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Solve the system of linear equations using the Gauss-Jordan elimination method. $$ \begin{array}{r} 5 x+3 y=9 \\ -2 x+y=-8 \end{array} $$

Short Answer

Expert verified
The solution of the given system of linear equations using the Gauss-Jordan elimination method is \(x = \frac{52}{11}\) and \(y = -\frac{7}{11}\).

Step by step solution

01

Write the augmented matrix

We will start by writing the given system of linear equations as an augmented matrix. This matrix consists of the coefficients of the variables, with an extra column for the constants on the right-hand side of the equations. $$ \left[\begin{array}{cc|c} 5 & 3 & 9 \\ -2 & 1 & -8 \end{array}\right] $$
02

Perform row operations to get a 1 in the top left corner

Our aim is to get a 1 in the top left corner. To achieve this, we can divide the first row by 5: $$ \left[\begin{array}{cc|c} 1 & \frac{3}{5} & \frac{9}{5} \\ -2 & 1 & -8 \end{array}\right] $$
03

Eliminate the -2 in the lower left corner

Now, we want to eliminate the -2 in the lower left corner to get a 0. We can do this by adding twice the first row to the second row: $$ \left[\begin{array}{cc|c} 1 & \frac{3}{5} & \frac{9}{5} \\ 0 & \frac{11}{5} & -\frac{7}{5} \end{array}\right] $$
04

Get a 1 in the bottom right non-corner entry

We want to get a 1 in the second row's non-corner entry (11/5). To do this, multiply the second row by 5/11: $$ \left[\begin{array}{cc|c} 1 & \frac{3}{5} & \frac{9}{5} \\ 0 & 1 & -\frac{7}{11} \end{array}\right] $$
05

Eliminate the 3/5 in the top right non-corner entry

Finally, our goal is to get a 0 in the first row's non-corner entry (3/5). We can do this by subtracting (3/5) times the second row from the first row: $$ \left[\begin{array}{cc|c} 1 & 0 & \frac{52}{11} \\ 0 & 1 & -\frac{7}{11} \end{array}\right] $$
06

Read the solution

Now that we have transformed the augmented matrix into reduced row echelon form, we can read the solution directly from the matrix. From the matrix, we have: $$ x = \frac{52}{11} \\ y = -\frac{7}{11} $$ Thus, the solution of the given system of linear equations is \(x = \frac{52}{11}\) and \(y = -\frac{7}{11}\).

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Key Concepts

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

Gauss-Jordan Elimination
Gauss-Jordan Elimination is a systematic method used to solve systems of linear equations. It simplifies matrices to what is known as a reduced row-echelon form (RREF) using a sequence of specific row operations. The ultimate goal of this method is to transform the matrix into a form that clearly displays the solutions to the equations.
This approach involves three primary row operations: swapping two rows, multiplying a row by a non-zero scalar, and adding or subtracting a multiple of one row to another. These operations help in systematically reducing the matrix until each variable can easily be solved.
The beauty of Gauss-Jordan Elimination is in its ability to provide a clear and direct solution by converting the entire system of equations into this simplified form. Once in reduced row-echelon form, determining the values of the variables becomes straightforward.
Augmented Matrix
An Augmented Matrix is a compact way of writing a system of linear equations. It consists of two main parts: a coefficient matrix on the left which includes all the coefficients of the variables, and an extra column on the right which includes the constants from the equations.
For example, the matrix for the system of equations \(5x + 3y = 9\) and \(-2x + y = -8\) is written as:
\[\begin{array}{cc|c} 5 & 3 & 9 \ -2 & 1 & -8 \end{array}\]
The vertical line in the notation separates the coefficients of the variables from the constants. This matrix representation offers a clearer overview of the system, allowing us to manipulate the rows in a more streamlined way as we apply row operations.
Row Operations
Row Operations are the heart of manipulating an augmented matrix to solve systems of equations. There are three types of row operations that can be used:
  • **Row Swapping**: This involves switching the positions of two rows.
  • **Row Multiplication**: This means multiplying all elements of a row by a non-zero scalar.
  • **Row Addition/Subtraction**: This entails adding or subtracting a multiple of one row to another row to create zeros in specific positions.

