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Chapter 4 · Determinants

Exercise 4.1 · Q5: Evaluate four 3×33\times3 determinants

Exercise 4.1Determinant of a matrix of order 3 × 3 · Section 4.2.3
The question

Evaluate the determinants

(i) ∣3−1−200−13−50∣\text{(i)}\ \begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix} (ii) ∣3−4511−2231∣\text{(ii)}\ \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix} (iii) ∣012−10−3−230∣\text{(iii)}\ \begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix} (iv) ∣2−1−202−13−50∣\text{(iv)}\ \begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}
Exam answer

(i) Expanding along R2R_2⁠:

Δ1=−0∣−1−2−50∣+0∣3−230∣−(−1)∣3−13−5∣=0+0+(−15+3)\begin{aligned} \Delta_1 &= -0\begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} + 0\begin{vmatrix} 3 & -2 \\ 3 & 0 \end{vmatrix} \\ &\quad - (-1)\begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix} \\ &= 0 + 0 + (-15 + 3) \end{aligned}

∴ Δ1=−12\therefore\ \Delta_1 = -12⁠.

(ii) Expanding along R1R_1⁠:

Δ2=3∣1−231∣−(−4)∣1−221∣+5∣1123∣=3(1+6)+4(1+4)+5(3−2)=21+20+5\begin{aligned} \Delta_2 &= 3\begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} - (-4)\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} \\ &\quad + 5\begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} \\ &= 3(1 + 6) + 4(1 + 4) \\ &\quad + 5(3 - 2) \\ &= 21 + 20 + 5 \end{aligned}

∴ Δ2=46\therefore\ \Delta_2 = 46⁠.

(iii) Expanding along R1R_1⁠:

Δ3=0∣0−330∣−1∣−1−3−20∣+2∣−10−23∣=0−1(0−6)+2(−3−0)=6−6\begin{aligned} \Delta_3 &= 0\begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} - 1\begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix} \\ &\quad + 2\begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix} \\ &= 0 - 1(0 - 6) + 2(-3 - 0) \\ &= 6 - 6 \end{aligned}

∴ Δ3=0\therefore\ \Delta_3 = 0⁠.

(iv) Expanding along C1C_1⁠:

Δ4=2∣2−1−50∣−0∣−1−2−50∣+3∣−1−22−1∣=2(0−5)−0+3(1+4)=−10+15\begin{aligned} \Delta_4 &= 2\begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} - 0\begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix} \\ &\quad + 3\begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix} \\ &= 2(0 - 5) - 0 + 3(1 + 4) \\ &= -10 + 15 \end{aligned}

∴ Δ4=5\therefore\ \Delta_4 = 5⁠.

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

The one idea this problem is built on

A 3×33\times3 determinant is reduced to three 2×22\times2 determinants by expanding along one row or one column: each element is multiplied by its sign (−1)i+j(-1)^{i+j} and by the 2×22\times2 determinant left when its row and column are deleted. Any row or column gives the same value, so we choose the one with the most zeros (zeros kill whole terms) and then expand along a second line to check.

Given, to find, and the plan
Given
Four determinants of order 3 with integer entries.
To find
The value of each.
Plan
For each part: (1) count zeros and choose the row or column to expand along, writing its signs; (2) write the three 2×22\times2 determinants (minors) and their values; (3) form the signed sum; (4) confirm by expanding along a different row or column.

