Exercises on Solving Linear Systems Using Row-Echelon Form

The following examples illustrate how to solve linear systems by reducing their augmented matrices to row-echelon form. Each step is shown explicitly to highlight the underlying algebraic structure.

Exercise 1

Consider the system

$$ \begin{cases} x - y = 3 \\ x + y = 9 \end{cases} $$

To solve it, I rewrite the system as an augmented matrix and then reduce it to row-echelon form.

$$ A|B = \begin{pmatrix} 1 & -1 & 3 \\ 1 & 1 & 9 \end{pmatrix} $$

The first two columns form the coefficient matrix A.

$$ A = \begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} $$

The last column is the vector of constant terms B.

$$ B = \begin{pmatrix} 3 \\ 9 \end{pmatrix} $$

I now carry out row operations to bring the matrix into row-echelon form. I subtract the first row from the second:

$$ R2 = R2 - R1 $$

$$ A|B = \begin{pmatrix} 1 & -1 & 3 \\ 1-1 & 1 - (-1) & 9 - 3 \end{pmatrix} $$

$$ A|B = \begin{pmatrix} 1 & -1 & 3 \\ 0 & 2 & 6 \end{pmatrix} $$

The matrix is now in row-echelon form. Translating it back into equations gives

$$ \begin{cases} x - y = 3 \\ 2y = 6 \end{cases} $$

which immediately yields

$$ y = 3. $$

Substituting this into the first equation,

$$ x - 3 = 3 $$

gives

$$ x = 6. $$

The solution to the system is therefore x = 6 and y = 3.

Exercise 2

Now consider the system

$$ \begin{cases} 2x - 5y = 7 \\ x - 3y = 1 \end{cases} $$

I proceed as before by writing its augmented matrix:

$$ A|B = \begin{pmatrix} 2 & -5 & 7 \\ 1 & -3 & 1 \end{pmatrix} $$

The coefficient matrix A is

$$ A = \begin{pmatrix} 2 & -5 \\ 1 & -3 \end{pmatrix} $$

and the vector of constant terms is

$$ B = \begin{pmatrix} 7 \\ 1 \end{pmatrix} $$

To simplify the reduction, I swap the two rows so that the leading coefficient in the first row is 1:

$$ R1 \Leftrightarrow R2 $$

$$ A|B = \begin{pmatrix} 1 & -3 & 1 \\ 2 & -5 & 7 \end{pmatrix} $$

I then eliminate the entry beneath the leading 1 by adding -2 times the first row to the second:

$$ R2 = R2 + R1 \cdot (-2) $$

$$ A|B = \begin{pmatrix} 1 & -3 & 1 \\ 2 - 2 & -5 + 6 & 7 - 2 \end{pmatrix} $$

$$ A|B = \begin{pmatrix} 1 & -3 & 1 \\ 0 & 1 & 5 \end{pmatrix} $$

This is row-echelon form, so the system becomes

$$ \begin{cases} x - 3y = 1 \\ y = 5 \end{cases} $$

The second equation gives y immediately. Substituting y = 5 into the first equation,

$$ x - 3 \cdot 5 = 1 $$

leads to

$$ x = 16. $$

The solution to the system is therefore x = 16 and y = 5.

Exercise 3

I now examine the following homogeneous linear system:

$$ \begin{cases} x_1 + 3x_3 - 3x_4 = 0 \\ 2x_2 -4x_3 +4x_4 = 0 \\ -x_1 + x_2 -5x_3 + x_4 = 0 \\ 2x_1 + x_2 + 4x_3 + 4x_4 = 0 \end{cases} $$

To solve it, I apply Gaussian elimination to reduce the augmented matrix to row-echelon form.

