Worked examples · Sheet 2

Determinants with Laplace expansion

Theory used: Minor and cofactor Determinant of order n Laplace's theorem

Exercise 1

Compute the determinant of

A = 204531 - 1 02 .

det A = a_{22} C_{22} = 3 \cdot (- 1)^{2 + 2} 24 - 1 2

24 - 1 2 = 4 - (- 4) = 8

det A = 3 \cdot 8 = 24

The second row contains two zeros, so by Laplace’s theorem it is convenient to expand along this row, since the terms $a_{21} C_{21}$ and $a_{23} C_{23}$ are zero and only the term $a_{22} C_{22}$ remains. The cofactor $C_{22}$ has sign $(- 1)^{4} = + 1$ and is obtained by deleting the second row and the second column of the matrix. The remaining determinant of order $2$ equals $8$ , therefore $det A = 24$ .

Exercise 2

Compute the determinant of

A = 130202002141 - 1 013 .

det A = a_{22} C_{22} = 2 \cdot (- 1)^{2 + 2} 102241 - 1 13

102241 - 1 13 = 1 \cdot (- 1)^{1 + 1} 4113 + 2 \cdot (- 1)^{3 + 1} 24 - 1 1 = 1 \cdot (12 - 1) + 2 \cdot (2 + 4) = 11 + 12 = 23

det A = 2 \cdot 23 = 46

The second column has only one nonzero entry, $a_{22} = 2$ , so the expansion along the second column reduces to a single determinant of order $3$ , with sign $(- 1)^{2 + 2} = + 1$ . Deleting the second row and the second column leaves rows $1$ , $3$ , $4$ and columns $1$ , $3$ , $4$ of $A$ . The determinant of order $3$ is in turn expanded along the first column, which contains a zero, so two terms remain, both with sign $+$ , since $1 + 1$ and $3 + 1$ are even. The sum of the two terms is $23$ , therefore $det A = 46$ .

Exercise 3

Compute the determinant of

A = 301212400050 20 - 1 1 .

det A = a_{22} C_{22} = 2 \cdot (- 1)^{2 + 2} 312050 2 - 1 1

312050 2 - 1 1 = 5 \cdot (- 1)^{2 + 2} 3221 = 5 \cdot (3 - 4) = - 5

det A = 2 \cdot (- 5) = - 10

The second row has three zeros, so the expansion along this row leaves only the term $a_{22} C_{22}$ , with sign $+$ . The matrix of order $3$ obtained by deleting the second row and the second column of $A$ in turn has two zeros in the second column, therefore it is expanded along that column and the entry $5$ remains, in position $(2, 2)$ of the reduced matrix, with sign $+$ . The final determinant of order $2$ equals $- 1$ , so $det A = - 10$ . Choosing a row or column with many zeros at each step reduces the computation of a determinant of order $4$ to a single determinant of order $2$ .