Question

Find all solutions to the following systems. a. \(\begin{aligned} 3 x_{1}+8 x_{2}-3 x_{3}-14 x_{4} &=2 \\ 2 x_{1}+3 x_{2}-x_{3}-2 x_{4} &=1 \\ x_{1}-2 x_{2}+x_{3}+10 x_{4} &=0 \\ x_{1}+5 x_{2}-2 x_{3}-12 x_{4} &=1 \end{aligned}\) b. \(\begin{aligned} & x_{1}-x_{2}+x_{3}-x_{4}=0 \\\\-& x_{1}+x_{2}+x_{3}+x_{4}=0 \\ & x_{1}+x_{2}-x_{3}+x_{4}=0 \\ & x_{1}+x_{2}+x_{3}+x_{4}=0 \end{aligned}\) c. \(\begin{aligned} x_{1}-x_{2}+x_{3}-2 x_{4} &=1 \\\\-x_{1}+x_{2}+x_{3}+x_{4} &=-1 \\\\-x_{1}+2 x_{2}+3 x_{3}-x_{4} &=2 \\ x_{1}-x_{2}+2 x_{3}+x_{4} &=1 \end{aligned}\) d. \(\begin{aligned} x_{1}+x_{2}+2 x_{3}-x_{4} &=4 \\ 3 x_{2}-x_{3}+4 x_{4} &=2 \\ x_{1}+2 x_{2}-3 x_{3}+5 x_{4} &=0 \\ x_{1}+x_{2}-5 x_{3}+6 x_{4} &=-3 \end{aligned}\)

Short answer

The solutions depend on each system; use Gaussian elimination and note systems with infinite solutions.

Step-by-step solution

Setup the Augmented Matrix for System (a)

First, we set up the augmented matrix for the given system of equations:\[\begin{bmatrix}3 & 8 & -3 & -14 & | & 2 \2 & 3 & -1 & -2 & | & 1 \1 & -2 & 1 & 10 & | & 0 \1 & 5 & -2 & -12 & | & 1\end{bmatrix}\]

Use Gaussian Elimination for System (a)

Apply Gaussian elimination to transform the augmented matrix into row-echelon form. Begin by eliminating the first entries in the second and third rows using row operations. Continue until you have upper triangular form.

Back Substitute for System (a)

After obtaining the row-echelon form, use back substitution to find the values for \(x_1, x_2, x_3,\) and \(x_4\). Solve the equations starting from the last row upward.

Setup the Augmented Matrix for System (b)

Construct the augmented matrix for system (b):\[\begin{bmatrix}1 & -1 & 1 & -1 & | & 0 \1 & 1 & 1 & 1 & | & 0 \1 & 1 & -1 & 1 & | & 0 \1 & 1 & 1 & 1 & | & 0\end{bmatrix}\]

Solve System (b) by Recognizing Dependent Rows

Observe that rows 2 and 4 are identical, indicating that the system is dependent. Also, row 3 can be created by a linear combination of the other rows. Simplify the system to find infinite solutions based on free variables.

Setup the Augmented Matrix for System (c)

Construct the augmented matrix for system (c):\[\begin{bmatrix}1 & -1 & 1 & -2 & | & 1 \-1 & 1 & 1 & 1 & | & -1 \-1 & 2 & 3 & -1 & | & 2 \1 & -1 & 2 & 1 & | & 1\end{bmatrix}\]

Apply Gaussian Elimination to System (c)

Use row operations to simplify the augmented matrix of system (c) to row-echelon form. Solve the resulting system by back substitution to determine the values of \(x_1, x_2, x_3,\) and \(x_4\).

Setup the Augmented Matrix for System (d)

Construct the augmented matrix for system (d):\[\begin{bmatrix}1 & 1 & 2 & -1 & | & 4 \0 & 3 & -1 & 4 & | & 2 \1 & 2 & -3 & 5 & | & 0 \1 & 1 & -5 & 6 & | & -3\end{bmatrix}\]

Transform System (d) Using Gaussian Elimination

Apply Gaussian elimination to convert the augmented matrix of system (d) to row-echelon or reduced row-echelon form. Solve for all variables using back substitution.

Key concepts

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. This is useful for applying techniques like Gaussian elimination efficiently. The augmented matrix of a system of equations includes the coefficients of the variables on the left and the constants from the other side of the equation on the right. For example, consider a simple system:

• Equation 1: \( 3x_1 + 2x_2 = 5 \)
• Equation 2: \( x_1 - 4x_2 = 7 \)

The augmented matrix for this system is:\[\begin{bmatrix} 3 & 2 & | & 5 \ 1 & -4 & | & 7 \end{bmatrix}\]The vertical bar separates the coefficients from the constants, making it clear which part of the matrix corresponds to the matrix of coefficients and which part represents the constants column vector. This setup is essential in simplifying the problem-solving process of linear systems using matrix operations.

Row-Echelon Form

The row-echelon form of a matrix is a type of matrix that results during Gaussian elimination. It is characterized by having zeros below the pivots, or leading coefficients, forming a staircase-like pattern from top-left to bottom-right.
In a row-echelon form:

• The leading entry (pivot) of each row is to the right of the leading entry of the row above it.
• All non-zero rows are above any rows of all zeros.
• Each leading entry is the only non-zero entry in its column.

An example of a row-echelon form for a 3x3 matrix is:\[\begin{bmatrix} 1 & 3 & 4 \ 0 & 2 & 5 \ 0 & 0 & 6 \end{bmatrix}\]This is not to be confused with reduced row-echelon form, where each leading entry is also 1 and is the only non-zero entry in its column. Achieving row-echelon form is a crucial step before applying back substitution to find the solutions to a system of linear equations.

Back Substitution

Back substitution is the process of solving for variables in a system of linear equations once the augmented matrix is in row-echelon form. This method begins with the last equation, which often contains only one variable, making it straightforward to compute its value.
Here’s how back substitution works:

• Start with the bottom row of the matrix. If there is only one variable, solve for it directly.
• Go to the second-to-last row and substitute the known value back into this row, then solve for the next variable.
• Continue this process upward to the first row, substituting the known values to solve for each variable step-by-step.

Ultimately, back substitution allows you to find specific solutions for each variable using the simplified equations created during Gaussian elimination.