Problem 22 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

(a) \( \begin{bmatrix} -1 & -3 & -5 \\ -1 & 3 & -6 \end{bmatrix} \); (b) \( \begin{bmatrix} -14 & -8 & -30 \\ -6 & 10 & -24 \end{bmatrix} \)."

Step 1: Determine the dimensions of matrices C and B

Matrix \( C \) has dimensions 2x3, and matrix \( B \) also has dimensions 2x3. Since both matrices have the same dimensions, the subtraction operation \( C - B \) is valid.

Step 2: Perform the subtraction for C - B

Subtract corresponding elements from matrix \( B \) from matrix \( C \): \[C - B = \begin{bmatrix} 2 & -\frac{5}{2} & 0 \ 0 & 2 & -3 \end{bmatrix} - \begin{bmatrix} 3 & \frac{1}{2} & 5 \ 1 & -1 & 3 \end{bmatrix}= \begin{bmatrix} 2-3 & -\frac{5}{2}-\frac{1}{2} & 0-5 \ 0-1 & 2+1 & -3-3 \end{bmatrix}= \begin{bmatrix} -1 & -3 & -5 \ -1 & 3 & -6 \end{bmatrix}\]

Step 3: Express and simplify 2C and 6B

Calculate \( 2C \) and \( 6B \) by multiplying the matrices \( C \) and \( B \) by 2 and 6 respectively:\[ 2C = 2 \cdot \begin{bmatrix} 2 & -\frac{5}{2} & 0 \ 0 & 2 & -3 \end{bmatrix} = \begin{bmatrix} 4 & -5 & 0 \ 0 & 4 & -6 \end{bmatrix} \]\[ 6B = 6 \cdot \begin{bmatrix} 3 & \frac{1}{2} & 5 \ 1 & -1 & 3 \end{bmatrix} = \begin{bmatrix} 18 & 3 & 30 \ 6 & -6 & 18 \end{bmatrix} \]

Step 4: Determine if 2C - 6B can be performed

Since \( 2C \) and \( 6B \) both have dimensions of 2x3, the subtraction \( 2C - 6B \) is valid.

Step 5: Perform the subtraction for 2C - 6B

Subtract the elements of \( 6B \) from \( 2C \): \[2C - 6B = \begin{bmatrix} 4 & -5 & 0 \ 0 & 4 & -6 \end{bmatrix} - \begin{bmatrix} 18 & 3 & 30 \ 6 & -6 & 18 \end{bmatrix}= \begin{bmatrix} 4-18 & -5-3 & 0-30 \ 0-6 & 4+6 & -6-18 \end{bmatrix}= \begin{bmatrix} -14 & -8 & -30 \ -6 & 10 & -24 \end{bmatrix}\]

Key Concepts

Matrix Subtraction

Matrix subtraction is an operation where you subtract one matrix from another. It's similar to subtracting numbers, but it happens element by element. This means that each element in one matrix is subtracted from the corresponding element in the other matrix.

For matrix subtraction to work, both matrices must have the same dimensions. If one matrix is a 2x3 matrix, the other must also be 2x3. You can think of these dimensions as rules that need to be followed for subtraction to take place.

In the exercise given, matrix subtraction looks like this:

• Given: two matrices with dimensions 2x3:
  • Matrix C: \( \begin{bmatrix} 2 & -\frac{5}{2} & 0 \ 0 & 2 & -3 \end{bmatrix} \)
  • Matrix B: \( \begin{bmatrix} 3 & \frac{1}{2} & 5 \ 1 & -1 & 3 \end{bmatrix} \)
• Subtract each corresponding element:
  • For example, the element in the first row, first column: \( 2 - 3 = -1 \)
  • Continue this process for each element to get the resulting matrix.

The resulting matrix is \( \begin{bmatrix} -1 & -3 & -5 \ -1 & 3 & -6 \end{bmatrix} \). As shown, subtraction is straightforward when the matrices are of the same dimensions.

Matrix Multiplication

Matrix multiplication is slightly more complex than addition or subtraction because it doesn’t happen element by element. Instead, you multiply rows by columns. This requires a specific rule to be followed: the number of columns in the first matrix must equal the number of rows in the second matrix.

The exercise involved multiplying matrices by a scalar (a single number), which is a form of matrix multiplication. In scalar multiplication, each element of the matrix is multiplied by the scalar. Let's look into the specific multiplications performed in the exercise:

• To find \(2C\): Multiply every element of Matrix C by 2:
  • For example, the first element: \(2 \times 2 = 4\)
  • Apply this to each element resulting in the matrix \( \begin{bmatrix} 4 & -5 & 0 \ 0 & 4 & -6 \end{bmatrix} \)
• To find \(6B\): Multiply every element of Matrix B by 6:
  • For example, the first element: \(6 \times 3 = 18\)
  • Repeat this for each element to get \( \begin{bmatrix} 18 & 3 & 30 \ 6 & -6 & 18 \end{bmatrix} \)

This operation is simple but essential, laying the groundwork for more complex forms of matrix multiplication.

2x3 Matrices

The concept of a 2x3 matrix is fundamental in understanding simple matrix operations as presented in the example. A 2x3 matrix has two rows and three columns. This structure is part of what defines how operations like addition, subtraction, and multiplication are carried out.

Here are some key points about working with 2x3 matrices:

• The numbers of rows and columns (2 rows, 3 columns) determine which operations can be done. A 2x3 matrix can only be added to or subtracted from another matrix of the same size.
• In terms of multiplication with other matrices, a 2x3 matrix can be multiplied by a 3xN matrix, where N can be any number of columns, but the inner dimensions (3 in this case) must match.
• When tasks involve scalar multiplication, such as multiplying by 2 or 6, each element is treated independently.

Understanding the dimensions of a matrix helps in identifying which operations are possible at a glance. Being familiar with these basic principles simplifies working with matrices in various mathematical tasks.