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Chapter 4 · Determinants

Miscellaneous · Q5: Evaluate a determinant in xx⁠, yy

Miscellaneous ExerciseExpanding a 3 × 3 determinant · Section 4.2.3
The question

Evaluate

∣xyx+yyx+yxx+yxy∣.\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}.
Exam answer

Let Δ\Delta be the given determinant.

Expanding along R1R_1⁠:

Δ=x∣x+yxxy∣−y∣yxx+yy∣+(x+y)∣yx+yx+yx∣=x(xy+y2−x2)−y(y2−x2−xy)+(x+y)(−x2−xy−y2)=x2y+xy2−x3−y3+x2y+xy2−x3−2x2y−2xy2−y3=−2x3−2y3\begin{aligned} \Delta &= x\begin{vmatrix} x+y & x \\ x & y \end{vmatrix} \\ &\quad - y\begin{vmatrix} y & x \\ x+y & y \end{vmatrix} \\ &\quad + (x+y)\begin{vmatrix} y & x+y \\ x+y & x \end{vmatrix} \\ &= x(xy + y^2 - x^2) \\ &\quad - y(y^2 - x^2 - xy) \\ &\quad + (x+y)(-x^2 - xy - y^2) \\ &= x^2y + xy^2 - x^3 \\ &\quad - y^3 + x^2y + xy^2 \\ &\quad - x^3 - 2x^2y - 2xy^2 - y^3 \\ &= -2x^3 - 2y^3 \end{aligned}

∴ Δ=−2(x3+y3)\therefore\ \Delta = -2(x^3 + y^3)⁠.

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

The one idea this problem is built on

Every row contains the same three expressions xx⁠, yy⁠, x+yx + y⁠, each shifted one place to the right. There is no zero to exploit, so we expand along the first row and organise the algebra: compute each minor as a polynomial, multiply by its entry, and collect like terms by type (x3x^3⁠, x2yx^2y⁠, xy2xy^2⁠, y3y^3⁠). All the mixed terms cancel, leaving −2x3−2y3-2x^3 - 2y^3⁠.

Given, to find, and the plan
Given
A 3×33\times3 determinant Δ\Delta with entries xx⁠, yy⁠, x+yx + y⁠.
To find
Δ\Delta as a simplified polynomial in xx and yy⁠.
Plan
(1) Expand along R1R_1 with signs +,−,++, -, +⁠. (2) Evaluate the three minors. (3) Multiply each by its entry. (4) Collect like terms. (5) Check with numbers.

Basics you need

1A 2×22\times2 determinant

Rows (a,b)(a, b) and (c,d)(c, d) give ad−bcad - bc⁠. Basics: where the determinant comes from

2Expansion along a row

Δ=a11M11−a12M12+a13M13\Delta = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} along R1R_1⁠. Lesson: expanding a 3 × 3 determinant

3Expanding brackets

(x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2⁠, and a product of brackets is the sum of every term of one times every term of the other. Basics: algebra the chapter leans on

4The sign pattern

Along the first row the signs are +,−,++, -, +⁠. Basics: the + − + sign pattern

Solution

  1. Set up the expansion

    No entry is zero, so every line is equally good. We use the first row (x, y, x+y)(x,\ y,\ x + y)⁠:

    Δ=x M11−y M12+(x+y) M13.\Delta = x\,M_{11} - y\,M_{12} + (x + y)\,M_{13}.
    The value is the same along any row or column (Section 4.2.3). R1R_1 has two single-letter entries, xx and yy⁠, which keeps two of the three products simple.
    Sign of position (1,j)(1, j) is (−1)1+j(-1)^{1+j}⁠. Basics: the + − + sign pattern

  2. The three minors

    M11=∣x+yxxy∣=(x+y)y−x⋅x=xy+y2−x2\begin{aligned} M_{11} &= \begin{vmatrix} x+y & x \\ x & y \end{vmatrix} = (x + y)y - x\cdot x \\ &= xy + y^2 - x^2 \end{aligned} M12=∣yxx+yy∣=y⋅y−x(x+y)=y2−x2−xy\begin{aligned} M_{12} &= \begin{vmatrix} y & x \\ x+y & y \end{vmatrix} = y\cdot y - x(x + y) \\ &= y^2 - x^2 - xy \end{aligned} M13=∣yx+yx+yx∣=xy−(x+y)2=xy−x2−2xy−y2=−x2−xy−y2\begin{aligned} M_{13} &= \begin{vmatrix} y & x+y \\ x+y & x \end{vmatrix} = xy - (x + y)^2 \\ &= xy - x^2 - 2xy - y^2 \\ &= -x^2 - xy - y^2 \end{aligned}
    Each minor comes from deleting row 1 and one column: column 1 for M11M_{11}⁠, column 2 for M12M_{12}⁠, column 3 for M13M_{13}⁠. In M13M_{13} the whole square (x+y)2(x+y)^2 is subtracted, so all three of its terms change sign.
    Minor MijM_{ij}⁠: Section 4.4, Definition 1. Lesson: minors and cofactors

  3. Multiply each minor by its entry, with its sign

    x M11=x2y+xy2−x3−y M12=−y3+x2y+xy2\begin{aligned} x\,M_{11} &= x^2y + xy^2 - x^3 \\ -y\,M_{12} &= -y^3 + x^2y + xy^2 \end{aligned}

