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Questions and Answers
What is the purpose of Gaussian elimination in the context of linear systems?
What is the purpose of Gaussian elimination in the context of linear systems?
Which type of elementary row operation involves multiplying a row by a nonzero constant?
Which type of elementary row operation involves multiplying a row by a nonzero constant?
What is the purpose of making the off-diagonal elements zero during Gaussian elimination?
What is the purpose of making the off-diagonal elements zero during Gaussian elimination?
What does an augmented matrix represent in the context of linear systems?
What does an augmented matrix represent in the context of linear systems?
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Which operation would be used to make the off-diagonal elements of the first column zero during Gaussian elimination?
Which operation would be used to make the off-diagonal elements of the first column zero during Gaussian elimination?
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What is the significance of transforming a matrix into upper triangular form during Gaussian elimination?
What is the significance of transforming a matrix into upper triangular form during Gaussian elimination?
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What is the first step in solving a linear system using Gaussian elimination?
What is the first step in solving a linear system using Gaussian elimination?
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After interchanging the 2nd and 3rd rows in Gaussian elimination, what is the next step?
After interchanging the 2nd and 3rd rows in Gaussian elimination, what is the next step?
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What is the purpose of converting a matrix into upper triangular form in Gaussian elimination?
What is the purpose of converting a matrix into upper triangular form in Gaussian elimination?
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What is the final step in solving a linear system using Gaussian elimination?
What is the final step in solving a linear system using Gaussian elimination?
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What is the correct order of operations when using Gaussian elimination to solve a linear system?
What is the correct order of operations when using Gaussian elimination to solve a linear system?
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Which operation is NOT typically involved in Gaussian elimination when solving linear systems?
Which operation is NOT typically involved in Gaussian elimination when solving linear systems?
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Study Notes
Gaussian Elimination: Linear Systems
Gaussian elimination is a classic technique used in the study of linear systems. It provides a systematic approach to solving a wide range of problems involving linear equations. In this article, we discuss how Gaussian elimination can be applied to solve linear systems.
Writing the Augmented Matrix
To start, we need to represent the linear system of equations in matrix form. This is done by writing the coefficients of the variables in a rectangular array known as the matrix. For example, consider the system of equations:
[ 3x + y = 1 ] [ 2x - y = 0 ] [ x - z = 2 ]
We can represent this as the augmented matrix:
[ \begin{bmatrix} 3 & 1 & | & 1 \ 2 & -1 & | & 0 \ 1 & 0 & | & 2 \end{bmatrix} ]
Performing Elementary Row Operations
The goal in Gaussian elimination is to transform the original matrix into upper triangular form. This involves performing a series of elementary row operations, which are defined as follows:
- Interchange any two rows.
- Multiply a row by a nonzero constant.
- Add a multiple of one row to another row.
By applying these operations, we aim to eliminate all but the diagonal entries below each element above it. For instance, we would first apply operation 3 to make the off-diagonal elements of the first column zero. Then, we would use operation 1 if necessary to move up a row with a non-zero entry. Next, we'd multiply any existing non-zero row by a suitable factor and then add or subtract multiples of other rows to reduce higher order terms. Finally, we'd perform back substitution to solve for the variables.
Solving Linear Systems
Let's illustrate this process using an example. Consider the following linear system of equations:
[ \begin{align*} 2x + y &= 6 \ 4x - y &= 8 \ x - y &= 7 \end{align*} ]
To solve this system using Gaussian elimination, we start by writing the equations as the augmented matrix:
[ \begin{bmatrix} 2 & 1 & | & 6 \ 4 & -1 & | & 8 \ 1 & -1 & | & 7 \end{bmatrix} ]
First, we use operation 1 and interchange rows 2 and 3 to get:
[ \begin{bmatrix} 2 & 1 & | & 6 \ 1 & -1 & | & 7 \ 4 & -1 & | & 8 \end{bmatrix} ]
Next, we multiply row 1 by 2 and subtract it from row 2:
[ \begin{bmatrix} 2 & 1 & | & 6 \ 0 & -3 & | & -14 \ 4 & -1 & | & 8 \end{bmatrix} ]
Now, divide row 2 by 3:
[ \begin{bmatrix} 2 & 1 & | & 6 \ 0 & 1 & | & \frac{-14}{3} \ 4 & -1 & | & 8 \end{bmatrix} ]
Finally, add the second row to the third row:
[ \begin{bmatrix} 2 & 1 & | & 6 \ 0 & 1 & | & \frac{-14}{3} \ 0 & 0 & | & \frac{-8}{3} \end{bmatrix} ]
At this point, our matrix is now upper triangular, so we can perform back substitution. From the bottom row, we have (\frac{-8}{3}). Substituting this into the second row, we get:
[ \frac{-14}{3} - \left(\frac{-8}{3}\right)\cdot\frac{1}{4} = \frac{-20}{12} = -\frac{5}{3} ]
Substituting (-\frac{5}{3}) into the top row, we find:
[ \frac{11}{4} - \left(-\frac{5}{3}\right)\cdot\frac{1}{2} = \frac{13}{6} ]
Thus, the solution to the system is (\left(\frac{13}{6}, -\frac{5}{3}, \frac{-8}{3}\right)).
In conclusion, Gaussian elimination provides a powerful tool for solving linear systems. By carefully applying elementary row operations, we can convert our matrix into upper triangular form, allowing us to easily solve for the unknown variables through back substitution.
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Description
Explore the concept of Gaussian elimination, a method used to solve linear systems by transforming matrices into upper triangular form through elementary row operations. Learn how to write augmented matrices, perform row operations, and solve equations through back substitution.
