Problem 20
Question
For the following exercises, solve the system by Gaussian elimination. $$ \left[\begin{array}{rr|r} -2 & 0 & 1 \\ 0 & 2 & -1 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The solution is \(x = -\frac{1}{2}\) and \(y = -\frac{1}{2}\).
1Step 1: Write Augmented Matrix
Start with the given system of equations and write it in augmented matrix form. The matrix given is already in augmented form, which is: \[ \left[\begin{array}{rr|r} -2 & 0 & 1 \ 0 & 2 & -1 \end{array}\right] \] This represents the system of equations: \[-2x + 0y = 1\] \[0x + 2y = -1\]
2Step 2: Simplify Each Row
Start by simplifying each row if necessary. Row 1 is \([-2, 0, |, 1]\), which can be simplified by dividing by \(-2\): \[R1 = \frac{1}{-2}R1\] This results in: \[R1: [1, 0, |, -\frac{1}{2}]\]. Row 2 is already simple: \[R2: [0, 2, |, -1].\]
3Step 3: Interpret the Solution
The transformed matrix is: \[ \left[\begin{array}{rr|r} 1 & 0 & -\frac{1}{2} \ 0 & 2 & -1 \end{array}\right] \] This corresponds to the equations: \[x = -\frac{1}{2}\] \[2y = -1\] Divide the second equation by 2 to solve for \(y\): \[y = -\frac{1}{2}.\]
4Step 4: Solution Verification
The final values obtained are \(x = -\frac{1}{2}\) and \(y = -\frac{1}{2}\). Substitute these back into the original equations to verify consistency: For the first equation: \(-2(-\frac{1}{2}) + 0(-\frac{1}{2}) = 1\) This simplifies to: \(1 = 1\) For the second equation: \(0(-\frac{1}{2}) + 2(-\frac{1}{2}) = -1\) This simplifies to: \(-1 = -1\) Both conditions satisfy the original equations, confirming the solution.
Key Concepts
Systems of EquationsAugmented MatrixSolution Verification
Systems of Equations
In mathematics, a **system of equations** is a set of two or more equations that share two or more unknowns. The goal is to find values for these unknowns that satisfy all equations in the system simultaneously. Typically, systems of equations can appear in different forms, such as linear or nonlinear.
Linear systems are the most common and are composed of linear equations. An example is:
Linear systems are the most common and are composed of linear equations. An example is:
- \(-2x + 0y = 1\)
- \(0x + 2y = -1\)
Augmented Matrix
An **augmented matrix** is a powerful tool used to solve systems of equations. It combines the coefficients and constants of a system into a matrix format.
The matrix used in this exercise is:\[\begin{array}{rr|r} -2 & 0 & 1 \ 0 & 2 & -1 \\end{array}\]Each row represents an equation, and each column corresponds to a variable's coefficient or constant term.
The vertical line in the matrix separates the coefficients of the variable from the constants on the right side of the equation.
The matrix used in this exercise is:\[\begin{array}{rr|r} -2 & 0 & 1 \ 0 & 2 & -1 \\end{array}\]Each row represents an equation, and each column corresponds to a variable's coefficient or constant term.
The vertical line in the matrix separates the coefficients of the variable from the constants on the right side of the equation.
- The left side: coefficients of variables.
- The right side: constant terms.
Solution Verification
**Solution verification** is a crucial step in solving systems of equations. After finding values for the unknowns, it’s important to ensure they satisfy all original equations.
This means substituing the found values back into the original set of equations.
In our solution, we obtained:
This means substituing the found values back into the original set of equations.
In our solution, we obtained:
- \(x = -\frac{1}{2}\)
- \(y = -\frac{1}{2}\)
Other exercises in this chapter
Problem 20
For the following exercises, find the determinant. $$ \left|\begin{array}{rrr} 6 & -1 & 2 \\ -4 & -3 & 5 \\ 1 & 9 & -1 \end{array}\right| $$
View solution Problem 20
For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$ \left[\begin{array}{ccc} 1 & 0 & 6 \\ -2 & 1 & 7 \\ 3 & 0 & 2 \end
View solution Problem 20
For the following exercises, use the matrices below to perform matrix multiplication. \(A=\left[\begin{array}{rr}-1 & 5 \\ 3 & 2\end{array}\right], B=\left[\beg
View solution Problem 20
For the following exercises, find the decomposition of the partial fraction for the repeating linear factors. $$ \frac{-5 x-19}{(x+4)^{2}} $$
View solution