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Chapter 4 · Determinants

Example 1: Evaluate a 2×22\times2 determinant

ExamplesDeterminant of a matrix of order two · Section 4.2.2
The question

Evaluate

∣24−12∣.\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix}.
Exam answer Δ=∣24−12∣=2(2)−(−1)(4)=4+4\begin{aligned} \Delta &= \begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} \\ &= 2(2) - (-1)(4) \\ &= 4 + 4 \end{aligned}

∴ Δ=8\therefore\ \Delta = 8⁠.

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

The one idea this problem is built on

A 2×22\times2 determinant is a single number made from the four entries by one fixed rule: (product of the main diagonal) minus (product of the other diagonal). The main diagonal runs from top left to bottom right. The only real danger is the sign: here one entry is negative, so the second product is negative, and subtracting a negative number adds.

Given, to find, and the plan
Given
A determinant with first row 2, 42,\ 4 and second row −1, 2-1,\ 2⁠.
To find
Its value (one number).
Plan
Name the four entries by position, multiply along each diagonal, then subtract the second product from the first, keeping the minus sign of −1-1 in brackets.

Basics you need

1Positions in a square array

aija_{ij} is the entry in row ii⁠, column jj⁠: row first, column second. In a 2×22\times2 array the entries are a11a_{11}⁠, a12a_{12} (top row) and a21a_{21}⁠, a22a_{22} (bottom row). Basics: rows, columns and positions

2The 2×22\times2 rule

For a matrix of order 2, the book defines (Section 4.2.2)

∣a11a12a21a22∣=a11a22−a21a12.\begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{21}a_{12}.

This number is exactly the quantity that decides whether two linear equations have a unique solution. Basics: where the determinant comes from

3Signed arithmetic

4×(−1)=−44\times(-1) = -4⁠, and subtracting −4-4 is the same as adding 4: 4−(−4)=4+4=84 - (-4) = 4 + 4 = 8⁠. Basics: algebra the chapter leans on

4Vertical bars mean "determinant"

The bars ∣  ∣|\ \ | around an array mean the determinant, a number. They do not mean absolute value here, so a determinant can come out negative (Remark (i), Section 4.2).

Solution

  1. Name the four entries

    Read the entries by position:

    a11=2,a12=4,a21=−1,a22=2.\begin{aligned} a_{11} &= 2, & a_{12} &= 4,\\ a_{21} &= -1, & a_{22} &= 2. \end{aligned}

    The array has 2 rows and 2 columns, so it is square of order 2 and the 2×22\times2 rule applies.

    The rule is stated in terms of a11,a12,a21,a22a_{11}, a_{12}, a_{21}, a_{22}⁠, so we must know which number sits where before we use it. Only square arrays have determinants (Remark (ii), Section 4.2), which is why the order is checked first.
    Row first, column second: a21a_{21} is in row 2, column 1, which is the −1-1⁠. Basics: rows, columns and positions

  2. Multiply along the main diagonal

    The main diagonal runs from top left to bottom right, through a11a_{11} and a22a_{22}⁠:

    a11a22=2×2=4.a_{11}a_{22} = 2\times 2 = 4.
    This is the first term of the rule. In the figure below it is the blue arrow.

  3. Multiply along the other diagonal

    The other diagonal runs from bottom left to top right, through a21a_{21} and a12a_{12}⁠:

    a21a12=(−1)×4=−4.a_{21}a_{12} = (-1)\times 4 = -4.
    Keep the sign of −1-1 with it. A negative times a positive is negative, so this product is −4-4⁠, not 4.
    Sign rule for products: (−)×(+)=(−)(-)\times(+) = (-)⁠. Basics: algebra the chapter leans on

  4. Subtract the second product from the first

    ∣24−12∣=a11a22−a21a12=2(2)−4(−1)=4−(−4)=4+4=8.\begin{aligned} \begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} &= a_{11}a_{22} - a_{21}a_{12}\\ &= 2(2) - 4(-1)\\ &= 4 - (-4)\\ &= 4 + 4 = 8. \end{aligned}
    The rule always subtracts the other-diagonal product. Because that product is −4-4⁠, subtracting it adds 4. Writing 4(−1)4(-1) with brackets, as the book does, keeps the two minus signs (the one in the rule and the one in the entry) visibly separate, so neither is lost.
    p−(−q)=p+qp - (-q) = p + q⁠. Lesson: determinant of a square matrix
2 4 −1 2 main diagonal: 2 × 2 = 4 other diagonal: (−1) × 4 = −4 determinant = 4 − (−4) = 8
Blue product minus gold product. Because the gold product is negative, subtracting it adds 4.
Result ∣24−12∣=8.\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = 8.

This agrees with the textbook's solution.

Check it with numbers

Swap rows and columns. Writing the columns as rows gives a different-looking determinant with the same value:

∣2−142∣=2(2)−4(−1)=8.\begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} = 2(2) - 4(-1) = 8.

Link to equations. The determinant is nonzero, so the system 2x+4y=c12x + 4y = c_1⁠, −x+2y=c2-x + 2y = c_2 has exactly one solution for any c1,c2c_1, c_2⁠. For example with c1=8c_1 = 8⁠, c2=0c_2 = 0⁠: from the second equation x=2yx = 2y⁠, then 4y+4y=84y + 4y = 8⁠, so y=1y = 1⁠, x=2x = 2⁠. Check: 2(2)+4(1)=82(2) + 4(1) = 8 and −2+2=0-2 + 2 = 0⁠.

Common mistakes
  1. Dropping the sign of −1-1⁠. Computing 4−4=04 - 4 = 0 treats a21a_{21} as +1+1⁠. Keep negative entries in brackets.
  2. Adding the diagonals. The rule is a difference, a11a22−a21a12a_{11}a_{22} - a_{21}a_{12}⁠, never a sum.
  3. Subtracting in the wrong order. a21a12−a11a22=−8a_{21}a_{12} - a_{11}a_{22} = -8 gives the right size with the wrong sign. The main diagonal comes first.
  4. Reading the bars as absolute value. A determinant can be negative; the bars only mean "determinant".
What to take away
  1. For order 2: (top left ×\times bottom right) −- (bottom left ×\times top right).
  2. Put every negative entry in brackets before multiplying; the minus in the rule and the minus in an entry are different minuses.
  3. A nonzero determinant means the matching pair of linear equations has a unique solution.

Every fact used, and where it comes from

FactUsed inSource
Only square matrices have determinants; ∣A∣|A| is not a modulusStep 1Section 4.2, Remarks (i) and (ii) (Lesson: determinant of a square matrix)
∣A∣=a11a22−a21a12|A| = a_{11}a_{22} - a_{21}a_{12} for order 2Steps 2–4Section 4.2.2
(−1)×4=−4(-1)\times4 = -4⁠; p−(−q)=p+qp-(-q) = p+qSteps 3, 4Rules of signs
Nonzero determinant gives a unique solutionCheckSection 4.1, Introduction
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