Worked examples · Sheet 2

Computing the rank of a matrix

Theory used: Minor of a matrix Rank Rank and reduction to echelon form Bordered minor Bordered minors theorem (Kronecker)

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Compute the rank of the matrix

A = 124215 - 1 1 - 1 .

124215 - 1 1 - 1 R_{2} \to R_{2} - 2 R_{1} R_{3} \to R_{3} - 4 R_{1} 100 2 - 3 - 3 - 1 33 R_{3} \to R_{3} - R_{2} 100 2 - 3 0 - 1 30

rk A = 2

Elementary row operations do not change the rank, so $A$ is reduced to echelon form and the pivots are counted, applying the theorem on rank and reduction to echelon form. After the first step the second and third rows coincide, and the subtraction $R_{3} \to R_{3} - R_{2}$ reduces them to a zero row. Two pivots remain, $1$ and $- 3$ , therefore $rk A = 2$ . The result is consistent with the definition: the third row of $A$ is $2 R_{1} + R_{2}$ , so $det A = 0$ , whereas the minor $1221 = - 3$ is nonzero.

Exercise 2

Compute the rank of the matrix

A = 120124030111 11 - 1 0 .

A R_{2} \to R_{2} - 2 R_{1} R_{4} \to R_{4} - R_{1} 100020010111 1 - 1 - 1 - 1 R_{2} \leftrightarrow R_{4} 100021000111 1 - 1 - 1 - 1

R_{4} \to R_{4} - R_{3} 100021000110 1 - 1 - 1 0 ⟹ rk A = 3

The first step makes the first column zero below the pivot, but it also makes the entry of $R_{2}$ in the second column zero; since the fourth row has the entry $1$ in that column, the swap $R_{2} \leftrightarrow R_{4}$ brings a pivot into the right position. At this point the third and fourth rows coincide, and $R_{4} \to R_{4} - R_{3}$ produces a zero row. The echelon matrix has three pivots, so by the theorem on rank and reduction to echelon form $rk A = 3$ . With the definition alone one would have had to verify that $det A = 0$ and then look for a nonzero minor of order $3$ among the sixteen possible ones; the reduction gives both pieces of information together.

Exercise 3

Compute with the bordered minors theorem the rank of the matrix

A = 213101314123 .

M = 2110 = - 1 \neq = 0 (rows 1, 2; columns 1, 2)

213101314 = 2 (0 - 1) - 1 (4 - 3) + 3 (1 - 0) = 0

213101123 = 2 (0 - 2) - 1 (3 - 6) + 1 (1 - 0) = 0

rk A = 2

The minor $M$ of the first two rows and the first two columns equals $- 1$ , so $rk A \geq 2$ . The bordered minors of $M$ are obtained by adding the only remaining row, the third, and one of the two remaining columns, the third or the fourth; there are therefore $(3 - 2) (4 - 2) = 2$ of them. Both, expanded along the first row, equal $0$ . By the bordered minors theorem one concludes $rk A = 2$ without computing the other two minors of order $3$ . The result is explained by observing that the third row of $A$ is the sum of the first two.

Exercise 4

Compute the rank of the matrix

A = 123213101 01 k

as the parameter $k \in R$ varies.

M = 1221 = - 3 \neq = 0 ⟹ rk A \geq 2 for every k

123213101 = 1 (1 - 0) - 2 (2 - 0) + 1 (6 - 3) = 0

123213 01 k = 1 (k - 3) - 2 (2 k - 3) + 0 = 3 - 3 k

rk A = {23 if k = 1 if k \neq = 1

The minor $M$ of the first two rows and columns does not contain the parameter and equals $- 3$ , so the rank is at least $2$ for every value of $k$ . The first bordered minor, with the third column, is zero for every $k$ . The second bordered minor, with the fourth column, equals $3 - 3 k$ and vanishes only for $k = 1$ . If $k \neq = 1$ , there exists a nonzero minor of order $3$ , which is the maximum order for a $3 \times 4$ matrix, therefore $rk A = 3$ . If $k = 1$ , both bordered minors of $M$ are zero and by the bordered minors theorem $rk A = 2$ ; indeed for $k = 1$ the third row $(3, 3, 1, 1)$ is the sum of the first two.