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Chapter 6: Problem 33

Describe what happens when Gaussian elimination is used to solve an inconsistent system.

Short Answer

Expert verified
When Gaussian elimination is used on an inconsistent system, the process generates a row in the augmented matrix that correlates to an equation like 0 = c (where c is non-zero). This signifies that the system of equations has no solution.

Step by step solution

01

Understanding Gaussian Elimination

Gaussian elimination is a method of solving a linear system. It involves performing row operations on an augmented matrix (which represents the system) to transform it into an upper triangular matrix or row-echelon form. The three row operations are: swapping two rows, multiplying a row by a nonzero scalar, and adding or subtracting one row from another.
02

Understanding Inconsistent System

In linear algebra, an inconsistent system is a system of linear equations that has no solution. This happens when there is at least one row in its augmented matrix form that represents an equation which is impossible, such as 0 = 1.
03

Gaussian Elimination on Inconsistent System

When Gaussian elimination is used on an inconsistent system, the process will generate a row in the augmented matrix that corresponds to an equation like 0 = c, where c is a nonzero number. In such cases, we can say that the original system of equations is inconsistent, meaning that it has no solution. So, in the process of Gaussian elimination, if we stumble upon a situation where we have a row that reads 0 = a nonzero number, it's a sign that the system is inconsistent.

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Most popular questions from this chapter

In Exercises \(27-36,\) find (if possible): \(\begin{array}{llll}\text { a. } A B & \text { and } & \text { b. } B A\end{array}\) $$ A=\left[\begin{array}{rrr} 1 & -1 & 1 \\ 5 & 0 & -2 \\ 3 & -2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rrr} 1 & 1 & 0 \\ 1 & -4 & 5 \\ 3 & -1 & 2 \end{array}\right] $$

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a. Evaluate: \(\left|\begin{array}{ll}a & a \\ 0 & a\end{array}\right|\) b. Evaluate: \(\left|\begin{array}{lll}a & a & a \\ 0 & a & a \\ 0 & 0 & a\end{array}\right|\) c. Evaluate: \(\left|\begin{array}{llll}a & a & a & a \\ 0 & a & a & a \\ 0 & 0 & a & a \\ 0 & 0 & 0 & a\end{array}\right|\) d. Describe the pattern in the given determinants. e. Describe the pattern in the evaluations.

In Exercises \(17-26,\) let $$ A=\left[\begin{array}{rr} -3 & -7 \\ 2 & -9 \\ 5 & 0 \end{array}\right] \text { and } B=\left[\begin{array}{rr} -5 & -1 \\ 0 & 0 \\ 3 & -4 \end{array}\right] $$ Solve each matrix equation for \(X\). $$ 2 X+A=B $$
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