Problem 3 - Active Calculus Multivariable

Question

Let \(\mathbf{a}=<-3,-4,-4>\) and \(\mathbf{b}=<2,2,4>\). Show that there are scalars \(\mathrm{s}\) and \(\mathrm{t}\) so that \(s \mathbf{a}+t \mathbf{b}=<20,24,32>\). You might want to sketch the vectors to get some intuition. \(s=\)_______ \(t=\)_______

Step-by-Step Solution

Answer

The scalars \(s\) and \(t\) that satisfy the equation \(s\mathbf{a} + t\mathbf{b} = <20, 24, 32>\) are \(s = -4\) and \(t = 4\).

Step 1: Set up the equation

We need to find scalars s and t such that: \(s\mathbf{a} + t\mathbf{b} = <20, 24, 32>\) where \(\mathbf{a} = <-3, -4, -4>\) and \(\mathbf{b} = <2, 2, 4>\). This equation can be represented as a system of linear equations as follows: \(-3s + 2t = 20\) \(-4s + 2t = 24\) \(-4s + 4t = 32\)

Step 2: Solve the system of linear equations

To solve the system of linear equations, we can use substitution, elimination or any other method. Using the elimination method, multiply the first equation by 2 to match the coefficients of t in the second equation: \(-6s + 4t = 40\) Now, subtract the second equation from the manipulated first equation: \(-6s + 4t = 40\) \(- (-4s + 2t = 24)\) This results in: \(-2s + 2t = 16\) Now, divide the equation by 2: \(-s + t = 8\) Next, from the first original equation, solve for t: \(t = \frac{20 + 3s}{2}\) Substitute this value of t into the last expression: \(-s + \frac{20 + 3s}{2} = 8\) Multiply both sides by 2 to eliminate the fraction: \(-2s + 20 + 3s = 16\)

Step 3: Find the values of s and t

Solve for s in the equation: \(s = 16 - 20 = -4\) Substitute the value of s back into the expression for t: \(t = \frac{20 + 3(-4)}{2} = \frac{20 - 12}{2} = \frac{8}{2} = 4\) The scalars s and t are found to be s = -4 and t = 4.

Key Concepts

Linear Algebra

Linear algebra is a significant branch of mathematics which deals with vectors, matrices, and linear transformations. It is foundational to understanding various concepts such as geometric transformations, system optimization, and even in advanced digital computations which include computer graphics and machine learning algorithms.

In the context of our exercise, linear algebra plays a role in understanding how to express one vector as a linear combination of others, which essentially involves finding the right scalars (or coefficients) that when multiplied with the given vectors and then added together, result in a specific target vector.

Systems of Linear Equations

A system of linear equations consists of two or more linear equations that share a set of variables. The goal is to find the values for each variable that will satisfy all the equations simultaneously. These systems can be represented in many forms, including matrices.

In practical scenarios, like in the exercise, we are often required to elucidate the relationship between multiple factors - which, in this scenario, are the vector components - and solve for their magnitudes. Understanding how to solve these systems is essential for various areas in both pure and applied mathematics.

Vector Operations

Vector operations include tasks such as adding, scaling, and dot product among vectors. Vectors are mathematical objects that have both magnitude and direction, and are often used to represent physical quantities like force, velocity, and displacement.

In our exercise, two operations were primarily used: scalar multiplication and vector addition. The scalar multiplication scales a vector by a certain factor, changing its magnitude but not its direction. Vector addition, on the other hand, combines two vectors to produce a third one which often represents the cumulative effect of the factors being considered in a given problem.

Elimination Method

The elimination method is a technique for solving systems of linear equations. This method involves manipulating the equations to eliminate one variable at a time, thus simplifying the system to a point where it can be solved easily. It is achieved by adding or subtracting equations from one another to cancel out one of the variables.

In the step-by-step solution of our given exercise, the elimination method allowed us to find the values of the scalars s and t by strategically eliminating variables and solving the resultant equations. Mastering this method is beneficial for efficiently solving more complex linear systems.