Problem 2
Question
Verify that the given triple of real numbers is a solution to the given system. (2,-3,1); $$\begin{aligned}x_{1}+x_{2}-2 x_{3} &=-3 \\\3 x_{1}-x_{2}-7 x_{3} &=2 \\ x_{1}+x_{2}+x_{3} &=0 \\\2 x_{1}+2 x_{2}-4 x_{3} &=-6\end{aligned}$$
Step-by-Step Solution
Verified Answer
The given triple of real numbers (2, -3, 1) is a solution to the given system of equations, as substituting these values into the expressions and verifying the right-hand sides result in true equations: (1) -3 = -3, (2) 2 = 2, (3) 0 = 0, and (4) -6 = -6.
1Step 1: Identify the given triple of real numbers and the system of equations
The given triple of real numbers is (2, -3, 1), and the system of equations is as follows:
\( \begin{aligned}x_{1}+x_{2}-2 x_{3} &=-3 \\\3 x_{1}-x_{2}-7 x_{3} &=2 \\\ x_{1}+x_{2}+x_{3} &=0 \\\2 x_{1}+2 x_{2}-4 x_{3} &=-6\end{aligned}\)
2Step 2: Substitute the given values into the equations
We will substitute the values from the given triple (2, -3, 1) for \(x_1\), \(x_2\), and \(x_3\) in each of the equations:
\( \begin{aligned}(1)\,\, 2+(-3)-2(1) &= -3 \\\(2)\,\,3(2)-(-3)-7(1) &=2 \\\(3)\,\, 2+(-3)+1 &=0 \\\(4)\,\,2(2)+2(-3)-4(1) &=-6\end{aligned}\)
3Step 3: Evaluate and simplify the expressions
Now, we will evaluate and simplify the expressions on the left-hand side of each equation:
\( \begin{aligned}(1)\,\, 2 - 3 - 2 &= -3 \\\(2)\,\,6 + 3 - 7 &=2 \\\(3)\,\, 2 - 3 + 1 &=0 \\\(4)\,\,4 - 6 - 4 &=-6\end{aligned}\)
4Step 4: Verify if the expressions match the right-hand side of the equations
Finally, we will verify if the simplified expressions match the right-hand side of the corresponding equations:
(1) -3 = -3 (True)
(2) 2 = 2 (True)
(3) 0 = 0 (True)
(4) -6 = -6 (True)
Thus, as all the equations are true, we can confirm that the given triple of real numbers (2, -3, 1) is indeed a solution to the given system of equations.
Key Concepts
Solving Linear SystemsSubstitution MethodLinear AlgebraVerification of Solutions
Solving Linear Systems
Linear systems consist of two or more linear equations with multiple variables that have one or more solutions. The objective is to find a set of values that satisfies all equations simultaneously. There are various methods to solve these systems, including the substitution method, elimination method, and graphing. An important concept is the type of solutions a linear system can have: it could be a unique solution, infinitely many solutions (when equations are dependent), or no solution at all (inconsistent system). The strategies for solving linear systems must be chosen based on the structure and complexity of the equations.
It's crucial for students to understand not just how to execute the methods, but also how to interpret the results and understand the implications of what it means for equations to have consistent and consistent solutions. Visualization techniques, such as graphing the equations, can provide additional insight into where solutions might lie.
It's crucial for students to understand not just how to execute the methods, but also how to interpret the results and understand the implications of what it means for equations to have consistent and consistent solutions. Visualization techniques, such as graphing the equations, can provide additional insight into where solutions might lie.
Substitution Method
The substitution method is a technique used in linear algebra to find the solution to a system of linear equations. This involves expressing one variable in terms of the others using one equation and then substituting this expression into the other equations. This method is particularly useful when an equation in the system can be easily solved for one variable. The steps to apply this method are identifying an equation with an easy-to-isolate variable, solving for that variable, and substituting its expression in the other equations.
Steps to Solve Using Substitution
- Pick an equation and solve for one variable.
- Substitute this variable's expression into the other equations.
- Solve the new equations, now in fewer variables.
- Continue the process until all variable values are found.
Linear Algebra
Linear algebra is a branch of mathematics that deals with vector spaces, matrices, and linear equations. It is fundamental to several areas of mathematics and applied sciences, especially those involving three-dimensional space. Linear algebra allows us to perform operations on sets of equations or vectors collectively, which can streamline problem-solving and lead to more elegant solutions.
The primary objects of study in linear algebra are vectors, which represent quantities with both magnitude and direction, and matrices, which can represent coefficients of linear equations or transformations. Understanding core concepts like vector addition, scalar multiplication, matrix multiplication, and the determinant of a matrix is essential for effectively working with linear systems.
The primary objects of study in linear algebra are vectors, which represent quantities with both magnitude and direction, and matrices, which can represent coefficients of linear equations or transformations. Understanding core concepts like vector addition, scalar multiplication, matrix multiplication, and the determinant of a matrix is essential for effectively working with linear systems.
Verification of Solutions
Once potential solutions are found for a system of linear equations, it is essential to verify these solutions to ensure they are correct. Verification involves substituting the solution back into the original equations and checking if the equations hold true. This step is critical; omitting it might lead to assuming an incorrect solution is valid.
Why Verification is Important
- Confirms the accuracy of the solution.
- Helps identify any computational mistakes.
- Ensures that the solution satisfies all equations in the system.
Other exercises in this chapter
Problem 2
Use Gaussian elimination to determine the solution set to the given system. $$\begin{aligned} &4 x_{1}-x_{2}=8\\\ &2 x_{1}+x_{2}=1 \end{aligned}$$
View solution Problem 2
determine whether the given matrices are in reduced row-echelon form, row- echelon form but not reduced row-echelon form, or neither. $$\left[\begin{array}{ll}
View solution Problem 2
$$^{165}=\left[\begin{array}{c}-1^{-1}-\frac{1}{-3} \\ -\frac{5}{-1} \\ -1 \\\ 6 \\ -\frac{4}{4}\end{array}\right]$$ (a) \(b_{12}, b_{33}, b_{41}, b_{43}, b_{51
View solution Problem 3
Determine elementary matrices that reduce the given matrix to row-echelon form. $$\left[\begin{array}{rr}3 & 5 \\\1 & -2\end{array}\right]$$
View solution