 2 \times 0 & 2 \times 2 & 2 \times -3 \end{bmatrix} = \begin{bmatrix} 4 & -5 & 0 \ 0 & 4 & -6 \end{bmatrix} \].

Step 2: Matrix Multiplication by a Scalar for Matrix B

Next, we multiply each entry of matrix \( B \) by 6 to perform the operation \( 6B \). The matrix \( B \) is given by:\[ B = \begin{bmatrix} 3 & \frac{1}{2} & 5 \ 1 & -1 & 3 \end{bmatrix} \]. The multiplication gives:\[ 6B = \begin{bmatrix} 6 \times 3 & 6 \times \frac{1}{2} & 6 \times 5 \ 6 \times 1 & 6 \times -1 & 6 \times 3 \end{bmatrix} = \begin{bmatrix} 18 & 3 & 30 \ 6 & -6 & 18 \end{bmatrix} \].

Step 3: Attempt to Subtract Matrices 2C and 6B

Now, we need to find \( 2C - 6B \). Subtraction of matrices can only be performed if they have the same dimensions. Matrix \( 2C \) has dimensions \( 2 \times 3 \) and matrix \( 6B \) also has dimensions \( 2 \times 3 \), so subtraction is possible. We perform the subtraction entry-wise:\[2C - 6B = \begin{bmatrix} 4 & -5 & 0 \ 0 & 4 & -6 \end{bmatrix} - \begin{bmatrix} 18 & 3 & 30 \ 6 & -6 & 18 \end{bmatrix} = \begin{bmatrix} 4-18 & -5-3 & 0-30 \ 0-6 & 4+6 & -6-18 \end{bmatrix} = \begin{bmatrix} -14 & -8 & -30 \ -6 & 10 & -24 \end{bmatrix} \].

Key Concepts

Matrix Operations

Matrix operations are a crucial part of understanding matrix algebra. These operations include addition, subtraction, multiplication, and scalar multiplication. Each operation has specific rules that must be followed. These rules are framed by the nature and structure of matrices themselves.

Matrices are organized as rectangular arrays of numbers arranged in rows and columns. The dimensions of a matrix are described as "m x n," where "m" is the number of rows, and "n" is the number of columns.

Matrix operations can only be performed on matrices that have compatible dimensions:

• Addition and subtraction require both matrices to have the same dimensions.
• Multiplication needs the number of columns in the first matrix to be equal to the number of rows in the second matrix.
• Scalar multiplication, however, can be performed on any matrix because it involves multiplying every element by a single number (the scalar).

Understanding these operations forms the foundation for more advanced topics in matrix and linear algebra.

Matrix Addition and Subtraction

Matrix addition and subtraction are straightforward operations, provided that the dimensions of the matrices involved are the same. These operations involve the element-wise addition or subtraction of corresponding entries in the matrices.

For two matrices, say, matrix A and matrix B to be added or subtracted, they must have the same size. This means if A is a 2x3 matrix, B must also be a 2x3 matrix for their addition or subtraction to be valid.

• Addition example: Given matrices A and B:
  If \( A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 5 & 6 \ 7 & 8 \end{bmatrix} \), then \( A + B = \begin{bmatrix} 1+5 & 2+6 \ 3+7 & 4+8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \ 10 & 12 \end{bmatrix} \).


• Subtraction example: If \( A = \begin{bmatrix} 5 & 7 \ 9 & 11 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & 4 \ 1 & 2 \end{bmatrix} \), then \( A - B = \begin{bmatrix} 5-3 & 7-4 \ 9-1 & 11-2 \end{bmatrix} = \begin{bmatrix} 2 & 3 \ 8 & 9 \end{bmatrix} \).

This simple element-wise handling allows these operations to be memorably easy to apply but emphasizes the need for strict compliance with matrix dimension rules.

Scalar Multiplication

Scalar multiplication of a matrix is an operation where each element of a matrix is multiplied by a scalar (a constant number). This is a basic but potent concept in matrix algebra that often serves as a building block for more complex matrix operations.

When a matrix is multiplied by a scalar, each element of the matrix undergoes this transformation.

• For example, consider matrix A:
• \( A = \begin{bmatrix} 3 & 5 \ 7 & 9 \end{bmatrix} \)
• If we multiply matrix A by a scalar \( k = 2 \), then \( 2A = \begin{bmatrix} 2 \times 3 & 2 \times 5 \ 2 \times 7 & 2 \times 9 \end{bmatrix} = \begin{bmatrix} 6 & 10 \ 14 & 18 \end{bmatrix} \).

This operation retains the structure of the original matrix while effectively scaling its values by the specified constant. Scalar multiplication is incredibly valuable, especially when manipulating matrices to simplify calculations or apply theoretical concepts practically. It lays the groundwork for more advanced topics such as eigenvectors and matrix factorization.