Problem 36 - Algebra and Trigonometry

Question

Determine whether the system of linear equations is inconsistent or dependent. If it is dependent, find the complete solution. $$ \left\\{\begin{aligned} 3 r+2 s-3 t &=10 \\ r-s-t &=-5 \\ r+4 s-t &=20 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is dependent. Solutions: \((t, 5, t)\).

Step 1: Write the system in matrix form

First, we'll write the system of equations as an augmented matrix, which is a matrix representation that includes the coefficients of the variables and the constants from the equations.\[ \begin{bmatrix} 3 & 2 & -3 & | & 10 \ 1 & -1 & -1 & | & -5 \ 1 & 4 & -1 & | & 20 \end{bmatrix} \]

Step 2: Transform the matrix to Row Echelon Form

We will use Gaussian elimination to transform the matrix into row echelon form. Start by using the first row to eliminate the first variable from the second and third rows.Subtract row 1 multiplied by \(\frac{1}{3}\) from row 2:\[ R_2 = R_2 - \frac{1}{3}R_1 \ \begin{bmatrix} 3 & 2 & -3 & | & 10 \ 0 & -\frac{5}{3} & 0 & | & -\frac{25}{3} \ 1 & 4 & -1 & | & 20 \end{bmatrix} \]Subtract row 1 multiplied by \(\frac{1}{3}\) from row 3:\[ R_3 = R_3 - \frac{1}{3}R_1 \ \begin{bmatrix} 3 & 2 & -3 & | & 10 \ 0 & -\frac{5}{3} & 0 & | & -\frac{25}{3} \ 0 & \frac{10}{3} & 0 & | & \frac{50}{3} \end{bmatrix} \]

Step 3: Further simplify the matrix

Now, simplify row 2 by multiplying it by \(-\frac{3}{5}\):\[ R_2 = -\frac{3}{5}R_2 \ \begin{bmatrix} 3 & 2 & -3 & | & 10 \ 0 & 1 & 0 & | & 5 \ 0 & \frac{10}{3} & 0 & | & \frac{50}{3} \end{bmatrix} \]Use row 2 to eliminate the second variable in row 3:\[ R_3 = R_3 - 10R_2 \ \begin{bmatrix} 3 & 2 & -3 & | & 10 \ 0 & 1 & 0 & | & 5 \ 0 & 0 & 0 & | & 0 \end{bmatrix} \]

Step 4: Analyze the row echelon form

The matrix now looks like this:\[\begin{bmatrix}3 & 2 & -3 & | & 10 \0 & 1 & 0 & | & 5 \0 & 0 & 0 & | & 0 \\end{bmatrix}\]We can see that the third row contains all zeroes. This indicates that the system of equations might be dependent, meaning there are infinitely many solutions. We need to express one variable in terms of the others to find the complete solution set.

Step 5: Express a variable in terms of others

From row 2, we have:\[ s = 5 \]Substitute \(s = 5\) into row 1:\[ 3r + 2(5) - 3t = 10 \Rightarrow 3r - 3t = 0 \ \Rightarrow r = t \]Therefore, we have:- \( r = t \)- \( s = 5 \)So the complete solution is in terms of \(t\):\[ (r, s, t) = (t, 5, t) \]

Key Concepts

Gaussian elimination

Gaussian elimination is a powerful technique used to solve systems of linear equations. The main goal of this method is to transform a given system of equations into a simpler form, called the Row Echelon Form (REF).
To achieve this, we use three types of row operations:

• Swapping two rows.
• Multiplying a row by a non-zero constant.
• Adding or subtracting a multiple of one row from another row.

Starting from the top left corner, these operations pivot each row to make the leading coefficients down the diagonal of the matrix.
This simplifies the system, making it easier to solve through back substitution.
Using the matrix from our original system of equations as an example, Gaussian elimination helped us identify a dependent system by transforming our matrix into a form where one row was all zeros.

dependent system

A dependent system describes a set of equations where all equations essentially describe the same geometric space. In these cases, instead of having one unique solution, there are infinitely many solutions.
This happens when one of the rows in the Row Echelon Form (REF) of the system of equations has all zero coefficients but also zero on the constant side (like in the exercise's final matrix row).
Consider our system of equations after applying Gaussian elimination:

• The third row becomes all zeros, indicating that any value of the variable "t" can satisfy the system.

By expressing the other variables in terms of "t," we notice the system is reliant upon a free parameter, leading to the infinite solutions. In this case, given the equations:

• The solution was expressed as \((r, s, t) = (t, 5, t)\).

Although this set seems complex on the surface, it simply means that as "t" changes, the values of the others pivot, forming a line of solutions in the given equation space.

augmented matrix

An augmented matrix is a concise way of representing a system of linear equations. It combines both the coefficients of the variables and the constants from each equation into a single matrix.
This matrix not only simplifies computations but also aids in visualization.
For example, consider the original system of equations provided. When expressed as an augmented matrix, it looks like this:

• \[ \begin{bmatrix} 3 & 2 & -3 & | & 10 \ 1 & -1 & -1 & | & -5 \ 1 & 4 & -1 & | & 20 \end{bmatrix} \]

The vertical line "|" separates the coefficients from the constants.
It’s crucial in applying Gaussian elimination, because it allows performing row operations uniformly on equations as a whole.
Watching how this matrix changes through each calculation step is key to understanding the transformation process and interpreting the system of solutions.