Worked examples · Sheet 1

Determinants of order 2 and 3 with Sarrus' rule

Theory used: Determinant of order 2 Determinant of order 3 and Sarrus' rule Geometric meaning

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Compute the determinants

42 - 3 5, \frac{1}{2} \frac{1}{3} 34, a a - 1 a + 1 a (a \in R) .

42 - 3 5 = 4 \cdot 5 - (- 3) \cdot 2 = 20 + 6 = 26

\frac{1}{2} \frac{1}{3} 34 = \frac{1}{2} \cdot 4 - 3 \cdot \frac{1}{3} = 2 - 1 = 1

a a - 1 a + 1 a = a \cdot a - (a + 1) (a - 1) = a^{2} - (a^{2} - 1) = 1

For a matrix of order $2$ the determinant is the product of the main diagonal minus the product of the antidiagonal. In the first case the second product is $(- 3) \cdot 2 = - 6$ , and subtracting it is the same as adding $6$ , so the determinant is $26$ . In the second case the products are $\frac{1}{2} \cdot 4 = 2$ and $3 \cdot \frac{1}{3} = 1$ , so the determinant is $1$ . In the third case the product of the antidiagonal is the special product $(a + 1) (a - 1) = a^{2} - 1$ ; since the $a^{2}$ terms cancel, the determinant equals $1$ and is independent of $a$ .

Exercise 2

Compute with Sarrus’ rule the determinant of

A = 201142315 .

201142315201142

diagonals to the right: 2 \cdot 4 \cdot 5 + 1 \cdot 1 \cdot 1 + 3 \cdot 0 \cdot 2 = 40 + 1 + 0 = 41

diagonals to the left: 3 \cdot 4 \cdot 1 + 2 \cdot 1 \cdot 2 + 1 \cdot 0 \cdot 5 = 12 + 4 + 0 = 16

det A = 41 - 16 = 25

The first two columns are copied to the right of the matrix, so that each diagonal contains three entries. The three diagonals going down to the right start from $a_{11} = 2$ , $a_{12} = 1$ , $a_{13} = 3$ and give the products $40$ , $1$ and $0$ , with sum $41$ . The three diagonals going down to the left start from $a_{13} = 3$ and from the two copied columns, and give the products $12$ , $4$ and $0$ , with sum $16$ . The determinant is the difference between the two sums, hence $det A = 25$ .

Exercise 3

Compute with Sarrus’ rule the determinant of

A = 12 - 3 - 2 14 3 - 1 2 .

12 - 3 - 2 14 3 - 1 2 12 - 3 - 2 14

to the right: 1 \cdot 1 \cdot 2 + (- 2) \cdot (- 1) \cdot (- 3) + 3 \cdot 2 \cdot 4 = 2 - 6 + 24 = 20

to the left: 3 \cdot 1 \cdot (- 3) + 1 \cdot (- 1) \cdot 4 + (- 2) \cdot 2 \cdot 2 = - 9 - 4 - 8 = - 21

det A = 20 - (- 21) = 41

The scheme is the same as in the previous exercise; with negative numbers it is convenient to determine the sign of each product before adding. The product $(- 2) \cdot (- 1) \cdot (- 3)$ has three negative factors, so it is negative and equals $- 6$ . The sum of the diagonals to the right is $20$ , that of the diagonals to the left is $- 21$ . Since a negative number is subtracted, the result is $20 + 21$ , that is, $det A = 41$ .

Exercise 4

Compute the area of the triangle with vertices $P = (1, 1)$ , $Q = (5, 2)$ , $R = (2, 4)$ .

u = Q - P = (4, 1), v = R - P = (1, 3)

det (4113) = 12 - 1 = 11

Area (P QR) = \frac{1}{2} ∣11∣ = \frac{11}{2}

The triangle

P QR

has two sides along the vectors

u = (4, 1)

and

v = (1, 3)

applied at

P

: its area is

\frac{11}{2}

, half the area

11

of the parallelogram with sides

u

and

v

The triangle has two sides leaving from $P$ , represented by the vectors $u = Q - P$ and $v = R - P$ . The parallelogram built on $u$ and $v$ has area equal to the absolute value of the determinant of the matrix with rows $u$ and $v$ , that is, $11$ . The diagonal $QR$ divides the parallelogram into two congruent triangles, so the area of the triangle $P QR$ is $\frac{11}{2}$ . The determinant is positive because the shortest rotation taking $u$ to $v$ is counterclockwise; the area, however, is computed with the absolute value, and the order of the vectors does not change the result.