Problem 25 - Algebra and Trigonometry

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} 2 x+3 y-z &=1 \\ x+2 y &=3 \\ x+3 y+z &=4 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is inconsistent, so there is no solution.

Step 1: Write the system in matrix form

The system of linear equations can be represented by an augmented matrix. The system is: \( \begin{align*} 2x + 3y - z &= 1 \ x + 2y &= 3 \ x + 3y + z &= 4 \end{align*} \). We can write this as the augmented matrix: \[ \begin{bmatrix} 2 & 3 & -1 & | & 1 \ 1 & 2 & 0 & | & 3 \ 1 & 3 & 1 & | & 4 \end{bmatrix} \]

Step 2: Perform row operations to transform the matrix

We aim to convert the matrix into upper triangular form. Start by eliminating the \(x\) from the third row. Subtract the first row (scaled appropriately) from the third row to obtain:Subtract Row 2 from Row 1 to get a new Row 1:\[ \begin{bmatrix} 1 & 1 & -1 & | & -2 \ 1 & 2 & 0 & | & 3 \ 1 & 3 & 1 & | & 4 \end{bmatrix} \]Now, subtract the updated Row 1 from Row 3:\[ \begin{bmatrix} 1 & 1 & -1 & | & -2 \ 1 & 2 & 0 & | & 3 \ 0 & 2 & 2 & | & 6 \end{bmatrix} \]

Step 3: Continue row operations

Now we aim to make the leading coefficient of the second row equal to 1 and eliminate the first column of other rows:Subtract Row 1 from Row 2:\[ \begin{bmatrix} 1 & 1 & -1 & | & -2 \ 0 & 1 & 1 & | & 5 \ 0 & 2 & 2 & | & 6 \end{bmatrix} \]Subtract twice of new Row 2 from Row 3:\[ \begin{bmatrix} 1 & 1 & -1 & | & -2 \ 0 & 1 & 1 & | & 5 \ 0 & 0 & 0 & | & -4 \end{bmatrix} \]

Step 4: Analyze the transformed matrix

The augmented matrix now has a row \((0, 0, 0 | -4)\) which shows an inconsistency; this means that there is no solution to the system. A row of zeros on the left with a non-zero entry on the right indicates a contradiction.

Key Concepts

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. Instead of writing out the individual equations, we use a matrix to capture both the coefficients of the variables and the constants from the right-hand side. The augmented matrix follows the structure:

• The left side of the matrix contains the coefficients of each variable for every equation.
• The right side, separated by a vertical line or appended as the last column, holds the constants from each equation.

For instance, when given equations like:\[\begin{align*}2x + 3y - z &= 1 \x + 2y &= 3 \x + 3y + z &= 4\end{align*}\]The augmented matrix becomes:\[\begin{bmatrix}2 & 3 & -1 & | & 1 \1 & 2 & 0 & | & 3 \1 & 3 & 1 & | & 4 \end{bmatrix}\]This matrix serves as a starting point for solving the system using row operations.

Row Operations

Row operations are used to simplify augmented matrices and solve the system of linear equations they represent. These operations are essential to creating a simpler matrix that is easier to interpret. There are three basic row operations:

• Row Switching: Swap two rows within the matrix. This is useful when rearranging the matrix for simplification.
• Row Multiplication: Multiply all entries of a row by a non-zero scalar. This adjusts the row without changing the solution set.
• Row Addition: Add or subtract a multiple of one row to another row. This is particularly useful for eliminating variables to form simpler equations.

By systematically applying these operations, we can transform the augmented matrix into reduced row-echelon form or another simplified state. For example, in our problem, subtracting multiples and swapping rows are used to achieve the simplest possible form. This process mirrors the elimination method seen in solving equations.

Inconsistent System

An inconsistent system of linear equations is one that has no solutions. This can often be identified during the row operations phase, where the augmented matrix is transformed. A clear sign of inconsistency is a row in the matrix where all the coefficients of the variables are zero, but the constant on the right is not zero, such as \[0x + 0y + 0z | = b\]where \(b\) is a non-zero number.This row translates back to a mathematical statement that's impossible, for instance, "0 = -4," which cannot be true.In the given exercise, the resulting augmented matrix contains the row:\[\begin{bmatrix}0 & 0 & 0 & | & -4\end{bmatrix}\]This shows inconsistency, indicating our system has no solutions. The presence of such a row means that the original set of equations contradicts itself, making it impossible to find a common solution.