Chapter 4 · Determinants

Exercise 4.1 · Q1: Evaluate a $2\times2$ determinant

Exercise 4.1Determinant of a matrix of order two · Section 4.2.2

The question

Evaluate the determinant

\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}

Exam answer

\begin{aligned} \Delta &= \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} \\ &= 2(-1) - (-5)(4) \\ &= -2 + 20 \end{aligned}

$\therefore\ \Delta = 18$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A $2\times2$ determinant is one number: the product down the main diagonal minus the product up the other diagonal, $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.$ The only danger here is the signs: two of the four entries are negative, and the formula itself contains a subtraction, so a minus is subtracted.

Given, to find, and the plan

Given: A determinant of order 2 with first row $2,\ 4$ and second row $-5,\ -1$ ⁠.
To find: Its value.
Plan: (1) Name the four entries $a_{11}, a_{12}, a_{21}, a_{22}$ ⁠. (2) Apply the definition $a_{11}a_{22} - a_{21}a_{12}$ ⁠, putting brackets round every negative entry. (3) Simplify, then check.

Basics you need

1Entries and their positions

$a_{ij}$ is the entry in row $i$ ⁠, column $j$ ⁠. In a $2\times2$ array the main diagonal runs from $a_{11}$ (top left) to $a_{22}$ (bottom right). Basics: rows, columns and positions

2The $2\times2$ determinant

For a matrix $A$ of order 2 with first row $a_{11}, a_{12}$ and second row $a_{21}, a_{22}$ ⁠, the book defines $|A| = a_{11}a_{22} - a_{21}a_{12}$ (Section 4.2.2). It is the number that decides whether the two equations $a_{11}x + a_{12}y = c_1$ ⁠, $a_{21}x + a_{22}y = c_2$ have exactly one solution. Basics: where the determinant comes from

3Sign rules

A negative times a positive is negative; a negative times a negative is positive; subtracting a negative number is the same as adding its size: $-2 - (-20) = -2 + 20$ ⁠. Basics: algebra the chapter leans on

Solution

Read off the four entries
$\begin{aligned} a_{11} &= 2, & a_{12} &= 4,\\ a_{21} &= -5, & a_{22} &= -1. \end{aligned}$
Writing the entries with their addresses first means the formula in the next step can be filled in mechanically, with no chance of using the wrong diagonal. The minus signs belong to the entries $-5$ and $-1$ ⁠; they are part of the numbers, not operations.

Row number first, column number second. Basics: rows, columns and positions
Apply the definition, brackets round negatives
$\begin{aligned} \begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} &= a_{11}a_{22} - a_{21}a_{12}\\ &= 2\cdot(-1) - (-5)\cdot 4. \end{aligned}$
The first product is the main diagonal, $2$ and $-1$ ⁠. The second product is the other diagonal, $-5$ and $4$ ⁠, and it is subtracted. Brackets keep each negative entry attached to its own number, so that " $-\,(-5)\cdot 4$ " is not misread as " $-5\cdot4$ " with one minus lost.

Definition of the determinant of order 2, Section 4.2.2. Lesson: determinant of a square matrix
Work out each product, then subtract
$\begin{aligned} 2\cdot(-1) &= -2,\\ (-5)\cdot 4 &= -20,\\ (-2) - (-20) &= -2 + 20 = 18. \end{aligned}$
So the determinant equals 18.

Positive times negative is negative, so both diagonal products are negative here. Subtracting $-20$ means adding 20, which is why the answer is positive even though both products are negative. This is exactly the pattern of the book's Example 1, where $2(2) - 4(-1) = 4 + 4$ ⁠.

Subtracting a negative number adds its size. Basics: algebra the chapter leans on

Main diagonal product minus other diagonal product:

-2-(-20)=18

⁠.

Result

$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} = 18.$ (Textbook answer: 18.)

Check it with numbers

See the 18 appear when solving the equations $2x + 4y = 18$ and $-5x - y = 0$ ⁠, whose coefficients are the entries. Multiply the first equation by $-1$ and the second by $4$ (the entries $a_{22}$ and $a_{12}$ ⁠):

\begin{aligned} -2x - 4y &= -18,\\ -20x - 4y &= 0. \end{aligned}

Subtracting the second from the first removes $y$ and leaves $18x = -18$ ⁠. The number multiplying $x$ is $(-2) - (-20) = 18$ ⁠, the determinant ✓. So $x = -1$ ⁠, and then $y = -5x = 5$ ⁠; in the first equation $2(-1) + 4(5) = 18$ ✓.

Common mistakes

Losing a minus: writing $-2 - 20 = -22$ ⁠. The second product is $(-5)\cdot4 = -20$ ⁠, and subtracting it gives $+20$ ⁠.
Adding the diagonals ( $-2 + (-20) = -22$ ⁠). The definition subtracts the second diagonal product.
Using the wrong diagonals, for example $2\cdot4 - (-5)(-1) = 3$ ⁠. The products are along the diagonals, not along the rows.
Reading $|\ \ |$ as absolute value. The bars mean "determinant"; a determinant can be negative (Remark (i), Section 4.2).

What to take away

A $2\times2$ determinant with rows $a, b$ and $c, d$ equals $ad - bc$ ⁠: main diagonal product minus the other diagonal product.
Bracket every negative entry before multiplying; most errors in this chapter are sign errors.

Every fact used, and where it comes from

Fact	Used in	Source
$a_{ij}$ is the entry in row $i$ ⁠, column $j$	Step 1	Section 4.2 and Chapter 3, Section 3.2
$\|A\| = a_{11}a_{22} - a_{21}a_{12}$ for order 2	Steps 2, 3	Section 4.2.2 (Lesson: determinant of a square matrix)
$\|A\|$ means determinant, not modulus	Common mistakes	Remark (i), Section 4.2
The same calculation pattern	Step 3	Example 1
Value 18 and the check-box system	Step 3, Check	Recomputed in the site's verification script

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Exercise 4.1 · Q1: Evaluate a 2×22\times22×2 determinant

Basics you need

Solution

Read off the four entries

Apply the definition, brackets round negatives

Work out each product, then subtract

Every fact used, and where it comes from

Exercise 4.1 · Q1: Evaluate a $2\times2$ determinant