Problem 13 - College Algebra

Question

In Exercises \(9-16,\) find: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{r} 2 \\ -4 \\ 1 \end{array}\right], \quad B=\left[\begin{array}{r} -5 \\ 3 \\ -1 \end{array}\right] $$

Step-by-Step Solution

Answer

a. \( A + B = \)\[\begin{array}{c} -3\ -1\ 0 \end{array}\]\nb. \( A - B = \)\[\begin{array}{c} 7\ -7\ 2 \end{array}\]\nc. \( -4A = \)\[\begin{array}{c} -8\ 16\ -4 \end{array}\]\nd. \( 3A + 2B = \)\[\begin{array}{c} -4\ -6\ 1 \end{array}\]

Step 1: Find A + B

Add the corresponding elements in each vector to find the sum. \n \( A + B = \)\[\begin{array}{c} 2 -5\ -4 + 3\ 1 -1 \end{array}]\( = \)\[\begin{array}{c} -3\ -1\ 0 \end{array}\]

Step 2: Find A - B

Subtract the corresponding elements in \( B \) from those in \( A \) to find the difference. \n \( A - B = \)\[\begin{array}{c} 2 - (-5)\ -4 - 3\ 1 - (-1) \end{array}]\( = \)\[\begin{array}{c} 7\ -7\ 2 \end{array}\]

Step 3: Find -4A

Multiply each element in \( A \) by -4. \n \( -4A = -4 \times \)\[\begin{array}{c} 2\ -4\ 1 \end{array}]\( = \)\[\begin{array}{c} -8\ 16\ -4 \end{array}\]

Step 4: Find 3A + 2B

First perform the scalar multiplication, then add the resulting vectors. \n \( 3A = 3 \times \)\[\begin{array}{c} 2\ -4\ 1 \end{array}]\( = \)\[\begin{array}{c} 6\ -12\ 3 \end{array}\]\n\( 2B = 2 \times \)\[\begin{array}{c} -5\ 3\ -1 \end{array}]\( = \)\[\begin{array}{c} -10\ 6\ -2 \end{array}\]\nTherefore, \( 3A + 2B = \)\[\begin{array}{c} 6 + (-10)\ -12 + 6\ 3 + (-2) \end{array}]\( = \)\[\begin{array}{c} -4\ -6\ 1 \end{array}\]

Key Concepts

Vector Addition

Understanding vector addition is pivotal in fields such as physics, engineering, and computer science. It's essentially about combining two vectors to find their sum, much like regular addition, but with a focus on direction and magnitude. For instance, consider vectors \( A \) and \( B \) from our exercise: \( A = \begin{bmatrix} 2 \ -4 \ 1 \end{bmatrix} \), and \( B = \begin{bmatrix} -5 \ 3 \ -1 \end{bmatrix} \). To get \( A + B \) we simply add each corresponding component of the vectors to find the resulting vector.
In layman terms, you're walking the steps indicated by the first vector and then continuing the journey by following the second vector's instructions to reach the final destination. This operation is commutative, meaning that \( A + B = B + A \). It's like saying, 'Whether you put on your socks first or second, you're still going to end up wearing them both.'

Vector Subtraction

If vector addition is akin to putting things together, then vector subtraction is like taking them apart. It's a way to find the difference between vectors. With our vectors \( A \) and \( B \) from the exercise above, we calculate \( A - B \) to move according to \( A \) and then backtracking according to \( B \). We do this by subtracting each element of \( B \) from the corresponding element of \( A \) to get the resulting vector which represents the difference between the two.
Imagine you've moved from point A to point B using the direction and magnitude given by the first vector. Now, to go back or undo just the movement dictated by the second vector, we do this reverse operation which shows us where we'd be if \( B \) was never a factor.

Scalar Multiplication

Ever wonder how you scale objects up or down? That's where scalar multiplication comes in—it's the operation that magnifies or diminishes a vector by a real number (called a scalar). For example, in our problem, multiplying \( A \) by \( -4 \) essentially means stretching or compressing the original vector \( 4 \) times in the opposite direction because the scalar is negative.
This operation is key in resizing vectors without changing their direction, unless the scalar is negative, which also flips the direction. The term 'scalar' simply refers to a quantity represented by a single value, unlike a vector that combines both magnitude and direction. It's like changing the volume of music—it doesn't change the melody, just how loud or soft it is.

Linear Combinations of Vectors

Picture painting with only two colors, but by mixing them, you can create a plethora of new shades. This is similar to linear combinations of vectors, where we mix vectors with scalar multiples to create a new vector. In our case, the combination \( 3A + 2B \) used in the exercise creates a new vector from the originals \( A \) and \( B \) by scaling and then adding them.
The power of linear combinations lies in their ability to span vector spaces, essentially building new vectors as needed. This is fundamental in solving systems of equations, understanding transformations, and even in computer graphics to model objects and animations. It's like having a basic recipe that you can tweak with different amounts of ingredients to bake a variety of cakes.