Problem 31 - Precalculus: Mathematical for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{array}{rcc} x+\quad\quad z+2 w & = & 6 \\ y-2 z \quad\quad& = & -3 \\ x+2 y-\quad z\quad\quad & = & -2 \\ 2 x+y+3 z-2 w & = & 0 \end{array}\right.$$

Step-by-Step Solution

Answer

The system has infinitely many solutions, parameterized by \(w\).

Step 1: Write the System in Matrix Form

First, express the given system of equations in augmented matrix form. The system is:\[\begin{align*}x + 0y + z + 2w &= 6, \y - 2z + 0w &= -3, \x + 2y - z + 0w &= -2, \2x + y + 3z - 2w &= 0.\end{align*}\]The corresponding augmented matrix is:\[\begin{bmatrix}1 & 0 & 1 & 2 & | & 6 \0 & 1 & -2 & 0 & | & -3 \1 & 2 & -1 & 0 & | & -2 \2 & 1 & 3 & -2 & | & 0\end{bmatrix}.\]

Step 2: Perform Row Operations for Row Echelon Form

Start transforming the matrix to upper triangular form using Gaussian elimination. First, subtract the first row from the third row:\[R3 = R3 - R1 \]The matrix becomes:\[\begin{bmatrix}1 & 0 & 1 & 2 & | & 6 \0 & 1 & -2 & 0 & | & -3 \0 & 2 & -2 & -2 & | & -8 \2 & 1 & 3 & -2 & | & 0\end{bmatrix}.\]Next, perform the operation to zero the first element of the fourth row:\[R4 = R4 - 2R1\]The matrix becomes:\[\begin{bmatrix}1 & 0 & 1 & 2 & | & 6 \0 & 1 & -2 & 0 & | & -3 \0 & 2 & -2 & -2 & | & -8 \0 & 1 & 1 & -6 & | & -12\end{bmatrix}.\]

Step 3: Further Simplify to Row Echelon Form

Proceed by eliminating the entry below the pivot in the second column by subtracting the second row from the fourth row:\[R4 = R4 - R2\]The matrix becomes:\[\begin{bmatrix}1 & 0 & 1 & 2 & | & 6 \0 & 1 & -2 & 0 & | & -3 \0 & 2 & -2 & -2 & | & -8 \0 & 0 & 3 & -6 & | & -9\end{bmatrix}.\]Now, scale the third row (for simplicity):\[R3 = \frac{1}{2}R3 \]The matrix becomes:\[\begin{bmatrix}1 & 0 & 1 & 2 & | & 6 \0 & 1 & -2 & 0 & | & -3 \0 & 1 & -1 & -1 & | & -4 \0 & 0 & 3 & -6 & | & -9\end{bmatrix}.\]

Step 4: Solve for Variables Using Back Substitution

With the matrix in row echelon form, begin back substitution from the last row upwards. Solve for variable in the last row:\[3z - 6w = -9 \ z - 2w = -3 \ z = 2w - 3.\]Plug this into the third row:\[y - (2w - 3) - 1w = -4 \ y - 2w + 3 - w = -4 \ y - 3w = -7 \ y = 3w - 7.\]Use the second row to find an expression:\[x + (2w - 3) + 2(3w - 7) = 6 \ x + 2w - 3 + 6w - 14 = 6 \ x + 8w = 23 \ x = -8w + 23.\]

Step 5: Write the General Solution

The system has a one-parametric family of solutions with free variable \(w\). Thus, the complete solution is:\[x = 23 - 8w, \y = 3w - 7, \z = 2w - 3, \w = w.\]This solution implies that for each value of \(w\), we can find corresponding values of \(x\), \(y\), and \(z\).

Key Concepts

Gaussian Elimination

Gaussian elimination is a fundamental technique used for solving systems of linear equations. It's a step-by-step method that transforms the original system into a simpler form, from which the solutions can be easily deduced. The primary goal is to convert the system's augmented matrix into a row-echelon form, which is an upper triangular matrix.

Here's a brief overview of the process:

• Perform row operations to create zeros below the pivot positions (the leading non-zero entry in each row).
• Focus on one column at a time, working from left to right. Create zeros below the pivot by adding or subtracting multiples of earlier rows from subsequent rows.
• Once the matrix is in row-echelon form, proceed to back substitution to find the solutions.

Gaussian elimination is powerful because it provides a systematic way to address any system of linear equations, ensuring no solution is overlooked.

Augmented Matrix

The augmented matrix is a critical concept when dealing with linear systems of equations. It streamlines the process of solving these systems by organizing all coefficients and constants concisely. By doing so, it allows for an efficient manipulation of the data using Gaussian elimination.

To construct an augmented matrix:

• Write the coefficients of the variables from each equation into rows, maintaining a consistent order of variables across all rows.
• Add the constant terms (that are on the other side of the equation) as an additional column, separated by a vertical line to indicate the division between variables and constants.

This form is advantageous because it simplifies large systems of equations, making the data easier to handle and visualize during row operations. Understanding augmented matrices is a crucial step toward mastering linear algebra.

Back Substitution

Back substitution is the process of solving a system of equations once the matrix is in row-echelon form after applying Gaussian elimination. It's like unraveling a puzzle backward, starting from the simplest equation and working your way up through the more complex ones.

The procedure involves:

• Starting with the bottom-most equation in the row-echelon form, solve for the variable in terms of the others.
• Substitute this solution back into the equations above it. This step reduces the number of unknowns in the earlier equations, gradually arriving at a complete solution.
• Continue substituting upward until all variables are resolved in terms of known quantities or parameters.

Back substitution is a satisfying conclusion to the elimination process, making the abstract problem tangible by providing explicit solutions to the variables involved.