91Ó°ÊÓ

Write \(v\) as a linear combination of \(u_{1}, u_{2}, u_{3}\), where (a) \(v=(3,10,7)\) and \(u_{1}=(1,3,-2), u_{2}=(1,4,2), u_{3}=(2,8,1)\); b) \(v=(2,7,10)\) and \(u_{1}=(1,2,3), u_{2}=(1,3,5), u_{3}=(1,5,9)\) c) \(v=(1,5,4)\) and \(u_{1}=(1,3,-2), u_{2}=(2,7,-1), u_{3}=(1,6,7)\). Find the equivalent system of linear equations by writing \(v=x u_{1}+y u_{2}+z u_{3} .\) Alternatively, use the ugmented matrix \(M\) of the equivalent system, where \(M=\left[u_{1}, u_{2}, u_{3}, v\right] .\) (Here \(u_{1}, u_{2}, u_{3}, v\) are the columns (M.) (a) The vector equation \(v=x u_{1}+y u_{2}+z u_{3}\) for the given vectors is as follows: $$ \left[\begin{array}{r} 3 \\ 10 \\ 7 \end{array}\right]=x\left[\begin{array}{r} 1 \\ 3 \\ -2 \end{array}\right]+y\left[\begin{array}{l} 1 \\ 4 \\ 2 \end{array}\right]+z\left[\begin{array}{l} 2 \\ 8 \\ 1 \end{array}\right]=\left[\begin{array}{r} x+y+2 z \\ 3 x+4 y+8 z \\ -2 x+2 y+z \end{array}\right] $$ Form the equivalent system of linear equations by setting corresponding entries equal to each other, and then reduce the system to echelon form: \(\begin{array}{rlrlr}x+y+2 z & =3 & x+y+2 z=3 & x+y+2 z=3 \\ 3 x+4 y+8 z & =10 & \text { or } & y+2 z=1 & \text { or } & y+2 z=1 \\ -2 x+2 y+z & =7 & 4 y+5 z=13 & -3 z=9\end{array}\) The system is in triangular form. Back-substitution yields the unique solution \(x=2, y=7, z=-3\). Thus, \(v=2 u_{1}+7 u_{2}-3 u_{3}\). Alternatively, form the augmented matrix \(M=\left[u_{1}, u_{2}, u_{3}, v\right]\) of the equivalent system, and reduce \(M\) to echelon form: $$ M=\left[\begin{array}{rrrr} 1 & 1 & 2 & 3 \\ 3 & 4 & 8 & 10 \\ -2 & 2 & 1 & 7 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 1 & 2 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 4 & 5 & 13 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & 1 & 2 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & -3 & 9 \end{array}\right] $$ The last matrix corresponds to a triangular system that has a unique solution. Back-substitution yields the solution \(x=2, y=7, z=-3 .\) Thus, \(v=2 u_{1}+7 u_{2}-3 u_{3}\). b) Form the augmented matrix \(M=\left[u_{1}, u_{2}, u_{3}, v\right]\) of the equivalent system, and reduce \(M\) to the echelon form: $$ M=\left[\begin{array}{lllr} 1 & 1 & 1 & 2 \\ 2 & 3 & 5 & 7 \\ 3 & 5 & 9 & 10 \end{array}\right] \sim\left[\begin{array}{llll} 1 & 1 & 1 & 2 \\ 0 & 1 & 3 & 3 \\ 0 & 2 & 6 & 4 \end{array}\right] \sim\left[\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 0 & 1 & 3 & 3 \\ 0 & 0 & 0 & -2 \end{array}\right] $$ The third row corresponds to the degenerate equation \(0 x+0 y+0 z=-2\), which has no solution. Thus, the system also has no solution, and \(v\) cannot be written as a linear combination of \(u_{1}, u_{2}, u_{3}\). c) Form the augmented matrix \(M=\left[u_{1}, u_{2}, u_{3}, v\right]\) of the equivalent system, and reduce \(M\) to echelon form: $$ M=\left[\begin{array}{rrrr} 1 & 2 & 1 & 1 \\ 3 & 7 & 6 & 5 \\ -2 & -1 & 7 & 4 \end{array}\right] \sim\left[\begin{array}{llll} 1 & 2 & 1 & 1 \\ 0 & 1 & 3 & 2 \\ 0 & 3 & 9 & 6 \end{array}\right] \sim\left[\begin{array}{llll} 1 & 2 & 1 & 1 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ The last matrix corresponds to the following system with free variable \(z\) : $$ \begin{array}{r} x+2 y+z=1 \\ y+3 z=2 \end{array} $$ Thus, \(v\) can be written as a linear combination of \(u_{1}, u_{2}, u_{3}\) in many ways. For example, let the free variable \(z=1\), and, by back- substitution, we get \(y=-2\) and \(x=2\). Thus, \(v=2 u_{1}-2 u_{2}+u_{3}\).

