Question

For the following system, find the augmented matrix; then, by reducing, determine whether the system has a solution. $$ \begin{aligned} 3 x-y+z &=1 \\ 7 x+y-z &=6 \\ 2 x+y-z &=2 \end{aligned} $$

Short answer

The augmented matrix for the given system of equations is: \[ \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \\ 0 & -2.5 & 2.5 & -1 \\ 0 & 0 & 0 & -1 \\ \end{array} \right] \] After applying Gaussian elimination, we found that the system is inconsistent and has no solution because the resulting matrix has a row with all zeros except for the last entry (-1), making the statement false: \(0 = -1\).

Step-by-step solution

Write the augmented matrix for the given system of equations

To find the augmented matrix, we need to write the coefficients of the variables and the constants in the matrix form. The given system of equations is: \( \begin{aligned} 3 x-y+z &=1 \\\ 7 x+y-z &=6 \\\ 2 x+y-z &=2 \end{aligned} \) The augmented matrix for this system of equations is: \[ \left[ \begin{array}{ccc|c} 3 & -1 & 1 & 1 \\ 7 & 1 & -1 & 6 \\ 2 & 1 & -1 & 2 \\ \end{array} \right] \]

Apply Gaussian elimination to reduce the augmented matrix to its simplest form

We want to get the matrix in the form of row echelon form or reduced row echelon form. To do this, we will perform elementary row operations, which include swapping rows, multiplying a row by a non-zero constant, and adding or subtracting rows. 1. Swap row 1 and row 3: \[ \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \\ 7 & 1 & -1 & 6 \\ 3 & -1 & 1 & 1 \\ \end{array} \right] \] 2. Subtract 3.5 times row 1 from row 2: \( \begin{aligned} R_2 &= R_2 - 3.5 R_1 \\\ 7 - 3.5(2) = 0 \quad & 1 - 3.5(1) = -2.5 \\\ -1 - 3.5(-1) = 2.5 \quad & 6 - 3.5(2) = -1 \end{aligned} \) setResulting in: \[ \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \\ 0 & -2.5 & 2.5 & -1 \\ 3 & -1 & 1 & 1 \\ \end{array} \right] \] 3. Subtract 1.5 times row 1 from row 3: \( \begin{aligned} R_3 &= R_3 - 1.5 R_1 \\\ 3 - 1.5(2) = 0 \quad & -1 - 1.5(1) = -2.5 \\\ 1 - 1.5(-1) = 2.5 \quad & 1 - 1.5(2) = -2 \end{aligned} \) setResulting in: \[ \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \\ 0 & -2.5 & 2.5 & -1 \\ 0 & -2.5 & 2.5 & -2 \\ \end{array} \right] \] 4. Since Row 2 and Row 3 are multiples, subtract row 2 from row 3: \[ \left[ \begin{array}{ccc|c} 2 & 1 & -1 & 2 \\ 0 & -2.5 & 2.5 & -1 \\ 0 & 0 & 0 & -1 \\ \end{array} \right] \]

Determine if the system has a solution

The matrix we obtained in step 2 is in a row echelon form. Notice that we have a row with all zeros except for the last entry, which means the corresponding equation is \(0 = -1\) Since this statement is false, the system is inconsistent, and it does not have a solution.

Key concepts

Augmented Matrix

An augmented matrix is an essential tool in linear algebra, especially when solving systems of linear equations like the one given. To construct it, you line up the coefficients of each variable from every equation in a tabular format. You also append the constants from each equation as a new column. This is called "augmented" because this extra column extends the matrix beyond the coefficient matrix. For example, given the system: \(3x - y + z = 1\), \(7x + y - z = 6\), and \(2x + y - z = 2\), the augmented matrix would be written as: \[\begin{bmatrix} 3 & -1 & 1 & | & 1 \ 7 & 1 & -1 & | & 6 \ 2 & 1 & -1 & | & 2 \end{bmatrix}\] The vertical bar partitions the coefficients from the constants, forming an easy-to-read format that facilitates further operations to solve the system.

Row Echelon Form

The Row Echelon Form (REF) is a special form of a matrix that makes solving systems of equations straightforward. It is achieved through Gaussian elimination. In REF, each leading entry of a row (other than zero) is 1 and is located to the right of any leading entries in the rows above. All the rows consisting entirely of zeros are at the bottom. This form helps reveal the structure of the system of equations, making it easier to identify solutions or determine if any exist. For instance, transforming the given system's augmented matrix into the following form is part of the process: \[\begin{bmatrix} 2 & 1 & -1 & | & 2 \ 0 & -2.5 & 2.5 & | & -1 \ 0 & 0 & 0 & | & -1 \end{bmatrix}\] This matrix shows one row of zeros followed by a non-zero constant, indicating an inconsistency in the system.

Elementary Row Operations

Elementary row operations are crucial for transforming a matrix into row echelon form. These operations are

• Row swapping: Exchanging two rows to reposition the non-zero entries strategically.
• Row multiplication: Multiplying a row by a non-zero scalar to simplify coefficients.
• Row addition/subtraction: Adding a multiple of another row to pivot or eliminate specific coefficients.

These operations aim to simplify the matrix while maintaining the equivalence of the original system of equations. For example, in the given exercise, swap and multiply operations are performed to progressively zero in on a solution, eventually revealing that the system is inconsistent due to an impossible equation appearing in the process.

Inconsistent System

An inconsistent system occurs when equations collectively form a contradiction, resulting in no available solutions. It happens when the resultant row echelon form of the matrix reveals a row where the leading coefficients are zero, but the constant term is non-zero, like \(0 = -1\). In our example, by reducing the augmented matrix, we reach a situation like: \[\begin{bmatrix} 2 & 1 & -1 & | & 2 \ 0 & -2.5 & 2.5 & | & -1 \ 0 & 0 & 0 & | & -1 \end{bmatrix}\] This highlights the inconsistency, where the third row row states an impossibility. Thus, the set of equations contradicts, indicating there can be no common solution that satisfies all equations simultaneously. Identifying such inconsistency early can save time and effort in solving an impossible setup.