Worked examples · Sheet 3

Row echelon reductions

Theory used: Elementary row operations Row echelon matrix and pivots Gaussian elimination

Exercise 1

Reduce the matrix

A = 12 - 1 - 1 14 21 - 3

to row echelon form and indicate the pivots.

12 - 1 - 1 14 21 - 3 R_{2} \to R_{2} - 2 R_{1} R_{3} \to R_{3} + R_{1} 100 - 1 33 2 - 3 - 1

R_{3} \to R_{3} - R_{2} 100 - 1 30 2 - 3 2

pivots: 1, 3, 2 in columns 1, 2, 3

The first pivot is the entry $a_{11} = 1$ , which is nonzero. To make the entries below it zero one subtracts twice the first row from the second and adds the first row to the third, since $2 - 2 \cdot 1 = 0$ and $- 1 + 1 = 0$ . In the submatrix formed by the last two rows the first nonzero entry is $3$ , in the second column, and it becomes the second pivot; the operation $R_{3} \to R_{3} - R_{2}$ makes the entry below it zero. The resulting matrix is in row echelon form, because each pivot lies strictly to the right of that of the previous row, and it has three pivots. A square matrix of order $3$ with three pivots is invertible by the Gauss-Jordan theorem.

Exercise 2

Reduce the matrix

A = 012214 - 2 22 404

to row echelon form and find its reduced row echelon form.

A R_{1} \leftrightarrow R_{2} 102124 2 - 2 2 044 R_{3} \to R_{3} - 2 R_{1} 100122 2 - 2 - 2 044

R_{3} \to R_{3} - R_{2} 100120 2 - 2 0 040

R_{2} \to \frac{1}{2} R_{2} 100110 2 - 1 0 020 R_{1} \to R_{1} - R_{2} 100010 3 - 1 0 - 2 20

The first column is nonzero but its first entry is $0$ , so the swap $R_{1} \leftrightarrow R_{2}$ brings a nonzero entry to the first row, which becomes the first pivot. After making the entry $2$ of the third row zero, the second and third rows coincide, therefore the difference $R_{3} - R_{2}$ is the zero row, which stays at the bottom. The row echelon form has two pivots, in columns $1$ and $2$ . To reach the reduced form one divides the second row by its pivot $2$ and then makes the entry above the second pivot zero with $R_{1} \to R_{1} - R_{2}$ . In the final matrix every pivot equals $1$ and is the only nonzero entry of its column, therefore the matrix is in reduced row echelon form.

Exercise 3

Reduce the matrix

A = 121363130252

to row echelon form and find its reduced row echelon form.

A R_{2} \to R_{2} - 2 R_{1} R_{3} \to R_{3} - R_{1} 100300 11 - 1 210 R_{3} \to R_{3} + R_{2} 100300110211

R_{2} \to R_{2} - R_{3} 100300110201 R_{1} \to R_{1} - R_{2} - 2 R_{3} 100300010001

After the first step the second column has zero entries below the first row, so no pivot lies in the second column and the second pivot is sought in the third: it is the entry $1$ of the second row. The operation $R_{3} \to R_{3} + R_{2}$ makes the entry $- 1$ below it zero and leaves as the first nonzero entry of the third row the number $1$ in the fourth column. The row echelon form therefore has three pivots, in columns $1$ , $3$ and $4$ . All pivots already equal $1$ , so for the reduced form it remains to make the entries above them zero, proceeding from the last pivot to the first. The operation $R_{2} \to R_{2} - R_{3}$ makes the entry above the third pivot in the second row zero; the last arrow combines the two operations $R_{1} \to R_{1} - R_{2}$ and $R_{1} \to R_{1} - 2 R_{3}$ , which make the entries $1$ and $2$ of the first row in columns $3$ and $4$ zero. The second column, which contains no pivot, remains $(3, 0, 0)$ .