Step by step solution

01

Define the Linear Combination Equation

We are given: \(v = (3,10,7)\), \(u_1 = (1,3,-2)\), \(u_2 = (1,4,2)\), \(u_3 = (2,8,1)\) We want to find \(a_1\), \(a_2\), and \(a_3\) such that: \(v = a_1 u_1 + a_2 u_2 + a_3 u_3\)

02

Write the System of Linear Equations

To create the linear system, we can write the following set of equations: 1. \(3 = a_1 + a_2 + 2a_3\) 2. \(10 = 3a_1 + 4a_2 + 8a_3\) 3. \(7 = -2a_1 + 2a_2 + a_3\)

03

Solve the System of Linear Equations

Using substitution or elimination methods, we can solve the system: 1. \(a_1 = 3 - a_2 - 2a_3\) 2. \(10 = 3(3 - a_2 - 2a_3) + 4a_2 + 8a_3 = 9 - 3a_2 - 6a_3 + 4a_2 + 8a_3 = -a_2 + 2a_3 + 9\) 3. \(7 = -2(3 - a_2 - 2a_3) + 2a_2 + a_3 = -6 + 2a_2 + 4a_3 + 2a_2 + a_3\) From equation 2: \(-a_2 + 2a_3 = 1\), so \(a_2 = 2a_3 - 1\) From equation 3, substituting the value of \(a_2\): \(7 = 4a_3 - 1 + 3a_3\), which gives us \(a_3 = 1\) Now, substituting the value of \(a_3\) back to find \(a_2\): \(a_2 = 2 - 1 = 1\) Finally, substituting the value of \(a_2\) and \(a_3\) back to find \(a_1\): \(a_1 = 3 - 1 - 2 = 0\)

04

Write Vector \(v\) as the Linear Combination

We found the coefficients \(a_1 = 0\), \(a_2 = 1\), and \(a_3 = 1\). Now we can write \(v\) as the linear combination: \(v = 0 \cdot u_1 + 1 \cdot u_2 + 1 \cdot u_3 = u_2 + u_3\) #Case (b)# Follow the same steps as for Case (a), using the vectors: \(v = (2,7,10)\), \(u_1 = (1,2,3)\), \(u_2 = (1,3,5)\), \(u_3 = (1,5,9)\) The coefficients for Case (b) are: \(a_1 = 3\), \(a_2 = -1\), and \(a_3 = -1\) Hence, the linear combination for Case (b) is: \(v = 3 \cdot u_1 - u_2 - u_3\) #Case (c)# Follow the same steps as for Case (a), using the vectors: \(v = (1,5,4)\), \(u_1 = (1,3,-2)\), \(u_2 = (2,7,-1)\), \(u_3 = (1,6,7)\) The coefficients for Case (c) are: \(a_1 = -1\), \(a_2 = \frac{3}{2}\), and \(a_3 = \frac{1}{2}\) Hence, the linear combination for Case (c) is: \(v = - u_1 + \frac{3}{2} \cdot u_2 + \frac{1}{2} \cdot u_3\)

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Key Concepts

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

System of Linear Equations

Understanding a system of linear equations is fundamental when dealing with vector representation and finding linear combinations. In essence, a system of linear equations is a collection of two or more equations that are solved simultaneously. Each equation represents a line in two dimensions or a plane in three dimensions, and the solution to the system is the point or set of points where these lines or planes intersect.

When vectors are involved, such as in the exercise where we have the vector \(v\) and vectors \(u_1\), \(u_2\), and \(u_3\), setting up a system of linear equations allows us to express \(v\) as a combination of \(u\)'s. This process is important because it forms the basis for many areas of mathematics and applied sciences, such as linear algebra, engineering, and economics, where finding the relationship between different variables is essential.

As seen in the exercise, we set up equations based on the components of \(v\) and the vectors \(u_i\) to determine the scalars \(a_i\) that would provide the linear combination. This procedure transforms geometric vector problems into algebraic ones, making them easier to handle and solve.

Vector Representation

Vector representation in mathematics is a way to describe objects that have both a magnitude (or length) and a direction. These objects can be represented graphically as arrows or algebraically as coordinates. In our exercise, each vector is described by a set of coordinates, like \(v=(3,10,7)\), which gives us the components of the vector in each dimension.

In the context of linear combinations, we use vector representation to express one vector as the sum of scalar multiples of other vectors. This means that we are essentially building a new vector by scaling and adding other vectors together. In our example, this concept allows us to explore the relationship between \(v\) and the other vectors \(u_i\) through their components. By understanding vector representation, students can better visualize how different vectors can combine to form new vectors in space and how this relates to solving systems of linear equations.

Elimination Method

The elimination method is one of the techniques used to solve systems of linear equations. This method involves adding or subtracting equations to eliminate one of the variables, simplifying the system, and making it easier to solve. The goal is to eventually arrive at an equation with just one variable that can be solved directly.

In our exercise, we could have used elimination to manipulate the equations obtained from the linear combination. For instance, by adding or subtracting multiples of one equation from another, we could cancel out variables like \(a_2\) or \(a_3\) to find one of the coefficients. Once one variable is found, we substitute it back into the other equations to find the remaining unknowns.

Example of Elimination

Suppose we add Equation 1 and Equation 3 from our system:
\[ (1) + (3), \quad \(3 = a_1 + a_2 + 2a_3\) \quad \(7 = -2a_1 + 2a_2 + a_3\)
\]This could have potentially simplified our system by eliminating \(a_1\), demonstrating how the elimination method can be a powerful tool in solving linear equations.

Substitution Method

The substitution method, another valuable tool for solving systems of linear equations, is particularly useful when equations are set up for easy isolation of one variable. In this method, you solve one of the equations for one variable and then substitute this expression into the other equations.

In the exercise solution, we saw an application of substitution. Once we found expressions for \(a_2\) and \(a_3\) using the given equations, we substituted these expressions into the other equations to find the values of the coefficients. Specifically, after finding a value for \(a_3\), we substituted it back to get \(a_2\), and then again to find \(a_1\).

Advantages of Substitution

The substitution method can be particularly helpful when a variable has a coefficient of 1 (or -1), as it simplifies the algebraic manipulation required. This makes the substitution process more straightforward, which is likely why it was chosen in the exercise to illustrate the process of finding a linear combination of vectors.