Problem 15 - Algebra and Trigonometry

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x+y+z &=4 \\ x+3 y+3 z &=10 \\ 2 x+y-z &=3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = 2 \), \( z = 1 \).

Step 1: Set Up Augmented Matrix

Convert the system of equations into an augmented matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \1 & 3 & 3 & | & 10 \2 & 1 & -1 & | & 3\end{bmatrix}\]

Step 2: Make Leading Coefficient in First Row 1

The first row already has a leading coefficient of 1, so no modification is necessary.

Step 3: Eliminate First Column Below the Leading 1

Subtract the first row from the second row and twice the first row from the third row:1. Row 2 = Row 2 - Row 12. Row 3 = Row 3 - 2 * Row 1Resulting matrix:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 2 & 2 & | & 6 \0 & -1 & -3 & | & -5\end{bmatrix}\]

Step 4: Make Leading Coefficient in Second Row 1

Divide the second row by 2 to make the leading coefficient 1:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 1 & 1 & | & 3 \0 & -1 & -3 & | & -5\end{bmatrix}\]

Step 5: Eliminate Second Column Below the Leading 1

Add the second row to the third row to eliminate the –1 below the leading 1 in the second row:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 1 & 1 & | & 3 \0 & 0 & -2 & | & -2\end{bmatrix}\]

Step 6: Make Leading Coefficient in Third Row 1

Divide the third row by -2 to get a leading 1:\[\begin{bmatrix}1 & 1 & 1 & | & 4 \0 & 1 & 1 & | & 3 \0 & 0 & 1 & | & 1\end{bmatrix}\]

Step 7: Back Substitution

Now use back substitution to find all the values:- Third row: \( z = 1 \)- Substitute \( z = 1 \) in the second row equation: \( y + 1 = 3 \Rightarrow y = 2 \)- Substitute \( y = 2 \) and \( z = 1 \) in the first row equation: \( x + 2 + 1 = 4 \Rightarrow x = 1 \)

Step 8: Verify the Solution

Substitute \( x = 1 \), \( y = 2 \), and \( z = 1 \) back into the original equations to verify that they satisfy all three equations.

Key Concepts

Augmented Matrix

An augmented matrix is a helpful tool for solving systems of linear equations. It combines the coefficients of each variable in the equations and the constants from the right-hand side into a single rectangular array. For example, consider the system of equations:

• \( x + y + z = 4 \)
• \( x + 3y + 3z = 10 \)
• \( 2x + y - z = 3 \)

This system can be transformed into the following augmented matrix:\[\begin{bmatrix} 1 & 1 & 1 & | & 4 \ 1 & 3 & 3 & | & 10 \ 2 & 1 & -1 & | & 3 \end{bmatrix}\]The vertical bar separates the coefficients of the variables from the constant terms. Essentially, this matrix represents the entire system's information in a compact form. With the augmented matrix, different methods like Gauss-Jordan elimination can be efficiently applied to solve the system.

Back Substitution

Back substitution is a method used to solve linear equations after transforming the system into an upper triangular matrix. In simpler terms, it's a way to find the unknowns in a step-by-step manner, starting from the last equation and moving upwards.Once the system is in the form:

• \( x + y + z = 4 \)
• \( 0 + y + z = 3 \)
• \( 0 + 0 + z = 1 \)

You can directly solve for the third variable first, which is \( z = 1 \). Moving upwards, substitute \( z = 1 \) into the second equation to find \( y = 2 \). Finally, use \( y = 2 \) and \( z = 1 \) in the first equation to determine \( x = 1 \).This sequential solving process ensures that all solutions are accurate and satisfy the original system of equations.

Gauss-Jordan Elimination

Gauss-Jordan elimination is an efficient method used to solve systems of linear equations by transforming the augmented matrix into a reduced row echelon form. The ultimate goal is to simplify the matrix so each row leads to a straightforward solution of the variables.This is accomplished through a series of operations:

• Swapping rows if needed to get a non-zero leading coefficient.
• Making the leading coefficient in each row 1, often by dividing the entire row by the value of the leading coefficient.
• Eliminating all other numbers in the column of the leading 1 by adding or subtracting multiples of other rows.

In the original problem, the steps showed successive elimination of coefficients below leading 1s, leading to a matrix form where each variable neatly lines up with only one non-zero coefficient:\[\begin{bmatrix} 1 & 1 & 1 & | & 4 \ 0 & 1 & 1 & | & 3 \ 0 & 0 & 1 & | & 1 \end{bmatrix}\]With this form, back substitution can be easily applied to find the solution for \( x, y, \) and \( z \), confirming the system of equations as consistent and solvable.