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Chapter 4 · Determinants

Miscellaneous · Q6: Evaluate another determinant in xx⁠, yy

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

Evaluate

∣1xy1x+yy1xx+y∣.\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}.
Exam answer

Let Δ\Delta be the given determinant.

Expanding along C1C_1⁠:

Δ=1∣x+yyxx+y∣−1∣xyxx+y∣+1∣xyx+yy∣=[(x+y)2−xy]−[x(x+y)−xy]+[xy−y(x+y)]=(x2+xy+y2)−x2−y2\begin{aligned} \Delta &= 1\begin{vmatrix} x+y & y \\ x & x+y \end{vmatrix} \\ &\quad - 1\begin{vmatrix} x & y \\ x & x+y \end{vmatrix} \\ &\quad + 1\begin{vmatrix} x & y \\ x+y & y \end{vmatrix} \\ &= \big[(x+y)^2 - xy\big] \\ &\quad - \big[x(x+y) - xy\big] \\ &\quad + \big[xy - y(x+y)\big] \\ &= (x^2 + xy + y^2) - x^2 - y^2 \end{aligned}

∴ Δ=xy\therefore\ \Delta = xy⁠.

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

The one idea this problem is built on

The first column is all 1s. Expanding along it, each term is just a sign times a minor, with no extra factor to multiply in. The three minors are short polynomials, and when they are added with the signs +,−,++, -, + everything cancels except xyxy⁠.

Given, to find, and the plan
Given
A 3×33\times3 determinant Δ\Delta whose first column is 1,1,11, 1, 1⁠.
To find
Δ\Delta in terms of xx and yy⁠.
Plan
(1) Expand along C1C_1⁠. (2) Evaluate M11M_{11}⁠, M21M_{21}⁠, M31M_{31}⁠. (3) Form the signed sum. (4) Check by expanding along R1R_1⁠, and 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 column

A determinant can be expanded along a column just as along a row: each entry times its minor, with the sign (−1)i+j(-1)^{i+j}⁠. Down the first column the signs are +,−,++, -, +⁠. Lesson: expanding a 3 × 3 determinant

3Expanding brackets

(x+y)2=x2+2xy+y2(x + y)^2 = x^2 + 2xy + y^2 and x(x+y)=x2+xyx(x + y) = x^2 + xy⁠. Basics: algebra the chapter leans on

Solution

  1. Expand along the first column

    Column 1 is (1,1,1)(1, 1, 1)⁠, with signs (−1)1+1=+(-1)^{1+1} = +⁠, (−1)2+1=−(-1)^{2+1} = -⁠, (−1)3+1=+(-1)^{3+1} = +⁠:

    Δ=1⋅M11−1⋅M21+1⋅M31=M11−M21+M31.\Delta = 1\cdot M_{11} - 1\cdot M_{21} + 1\cdot M_{31} = M_{11} - M_{21} + M_{31}.
    There is no zero to remove a term, but a column of 1s is the next best thing: the entries multiply the minors by 1, so all the work is in three 2×22\times2 determinants. The book shows expansion along C1C_1 in Section 4.2.3 and states that every row and column gives the same value.
    Sign of position (i,1)(i, 1) is (−1)i+1(-1)^{i+1}⁠. Basics: the + − + sign pattern

  2. The three minors

    Delete column 1 each time, and the row of the entry:

    M11=∣x+yyxx+y∣=(x+y)2−xy=x2+2xy+y2−xy=x2+xy+y2\begin{aligned} M_{11} &= \begin{vmatrix} x+y & y \\ x & x+y \end{vmatrix} = (x + y)^2 - xy \\ &= x^2 + 2xy + y^2 - xy \\ &= x^2 + xy + y^2 \end{aligned} M21=∣xyxx+y∣=x(x+y)−xy=x2\begin{aligned} M_{21} &= \begin{vmatrix} x & y \\ x & x+y \end{vmatrix} = x(x + y) - xy \\ &= x^2 \end{aligned} M31=∣xyx+yy∣=xy−y(x+y)=−y2\begin{aligned} M_{31} &= \begin{vmatrix} x & y \\ x+y & y \end{vmatrix} = xy - y(x + y) \\ &= -y^2 \end{aligned}
    M11M_{11} keeps rows 2 and 3, M21M_{21} keeps rows 1 and 3, M31M_{31} keeps rows 1 and 2, each without column 1. In M21M_{21}⁠, x(x+y)=x2+xyx(x+y) = x^2 + xy and subtracting xyxy leaves x2x^2⁠. In M31M_{31}⁠, y(x+y)=xy+y2y(x + y) = xy + y^2 and xy−xy−y2=−y2xy - xy - y^2 = -y^2⁠.
    Minor MijM_{ij}⁠: Section 4.4, Definition 1. Lesson: minors and cofactors

  3. Form the signed sum

    Δ=(x2+xy+y2)−x2+(−y2)=xy.\begin{aligned} \Delta &= (x^2 + xy + y^2) - x^2 + (-y^2) \\ &= xy. \end{aligned}

    So Δ=xy\Delta = xy.

    The x2x^2 from M11M_{11} cancels the −x2-x^2 from −M21-M_{21}⁠, and the y2y^2 from M11M_{11} cancels M31=−y2M_{31} = -y^2⁠. Only the middle term xyxy of M11M_{11} survives.
    Expansion along C1C_1⁠, Section 4.2.3. Lesson: expanding a 3 × 3 determinant
Result ∣1xy1x+yy1xx+y∣=xy\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix} = xy

(Textbook answer: xyxy⁠.)

Check it with numbers

Another line. Along R1=(1,x,y)R_1 = (1, x, y) with signs +,−,++, -, +⁠:

Δ=1⋅[(x+y)2−xy]−x [(x+y)−y]+y [x−(x+y)]=(x2+xy+y2)−x2−y2=xy ✓\begin{aligned} \Delta &= 1\cdot[(x+y)^2 - xy] - x\,[(x + y) - y] \\ &\quad + y\,[x - (x + y)] \\ &= (x^2 + xy + y^2) - x^2 - y^2 = xy \ ✓ \end{aligned}

Numbers. With x=2x = 2⁠, y=3y = 3 the rows are (1,2,3)(1, 2, 3)⁠, (1,5,3)(1, 5, 3)⁠, (1,2,5)(1, 2, 5)⁠. Along R1R_1⁠: 1(25−6)−2(5−3)+3(2−5)1(25 - 6) - 2(5 - 3) + 3(2 - 5) =19−4−9=6=2⋅3= 19 - 4 - 9 = 6 = 2\cdot3 ✓.

Common mistakes
  1. Using +,+,++, +, + down the column. The middle position (2,1)(2, 1) has sign −-⁠.
  2. Writing (x+y)2=x2+y2(x + y)^2 = x^2 + y^2: the cross term 2xy2xy is exactly what produces the answer.
  3. Mixing up which rows a minor keeps: M21M_{21} uses rows 1 and 3, not rows 2 and 3.
What to take away
  1. Expanding along a column is just as valid as along a row; pick whichever line makes the products simplest (zeros first, then 1s).
  2. A second expansion along a different line is an independent check.

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 C1C_1 or R1R_1⁠; same value along every lineSteps 1, 3, 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^2Step 2Algebraic identity (Basics: algebra the chapter leans on)
Δ=xy\Delta = xy⁠, the minors, and the numerical checkSteps 2, 3, CheckRecomputed with sympy in the site's verification script
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