Question

Solve each system of equations using matrices (row operations). If the system has no solution, say that it is inconsistent. $$ \left\\{\begin{array}{r} 2 x-3 y-z=0 \\ -x+2 y+z=5 \\ 3 x-4 y-z=1 \end{array}\right. $$

Short answer

The system is inconsistent and has no solution.

Step-by-step solution

Write the System of Equations as an Augmented Matrix

Rewrite the given system as an augmented matrix: \[\begin{bmatrix} 2 & -3 & -1 & | & 0 \ -1 & 2 & 1 & | & 5 \ 3 & -4 & -1 & | & 1 \end{bmatrix}\]

Perform Row Operations to Obtain a Row-Echelon Form

For the augmented matrix, perform row operations to make the leading coefficient of the first row 1:\( R_1 \leftrightarrow R_2 \)New matrix: \[\begin{bmatrix} -1 & 2 & 1 & | & 5 \ 2 & -3 & -1 & | & 0 \ 3 & -4 & -1 & | & 1 \end{bmatrix}\]Multiply the first row by -1: \( R_1 \rightarrow -R_1 \)New matrix: \[\begin{bmatrix} 1 & -2 & -1 & | & -5 \ 2 & -3 & -1 & | & 0 \ 3 & -4 & -1 & | & 1 \end{bmatrix}\]

Eliminate the x-Terms in the Second and Third Rows

Use the first row to eliminate the x-terms in the second and third rows:\( R_2 \rightarrow R_2 - 2R_1 \) and \( R_3 \rightarrow R_3 - 3R_1 \)Results in:\[\begin{bmatrix} 1 & -2 & -1 & | & -5 \ 0 & 1 & 1 & | & 10 \ 0 & 2 & 2 & | & 16 \end{bmatrix}\]

Eliminate the y-Terms in the Third Row

Use the second row to eliminate the y-term in the third row:\( R_3 \rightarrow R_3 - 2R_2 \)New matrix:\[\begin{bmatrix} 1 & -2 & -1 & | & -5 \ 0 & 1 & 1 & | & 10 \ 0 & 0 & 0 & | & -4 \end{bmatrix}\]

Determine Consistency

Observe the last row: \[ 0x + 0y + 0z = -4 \]This is an inconsistent system because it represents a contradiction (no solution).

Key concepts

Row Operations

Row operations are mathematical actions we perform on the rows of a matrix to solve a system of linear equations. There are three main types of row operations: exchanging two rows, multiplying a row by a non-zero scalar, and adding or subtracting a multiple of one row to another row. These operations help convert a matrix into a more manageable form, commonly the row-echelon form, making it easier to solve the system. For example, in our exercise, we exchanged rows and multiplied rows by a scalar to change our augmented matrix step-by-step.

Augmented Matrix

An augmented matrix is a compact way to represent a system of linear equations. It includes all the coefficients of the variables and the constants from the equations in a single matrix. This makes it easier to apply row operations and solve the system. In our exercise, the augmented matrix is:\[\begin{bmatrix} 2 & -3 & -1 & | & 0 \ -1 & 2 & 1 & | & 5 \ 3 & -4 & -1 & | & 1 \ \ \end{bmatrix}\]The vertical bar separates the coefficients from the constants. By manipulating this matrix, we can find solutions to the system or determine if it is inconsistent.

Row-Echelon Form

Row-echelon form is a special form of a matrix in which each subsequent row has an increasing number of leading zeros. The leading entry of each non-zero row is 1, and is the only non-zero entry in its column. Achieving this form simplifies solving a system of equations. In our exercise, we transformed the augmented matrix step-by-step using row operations until we reached: \[\begin{bmatrix} 1 & -2 & -1 & | & -5 \ 0 & 1 & 1 & | & 10 \ 0 & 0 & 0 & | & -4 \ \end{bmatrix}\]Reaching row-echelon form allowed us to see if the system was consistent or inconsistent.

Inconsistent System

An inconsistent system of equations is one that has no solutions. This happens when the equations contradict each other. You can identify an inconsistent system by looking for a row in the augmented matrix that leads to an impossible equation, like \(0x + 0y + 0z = -4\). In our exercise, we found such a row: \[\begin{bmatrix} 0 & 0 & 0 & | & -4 \ \end{bmatrix}\]This tells us there's no set of values for \(x, y, z\) that satisfies all the equations simultaneously, meaning the system is inconsistent.