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Solve each of the following systems: \(\begin{array}{rrr}x+2 y-4 z=-4 & x+2 y-3 z=-1 & x+2 y-3 z=1 \\ 2 x+5 y-9 z=-10 & -3 x+y-2 z=-7 & 2 x+5 y-8 z=4 \\ 3 x-2 y+3 z=11 & 5 x+3 y-4 z=2 & 3 x+8 y-13 z=7 \\ \text { (a) } & \text { (b) } \\ \begin{array}{rrr} & & \end{array}\end{array}\) Reduce each system to triangular or echelon form using Gaussian elimination: a) Apply "Replace \(L_{2}\) by \(-2 L_{1}+L_{2}^{\prime \prime}\) and "Replace \(L_{3}\) by \(-3 L_{1}+L_{3}^{n}\) to eliminate \(x\) from the second and third equations, and then apply "Replace \(L_{3}\) by \(8 L_{2}+L_{3}\) " to eliminate \(y\) from the third equation. These operations yield $$ \begin{array}{rlrl} x+2 y-4 z & =-4 & x+2 y-4 z & =-4 \\ y-z & =-2 & \text { and then } & y-z & =-2 \\ -8 y+15 z & =23 & 7 z & =7 \end{array} $$ The system is in triangular form. Solve by back-substitution to obtain the unique solution \(u=(2,-1,1)\). b) Eliminate \(x\) from the second and third equations by the operations "Replace \(L_{2}\) by \(3 L_{1}+L_{2} "\) and "Replace \(L_{3}\) by \(-5 L_{1}+L_{3} . "\) This gives the equivalent system $$ \begin{array}{r} x+2 y-3 z=-1 \\ 7 y-11 z=-10 \\ -7 y+11 z=7 \end{array} $$ The operation "Replace \(L_{3}\) by \(L_{2}+L_{3}\) " yields the following degenerate equation with a nonzero constant: $$ 0 x+0 y+0 z=-3 $$ This equation and hence the system have no solution. c) Eliminate \(x\) from the second and third equations by the operations "Replace \(L_{2}\) by \(-2 L_{1}+L_{2} "\) "and "Replace \(L_{3}\) by \(-3 L_{1}+L_{3}\)." This yields the new system $$ \begin{aligned} x+2 y-3 z &=1 \\ y-2 z &=2 \\ 2 y-4 z &=4 \end{aligned} \quad \text { or } \quad \begin{aligned} x+2 y-3 z=1 \\ y-2 z=2 \end{aligned} $$ (The third equation is deleted, because it is a multiple of the second equation.) The system is in echelon form with pivot variables \(x\) and \(y\) and free variable \(z\). To find the parametric form of the general solution, set \(z=a\) and solve for \(x\) and \(y\) by backsubstitution. Substitute \(z=a\) in the second equation to get \(y=2+2 a\). Then substitute \(z=a\) and \(y=2+2 a\) in the first equation to get $$ x+2(2+2 a)-3 a=1 \quad \text { or } \quad x+4+a=1 \quad \text { or } \quad x=-3-a $$ Thus, the general solution is $$ x=-3-a, \quad y=2+2 a, \quad z=a \quad \text { or } \quad u=(-3-a, \quad 2+2 a, \quad a) $$ where \(a\) is a parameter.

Step by step solution

01

Eliminate x from the second and third equations

To do this, apply "Replace L鈧� by -2L鈧� + L鈧�" and "Replace L鈧� by -3L鈧� + L鈧�": \( \begin{cases} x + 2y - 4z = -4 \\ 2x + 5y - 9z = -10 \\ 3x - 2y + 3z = 11 \end{cases} \) 鈫� \( \begin{cases} x + 2y - 4z = -4 \\ y - z = -2 \\ -8y + 15z = 23 \end{cases}\)

02

Eliminate y from the third equation

To do this, apply "Replace L鈧� by 8L鈧� + L鈧�": \( \begin{cases} x + 2y - 4z = -4 \\ y - z = -2 \\ 7z = 7 \end{cases}\) System is now in triangular form. #Phase 2: Solve the triangular system#

03

Solve for z

From the last equation, we have \(7z=7\), so z=1.

04

Solve for y

Substitute z=1 in the second equation: \(y - 1 = -2\), so y=-1.

