Question

Solve the given set of equations, or else show that there is no solution. $$ \begin{aligned} x_{1} -x_{3}=0 \\ 3 x_{1}+x_{2}+x_{3} &=0 \\\\-x_{1}+x_{2}+2 x_{3} &=0 \end{aligned} $$

Short answer

Answer: The unique solution to the given system of linear equations is x_1 = 0, x_2 = 0, and x_3 = 0.

Step-by-step solution

Write Down the Coefficient and Augmented Matrices

For the given system of linear equations, we first write down the coefficient matrix \(A\) and the augmented matrix \([A|B]\): $$ A = \begin{bmatrix} 1 & 0 & -1 \\ 3 & 1 & 1 \\ -1 & 1 & 2 \end{bmatrix}, \quad [A|B] = \begin{bmatrix} 1 & 0 & -1 & 0\\ 3 & 1 & 1 & 0\\ -1 & 1 & 2 & 0 \end{bmatrix} $$

Check the Rank of the Coefficient and Augmented Matrices

Next, we calculate the rank of the coefficient matrix \(A\) and the augmented matrix \([A|B]\). Since each row of the augmented matrix has a zero on the right-hand side, the rank \(r(A) = r([A|B])\). As we have a 3x3 matrix with nonzero determinant, the rank turns out to be 3.

Perform Gaussian Elimination (Row Reduction) on the Augmented Matrix

Now, let's perform Gaussian elimination (row reduction) on the augmented matrix \([A|B]\) to obtain an equivalent system in row echelon form: $$ \begin{bmatrix} 1 & 0 & -1 & 0\\ 3 & 1 & 1 & 0\\ -1 & 1 & 2 & 0 \end{bmatrix} \xrightarrow{R_2-3R_1} \begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 1 & 4 & 0\\ -1 & 1 & 2 & 0 \end{bmatrix} \xrightarrow{R_3+R_1} \begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 1 & 4 & 0\\ 0 & 1 & 1 & 0\end{bmatrix} \xrightarrow{R_3-R_2} \begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 1 & 4 & 0\\ 0 & 0 & -3 & 0\end{bmatrix} $$ After these row operations, we have an equivalent system in row echelon form.

Perform Back Substitution to Find the Unique Solution

Finally, we perform back substitution to find the values of the unknowns (variables): From the row echelon form, we have the following system of equations: $$ \begin{aligned} x_1 - x_3 &= 0 \\ x_2 + 4x_3 &= 0 \\ -3x_3 &= 0 \end{aligned} $$ Using back substitution, we can solve these equations from bottom to top: $$ \begin{aligned} -3x_3 &= 0 \implies x_3 = 0 \\ x_2 + 4x_3 &= 0 \implies x_2 + 4(0) = 0 \implies x_2 = 0 \\ x_1 - x_3 &= 0 \implies x_1 - 0 = 0 \implies x_1 = 0 \end{aligned} $$ Therefore, the unique solution to the given system of linear equations is: $$ x_1 = 0, x_2 = 0, x_3 = 0 $$

Key concepts

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations. It involves performing operations on the rows of an augmented matrix, which includes both the coefficients of the variables and the constants from the right-hand side of the equations, to achieve a simpler equivalent system.

The process starts by arranging the coefficients into a matrix, and then performing row operations to form what is known as row echelon form. The goal is to create zeroes below the leading coefficients (the first nonzero number from the left in a row), which makes the system easier to solve. These operations include swapping two rows, multiplying a row by a nonzero scalar, and adding or subtracting multiples of rows from each other.

An important aspect of this method is that it can reveal whether a system is consistent (has one or more solutions), inconsistent (has no solution), or has infinitely many solutions. The augmented matrix also plays a crucial role in understanding the system's properties by aligning with its coefficient matrix.

Row Echelon Form

Row echelon form is achieved through the process of Gaussian elimination and is defined by specific characteristics. Each row must begin with a leading 1 (called a pivot) if not all zeroes, and this pivot must be to the right of the pivot in the row above. Furthermore, any rows consisting entirely of zeroes should be at the bottom of the matrix.

To explain further, here's an example using the example given in the exercise:

• The first row starts with a leading 1, and no operations are needed here.
• In the second and third rows, operations are performed to create zeros below this leading 1.
• When the matrix is in row echelon form, it can be easily used for back substitution to find the solution, thus simplifying the original problem significantly.

Back Substitution

Back substitution is the process used after converting the augmented matrix to row echelon form. Starting from the last row, where the number of unknowns matches the number of non-zero coefficients, we solve for variables one at a time and substitute back into the above equations to get the values for all unknowns.

In our example, the system has been reduced to a form where each equation has one more variable than the previous one starting from the bottom. This allows for an easy determination of the values of the variables, working our way up and