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Chapter 4 · Determinants

Example 2: Evaluate a 2×22\times2 determinant in xx

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

Evaluate

∣xx+1x−1x∣.\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix}.
Exam answer Δ=∣xx+1x−1x∣=x(x)−(x−1)(x+1)=x2−(x2−1)\begin{aligned} \Delta &= \begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix} \\ &= x(x) - (x-1)(x+1) \\ &= x^2 - (x^2 - 1) \end{aligned}

∴ Δ=1\therefore\ \Delta = 1⁠.

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

The one idea this problem is built on

The 2×22\times2 rule works the same way when the entries are expressions instead of numbers: main-diagonal product minus other-diagonal product. The other diagonal here is (x+1)(x−1)(x+1)(x-1)⁠, a difference of squares, equal to x2−1x^2 - 1⁠. Subtracting it as a whole (in brackets) makes the x2x^2 terms cancel and leaves the constant 1. So the determinant is 1 whatever xx is.

Given, to find, and the plan
Given
A 2×22\times2 determinant whose entries are xx⁠, x+1x+1 (row 1) and x−1x-1⁠, xx (row 2), where xx is any real number.
To find
Its value, simplified.
Plan
Apply a11a22−a21a12a_{11}a_{22} - a_{21}a_{12}⁠, expand (x+1)(x−1)(x+1)(x-1)⁠, keep it in brackets while subtracting, then simplify.

Basics you need

1The 2×22\times2 rule

For order 2 (Section 4.2.2),

∣abcd∣=ad−cb.\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - cb.

Entries may be numbers or algebraic expressions. Basics: where the determinant comes from

2Difference of squares

(p+q)(p−q)=p2−q2(p+q)(p-q) = p^2 - q^2⁠. With p=xp = x⁠, q=1q = 1⁠: (x+1)(x−1)=x2−1(x+1)(x-1) = x^2 - 1⁠. Basics: algebra the chapter leans on

3A minus sign in front of a bracket

−(x2−1)=−x2+1-(x^2 - 1) = -x^2 + 1⁠: the minus changes the sign of every term inside. Basics: algebra the chapter leans on

Solution

  1. Name the four entries

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

    The array is square of order 2, so the 2×22\times2 rule applies.

    The rule is written in terms of positions. Here the two diagonals are easy to confuse because the expressions look alike, so naming them first avoids multiplying the wrong pair.
    aija_{ij}⁠: row ii⁠, column jj⁠. Basics: rows, columns and positions

  2. Write the rule with brackets

    Δ=a11a22−a21a12=x(x)−(x+1)(x−1).\begin{aligned} \Delta &= a_{11}a_{22} - a_{21}a_{12}\\ &= x(x) - (x+1)(x-1). \end{aligned}
    Each entry with two terms goes in its own bracket, so that the whole of x+1x+1 multiplies the whole of x−1x-1⁠. The minus sign then applies to the entire product, not just to its first term.
    ∣A∣=a11a22−a21a12|A| = a_{11}a_{22} - a_{21}a_{12}⁠, Section 4.2.2. Lesson: determinant of a square matrix

  3. Work out each product

    Main diagonal: x(x)=x2x(x) = x^2⁠.

    Other diagonal, by the difference of squares:

    (x+1)(x−1)=x2−x+x−1=x2−1.\begin{aligned} (x+1)(x-1) &= x^2 - x + x - 1\\ &= x^2 - 1. \end{aligned}
    Expanding term by term, x⋅x=x2x\cdot x = x^2⁠, x⋅(−1)=−xx\cdot(-1) = -x⁠, 1⋅x=x1\cdot x = x and 1⋅(−1)=−11\cdot(-1) = -1⁠. The middle terms −x-x and +x+x cancel, which is exactly why (p+q)(p−q)(p+q)(p-q) always gives p2−q2p^2 - q^2⁠.
    (p+q)(p−q)=p2−q2(p+q)(p-q) = p^2 - q^2 with p=xp = x⁠, q=1q = 1⁠. Basics: algebra the chapter leans on

  4. Subtract and simplify

    Δ=x2−(x2−1)=x2−x2+1=1.\begin{aligned} \Delta &= x^2 - (x^2 - 1)\\ &= x^2 - x^2 + 1\\ &= 1. \end{aligned}
    Removing the bracket after a minus sign changes the sign of each term inside: −(x2−1)=−x2+1-(x^2 - 1) = -x^2 + 1⁠. Then x2−x2=0x^2 - x^2 = 0⁠, leaving 1. No xx is left, so the value does not depend on xx⁠.
    −(u−v)=−u+v-(u - v) = -u + v⁠. Basics: algebra the chapter leans on
Result ∣xx+1x−1x∣=1\begin{vmatrix} x & x+1 \\ x-1 & x \end{vmatrix} = 1

for every real number xx⁠.

This agrees with the textbook's solution.

Check it with numbers

Put in a few values of xx and compute the plain-number determinant directly:

xxEntries (row 1; row 2)a11a22a_{11}a_{22}a21a12a_{21}a_{12}Value
000, 1; −1, 00,\ 1;\ -1,\ 000−1-10−(−1)=10-(-1) = 1
333, 4; 2, 33,\ 4;\ 2,\ 399889−8=19 - 8 = 1
−5-5−5, −4; −6, −5-5,\ -4;\ -6,\ -52525242425−24=125-24 = 1
2.52.52.5, 3.5; 1.5, 2.52.5,\ 3.5;\ 1.5,\ 2.56.256.255.255.2511

Every value is 1, as Step 4 says.

Common mistakes
  1. Losing the bracket. Writing x2−x2−1=−1x^2 - x^2 - 1 = -1 subtracts only the x2x^2⁠. The whole product x2−1x^2 - 1 must be subtracted.
  2. Expanding (x+1)(x−1)(x+1)(x-1) as x2+1x^2 + 1 or x2−2x−1x^2 - 2x - 1⁠. The middle terms cancel exactly; the result is x2−1x^2 - 1⁠.
  3. Taking the wrong diagonal. The main diagonal is xx and xx⁠; the other diagonal is x−1x-1 and x+1x+1⁠.
What to take away
  1. The determinant rule works for expressions exactly as for numbers; bracket every entry that has more than one term.
  2. Look out for standard identities, such as the difference of squares, in the diagonal products.
  3. A determinant with variable entries can turn out to be a constant. Testing a few values is a quick way to catch slips.

Every fact used, and where it comes from

FactUsed inSource
∣A∣=a11a22−a21a12|A| = a_{11}a_{22} - a_{21}a_{12} for order 2Steps 1, 2Section 4.2.2 (Lesson: determinant of a square matrix)
(p+q)(p−q)=p2−q2(p+q)(p-q) = p^2 - q^2Step 3Algebraic identity (earlier classes)
−(u−v)=−u+v-(u - v) = -u + vStep 4Distributive law with −1-1
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