Problem 3
Question
The following matrix is the augmented matrix of a system of linear equations in the variables \(x, y,\) and \(z .\) (It is given in reduced row-echelon form.) $$ \left[\begin{array}{rrrr}{1} & {0} & {-1} & {3} \\ {0} & {1} & {2} & {5} \\\ {0} & {0} & {0} & {0}\end{array}\right] $$ (a) The leading variables are _____. (b) Is the system inconsistent or dependent? _____. (c) The solution of the system is: x=_____, y=_____, z=_____.
Step-by-Step Solution
Verified Answer
(a) x, y; (b) dependent; (c) x = 3 + z, y = 5 - 2z, z = z.
1Step 1: Identify Leading Variables
Look at the leading coefficients (1s) of each row in the augmented matrix. The first row has a leading 1 in the first column corresponding to the variable \(x\). The second row has a leading 1 in the second column corresponding to the variable \(y\). Thus, the leading variables are \(x\) and \(y\).
2Step 2: Determine System Consistency
Notice that the last row consists entirely of zeros (\(0 = 0\)). This indicates that the system does not have any conflicting equations, hence it is consistent. Since there is at least one variable without a leading 1 (\(z\) is the free variable), the system has infinitely many solutions and is dependent.
3Step 3: Express Solution in Terms of Free Variable
Since \(z\) is a free variable, we need to express \(x\) and \(y\) in terms of \(z\). Using the first row of the matrix: \(x - z = 3 \rightarrow x = 3 + z\). Using the second row: \(y + 2z = 5 \rightarrow y = 5 - 2z\). Therefore, the solutions are \(x = 3 + z\), \(y = 5 - 2z\), and \(z = z\).
Key Concepts
Reduced Row-Echelon FormFree VariablesDependent System
Reduced Row-Echelon Form
In solving systems of linear equations, reaching the reduced row-echelon form of a matrix is a key step. This form simplifies the matrix to make solving it more straightforward. In reduced row-echelon form, the matrix has the following characteristics:
- Each row that contains non-zero elements begins with 1, which is known as the leading 1.
- The leading 1 in any row is to the right of the leading 1 in the row above it.
- All rows consisting entirely of zeros are at the bottom.
- Each leading 1 is the only non-zero number in its column.
Free Variables
Free variables are an important concept when dealing with systems of linear equations. They come into play when you have more variables than unique equations. In our example, the variable \(z\) is a free variable because it does not have a leading 1 in its column in the reduced row-echelon form. Here's why free variables matter:
- They indicate that the system has infinitely many solutions.
- Free variables can take any value, making them selectable parameters in the solution.
Dependent System
A dependent system is one where the equations are linked in such a way that they do not provide distinct, unique solutions. In other words, some equations in the system are not providing any new information, which happens when the equations are multiples or combinations of each other. As is evident from the last row of all zeros, these equations translate into true but trivial identities like \(0=0\).
This kind of system characteristically has infinitely many solutions since the dependent nature introduces free variables, as discussed earlier. In our given system, \(z\) can take any value, so the solutions for \(x\) and \(y\) depend on \(z\). Consequently, for different values of \(z\), there are different solutions, confirming the system is dependent. This contrasts with independent systems, which have a single, unique solution, or inconsistent systems, which have no solutions.
This kind of system characteristically has infinitely many solutions since the dependent nature introduces free variables, as discussed earlier. In our given system, \(z\) can take any value, so the solutions for \(x\) and \(y\) depend on \(z\). Consequently, for different values of \(z\), there are different solutions, confirming the system is dependent. This contrasts with independent systems, which have a single, unique solution, or inconsistent systems, which have no solutions.
Other exercises in this chapter
Problem 3
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