Worked examples · Sheet 2

Computing determinants

Theory used: Determinant of a 2 × 2 matrix Sarrus's rule Complementary minor and cofactor Laplace expansion Properties of the determinant Determinant of a triangular matrix

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Compute the determinants

\begin{vmatrix} 3 & -2 \\ 5 & 4 \end{vmatrix}, \qquad \begin{vmatrix} 1 & 2 & -1 \\ 3 & 0 & 2 \\ -2 & 1 & 4 \end{vmatrix} .

\begin{vmatrix} 3 & -2 \\ 5 & 4 \end{vmatrix} = 3 \cdot 4 - (-2) \cdot 5 = 12 + 10 = 22

\begin{aligned} \begin{vmatrix} 1 & 2 & -1 \\ 3 & 0 & 2 \\ -2 & 1 & 4 \end{vmatrix} &= (1 \cdot 0 \cdot 4 + 2 \cdot 2 \cdot (-2) + (-1) \cdot 3 \cdot 1) \\[6pt] &\quad - \big( (-1) \cdot 0 \cdot (-2) + 1 \cdot 2 \cdot 1 + 2 \cdot 3 \cdot 4 \big) \\[6pt] &= (0 - 8 - 3) - (0 + 2 + 24) \\[6pt] &= -11 - 26 = -37 \end{aligned}

The solid diagonals give the products

0

- 8

- 3

, the dashed ones

0

2

24

For the matrix of order $2$ the determinant is the product of the main diagonal minus the product of the other diagonal; since $-2$ is negative, the second product equals $-10$ and the determinant is $12 + 10 = 22$ .

For the matrix of order $3$ we apply Sarrus’s rule. The first two columns are copied to the right, as in the diagram. The products along the three diagonals going down to the right are $1 \cdot 0 \cdot 4 = 0$ , $2 \cdot 2 \cdot (-2) = -8$ and $(-1) \cdot 3 \cdot 1 = -3$ , with sum $-11$ . The products along the three diagonals going up to the right are $(-2) \cdot 0 \cdot (-1) = 0$ , $1 \cdot 2 \cdot 1 = 2$ and $4 \cdot 3 \cdot 2 = 24$ , with sum $26$ . The determinant is the first sum minus the second, so it equals $-37$ .

Exercise 2

Compute using the Laplace expansion

\begin{vmatrix} 2 & 0 & 1 \\ 3 & -1 & 0 \\ 1 & 0 & 4 \end{vmatrix} .

\begin{aligned} \begin{vmatrix} 2 & 0 & 1 \\ 3 & -1 & 0 \\ 1 & 0 & 4 \end{vmatrix} &= a_{22}\, A_{22} = (-1) \cdot (-1)^{2 + 2} \begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} \\[6pt] &= -(8 - 1) = -7 \end{aligned}

The Laplace expansion can be carried out along any row or column, and it is convenient to choose the one with the most zeros. The second column contains two zeros, so in the expansion along this column only one term remains, that of the entry $a_{22} = -1$ . Its cofactor has sign $(-1)^{2 + 2} = +1$ , and the complementary minor is obtained by deleting the second row and the second column, that is, $\begin{vmatrix} 2 & 1 \\ 1 & 4 \end{vmatrix} = 7$ . Hence the determinant equals $(-1) \cdot 7 = -7$ . As a check, Sarrus’s rule gives $(-8 + 0 + 0) - (-1 + 0 + 0) = -7$ .

Exercise 3

Compute

\det A = \begin{vmatrix} 1 & 2 & 0 & 1 \\ 2 & 5 & 1 & 3 \\ 0 & 1 & 3 & 2 \\ 1 & 2 & 1 & 4 \end{vmatrix} .

\begin{vmatrix} 1 & 2 & 0 & 1 \\ 2 & 5 & 1 & 3 \\ 0 & 1 & 3 & 2 \\ 1 & 2 & 1 & 4 \end{vmatrix} = \begin{vmatrix} 1 & 2 & 0 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 3 & 2 \\ 0 & 0 & 1 & 3 \end{vmatrix} \qquad \begin{aligned} &R_2 \to R_2 - 2R_1 \\ &R_4 \to R_4 - R_1 \end{aligned}

\begin{aligned} \det A &= 1 \cdot \begin{vmatrix} 1 & 1 & 1 \\ 1 & 3 & 2 \\ 0 & 1 & 3 \end{vmatrix} \\[6pt] &= 1 \cdot (9 - 2) - 1 \cdot (3 - 0) + 1 \cdot (1 - 0) \\[6pt] &= 7 - 3 + 1 = 5 \end{aligned}

For a determinant of order $4$ it is convenient to first create zeros using the properties of the determinant. Adding to a row a multiple of another row does not change the determinant; we therefore subtract twice the first row from the second row ( $R_2 \to R_2 - 2R_1$ ) and the first row from the fourth row ( $R_4 \to R_4 - R_1$ ). The third row already has a zero in the first column.

In the new matrix the first column has as its only nonzero entry $a_{11} = 1$ , with sign $+$ . The Laplace expansion along the first column therefore reduces the computation to the minor of order $3$ obtained by deleting the first row and the first column. This minor is expanded along its first row, with signs $+$ , $-$ , $+$ , and equals $7 - 3 + 1 = 5$ . Hence $\det A = 5$ .

Exercise 4

Solve the equation

\begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} = 0 .

\begin{aligned} \begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} &= x\,(x^2 - 1) - 1 \cdot (x - 0) + 0 \\[6pt] &= x^3 - 2x = x\,(x^2 - 2) \end{aligned}

x\,(x^2 - 2) = 0 \iff x = 0 \ \text{ or } \ x = \pm\sqrt{2}

The determinant is a polynomial in $x$ , and we compute it by expanding along the first row. The first term is $x$ times the minor $\begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix} = x^2 - 1$ ; the second has sign $-$ and is $1$ times the minor $\begin{vmatrix} 1 & 1 \\ 0 & x \end{vmatrix} = x$ ; the third is zero because $a_{13} = 0$ . We obtain $x^3 - x - x = x^3 - 2x$ . Collecting $x$ , the polynomial factors as $x\,(x^2 - 2)$ , and a product is zero when at least one factor is. Therefore the solutions are $x = 0$ , $x = \sqrt{2}$ and $x = -\sqrt{2}$ , and for these three values the matrix is not invertible.