Problem 32 - Algebra and Trigonometry

Question

Find the complete solution of the linear system, or show that it is inconsistent. $$ \left\\{\begin{aligned} x+y+z+w &=0 \\ x+y+2 z+2 w &=0 \\ 2 x+2 y+3 z+4 w &=1 \\ 2 x+3 y+4 z+5 w &=2 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is consistent with solution \(x = -1\), \(y = 1\), \(z = -1\), \(w = 1\).

Step 1: Write the Augmented Matrix

First, we need to write the system of equations as an augmented matrix. This will help us to perform row operations more easily.The augmented matrix is: \[\begin{bmatrix}1 & 1 & 1 & 1 & | & 0 \1 & 1 & 2 & 2 & | & 0 \2 & 2 & 3 & 4 & | & 1 \2 & 3 & 4 & 5 & | & 2\end{bmatrix}\]

Step 2: Perform Row Operations

We begin by trying to simplify the matrix using elementary row operations to get a row-echelon form.- Subtract Row 1 (R1) from Row 2 (R2): \[ R2 = R2 - R1 \ 1 - 1 = 0, \ 1 - 1 = 0, \ 2 - 1 = 1, \ 2 - 1 = 1, \ 0 - 0 = 0 \ \rightarrow \begin{bmatrix} 0 & 0 & 1 & 1 & | & 0 \end{bmatrix} \]- Subtract 2 times Row 1 (2*R1) from Row 3 (R3):\[ R3 = R3 - 2R1 \ 2 - 2 = 0, \ 2 - 2 = 0, \ 3 - 2 = 1, \ 4 - 2 = 2, \ 1 - 0 = 1 \ \rightarrow \begin{bmatrix} 0 & 0 & 1 & 2 & | & 1 \end{bmatrix} \]- Subtract 2 times Row 1 (2*R1) from Row 4 (R4):\[ R4 = R4 - 2R1 \ 2 - 2 = 0, \ 3 - 2 = 1, \ 4 - 2 = 2, \ 5 - 2 = 3, \ 2 - 0 = 2 \ \rightarrow \begin{bmatrix} 0 & 1 & 2 & 3 & | & 2 \end{bmatrix} \]

Step 3: Further Simplify the Matrix

Continue simplifying the matrix to obtain a clearer understanding.- Subtract Row 2 (R2) from Row 3 (R3):\[ R3 = R3 - R2 \ 0 - 0 = 0, \ 0 - 0 = 0, \ 1 - 1 = 0, \ 2 - 1 = 1, \ 1 - 0 = 1 \ \rightarrow \begin{bmatrix} 0 & 0 & 0 & 1 & | & 1 \end{bmatrix} \]- Subtract 2 times Row 2 (2*R2) from Row 4 (R4):\[ R4 = R4 - 2R2 \ 0 - 2 \cdot 0 = 0, \ 1 - 2 \cdot 0 = 1, \ 2 - 2 \cdot 1 = 0, \ 3 - 2 \cdot 1 = 1, \ 2 - 2 \cdot 0 = 2 \ \rightarrow \begin{bmatrix} 0 & 1 & 0 & 1 & | & 2 \end{bmatrix} \]

Step 4: Examine the Row-Echelon Form

With the matrix now in a form close to row-echelon form, examine the last row. From the augmented part of the matrix row, we find:\[0x + 0y + 0z + 1w = 1 \]which corresponds to \(w = 1\).Similarly, another row\[0x + 1y + 0z + 1w = 2\]gives \(y + w = 2\). Since \(w = 1\), this implies \(y = 1\).

Step 5: Determine the System's Consistency

Since our simplification did not lead to any contradictions (such as a statement like \(0 = 1\)), the system is consistent. We can check the consistency from the row where all coefficients are 0 and the constant is non-zero, which is not the case here.

Step 6: Solve for Variables x and z

With known values for \(y\) and \(w\), substitute into remaining equations to solve for \(x\) and \(z\):From Row 2:\[0x + 0y + 1z + 1w = 0 \ z + 1 = 0 \ z = -1\]From original equations using values collected:- Use \(x + y + 2z + 2w = 0\): \[x + 1 - 2 + 2 = 0 \x + 1 = 0 \ herefore x = -1\]}],

Step 7: Short Answer

The system is consistent with solution \(x = -1\), \(y = 1\), \(z = -1\), \(w = 1\).

Key Concepts

Augmented Matrix

An augmented matrix is a crucial tool for solving systems of linear equations. It combines the coefficients of variables and the constants from the equations into a single matrix format. This allows for efficient manipulation, especially when performing row operations.
To form the augmented matrix, we take the coefficients of the variables from each equation and align them into columns. Then, we add a vertical line to separate these coefficients from the constants on the right side of the equations.
This setup makes it easier to visualize and perform operations that can simplify the system, such as getting it into a row-echelon form.

Row Operations

Row operations are the steps we perform on the augmented matrix to simplify it. They are designed to make the matrix easier to work with and help us find solutions to the system of linear equations. There are three main types of row operations:

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

These operations aim to create zeros in strategic positions, making it easier to isolate variables. By performing row operations, we gradually transform the matrix into a more manageable form, eventually reaching the row-echelon form.

Row-Echelon Form

The row-echelon form is a simplified form of a matrix that makes solving systems of equations straightforward. In this form, each row has more leading zeros than the previous one, and the first non-zero number in a row, called the leading coefficient, is always to the right of the leading coefficient in the row above.
The row-echelon form typically looks like a stair-step pattern, making it easier to back-substitute and solve for variables. The goal is to get as many zeros below each leading coefficient as possible. Achieving row-echelon form is a major step in solving linear systems and determining their consistency.

System Consistency

System consistency refers to whether a system of equations has at least one solution. A consistent system means there is at least one set of values for the variables that satisfies all the equations simultaneously. Conversely, an inconsistent system has no solutions.
In the context of the augmented matrix and row-echelon form, we check for consistency by examining the rows of the matrix. If we encounter a row where all the coefficients of the variables are zero but the constant is non-zero, it indicates a contradiction. Such a system is inconsistent.
In our example, since we reached a row-echelon form without such contradictions, the system is consistent, indicating that there is a solution. We found this solution through back-substitution, confirming the system's consistency.