    For the third term, multiply out (x+y)(−x2−xy−y2)(x + y)(-x^2 - xy - y^2) one part at a time:

    x(−x2−xy−y2)=−x3−x2y−xy2y(−x2−xy−y2)=−x2y−xy2−y3\begin{aligned} x(-x^2 - xy - y^2) &= -x^3 - x^2y - xy^2 \\ y(-x^2 - xy - y^2) &= -x^2y - xy^2 - y^3 \end{aligned} (x+y) M13=−x3−2x2y−2xy2−y3.(x + y)\,M_{13} = -x^3 - 2x^2y - 2xy^2 - y^3.
    For the second line, −y⋅y2=−y3-y\cdot y^2 = -y^3⁠, −y⋅(−x2)=+x2y-y\cdot(-x^2) = +x^2y and −y⋅(−xy)=+xy2-y\cdot(-xy) = +xy^2⁠: the minus sign in front of yy flips every sign in M12M_{12}⁠. Splitting (x+y)(x + y) into xx and yy for the third term avoids losing any of its six products.
    Distributive law. Basics: algebra the chapter leans on

  4. Collect like terms

    Add the three results of Step 3, sorting by type of term:

    x3:−1+0−1=−2x2y:1+1−2=0xy2:1+1−2=0y3:0−1−1=−2\begin{aligned} x^3&: -1 + 0 - 1 = -2 \\ x^2y&: 1 + 1 - 2 = 0 \\ xy^2&: 1 + 1 - 2 = 0 \\ y^3&: 0 - 1 - 1 = -2 \end{aligned}

    So

    Δ=−2x3−2y3=−2(x3+y3).\Delta = -2x^3 - 2y^3 = -2(x^3 + y^3).
    Every term of a 3×33\times3 determinant with these entries has degree 3, so only the four types x3x^3⁠, x2yx^2y⁠, xy2xy^2⁠, y3y^3 can occur. Listing the coefficient of each type from each of the three products (in order xM11xM_{11}⁠, −yM12-yM_{12}⁠, (x+y)M13(x+y)M_{13}⁠) makes the cancellation easy to see and easy to check. If you like, x3+y3=(x+y)(x2−xy+y2)x^3 + y^3 = (x + y)(x^2 - xy + y^2)⁠, so Δ=−2(x+y)(x2−xy+y2)\Delta = -2(x + y)(x^2 - xy + y^2) as well.
    Signed sum along R1R_1⁠, Section 4.2.3. Lesson: expanding a 3 × 3 determinant
Result ∣xyx+yyx+yxx+yxy∣=−2(x3+y3)\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix} = -2(x^3 + y^3)

(Textbook answer: −2(x3+y3)-2(x^3 + y^3)⁠.)

Check it with numbers

Take x=1x = 1⁠, y=2y = 2⁠. The rows become (1,2,3)(1, 2, 3)⁠, (2,3,1)(2, 3, 1)⁠, (3,1,2)(3, 1, 2)⁠. Along R1R_1⁠:

1(3⋅2−1⋅1)−2(2⋅2−1⋅3)+3(2⋅1−3⋅3)=5−2−21=−18.\begin{aligned} &1(3\cdot2 - 1\cdot1) - 2(2\cdot2 - 1\cdot3) \\ &\quad + 3(2\cdot1 - 3\cdot3) \\ &= 5 - 2 - 21 = -18. \end{aligned}

The formula gives −2(1+8)=−18-2(1 + 8) = -18 ✓. With x=3x = 3⁠, y=−1y = -1⁠: rows (3,−1,2)(3, -1, 2)⁠, (−1,2,3)(-1, 2, 3)⁠, (2,3,−1)(2, 3, -1)⁠, and the determinant is 3(−2−9)+1(1−6)+2(−3−4)3(-2 - 9) + 1(1 - 6) + 2(-3 - 4) =−33−5−14=−52= -33 - 5 - 14 = -52⁠, while −2(27−1)=−52-2(27 - 1) = -52 ✓.

Common mistakes
  1. Forgetting the middle minus: the term is −y M12-y\,M_{12}⁠, which flips every sign of M12M_{12}⁠.
  2. Subtracting only part of (x+y)2(x + y)^2 in M13M_{13}⁠: xy−(x+y)2=xy−x2−2xy−y2xy - (x + y)^2 = xy - x^2 - 2xy - y^2⁠.
  3. Dropping products when multiplying (x+y)(x + y) by a three-term bracket: there must be six products before collecting.
  4. Writing −2(x+y)3-2(x + y)^3 or −2(x3−y3)-2(x^3 - y^3) by mis-collecting signs. The numbers check catches this at once.
What to take away
  1. With no zeros, expand along any row, and keep each minor as a tidy polynomial before multiplying.
  2. Collect terms by type in a small table; it turns a messy expansion into four short sums.
  3. Always test a symbolic answer with one or two pairs of numbers.

Every fact used, and where it comes from

FactUsed inSource
Value of a 2×22\times2 determinantStep 2, CheckSection 4.2.2 (Lesson: determinant of a square matrix)
Expansion along R1R_1⁠; same value along every lineSteps 1, 4, CheckSection 4.2.3 (Lesson: expanding a 3 × 3 determinant)
MinorsStep 2Section 4.4, Definition 1 (Lesson: minors and cofactors)
(x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2⁠; x3+y3=(x+y)(x2−xy+y2)x^3 + y^3 = (x+y)(x^2 - xy + y^2)Steps 2, 4Algebraic identities (Basics: algebra the chapter leans on)
Δ=−2(x3+y3)\Delta = -2(x^3 + y^3)⁠, the minors, and the numerical checksSteps 2–4, CheckRecomputed with sympy in the site's verification script
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