05

Solve for x

Substitute y=-1 and z=1 in the first equation: \(x + 2(-1) - 4(1) = -4\), so x=2. The unique solution is (2, -1, 1). System (b):

06

Eliminate x from the second and third equations

To do this, apply "Replace L鈧� by 3L鈧� + L鈧�" and "Replace L鈧� by -5L鈧� + L鈧�": \( \begin{cases} x + 2y - 3z = -1 \\ -3x + y - 2z = -7 \\ 2x + 5y - 8z = 4 \end{cases} \) 鈫� \( \begin{cases} x + 2y - 3z = -1 \\ 7y - 11z = -10 \\ -7y + 11z = 7 \end{cases}\)

07

Eliminate y from the third equation

To do this, apply "Replace L鈧� by L鈧� + L鈧�": \( \begin{cases} x + 2y - 3z = -1 \\ 7y - 11z = -10 \\ 0x + 0y + 0z = -3 \end{cases}\) The last equation has no solution. Thus, the system has no solution. System (c):

08

Eliminate x from the second and third equations

To do this, apply "Replace L鈧� by -2L鈧� + L鈧�" and "Replace L鈧� by -3L鈧� + L鈧�": \( \begin{cases} x + 2y - 3z = 1 \\ 2x + 5y - 9z = -10 \\ 3x + 8y - 13z = 7 \end{cases} \) 鈫� \( \begin{cases} x + 2y - 3z = 1 \\ y - 2z = 2 \\ 2y - 4z = 4 \end{cases}\) The third equation is a multiple of the second equation. The system is now in echelon form with pivot variables x and y, and free variable z. #Phase 2: Parametric form of the general solution#

09

Solve for x and y

Set z=a, then use back-substitution: Substitute z=a in the second equation: y = 2 + 2a. Substitute z=a and y=2+2a in the first equation: x + 4 + a = 1, so x = -3 - a. Thus, the general solution is u=(-3-a, 2+2a, a), where a is a parameter.

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Key Concepts

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

Linear Equations

Linear equations form the backbone of many real-world problems in mathematics and science. In essence, these are equations that involve only linear terms, meaning each variable in the equation is raised to the power of one. A typical linear equation looks like this: \(ax + by + cz = d\). Here, \(a\), \(b\), and \(c\) are constants and \(x\), \(y\), and \(z\) are variables. A system of linear equations is a collection of one or more linear equations involving the same set of variables. The solution to such a system is a set of values for the variables that satisfies all of the equations simultaneously. Dealing with multiple equations together is common in solving real-world problems like finding the intersection point of lines or modeling various phenomena across science and engineering fields. A primary method to tackle these systems is Gaussian elimination, which simplifies the system to find solutions efficiently.

Triangular Form

Triangular form is a specific arrangement of a system of linear equations achieved through a series of operations. In this form, equations are aligned such that the first equation contains all unknown variables, the second equation contains all unknowns except the first, the third equation leaves out the first two variables, and so forth. Taking a system to its triangular form involves manipulating the equations using operations such as row swaps, scalar multiplication, and addition or subtraction of equations. These are performed so that each leading coefficient (or pivot) appears as a 1 in a downward staircase pattern across the equations.For example, a system in triangular form might look like:

• Equation 1: \(x + 2y - 4z = -4\)
• Equation 2: \(y - z = -2\)
• Equation 3: \(7z = 7\)

This configuration makes it straightforward to solve for these variables starting from the last equation upwards, eventually revealing every variable's value.

Echelon Form

The echelon form is an organized way of presenting linear systems that facilitates easier solutions. Similar to the triangular form, it involves aligning systems so that non-zero elements form a 'stair-step' pattern descending from left to right.Echelon form allows for easy identification of pivot variables (i.e., essential variables that determine the system's structure) and free variables (i.e., variables that can take any value without disrupting the system's validity). For example, in a system turned into echelon form, it might look like:

• Equation 1: \(x + 2y - 3z = 1\)
• Equation 2: \(y - 2z = 2\)
• Equation 3: This may be a multiple of another equation, thus removed or consolidated

Echelon forms help to recognize how many solutions exist for the system鈥攅ither a unique solution, an infinite number, or none鈥攁s observed in the provided exercise outcomes.

Back-Substitution

Back-substitution is the method used to find solutions from a system of linear equations that has already been simplified into either triangular or echelon form. The process begins with solving the last equation first, moving upwards through the previously aligned system.In triangular form, the last equation usually has only one variable, which can be solved directly. For instance, if the equation is \(7z = 7\), the solution is simply \(z = 1\). Once this value is identified, it is substituted back into the previous equation, allowing the next variable to be solved. Continuing this pattern, each solved variable is used to simplify and solve others until all variables are confidently determined.

• Start with the simplest equation (where only one variable remains).
• Substitute known values back into earlier equations.
• Continue until all unknowns are resolved.

Back-substitution efficiently uncovers the unique solution in systems like the one given in the exercise, demonstrating its vital role in solving linear equations.