Each of these operations helps in progressively simplifying the matrix, moving towards a reduced form that reveals the solutions. The key is to use these operations to strategically create leading 1s and zeros as needed without altering the solution scope of the original system.
Reduced Row Echelon Form
Reduced Row Echelon Form (RREF) is the goal of the Gauss-Jordan Elimination process. When a matrix is transformed into this form, it becomes extremely simple to read off the solutions to the system of equations it represents.
In RREF, each leading entry in a row is 1, and it is the only non-zero entry in its column. Rows of zeros, if any, are located at the bottom of the matrix. By achieving this form, each column corresponds directly to a single variable, making it easy to understand the solution.
In our example, we transformed the matrix into:
\[\begin{array}{cc|c} 1 & 0 & \frac{52}{11} \ 0 & 1 & -\frac{7}{11} \end{array}\]
From this matrix, it is clear that \(x = \frac{52}{11}\) and \(y = -\frac{7}{11}\), providing a clear, neat, and final answer to the system.

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Most popular questions from this chapter

Write the given system of linear equations in matrix form. $$ \begin{array}{rr} 3 x_{1}-5 x_{2}+4 x_{3}= & 10 \\ 4 x_{1}+2 x_{2}-3 x_{3}= & -12 \\ -x_{1}+x_{3}= & -2 \end{array} $$

Let $$ A=\left[\begin{array}{ll} 3 & 0 \\ 8 & 0 \end{array}\right] \text { and } B=\left[\begin{array}{ll} 0 & 0 \\ 4 & 5 \end{array}\right] $$ Show that \(A B=0\), thereby demonstrating that for matrix multiplication the equation \(A B=0\) does not imply that one or both of the matrices \(A\) and \(B\) must be the zero matrix.

Three network consultants, Alan, Maria, and Steven, each received a year-end bonus of \(\$ 10,000\), which they decided to invest in a \(401(\mathrm{k})\) retirement plan sponsored by their employer. Under this plan, employees are allowed to place their investments in three funds: an equity index fund (I), a growth fund (II), and a global equity fund (III). The allocations of the investments (in dollars) of the three employees at the beginning of the year are summarized in the matrix $$ \begin{array}{l} \text { II }\\\ \begin{array}{c} \text { Alan } \\ A=\text { Maria } \\ \text { Steven } \end{array}\left[\begin{array}{lll} 4000 & 3000 & 3000 \\ 2000 & 5000 & 3000 \\ 2000 & 3000 & 5000 \end{array}\right] \end{array} $$ The returns of the three funds after 1 yr are given in the matrix $$ \begin{array}{r} \mathrm{I} \\ B=\mathrm{II} \\ \mathrm{III} \end{array}\left[\begin{array}{l} 0.18 \\ 0.24 \\ 0.12 \end{array}\right] $$ Which employee realized the best return on his or her investment for the year in question? The worst return?

Mr. and Mrs. Garcia have a total of $$\$ 100,000$$ to be invested in stocks, bonds, and a money market account. The stocks have a rate of return of \(12 \% /\) year, while the bonds and the money market account pay \(8 \% /\) year and \(4 \%\) year, respectively. The Garcias have stipulated that the amount invested in the money market account should be equal to the sum of \(20 \%\) of the amount invested in stocks and \(10 \%\) of the amount invested in bonds. How should the Garcias allocate their resources if they require an annual income of $$\$$ 10,000$ from their investments?

(a) write each system of equations as a matrix equation and (b) solve the system of equations by using the inverse of the coefficient matrix. $$ \begin{array}{l} 3 x-2 y=b_{1} \\ 4 x+3 y=b_{2} \\ \text { where } & \text { (i) } b_{1}=-6, b_{2}=10 \\ \text { and } & \text { (ii) } b_{1}=3, b_{2}=-2 \end{array} $$

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