Problem 10 - College Algebra and Calculus: An Applied Approach

Question

Find (a) \(A+B\), (b) \(A-B\), (c) \(3 A\), and (d) \(3 A-2 B\). $$ A=\left[\begin{array}{r} 3 \\ 2 \\ -1 \end{array}\right], B=\left[\begin{array}{r} -4 \\ 6 \\ 2 \end{array}\right] $$

Step-by-Step Solution

Answer

(a) \(A + B = \[ \begin{array}{r} -1 \ 8 \ 1 \end{array} \]\), (b) \(A - B = \[ \begin{array}{r} 7 \ -4 \ -3 \end{array} \]\), (c) \(3A = \[ \begin{array}{r} 9 \ 6 \ -3 \end{array} \]\), (d) \(3A - 2B = \[ \begin{array}{r} 17 \ -6 \ -7 \end{array} \]\)

Step 1: Find \(A + B\)

To add two matrices or vectors \(A\) and \(B\), we add corresponding elements in the two matrices. Hence, \(A + B = \[ \begin{array}{r} 3 + (-4) \ 2 + 6 \ -1 + 2 \end{array} \] = \[ \begin{array}{r} -1 \ 8 \ 1 \end{array} \]\).

Step 2: Find \(A - B\)

Subtracting one vector from another involves subtracting corresponding elements. Hence, \(A - B = \[ \begin{array}{r} 3 - (-4) \ 2 - 6 \ -1 - 2 \end{array} \] = \[ \begin{array}{r} 7 \ -4 \ -3 \end{array} \]\).

Step 3: Find \(3A\)

Multiplying a matrix by a scalar involves multiplying each element in the matrix by the scalar. Therefore, \(3A = \[ \begin{array}{r} 3 * 3 \ 3 * 2 \ 3 * -1 \end{array} \] = \[ \begin{array}{r} 9 \ 6 \ -3 \end{array} \]\).

Step 4: Find \(3A - 2B\)

This is done by first multiplying each matrix by their respective scalars and then subtracting the second resultant matrix from the first. Thus, \(3A - 2B = (3A) - (2B) = \[ \begin{array}{r} 9 - -8 \ 6 - 12 \ -3 - 4 \end{array} \] = \[ \begin{array}{r} 17 \ -6 \ -7 \end{array} \]\).

Key Concepts

Matrix Addition

Matrix addition is a fundamental concept in linear algebra, where we sum two matrices or vectors by adding their corresponding elements. Each matrix must have the same dimensions for addition to be possible.
In our given example, the matrices \( A \) and \( B \) are vectors of size 3x1, meaning we simply add each corresponding element.

• The first element of \( A \) is 3 and the first element of \( B \) is -4. So, 3 + (-4) = -1.
• For the second elements: 2 from \( A \) and 6 from \( B \). 2 + 6 = 8.
• Finally, for the third elements: -1 from \( A \) and 2 from \( B \). -1 + 2 = 1.

Therefore, the sum \( A + B \) results in a new vector: \( \[ \begin{array}{r} -1 \ 8 \ 1 \end{array} \] \).
This process illustrates how simple and intuitive matrix addition is once you understand the basic rule of adding corresponding elements.

Matrix Subtraction

Matrix subtraction is quite similar to matrix addition, except that we subtract corresponding elements instead of adding them. Just like addition, subtraction can only be performed if the matrices involved are of the same dimensions.
Returning to our example, for the subtraction of vector \( B \) from vector \( A \):

• Subtract the first elements: \( 3 - (-4) = 7 \).
• Subtract the second elements: \( 2 - 6 = -4 \).
• Subtract the third elements: \( -1 - 2 = -3 \).

So, the difference \( A - B \) results in the vector \( \[ \begin{array}{r} 7 \ -4 \ -3 \end{array} \] \).
This operation is straightforward as long as you remember to align elements properly and keep track of negative signs.

Scalar Multiplication

Scalar multiplication involves multiplying every element in a matrix by a single number, known as a scalar. This operation is widely used in linear algebra, particularly in scaling matrices.
With our example, let's find \( 3A \):

• The first element is multiplied by 3: \( 3 \times 3 = 9 \).
• The second element is also multiplied by 3: \( 3 \times 2 = 6 \).
• The third element is \( 3 \times (-1) = -3 \).

Therefore, \( 3A \) results in \( \[ \begin{array}{r} 9 \ 6 \ -3 \end{array} \] \).
Scalar multiplication is a simple yet powerful operation that stretches or shrinks the original matrix, maintaining its direction in the vector space.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, matrices, and operations on them. It is essential for solving equations, optimizing processes, and much more in mathematical studies and applications.
Consider the combination operation \( 3A - 2B \): we first compute scalar multiplications \( 3A \) and \( 2B \), then perform subtraction.

• For the first element: Scale \( 3A = 9 \) and \( -2B = -8 \), then subtract: \( 9 + 8 = 17 \).
• For the second element: Scale \( 6 \) from \( 3A \) and \( 12 \) from \( 2B \), then subtract: \( 6 - 12 = -6 \).
• For the third: Scale \( -3 \) and \( 4 \), then subtract: \( -3 - 4 = -7 \).

This yields the resultant vector \( \[ \begin{array}{r} 17 \ -6 \ -7 \end{array} \] \).
Fundamentally, linear algebra provides tools to transform vectors and matrices, which are necessary in many fields like computer science, engineering, and physics.