Question

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{rr} x-y-z= & 1 \\ -x+2 y-3 z= & -4 \\ 3 x-2 y-7 z= & 0 \end{array}\right. $$

Short answer

The system has infinitely many solutions: \[ (x, y, z) = (5z - 2, 4z - 3, z) \]

Step-by-step solution

Write the system of equations in matrix form

Convert the given system of equations into an augmented matrix. The system is: \[ \begin{cases} x - y - z = 1 \ -x + 2y - 3z = -4 \ 3x - 2y - 7z = 0 d{cases}\] In matrix form, this becomes: \[ \begin{pmatrix} 1 & -1 & -1 & | & 1 \ -1 & 2 & -3 & | & -4 \ 3 & -2 & -7 & | & 0 \ d{pmatrix} \]

Use row operations to get zeros below the first entry of the first column

Perform row operations to create zeros below the pivot in the first column. Add Row 1 to Row 2: \[ R2 \rightarrow R2 + R1 \] This gives the new matrix: \[ \begin{pmatrix} 1 & -1 & -1 & | & 1 \ 0 & 1 & -4 & | & -3 \ 3 & -2 & -7 & | & 0 \ d{pmatrix} \] Next, subtract 3 times Row 1 from Row 3: \[ R3 \rightarrow R3 - 3R1 \] This results in: \[ \begin{pmatrix} 1 & -1 & -1 & | & 1 \ 0 & 1 & -4 & | & -3 \ 0 & 1 & -4 & | & -3 \ d{pmatrix} \]

Eliminate the second entry in Row 3

Subtract Row 2 from Row 3: \[ R3 \rightarrow R3 - R2 \] This simplifies the matrix to: \[ \begin{pmatrix} 1 & -1 & -1 & | & 1 \ 0 & 1 & -4 & | & -3 \ 0 & 0 & 0 & | & 0 \ d{pmatrix} \]

Interpret the resulting matrix

The resulting matrix from the row operations is equivalent to the system: \[ \begin{cases} x - y - z = 1 \ y - 4z = -3 \ 0 = 0 d{cases}\] The third equation is always true, indicating a dependent system with infinitely many solutions. Express the solutions in terms of the parameter.

Express the solutions in parametric form

From the second row: \[ y = 4z - 3 \] Now substitute this into the first row: \[ x - (4z - 3) - z = 1 \] Simplify to find: \[ x - 4z + 3 - z = 1 \] \[ x - 5z + 3 = 1 \] \[ x = 5z - 2 \] Thus, the solutions are: \[ (x, y, z) = (5z - 2, 4z - 3, z) \]

Key concepts

Matrix Form

Matrix form is a way of representing a system of linear equations in a compact form using matrices.
It involves translating the system into an augmented matrix that incorporates the coefficients of the variables and the constants from the equations.
For example, in our system:

• \(x - y - z = 1\)
• \(-x + 2y - 3z = -4\)
• \(3x - 2y - 7z = 0\)

we build the augmented matrix like this: \begin{pmatrix} 1 & -1 & -1 & | & 1 \ -1 & 2 & -3 & | & -4 \ 3 & -2 & -7 & | & 0 \end{pmatrix}
The vertical bar separates the coefficients from the constants of the equations, making it easier to apply systematic methods like Gaussian elimination.

Row Operations

Row operations help us simplify the matrix in order to solve for the variables.
These operations include swapping rows, multiplying a row by a non-zero scalar, and adding or subtracting rows.
For example:

• Adding Row 1 to Row 2: \(R2 \rightarrow R2 + R1\)
• Subtracting 3 times Row 1 from Row 3: \(R3 \rightarrow R3 - 3R1\)
• Subtracting Row 2 from Row 3: \(R3 \rightarrow R3 - R2\)

These steps work together to create zeros in strategic positions, which makes the matrix easier to solve. We perform these steps to reach a simpler form of the matrix: \begin{pmatrix} 1 & -1 & -1 & | & 1 \ 0 & 1 & -4 & | & -3 \ 0 & 0 & 0 & | & 0 \end{pmatrix}.

Parametric Form

Parametric form is used when a system has infinitely many solutions.
Instead of finding one unique solution, we express the solution in terms of free variables known as parameters.
In our simplified system:

• \(x - y - z = 1\)
• \(y - 4z = -3\)
• \(0 = 0\)

Since the third equation is always true, we have a dependent system allowing infinite solutions.
To express this in parametric form, set \(z = t\) (a parameter). From the second row: \(y = 4z - 3 = 4t - 3\) Now substitute into the first row: \(x - (4t - 3) - t = 1\) Simplifying gives: \(x = 5t - 2\) This gives us: \((x, y, z) = (5t - 2, 4t - 3, t)\), showing the solutions in parametric form.

Dependent System

A dependent system has infinitely many solutions because the equations depend on each other.
This happens when one of the equations can be derived from others or a row of zeros appears after row operations.
In our case, the system \begin{cases} \(x - y - z = 1\) \( y - 4z = -3\) \(0 = 0\)\begin{cases} is dependent as evidenced by the third equation becoming \(0 = 0\).
There are infinite solutions all lying on a line or plane described in parametric form. When encountering \(0 = 0\), it indicates we should find the remaining variable solutions in terms of free variables (parameters). This type of insight helps us identify dependencies easily without solving each part individually.