Question

Let \(\mathbf{u}=\left[\begin{array}{r}3 \\ -1 \\ 0\end{array}\right], \mathbf{v}=\left[\begin{array}{l}4 \\ 0 \\ 1\end{array}\right],\) and \(\mathbf{z}=\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right] .\) In each case, show that there are no numbers \(a, b,\) and \(c\) such that: a. \(a \mathbf{u}+b \mathbf{v}+c \mathbf{z}=\left[\begin{array}{l}1 \\ 2 \\\ 1\end{array}\right]\) b. \(a \mathbf{u}+b \mathbf{v}+c \mathbf{z}=\left[\begin{array}{r}5 \\ 6 \\\ -1\end{array}\right]\)

Short answer

In both cases, there are no solutions for \(a, b, c\) as the systems are inconsistent.

Step-by-step solution

Set up the Equation System for Case a

To find if there exist numbers \(a, b, c\) such that \(a \mathbf{u} + b \mathbf{v} + c \mathbf{z} = \left[\begin{array}{c}1 \ 2 \ 1\end{array}\right]\), we set up the following system of linear equations by substituting the values of \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{z}\):\[\begin{aligned}&a(3) + b(4) + c(1) = 1, \ &a(-1) + b(0) + c(1) = 2, \ &a(0) + b(1) + c(1) = 1.\end{aligned}\]

Solve the System of Equations for Case a

Solve the system: \[\begin{aligned}&3a + 4b + c = 1, \ &-a + c = 2, \ &b + c = 1.\end{aligned}\] Substituting \(c = a + 2\) from the second equation into the third gives \(b + (a + 2) = 1\), or \(b + a = -1\). Also, substitute \(c = a + 2\) in the first equation: \(3a + 4b + (a + 2) = 1\), simplifying to \(4a + 4b = -1\), leading to \(a + b = -\frac{1}{4}\). These two equations (\(b + a = -1\) and \(a + b = -\frac{1}{4}\)) are inconsistent.

Set up the Equation System for Case b

Similarly, to find if there exist numbers \(a, b, c\) such that \(a \mathbf{u} + b \mathbf{v} + c \mathbf{z} = \left[\begin{array}{c}5 \ 6 \ -1\end{array}\right]\), set up the system: \[\begin{aligned}&3a + 4b + c = 5, \ &-a + c = 6, \ &b + c = -1.\end{aligned}\]

Solve the System of Equations for Case b

Substitute \(c = a + 6\) from the second equation into the third gives \(b + (a + 6) = -1\), simplifying to \(b + a = -7\). Substitute \(c = a + 6\) in the first equation: \(3a + 4b + (a + 6) = 5\), leading to the equation \(4a + 4b = -1\), or \(a + b = -\frac{1}{4}\). Comparing the two equations (\(b + a = -7\) and \(a + b = -\frac{1}{4}\)), we find they are inconsistent.

Key concepts

Vector Equations

Vector equations are essential in linear algebra and help us express relationships between vectors using simple algebraic structures. In a vector equation like \(a \mathbf{u} + b \mathbf{v} + c \mathbf{z} = \mathbf{b}\), we are trying to find scalar multiples \(a, b,\) and \(c\) that transform our given vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{z}\) into the vector \(\mathbf{b}\).
This operation involves scaling each vector by its respective coefficient and then adding the resulting vectors together. This process is comparable to solving equations in arithmetic, but instead, we're working with entire vectors. To solve a vector equation, we often break it down into a system of linear equations, reflecting the coordinate positions of the vectors. This yields an intuitive way to handle computations and visualize vector transformations.

System of Linear Equations

A system of linear equations is a collection of one or more linear equations involving the same set of variables. Solving a system means finding values for these variables that satisfy all equations simultaneously.
In linear algebra, these systems can be represented as matrices, which makes them easier to manipulate and solve. For example, in Case a of the exercise, the vector equation morphs into a system of linear equations:

• \(3a + 4b + c = 1\)
• \(-a + c = 2\)
• \(b + c = 1\)

In this exercise, these equations originate from equating the coordinates of the transformed vectors to the target vector \(\left[\begin{array}{c}1 \ 2 \ 1\end{array}\right]\). Each equation represents a direct comparison of the same coordinate across all involved vectors, laying out a clear pathway for solving the problem by either substitution or elimination methods.

Inconsistency of Equations

An inconsistency in a system of equations arises when there is no possible set of solutions that can satisfy all equations simultaneously. For a system of linear equations, it means that there is no common intersection point in a graphical interpretation of these equations, usually resulting in parallel lines that never meet.
In the context of the exercise, both Case a and Case b demonstrate inconsistency. For instance, in Case a, the system yields:

• \(a + b = -\frac{1}{4}\)
• \(b + a = -1\)

These equations contradict each other, implying that no real numbers exist for \(a\) and \(b\) that will satisfy both equations simultaneously. The result is that the system has no solutions, or is inconsistent, meaning the desired transformation to reach the target vector \([1, 2, 1]\) is not possible with the given vectors.