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Find values for the scalars \(a\) and \(b\) that satisfy the given equation. $$ a\left[\begin{array}{l} 1 \\ 3 \end{array}\right]+b\left[\begin{array}{l} -3 \\ -9 \end{array}\right]=\left[\begin{array}{c} 4 \\ -12 \end{array}\right] $$

Short Answer

Expert verified
There are no values for \(a\) and \(b\) that satisfy the equation.

Step by step solution

01

Equation Setup

To solve for the scalars \(a\) and \(b\), start by setting up the vector equation into two separate component equations. We have \[ a \begin{bmatrix} 1 \ 3 \end{bmatrix} + b \begin{bmatrix} -3 \ -9 \end{bmatrix} = \begin{bmatrix} 4 \ -12 \end{bmatrix} \]This breaks down to two equations: 1. \( a - 3b = 4 \)2. \( 3a - 9b = -12 \).
02

Equation Simplification

Notice that the second equation \(3a - 9b = -12\) can be simplified by dividing the entire equation by 3, resulting in: \[ a - 3b = -4 \]Now, we have the system of equations:1. \( a - 3b = 4 \)2. \( a - 3b = -4 \)
03

Compare Simplified Equations

The simplified system shows inconsistent equations: both \( a - 3b = 4 \) and \( a - 3b = -4 \) cannot simultaneously be true. This indicates there is no solution for scalars \(a\) and \(b\) that satisfy the given vector equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Equations
Vector equations help us represent mathematical equations in a form involving vectors. A vector equation is established by expressing the combination of scalar multiples of vectors equated to another vector. This is crucial for understanding linear algebra because it portrays how different vectors relate in a space. In our previous exercise, we worked with the vector equation:
  • \( a\begin{bmatrix} 1 \ 3 \end{bmatrix} + b\begin{bmatrix} -3 \ -9 \end{bmatrix} = \begin{bmatrix} 4 \ -12 \end{bmatrix} \)
Here, \(a\) and \(b\) are scalars, and they determine how much of each vector \(\begin{bmatrix} 1 \ 3 \end{bmatrix}\) and \(\begin{bmatrix} -3 \ -9 \end{bmatrix}\) is utilized to "add up" to the target vector \(\begin{bmatrix} 4 \ -12 \end{bmatrix}\).
The beauty of vector equations is their ability to simplify complex problems into solvable components, where vectors can describe lines, planes, and other geometrical shapes in space. This process allows us to handle real-world problems in physics, engineering, and computer graphics effectively.
Systems of Linear Equations
Systems of linear equations consist of two or more linear equations that are solved simultaneously. They are a critical part of understanding relationships between multiple expressions that involve one or more variables. The results reflect common solutions that satisfy all participating equations. In the exercise provided, the vector equation was decomposed into the following system:
  • Equation 1: \( a - 3b = 4 \)
  • Equation 2: \( 3a - 9b = -12 \)
These equations share the same set of unknowns - in this case, scalar values \(a\) and \(b\).
To simplify, we divided the second equation by 3, aligning its coefficients with the first one. Such a system often requires methods such as substitution or elimination to find solutions. Solving these predicts any existing combination of variable values that fulfill all conditions presented in the system.
Linear systems appear frequently in various fields like economics (to predict market equilibria) and science (where they are indispensable for modeling reactions and predicting outcomes).
Inconsistent Systems
Inconsistent systems are those where no solution exists due to contradictory equations. This means the set of equations has no common solution; not even one single combination of values satisfies every equation simultaneously. In the step-by-step solution, simplifying both equations resulted in:
  • Equation 1: \( a - 3b = 4 \)
  • Equation 2: \( a - 3b = -4 \)
Clearly, these two statements are at odds—hypothetically asking for \(a - 3b\) to equal two distinct numbers at the same time, which is impossible.
When such inconsistencies arise, it indicates there is a deeper issue with expectations or constraints. An inconsistent system could signal errors in problem setup, or constraints needing to be re-evaluated. In practical scenarios, recognizing inconsistency is vital as it prevents false conclusions in predictive models or decision-making processes.

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Most popular questions from this chapter

A matrix \(A\) and vectors \(\vec{b}, \vec{u}\) and \(\vec{v}\) are given. Verify that \(\vec{u}\) and \(\vec{v}\) are both solutions to the equation \(A \vec{x}=\vec{b} ;\) that is, show that \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cccc} 0 & -3 & -1 & -3 \\ -4 & 2 & -3 & 5 \end{array}\right], \\ \vec{b}=\left[\begin{array}{c} 0 \\ 0 \end{array}\right], \vec{u}=\left[\begin{array}{c} 11 \\ 4 \\ -12 \\ 0 \end{array}\right], \\ \vec{v}=\left[\begin{array}{c} 9 \\ -12 \\ 0 \\ 12 \end{array}\right] \end{array} $$

A matrix \(A\) and vectors \(\vec{b}, \vec{u}\) and \(\vec{v}\) are given. Verify that \(\vec{u}\) and \(\vec{v}\) are both solutions to the equation \(A \vec{x}=\vec{b} ;\) that is, show that \(A \vec{u}=A \vec{v}=\vec{b}\). $$ \begin{array}{l} A=\left[\begin{array}{cc} 1 & -2 \\ -3 & 6 \end{array}\right] \\ \vec{b}=\left[\begin{array}{l} 0 \\ 0 \end{array}\right], \vec{u}=\left[\begin{array}{l} 2 \\ 1 \end{array}\right], \vec{v}=\left[\begin{array}{c} -10 \\ -5 \end{array}\right] \end{array} $$

A matrix \(A\) and vector \(\vec{b}\) are given. (a) Solve the equation \(A \vec{x}=\vec{O}\) (b) Solve the equation \(A \vec{x}=\vec{b}\). In each of the above, be sure to write your answer in vector format. Also, when possible, give 2 particular solutions to each equation. $$ A=\left[\begin{array}{cc} 0 & 2 \\ -1 & 3 \end{array}\right], \vec{b}=\left[\begin{array}{l} -2 \\ -1 \end{array}\right] $$

A matrix \(A\) and a vector \(\vec{x}\) are given. Find the product \(A \vec{x}\). $$ A=\left[\begin{array}{ccc} -2 & 0 & 3 \\ 1 & 1 & -2 \\ 4 & 2 & -1 \end{array}\right], \quad \vec{x}=\left[\begin{array}{l} 4 \\ 3 \\ 1 \end{array}\right] $$

Row and column vectors \(\vec{u}\) and \(\vec{v}\) are defined. Find the product \(\vec{u} \vec{v},\) where possible. $$ \vec{u}=\left[\begin{array}{ll} 1 & -4 \end{array}\right] \quad \vec{v}=\left[\begin{array}{c} -2 \\ 5 \end{array}\right] $$

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