Question

Solve each system of equations. If the system has no solution, state 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 solution is \(x = 11\), \(y = 8\), \(z = -2\).

Step-by-step solution

- Write the system of equations in matrix form

Transform the system of linear equations into an augmented matrix. The given system is \(\begin{cases} 2x - 3y - z = 0 \ -x + 2y + z = 5 \ 3x - 4y - z = 1 \end{cases}\). The corresponding augmented matrix is: \[ \begin{pmatrix} 2 & -3 & -1 &| & 0 \ -1 & 2 & 1 &| & 5 \ 3 & -4 & -1 &| & 1 \end{pmatrix} \]

- Perform row operations to create zeros below the first pivot

We want to eliminate the coefficients below the first pivot element (2 in the first row).First, perform the operation \(R2 + \frac{1}{2}R1 \rightarrow R2\) and \(R3 - \frac{3}{2}R1 \rightarrow R3\).The new matrix is: \[ \begin{pmatrix} 2 & -3 & -1 &| & 0 \ 0 & \frac{1}{2} & \frac{-1}{2} &| & 5 \ 0 & \frac{-1}{2} & \frac{-5}{2} &| & 1 \end{pmatrix} \]

- Create zeros below the second pivot

Next, we need to create a zero below the second pivot element (\(\frac{1}{2}\) in the second row).Perform the operation \(R3 + R2 \rightarrow R3\).The new matrix is: \[ \begin{pmatrix} 2 & -3 & -1 &| & 0 \ 0 & \frac{1}{2} & \frac{-1}{2} &| & 5 \ 0 & 0 & -3 &| & 6 \end{pmatrix} \]

- Solve for variables

Now, we can solve for the variables starting from the last row upwards.From the last row, solve for \(z\): \(-3z = 6\) which gives \(z = -2\).Substitute \(z = -2\) into the second row equation: \(\frac{1}{2}y - \frac{1}{2}(-2) = 5\) which simplifies to \(\frac{1}{2}y + 1 = 5\), thus \(y = 8\).Finally, substitute \(y = 8\) and \(z = -2\) into the first row equation: \(2x - 3(8) - (-1)(-2) = 0\) which simplifies to \(2x - 24 + 2 = 0\), thus \(2x = 22\) and \(x = 11\).

Key concepts

augmented matrix

An augmented matrix is a way to represent a system of linear equations. First, we combine the coefficient matrix and the constants into a single matrix. This helps us use matrix operations to solve the system. Here's how to form an augmented matrix:

- Write down the coefficients of each variable.
- Place these coefficients into rows for each equation.
- Add a separator, usually a vertical line, to separate the coefficients from the constant terms.

For example, for the system:\[ \begin{cases} 2x - 3y - z = 0 \ -x + 2y + z = 5 \ 3x - 4y - z = 1 \ \end{cases} \]
The augmented matrix is:\[ \begin{pmatrix} 2 & -3 & -1 & | & 0 \ -1 & 2 & 1 & | & 5 \ 3 & -4 & -1 & | & 1 \end{pmatrix} \]

This matrix format makes it easier to perform calculations to solve the equations.

row operations

Row operations are used to simplify matrices, making the equations easier to solve. There are three main types:

- Swapping two rows (Row Interchange).
- Multiplying a row by a nonzero scalar (Scaling).
- Adding or subtracting the multiples of rows (Row Addition/Subtraction).

Consider our example. To eliminate coefficients below the first pivot, we perform:
\( R2 + \frac{1}{2}R1 \rightarrow R2 \) and \( R3 - \frac{3}{2}R1 \rightarrow R3 \). The new matrix is:\[ \begin{pmatrix} 2 & -3 & -1 & | & 0 \ 0 & \frac{1}{2} & \frac{-1}{2} & | & 5 \ 0 & \frac{-1}{2} & \frac{-5}{2} & | & 1 \end{pmatrix} \]

These operations transform the system step-by-step to an easier form to solve.

Gaussian elimination

Gaussian elimination is a method to solve systems of linear equations. It involves using row operations to convert the matrix into Row Echelon Form (REF), where the goal is to form a triangular matrix. Here is a step-by-step approach:

1. Convert the system into an augmented matrix.
2. Use row operations to make zeros below each pivot.

For example, after the first steps, our augmented matrix is:\[ \begin{pmatrix} 2 & -3 & -1 & | & 0 \ 0 & \frac{1}{2} & \frac{-1}{2} & | & 5 \ 0 & \frac{-1}{2} & \frac{-5}{2} & | & 1 \end{pmatrix} \]

To create zeros below the second pivot, we perform:\( R3 + R2 \rightarrow R3 \).
The new matrix becomes:\[ \begin{pmatrix} 2 & -3 & -1 & | & 0 \ 0 & \frac{1}{2} & \frac{-1}{2} & | & 5 \ 0 & 0 & -3 & | & 6 \end{pmatrix} \]

This makes solving for variables straightforward, starting from the last row upwards.

solving equations

Once the matrix is in Row Echelon Form (or better, Reduced Row Echelon Form), solving the equations becomes simple. We start from the last equation and work our way up.

For example, from:\[ 0x + 0y - 3z = 6 \implies -3z = 6 \rightarrow z = -2 \]
Next, substitute \( z = -2 \) into the second row to solve for \( y \):\[ \frac{1}{2}y - \frac{1}{2}(-2) = 5 \rightarrow \frac{1}{2}y + 1 = 5 \rightarrow y = 8 \]
Finally, use \( y = 8 \) and \( z = -2 \) to solve for \( x \) using the first row:\[ 2x - 3(8) - (-1)(-2) = 0 \rightarrow 2x - 24 + 2 = 0 \rightarrow 2x = 22 \rightarrow x = 11 \]

Thus, the solution to the system of equations is:\( x = 11, y = 8, z = -2 \)