Problem 21 - Precalculus: Mathematical for Calculus

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$\left\\{\begin{aligned} y-2 z &=0 \\ 2 x+3 y\quad\quad &=2 \\ -x-2 y+\quad z &=-1 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system has infinitely many solutions: \( x = -1 - 3t, y = 2t, z = t \), where \( t \) is any real number.

Step 1: Writing the System in Matrix Form

First, let's write the system of equations in matrix form. The given equations are:\[\begin{aligned} y - 2z &= 0 \ 2x + 3y &= 2 \ -x - 2y + z &= -1 \end{aligned}\] The corresponding augmented matrix is:\[\begin{bmatrix} 0 & 1 & -2 & | & 0 \ 2 & 3 & 0 & | & 2 \ -1 & -2 & 1 & | & -1 \end{bmatrix}\]

Step 2: Applying Gaussian Elimination

Our goal is to transform the matrix into row-echelon form. Start with the first row as it is. Swap the first and third rows to bring a non-zero element to the top of the first column, getting:\[\begin{bmatrix} -1 & -2 & 1 & | & -1 \ 2 & 3 & 0 & | & 2 \ 0 & 1 & -2 & | & 0 \end{bmatrix}\] Next, eliminate the first element of the second row. Add twice the first row to the second:\[\begin{bmatrix} -1 & -2 & 1 & | & -1 \ 0 & -1 & 2 & | & 0 \ 0 & 1 & -2 & | & 0 \end{bmatrix}\]

Step 3: Simplifying the Second Row

To simplify, multiply the second row by -1:\[ \begin{bmatrix} -1 & -2 & 1 & | & -1 \ 0 & 1 & -2 & | & 0 \ 0 & 1 & -2 & | & 0 \end{bmatrix} \] Now, subtract the second row from the third row to eliminate redundancy:\[ \begin{bmatrix} -1 & -2 & 1 & | & -1 \ 0 & 1 & -2 & | & 0 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \] Since the third row is now all zeros with zero as the augmented value, it doesn't provide new information, simplifying our analysis.

Step 4: Back Substitution

With the matrix in this form, back substitution can be applied. Starting from the second row: \( y - 2z = 0 \), it means \( y = 2z \). Now, use the first row: \( -x - 2y + z = -1 \), and substitute \( y = 2z \):\(-x - 2(2z) + z = -1 \Rightarrow -x - 4z + z = -1 \Rightarrow -x - 3z = -1 \Rightarrow x = -1 - 3z \).

Step 5: Writing the General Solution

The solution is expressed in terms of the parameter \( z \) (denote \( z = t \), where \( t \) can be any real number): \[ \begin{align*} x &= -1 - 3t, \ y &= 2t, \ z &= t \end{align*} \] We have infinitely many solutions parameterized by \( t \).

Key Concepts

Gaussian elimination

Gaussian elimination is a systematic method used for solving systems of linear equations. It's a way of reducing a matrix to an easier form called row-echelon form.

The method involves two main operations:

• Row switching: Swap two rows to position a heavier numerical element if needed.
• Row addition and subtraction: Change a row by adding or subtracting multiples of another row to eliminate certain variables.

By applying these operations, the matrix simplifies, allowing us to easily find the variables in the system.

In our example, we started with an augmented matrix that represented the given system of equations. The primary goal was to create zeros in all positions below the main diagonal of the matrix (upper left to lower right) in order to simplify the matrix. Once simplified, this form helps identify the critical structure needed for back substitution.

matrix form

A system of linear equations can be conveniently rewritten using matrices. The matrix form separates the numerical coefficients of the variables from the constants, making complex mathematical operations straightforward and less error-prone.

For a system of equations:

• Each equation corresponds to a row in the matrix.
• Each variable has its column.
• The right-side values create an additional column, forming the augmented matrix.

Using matrix representation, calculations such as Gaussian elimination become manageable.
For instance, our example system was converted to an augmented matrix consisting of coefficients and constant sums from the equations, simplifying further operations needed for solving the system. This transformation is crucial as a systematic approach to handling more complex systems easily.

back substitution

After Gaussian elimination, the system concludes with a row-echelon form matrix.

Back substitution is the reverse process where we determine the values of the variables by starting from the last row and working upwards.

• In this method, the bottom row typically simplifies such that one variable or none can directly be solved.
• We use this solution to solve for variables in preceding rows, gradually determining all variable values.

In the original problem, after reaching row-echelon form, some variables were directly expressed in terms of another. This systematic breakdown allows deriving a general solution in terms of one free variable, often using parameters. This step ensures that you can verify the consistency of the solutions or identify any that don't fit, guaranteeing a comprehensive understanding of the problem's solution.