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Use Gaussian elimination to solve the system of linear equations. If there is no solution, state that the system is inconsistent. $$\left\\{\begin{aligned} 4 x-y &=2 \\ y-2 z &=1 \\ 6 x-z &=3 \end{aligned}\right.$$

Short Answer

Expert verified
The solution to the system of equations is: \(x = 1.22\), \(y = 0.72\), \(z = 0.47\).

Step by step solution

01

Write the Augmented Matrix

Let's first write the given system of equations in the augmented matrix form. \[ \left[ \begin{array}{ccc|c} 4 & -1 & 0 & 2 \ 0 & 1 & -2 & 1 \ 6 & 0 & -1 & 3 \end{array} \right] \]. Each column of the matrix correspond to coefficients of \(x\), \(y\), and \(z\) respectively and the last column represents the constants on the right-hand side of the equations.
02

Apply Gaussian Elimination

The next step is to transform the given matrix into row-reduced form. For that purpose, we must subtract \(1.5\) times the 1st row from the 3rd row. It gives us \[ \left[ \begin{array}{ccc|c} 4 & -1 & 0 & 2 \ 0 & 1 & -2 & 1 \ 0 & 1.5 & -1 & -0.5 \end{array} \right] \]. Now, add 1st row to the 2nd row to yield \[ \left[ \begin{array}{ccc|c} 4 & -1 & 0 & 2 \ 4 & 0 & -2 & 3 \ 0 & 1.5 & -1 & -0.5 \end{array} \right] \], and then simplify by dividing through the 2nd row by 4 and 3rd row by 1.5 which gives \[ \left[ \begin{array}{ccc|c} 4 & -1 & 0 & 2 \ 1 & 0 & -0.5 & 0.75 \ 0 & 1 & -0.67 & -0.33 \end{array} \right] \]. And finally, multiply the 2nd row by 4 to yield \[ \left[ \begin{array}{ccc|c} 4 & -1 & 0 & 2 \ 4 & 0 & -2 & 3 \ 0 & 1 & -0.67 & -0.33 \end{array} \right] \].
03

Solve for Variables

The row-reduced matrix can be rewritten into a system of equations from which the variables can be solved. The system is: \begin{aligned} 4x - y &= 2 \ x - z &= 0.75 \ y - 0.67z &= 0.33 \end{aligned}. From the second equation, it can be observed that \(x = z + 0.75\). Substituting \(x\) in the 1st equation gives \(y = 0.25 + z\). Substituting y in third equation results in \(z = 0.47\). Hence its observed that \(x = 1.22\), \(y = 0.72\), \(z = 0.47\).

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Key Concepts

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

System of Linear Equations
The system of linear equations is a collection of one or more equations involving the same set of variables. A linear equation in three variables, such as the ones in the exercise, can be visually interpreted as a plane in three-dimensional space. When we have multiple linear equations, we are looking for a point or points where these planes intersect, which represent the solutions to our system.

For instance, we may have one unique solution where all three planes intersect at a single point, or we may have infinitely many solutions if the planes overlap along a line or coincide completely. There is also a possibility that there is no solution, which occurs if the planes are parallel and never intersect, meaning the system is inconsistent.

The challenge in solving these systems is to find a systematic way to determine whether solutions exist and if they do, what they are. This is where Gaussian elimination, a method used in the provided exercise, comes into play. It transforms the system into an equivalent one that is easier to solve.
Augmented Matrix
To apply Gaussian elimination efficiently, we utilize what is known as an augmented matrix. This compact form of representation includes all the coefficients of the variables along with the constants from the right-hand side of the equations placed into one single matrix. Each row corresponds to an equation and each column, up to the last one, corresponds to a variable's coefficient in the equations.

In the solved exercise, the augmented matrix represents the system using less space and more structure, enabling clear manipulation of the equations. It is formed by aligning the coefficients of the variables in columns and appending the constants on the right, separated by a line to indicate the equality sign from the equations.

Augmented matrices are crucial for visual simplification and for performing row operations that help us reach a solution more systematically. By manipulating rows, which correspond to equations, we work towards simplifying the system without changing its solutions.
Row-Reduced Form
The row-reduced form of a matrix, often referred to as row-echelon form, is when it has been simplified through a series of row operations - namely row swapping, multiplying rows by non-zero scalars, and adding or subtracting row multiples to or from other rows.

Our aim with Gaussian elimination is to transform the augmented matrix into this form, where we have a staircase-like pattern with leading coefficients (typically '1's) and zeros beneath them in each column leading up to the last. In the ideal case where we reach a reduced row-echelon form, each leading '1' is the only non-zero entry in its column.

This simplification allows us to interpret the row-reduced matrix as a set of simpler, equivalent equations that can be solved with back-substitution. The process demonstrated in the exercise converts a complex system into one where the solutions to the variables are more apparent and accessible, revealing whether there is a unique solution, infinitely many solutions, or no solution at all.

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Most popular questions from this chapter

If \(A=\left[\begin{array}{rrr}1 & 0 & 1 \\ 0 & 0 & 1 \\ 2 & -1 & 0\end{array}\right]\) and \(B=\left[\begin{array}{rrr}0 & 3 & -1 \\ -1 & 2 & 0 \\\ 0 & 0 & 1\end{array}\right],\) for what values of \(a\) and \(b\) does \(A B=\left[\begin{array}{rrr}0 & 2 a+2 b+1 & 0 \\ 3 a+4 b & 0 & 1 \\ 1 & 4 & -2\end{array}\right] ?\)

If \(A=\left[\begin{array}{cc}a^{2}-3 a+3 & 1 \\ 0 & 2 b+5\end{array}\right]\) and \(B=\left[\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right],\) for what values of \(a\) and \(b\) does \(A B=\left[\begin{array}{ll}1 & 1 \\ 1 & 0\end{array}\right] ?\)

Find \(A^{2}\) (the product \(A A\) ) and \(A^{3}\) (the prod\(\left.u c t\left(A^{2}\right) A\right)\). $$A=\left[\begin{array}{rr}-4 & 0 \\\0 & 3\end{array}\right]$$

An electronics firm makes a clock radio in two different models: one (model 380 ) with a battery backup feature and the other (model 360 ) without. It takes 1 hour and 15 minutes to manufacture each unit of the model 380 radio, and only 1 hour to manufacture each unit of the model \(360 .\) At least 500 units of the model 360 radio are to be produced. The manufacturer realizes a profit per radio of \(\$ 15\) for the model 380 and only \(\$ 10\) for the model \(360 .\) If at most 2000 hours are to be allocated to the manufacture of the two models combined, how many of each model should be made to maximize the total profit?

Let \(A=\left[\begin{array}{ll}1 & -1 \\ 1 & -1\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] .\) What is the prod- uct \(A B ?\) Is it true that if \(A\) and \(B\) are matrices such that \(A B\) is defined and all the entries of \(A B\) are zero, then either all the entries of \(A\) must be zero or all the entries of \(B\) must be zero? Explain.

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