Properties of determinants

Expand­ing a $3 \times 3$ deter­mi­nant entry by entry works, but it is slow and easy to get wrong. The prop­er­ties in this les­son let you sim­plify a deter­mi­nant before expand­ing it, often until the answer is obvi­ous. They are writ­ten for rows, but every one of them is equally true for columns.

The prop­er­ties#

Prop­erty 1. A matrix and its trans­pose have the same deter­mi­nant: $|A'| = |A|$. So any­thing true for rows is true for columns.

Prop­erty 2. Inter­chang­ing two rows (or two columns) changes the sign of the deter­mi­nant.

Prop­erty 3. If two rows (or two columns) are iden­ti­cal, the deter­mi­nant is $0$. (Swap­ping them changes noth­ing, yet by Prop­erty 2 it changes the sign; the only num­ber equal to its own neg­a­tive is $0$.)

Prop­erty 4. If every entry of one row (or col­umn) is mul­ti­plied by $k$, the deter­mi­nant is mul­ti­plied by $k$. So a com­mon fac­tor of a row can be taken out­side the deter­mi­nant. It fol­lows that if two rows are pro­por­tional, the deter­mi­nant is $0$.

Prop­erty 5. For a square matrix of order $n$, $|kA| = k^n|A|$, because each of the $n$ rows con­tributes a fac­tor $k$.

Prop­erty 6. If each entry of a row is a sum of two terms, the deter­mi­nant splits into the sum of two deter­mi­nants:

$$\begin{vmatrix} a_1 + b_1 & c_1 \\ a_2 + b_2 & c_2 \end{vmatrix} = \begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix} + \begin{vmatrix} b_1 & c_1 \\ b_2 & c_2 \end{vmatrix}$$

Prop­erty 7. Adding a mul­ti­ple of one row to another row does not change the deter­mi­nant. This is writ­ten, for exam­ple, $R_2 \to R_2 - 2R_1$ (replace row 2 by row 2 minus twice row 1), or $C_2 \to C_2 - C_1$ for columns. It is the most use­ful prop­erty of all: it cre­ates zeros with­out chang­ing the value.

Prop­erty 8. The deter­mi­nant of a tri­an­gu­lar matrix (upper or lower) is the prod­uct of its diag­o­nal entries.

Prop­erty 9. For square matri­ces of the same order, $|AB| = |A||B|$.

Exam­ple 1#

With­out expand­ing, show that $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 7 & 8 & 9 \end{vmatrix} = 0$.

Row 2 is twice row 1, so the two rows are pro­por­tional. By Prop­erty 4 the deter­mi­nant is $0$.

Exam­ple 2#

Eval­u­ate $\begin{vmatrix} 101 & 102 \\ 103 & 104 \end{vmatrix}$.

Apply $C_2 \to C_2 - C_1$ (Prop­erty 7):

$$\begin{vmatrix} 101 & 1 \\ 103 & 1 \end{vmatrix} = 101 - 103 = -2$$

Expand­ing directly gives $101 \times 104 - 102 \times 103 = 10504 - 10506 = -2$, the same, but with much big­ger num­bers.

Exam­ple 3#

Show that $\begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{vmatrix} = 0$.

Apply $C_3 \to C_3 + C_2$. Each entry of the third col­umn becomes $a + b + c$:

$$\begin{vmatrix} 1 & a & a + b + c \\ 1 & b & a + b + c \\ 1 & c & a + b + c \end{vmatrix} = (a + b + c)\begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix} = 0$$

since, after tak­ing out the com­mon fac­tor, columns 1 and 3 are iden­ti­cal (Prop­erty 3).

Exam­ple 4#

Show that $\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} = (a - b)(b - c)(c - a)$.

Apply $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$ to cre­ate zeros in the first row:

$$\begin{vmatrix} 1 & 0 & 0 \\ a & b - a & c - a \\ a^2 & b^2 - a^2 & c^2 - a^2 \end{vmatrix}$$

Take $(b - a)$ out of col­umn 2 and $(c - a)$ out of col­umn 3, using $b^2 - a^2 = (b - a)(b + a)$:

$$(b - a)(c - a)\begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 1 \\ a^2 & b + a & c + a \end{vmatrix} = (b - a)(c - a)\left[(c + a) - (b + a)\right] = (b - a)(c - a)(c - b)$$

Rear­rang­ing the signs, $(b - a)(c - a)(c - b) = (a - b)(b - c)(c - a)$.

For exam­ple, with $a = 2$, $b = 3$, $c = 5$: $\begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 5 \\ 4 & 9 & 25 \end{vmatrix} = (2 - 3)(3 - 5)(5 - 2) = (-1)(-2)(3) = 6$.

Exam­ple 5#

For $A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -2 \\ 5 & 3 \end{bmatrix}$, check that $|AB| = |A||B|$.

$|A| = 8 - 3 = 5$ and $|B| = 3 + 10 = 13$, so $|A||B| = 65$.

$AB = \begin{bmatrix} 2 + 5 & -4 + 3 \\ 3 + 20 & -6 + 12 \end{bmatrix} = \begin{bmatrix} 7 & -1 \\ 23 & 6 \end{bmatrix}$, and $|AB| = 42 + 23 = 65$.

Exam­ple 6#

If $A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 3 & 2 \\ 0 & 1 & 4 \end{bmatrix}$, find $|A|$ and $|3A|$.

$|A| = 2(12 - 2) - 1(4 - 0) + 0 = 20 - 4 = 16$. By Prop­erty 5 with $n = 3$, $|3A| = 3^3 \times 16 = 27 \times 16 = 432$.

Prac­tice#

1. With­out expand­ing, find $\begin{vmatrix} 2 & 3 & 5 \\ 4 & 6 & 10 \\ 1 & 7 & 8 \end{vmatrix}$.
2. Eval­u­ate $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$ using $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$.
3. Eval­u­ate $\begin{vmatrix} 201 & 202 \\ 203 & 204 \end{vmatrix}$.
4. Eval­u­ate $\begin{vmatrix} 5 & 0 & 0 \\ 3 & -2 & 0 \\ 7 & 1 & 4 \end{vmatrix}$.
5. If $\begin{vmatrix} 3 & 1 & 4 \\ 1 & 5 & 9 \\ 2 & 6 & 5 \end{vmatrix} = -90$, find $\begin{vmatrix} 1 & 5 & 9 \\ 3 & 1 & 4 \\ 2 & 6 & 5 \end{vmatrix}$.
6. Eval­u­ate $\begin{vmatrix} 1 & 1 & 1 \\ 1 & 3 & 4 \\ 1 & 9 & 16 \end{vmatrix}$ using Exam­ple 4.
7. Show that $\begin{vmatrix} x + 1 & x + 2 \\ x + 3 & x + 4 \end{vmatrix}$ has the same value for every $x$, and find it.
8. Eval­u­ate $\begin{vmatrix} 10 & 11 & 12 \\ 13 & 14 & 15 \\ 16 & 17 & 18 \end{vmatrix}$.
9. If $|A| = 4$ for a $2 \times 2$ matrix $A$, find $|5A|$.
10. If $|A| = -3$ and $|B| = 7$ for $3 \times 3$ matri­ces, find $|AB|$ and $|2A|$.

Answers#