Worked examples · Sheet 3

Consistency and number of solutions with the Rouché-Capelli theorem

Theory used: Rank Bordered minors theorem (Kronecker)Rouché-Capelli theorem Choice of the free unknowns

Exercise 1

Determine whether the system

⎩ ⎨ ⎧ x + y + z = 3 x + 2 y - z = 2 2 x + 3 y = 5

is consistent and, if so, find its solutions.

det A = 112123 1 - 1 0 = 1 (0 + 3) - 1 (0 + 2) + 1 (3 - 4) = 0, 1112 = 1 \neq = 0

112123325 = 1 (10 - 6) - 1 (5 - 4) + 3 (3 - 4) = 0

rk A = rk (A ∣ b) = 2, n - r = 3 - 2 = 1

z = t ⟹ {x + y = 3 - t x + 2 y = 2 + t ⟹ y = - 1 + 2 t, x = 4 - 3 t

(x, y, z) = (4 - 3 t, - 1 + 2 t, t), t \in R

The determinant of $A$ is zero, so $rk A < 3$ ; the minor of the first two rows and the first two columns equals $1$ , therefore $rk A = 2$ . In the augmented matrix there are two bordered minors of this minor: with the column of $z$ one obtains $det A = 0$ , with the column of constant terms one again obtains $0$ . By the bordered minors theorem $rk (A ∣ b) = 2$ , and by the Rouché-Capelli theorem the system is consistent with $\infty^{3 - 2} = \infty^{1}$ solutions. To solve it, the two equations of the rows of the minor are kept, while the third, which is their sum, is discarded. The unknown $z$ does not appear in the minor, so we set $z = t$ and solve the $2 \times 2$ system in $x$ and $y$ : subtracting the first equation from the second gives $y = - 1 + 2 t$ , and the first gives $x = 3 - t - y = 4 - 3 t$ .

Exercise 2

Determine whether the system of four equations in three unknowns

⎩ ⎨ ⎧ x + y + z = 6 x - y + z = 2 2 x + y - z = 1 3 x + y + z = 8

is consistent and, if so, find its solutions.

112 1 - 1 1 11 - 1 = 1 (1 - 1) - 1 (- 1 - 2) + 1 (1 + 2) = 6 \neq = 0 ⟹ rk A = 3

1123 1 - 1 11 11 - 1 1 6218 R_{2} \to R_{2} - R_{1} R_{3} \to R_{3} - 2 R_{1}, R_{4} \to R_{4} - 3 R_{1} 1000 1 - 2 - 1 - 2 10 - 3 - 2 6 - 4 - 11 - 10

y = 2, - 2 - 3 z = - 11 ⟹ z = 3, - 4 - 2 z = - 10 ⟹ z = 3

x = 6 - y - z = 1, (x, y, z) = (1, 2, 3)

The matrix $A$ is $4 \times 3$ , so $rk A \leq 3$ ; the minor of the first three rows equals $6$ , therefore $rk A = 3$ . The augmented matrix is square of order $4$ and the system is consistent if and only if its rank is still $3$ . Instead of $det (A ∣ b)$ it is convenient to reduce: after the first step the second row gives $y = 2$ , the third gives $z = 3$ and the fourth, with $y = 2$ , again gives $z = 3$ . The four equations therefore have a common solution, and by the Rouché-Capelli theorem the ranks are equal; since the common rank is $3$ , equal to the number of unknowns, the solution is unique: $(1, 2, 3)$ . The check in the fourth equation gives $3 + 2 + 3 = 8$ .

Exercise 3

Determine whether the system

⎩ ⎨ ⎧ x + 2 y - z = 1 2 x + 4 y + z = 5 x + 2 y + 2 z = 3

is consistent.

det A = 0 (the column of y is twice that of x), 12 - 1 1 = 3 \neq = 0

rk A = 2

121 - 1 12 153 = 1 (3 - 10) + 1 (6 - 5) + 1 (4 - 1) = - 7 + 1 + 3 = - 3 \neq = 0

rk (A ∣ b) = 3 \neq = rk A

In every equation the coefficient of $y$ is twice that of $x$ , so the first two columns of $A$ are proportional and $det A = 0$ . The minor of the first two rows and of the columns of $x$ and $z$ equals $3$ , therefore $rk A = 2$ . In the augmented matrix the bordered minor of this minor with the third row and the column of constant terms equals $- 3$ , so the augmented matrix has a nonzero minor of order $3$ and $rk (A ∣ b) = 3$ . Since the two ranks are different, by the Rouché-Capelli theorem the system is inconsistent. The computation of the ranks settled the question without looking for the solutions.