Problem 51
Question
(Refer to Example \(6 .\) ) The augmented matrix is in reduced row-echelon form and represents a system of linear equations. If possible, solve the system. $$ \left[\begin{array}{ll|r} 1 & 0 & 12 \\ 0 & 1 & 3 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The solution is \(x = 12\) and \(y = 3\).
1Step 1: Understand the Augmented Matrix
The given augmented matrix is \( \begin{bmatrix} 1 & 0 & | & 12 \ 0 & 1 & | & 3 \end{bmatrix} \). This matrix corresponds to a system of linear equations, where each row represents an equation.
2Step 2: Interpret Each Row as an Equation
The first row \([1, 0 | 12]\) can be interpreted as \(x = 12\). Similarly, the second row \([0, 1 | 3]\) represents \(y = 3\).
3Step 3: Write Down the Solutions
Based on the interpretations from Step 2, the solution for the variables is \(x = 12\) and \(y = 3\).
4Step 4: Verify the Solution
Verify that these solutions satisfy each individual equation. These solutions directly come from a reduced row-echelon form, so they are consistent and check out.
Key Concepts
Row-Echelon FormAugmented MatrixSystem of Equations
Row-Echelon Form
Row-Echelon Form is a specific way of arranging a matrix that makes it easier to solve linear systems. A matrix in row-echelon form has zeros below the leading 1s (called pivot elements) in each row. Each leading 1 must be to the right of any leading 1s in the rows above. If a row is entirely zeros, it should be at the bottom. This structured format simplifies solving equations.
In a reduced row-echelon form, like the one in the original problem, additional rules apply. It's the form where every leading entry (the first non-zero number from the left in a row) is 1, and it is the only nonzero entry in its column. This makes it straightforward to see the solutions directly.
In a reduced row-echelon form, like the one in the original problem, additional rules apply. It's the form where every leading entry (the first non-zero number from the left in a row) is 1, and it is the only nonzero entry in its column. This makes it straightforward to see the solutions directly.
- Leading 1s: They make the solutions apparent as they isolate variables.
- Zeros below leading 1s: It helps simplify calculations by eliminating extra elements.
- No back and forth in calculations: The form is straightforward, clearly revealing the solution.
Augmented Matrix
An augmented matrix is a powerful tool in linear algebra to solve systems of equations. It combines the coefficients of variables and the constants they equate to in a single matrix. Instead of writing out equations separately, you condense everything into a compact form.
Consider how the matrix is set up. Each row in the augmented matrix represents one equation. The first part of the row corresponds to the coefficients of the variables, and the part after the vertical bar (often called the "augmented" part) represents the constants from each equation.
Consider how the matrix is set up. Each row in the augmented matrix represents one equation. The first part of the row corresponds to the coefficients of the variables, and the part after the vertical bar (often called the "augmented" part) represents the constants from each equation.
- Each row corresponds to one equation in the system.
- The use of the vertical bar separates coefficients from constant terms.
- It simplifies operations like elimination or substitution used in solving the system.
System of Equations
A system of equations is a collection of equations with common variables. They are addressed together because you often seek a common solution for them all. In our example, the equations derived from the augmented matrix using row-echelon form have a unique solution.
Here's how it works:
Here's how it works:
- Each equation depicts a constraint on the values the variables can take.
- Solving involves finding values of the variables that satisfy all the equations simultaneously.
- These systems can often be solved by substitution, elimination, or using matrices like augmented matrices.
Other exercises in this chapter
Problem 51
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