The augmented matrix is

$$ A|B = \begin{pmatrix} 1 & 0 & 3 & -3 & 0 \\ 0 & 2 & -4 & 4 & 0 \\ -1 & 1 & -5 & 1 & 0 \\ 2 & 1 & 4 & 4 & 0 \end{pmatrix} $$

I begin by eliminating the entry in the first column of the third row:

R3 = R3 + R1

$$ A|B = \begin{pmatrix} 1 & 0 & 3 & -3 & 0 \\ 0 & 2 & -4 & 4 & 0 \\ 0 & 1 & -2 & -2 & 0 \\ 2 & 1 & 4 & 4 & 0 \end{pmatrix} $$

I continue by clearing the first column of the fourth row:

R4 = R4 - 2R1

$$ A|B = \begin{pmatrix} 1 & 0 & 3 & -3 & 0 \\ 0 & 2 & -4 & 4 & 0 \\ 0 & 1 & -2 & -2 & 0 \\ 0 & 1 & -2 & 10 & 0 \end{pmatrix} $$

Next I eliminate the second-column entry in the second row using the fourth row:

R2 = R2 - 2R4

$$ A|B = \begin{pmatrix} 1 & 0 & 3 & -3 & 0 \\ 0 & 0 & 0 & -16 & 0 \\ 0 & 1 & -2 & -2 & 0 \\ 0 & 1 & -2 & 10 & 0 \end{pmatrix} $$

I then eliminate the entry beneath the pivot in the second column:

R4 = R4 - R3

$$ A|B = \begin{pmatrix} 1 & 0 & 3 & -3 & 0 \\ 0 & 0 & 0 & -16 & 0 \\ 0 & 1 & -2 & -2 & 0 \\ 0 & 0 & 0 & 12 & 0 \end{pmatrix} $$

To place the pivot row in the proper position, I swap rows 2 and 3:

R2 ⇔ R3

$$ A|B = \begin{pmatrix} 1 & 0 & 3 & -3 & 0 \\ 0 & 1 & -2 & -2 & 0 \\ 0 & 0 & 0 & -16 & 0 \\ 0 & 0 & 0 & 12 & 0 \end{pmatrix} $$

I scale the third row to make the pivot more evident:

R3 = 3R3

$$ A|B = \begin{pmatrix} 1 & 0 & 3 & -3 & 0 \\ 0 & 1 & -2 & -2 & 0 \\ 0 & 0 & 0 & -48 & 0 \\ 0 & 0 & 0 & 12 & 0 \end{pmatrix} $$

I also scale the fourth row for consistency:

R4 = 4R4

$$ A|B = \begin{pmatrix} 1 & 0 & 3 & -3 & 0 \\ 0 & 1 & -2 & -2 & 0 \\ 0 & 0 & 0 & -48 & 0 \\ 0 & 0 & 0 & 48 & 0 \end{pmatrix} $$

I add the third row to the fourth to eliminate the pivot entry:

R4 = R4 + R3

$$ A|B = \begin{pmatrix} 1 & 0 & 3 & -3 & 0 \\ 0 & 1 & -2 & -2 & 0 \\ 0 & 0 & 0 & -48 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix} $$

The matrix is now in row-echelon form.

Note. The absence of a pivot in the third column shows that \(x_3\) is a free variable. The pivot positions occur in three rows, so the rank is \(r = 3\).

Since the system has four variables (\(n = 4\)), the Rouché - Capelli theorem implies that the solution set has dimension \(n - r = 1\), so there are infinitely many solutions.

$$ \infty^{n-r} = \infty^{4-3} = \infty $$

Translating the row-echelon matrix back into equations gives

$$ \begin{cases} x_1 + 3x_3 - 3x_4 = 0 \\ x_2 - 2x_3 - 2x_4 = 0 \\ -48x_4 = 0 \end{cases} $$

From the last equation I obtain \(x_4 = 0\). Substituting into the first two equations yields

$$ \begin{cases} x_1 + 3x_3 = 0 \\ x_2 - 2x_3 = 0 \\ x_4 = 0 \end{cases} $$

I now solve for \(x_1\) and \(x_2\):

$$ \begin{cases} x_1 = -3x_3 \\ x_2 = 2x_3 \\ x_4 = 0 \end{cases} $$

The variable \(x_3\) is free and may take any real value. Let \(x_3 = k\).

The general solution is therefore

$$ \begin{cases} x_1 = -3k \\ x_2 = 2k \\ x_4 = 0 \\ x_3 = k \end{cases} $$

This describes a one-parameter family of solutions, one for each real value of \(k\).