Question

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

Short answer

The system has no solution and is inconsistent.

Step-by-step solution

- Represent the system of equations as an augmented matrix

Write the coefficients of the variables and constants in matrix form. The system of equations can be represented as:\resizebox{0.9\textwidth}{!}{\(\begin{bmatrix} 2 & 3 & -1 & | & 3 \ 1 & -1 & -1 & | & 0 \ -1 & 1 & 1 & | & 0 \ 1 & 1 & 3 & | & 5 \ \rowlines \end{bmatrix}\)}

- Create zeroes in the first column

Perform row operations to make the elements below the pivot in the first column zero. Subtract suitable multiples of the first row (\text{R1}):\resizebox{0.9\textwidth}{!}{\(\begin{bmatrix} 2 & 3 & -1 & | & 3 \ 1 & -1 & -1 & | & 0 \ (-1) \ -1 & 1 & 1 & | & 0 \ 1 & 1 & 3 & | & 5 \end{bmatrix} => \begin{bmatrix} 2 & 3 & -1 & | & 3 \ 0 & -\frac{7}{2} & -\frac{1}{2} & | & -\frac{3}{2} \ 0 & 2.5 & -0.5 & | & 1.5 \ 1 & 1 & 3 & | & 5 \end{bmatrix}\)}

- Create zeroes in the second column

Use the second row (\text{R2}) to create zeroes below the pivot in the second column:\resizebox{0.9\textwidth}{!}{\(\begin{bmatrix} 2 & 3 & -1 & | & 3 \ 0 & -3.5 & -0.5 & | & -1.5 \ 0 & 2.5 & -0.5 & | & 1.5 \ 0 & -0.5 & 4.5 & | & 4.5 \end{bmatrix}\)}

- Create zeroes in the third column

We need to form zeroes below and above the third pivot term across the rows for simplification:\resizebox{0.9\textwidth}{!}{\(\begin{bmatrix} 2 & 3 & -1 & | & 3 \ 0 & -3.5 & -0.5 & | & -1.5 \ 0 & 0 & 1.5 & | & 4.5 \ 0 & 0 & 2.0 & | & 6 \end{bmatrix}\)}

- Back-substitution and solve

\resizebox{0.9\textwidth}{!}{\(\begin{bmatrix} 2 & 3 & -1 & | & 3 \ 0 & -3.5 & -0.5 & | & -1.5 \ 0 & 0 & 1.5 & | & 4.5 \ \rowlines \ \begin{matrix} & & & 0 & \ 0 & 0 & 0.667 x = 2\ & x_1,x_2,x_3 \end{matrix} y \begin{bmatrix} => \begin{bvalues}x = 3 y = 6 \3z->4 \y = 2 x3kz\)}$}

Key concepts

system of linear equations

A system of linear equations consists of multiple linear equations that share common variables. The objective is to find values for these variables that satisfy all the equations simultaneously. For example, in the system of equations provided:

$$\[\[\begin{align*} 2x + 3y - z &= 3 \ x - y - z &= 0 \ -x + y + z &= 0 \ x + y + 3z &= 5 \end{align*}\]\]$$

we aim to find values for x, y, and z that make all the equations true at the same time.

Systems of linear equations can have:

• A single unique solution
• Multiple solutions
• No solution (inconsistent system)

One effective method to solve such systems is using matrices (row operations). This approach converts the system into a more manageable format, eventually leading to a straightforward solution or revealing that no solution exists.

augmented matrix

To solve systems of linear equations efficiently, we can represent them as an augmented matrix. An augmented matrix includes the coefficients of the variables and the constants from the right side of the equations combined into a single matrix. For our example, the augmented matrix is formed as follows:

$$\begin{bmatrix} 2 & 3 & -1 & | & 3 \ 1 & -1 & -1 & | & 0 \ -1 & 1 & 1 & | & 0 \ 1 & 1 & 3 & | & 5 \end{bmatrix}$$


In this matrix, each row represents a linear equation, and the elements of the columns represent the coefficients of the variables and the constants. The vertical bar separates the coefficients from the constants.

By performing row operations (adding, subtracting, or multiplying rows), we can simplify the augmented matrix step-by-step. The goal is to manipulate this matrix to make it simpler (like forming zeroes beneath pivot elements), which will help in solving the system easily.

back-substitution

Back-substitution is a method used to solve a system of equations once the augmented matrix has been simplified to an upper triangular form. In this form, all the elements below the main diagonal are zeros. For instance, after performing row operations on our example, we may get an upper triangular matrix like:

$$\begin{bmatrix} 2 & 3 & -1 & | & 3 \ 0 & -3.5 & -0.5 & | & -1.5 \ 0 & 0 & 1.5 & | & 4.5 \ 0 & 0 & 0 & | & 0 \end{bmatrix}$$

Now, we can solve the equations starting from the last non-zero row. For example, from the third row, we can solve for z:
$$1.5z = 4.5 \rightarrow z = 3$$

Then, substitute this value back into the rows above to find y and x:

$$-3.5y - 0.5(3) = -1.5 \rightarrow y = 2$$

Substitute both into the first row:

$$2x + 3(2) - 3 = 3 \rightarrow x = 0.5$$

By back-substituting step-by-step, we find the solution to the system efficiently. This method ensures that linear systems that have a solution can be solved systematically and accurately.