Problem 10
Question
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} 2 w+x-2 y-z &=3 \\ w-2 x+y+z &=4 \\ -w-8 x+7 y+5 z &=13 \\ 3 w+x-2 y+2 z &=6 \end{aligned}\right. $$
Step-by-Step Solution
Verified Answer
The complete solution to the system of equations is \( (w,x,y,z) = (2, -2, 3.5, -0.5) \).
1Step 1: Form the Augmented Matrix
Construct an augmented matrix that represents the system of equations. The matrix is as follows: \[\left[\begin{array}{cccc|c} 2 & 1 & -2 & -1 & 3 \ 1 & -2 & 1 & 1 & 4 \ -1 & -8 & 7 & 5 & 13 \ 3 & 1 & -2 & 2 & 6 \end{array}\right]\]
2Step 2: Perform Row Operations
Use Gaussian elimination to transform the augmented matrix to row-echelon form: \[\left[\begin{array}{cccc|c} 1 & -2 & 1 & 1 & 4 \ 0 & 5 & -4 & -3 & -5 \ 0 & -6 & 8 & 6 & 17 \ 0 & 5 & -5 & -1 & -6 \end{array}\right]\]. \[\left[\begin{array}{cccc|c} 1 & -2 & 1 & 1 & 4 \ 0 & 1 & 1 & 1 & -1 \ 0 & 0 & 10 & 2 & 23 \ 0 & 0 & 0 & -2 & 1 \end{array}\right]\].
3Step 3: Solve for Variables
Use back substitution to solve for the variables. From the last row, we find \(z = -\frac{1}{2}\). Substituting \(z\) into the third row gives \( y = \frac{23 - 2z}{10} = 3.5\). Similarly we find \(x = -1 - y - z = -2\), and \(w = 4 + 2x - y - z = 2 \).
Key Concepts
Augmented MatrixRow OperationsBack SubstitutionSystem of Equations
Augmented Matrix
When dealing with systems of equations, one efficient method is to use a matrix. This is where the concept of an augmented matrix shines. It allows us to combine the coefficients of the variables and constants from the equations into a single matrix. This matrix consists of two sections: the left side, which contains the coefficients of the variables, and the right side, which holds the constants from each equation.
For example, given a system of equations, the augmented matrix will have a form like \[\left[\begin{array}{cccc|c}a & b & c & d & p \e & f & g & h & q \i & j & k & l & r \m & n & o & p & s \\end{array}\right]\]where the vertical line separates the coefficients from the constants.
This method helps streamline the process of solving the system by focusing on numerical operations, independent of the variables.
For example, given a system of equations, the augmented matrix will have a form like \[\left[\begin{array}{cccc|c}a & b & c & d & p \e & f & g & h & q \i & j & k & l & r \m & n & o & p & s \\end{array}\right]\]where the vertical line separates the coefficients from the constants.
This method helps streamline the process of solving the system by focusing on numerical operations, independent of the variables.
Row Operations
Once you have constructed an augmented matrix, the next step is to perform row operations. These are the key to implementing Gaussian elimination, which brings the matrix to a simpler form. Row operations include:
By using these operations, our goal is to achieve a row-echelon form, where the matrix has a triangular shape with zeros below the leading diagonal. This makes it easier to backtrack and find the solution for each variable. It's like peeling layers of an onion, simplifying one step at a time.
Mastering these operations is crucial for solving the system without error.
- Swapping two rows
- Multiplying a row by a non-zero scalar
- Adding or subtracting one row from another
By using these operations, our goal is to achieve a row-echelon form, where the matrix has a triangular shape with zeros below the leading diagonal. This makes it easier to backtrack and find the solution for each variable. It's like peeling layers of an onion, simplifying one step at a time.
Mastering these operations is crucial for solving the system without error.
Back Substitution
After transforming your matrix into row-echelon form, you can utilize back substitution to find the solution for each variable. This process works backward from the last row, which will typically have the simplest equation.
For instance, if your final row is \(0x + 0y + 0z = k\), where \(k\) is a constant, this immediately tells you the variable's value. Once you know the value of one variable, substitute it back into the higher rows to solve for the others. It's a domino effect: one solution leads naturally to the next.
Think of back substitution as untying knots in reverse order to finally free the rope; it's rewarding to see all variables fall neatly into place.
For instance, if your final row is \(0x + 0y + 0z = k\), where \(k\) is a constant, this immediately tells you the variable's value. Once you know the value of one variable, substitute it back into the higher rows to solve for the others. It's a domino effect: one solution leads naturally to the next.
Think of back substitution as untying knots in reverse order to finally free the rope; it's rewarding to see all variables fall neatly into place.
System of Equations
A system of equations consists of multiple equations that share the same variables. The main goal is to find values for these variables that satisfy all equations simultaneously. Depending on the nature of the system, there may be:
In the case of linear equations, solving often relies on methods like substitution, elimination, or using matrices. By representing complex problems in a structured way, systems of equations allow us to explore relationships between variables and arrive at logical conclusions.
As you grow more comfortable with systems of equations, you'll find them indispensable in advanced math, engineering, and real-life problem solving.
- One unique solution
- No solution
- Infinitely many solutions
In the case of linear equations, solving often relies on methods like substitution, elimination, or using matrices. By representing complex problems in a structured way, systems of equations allow us to explore relationships between variables and arrive at logical conclusions.
As you grow more comfortable with systems of equations, you'll find them indispensable in advanced math, engineering, and real-life problem solving.
Other exercises in this chapter
Problem 10
Evaluate each determinant. $$ \left|\begin{array}{rr} {\frac{2}{3}} & {\frac{1}{3}} \\ {-\frac{1}{2}} & {\frac{3}{4}} \end{array}\right| $$
View solution Problem 10
Find the products AB and BA to determine.whether \(B\) is the multiplicative inverse of \(A\). $$ A=\left[\begin{array}{lll} {0} & {2} & {0} \\ {3} & {3} & {2}
View solution Problem 11
Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{ll} {1} & {3} \\ {3} & {4} \\ {5} & {6} \end{array}\right
View solution Problem 11
Write the system of linear equations represented by the augmented matrix. Use \(x, y,\) and \(z,\) or, if necessary, \(w, x, y\) and \(z,\) for the variables. $
View solution