Basics you need

1Expansion along a row or column

For the element aija_{ij}⁠, delete row ii and column jj⁠; the 2×22\times2 determinant that remains is called the minor MijM_{ij} (the name is given in Section 4.4). Then Δ=∑j(−1)i+jaijMij\Delta = \sum_{j} (-1)^{i+j} a_{ij} M_{ij} along row ii⁠, and similarly down a column. All six rows and columns give the same value (Section 4.2.3). Lesson: expanding a 3 × 3 determinant

2The sign pattern

(−1)i+j(-1)^{i+j} is ++ when i+ji+j is even and −- when it is odd (Remark (ii), Section 4.2.3): +−+−+−+−+\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix} Basics: the + − + sign pattern

3The 2×22\times2 determinant

∣abcd∣=ad−bc\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc With negative entries, bracket them: (−1)(0)−(−5)(−2)=−10(-1)(0) - (-5)(-2) = -10⁠. Lesson: determinant of a square matrix

4Choose the line with most zeros

A zero element contributes 0×(±1)×Mij=00\times(\pm1)\times M_{ij} = 0 whatever the minor is (Remark (i), Section 4.2.3).

Part (i): expand along the row with two zeros

  1. Choose row 2

    Row 2 is 0,0,−10, 0, -1⁠: two zeros, more than any other row or column. Its signs are −,+,−-, +, - (from (−1)2+1,(−1)2+2,(−1)2+3(-1)^{2+1}, (-1)^{2+2}, (-1)^{2+3}⁠).

    Two of the three terms will be 0, so only one 2×22\times2 determinant really matters. The middle row starts with a minus sign, because 2+1=32 + 1 = 3 is odd. Forgetting this is the classic error when expanding along row 2.
    Basics: the sign pattern

  2. The three minors along row 2

    M21=∣−1−2−50∣=(−1)(0)−(−5)(−2)=−10M22=∣3−230∣=3⋅0−3(−2)=6M23=∣3−13−5∣=3(−5)−3(−1)=−12\begin{aligned} M_{21} &= \begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix}\\ &= (-1)(0) - (-5)(-2) = -10\\ M_{22} &= \begin{vmatrix} 3 & -2 \\ 3 & 0 \end{vmatrix}\\ &= 3\cdot0 - 3(-2) = 6\\ M_{23} &= \begin{vmatrix} 3 & -1 \\ 3 & -5 \end{vmatrix}\\ &= 3(-5) - 3(-1) = -12 \end{aligned}
    For M21M_{21} delete row 2 and column 1, leaving rows −1,−2-1, -2 and −5,0-5, 0⁠. For M22M_{22} delete row 2 and column 2, leaving 3,−23, -2 and 3,03, 0⁠. For M23M_{23} delete row 2 and column 3, leaving 3,−13, -1 and 3,−53, -5⁠. In M23M_{23}⁠: 3(−5)=−153(-5) = -15 and 3(−1)=−33(-1) = -3⁠, so −15−(−3)=−12-15 - (-3) = -12⁠.

  3. Signed sum

    Δ=(−1)(0)(−10)+(+1)(0)(6)+(−1)(−1)(−12)=0+0−12=−12.\begin{aligned} \Delta &= (-1)(0)(-10) + (+1)(0)(6)\\ &\quad + (-1)(-1)(-12)\\ &= 0 + 0 - 12 = -12. \end{aligned}
    Each term is sign ×\times element ×\times minor. The first two vanish because their elements are 0. In the last term the sign −1-1 and the element −1-1 multiply to +1+1⁠, leaving M23=−12M_{23} = -12⁠.
    Expansion along a row, Section 4.2.3. Lesson: expanding a 3 × 3 determinant

  4. Confirm along row 1

    Row 1 is 3,−1,−23, -1, -2 with signs +,−,++, -, +⁠:

    M11=∣0−1−50∣=0−(−5)(−1)=−5M12=∣0−130∣=0−3(−1)=3M13=∣003−5∣=0−0=0\begin{aligned} M_{11} &= \begin{vmatrix} 0 & -1 \\ -5 & 0 \end{vmatrix} = 0 - (-5)(-1) = -5\\ M_{12} &= \begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} = 0 - 3(-1) = 3\\ M_{13} &= \begin{vmatrix} 0 & 0 \\ 3 & -5 \end{vmatrix} = 0 - 0 = 0 \end{aligned} Δ=3(−5)−(−1)(3)+(−2)(0)=−15+3+0=−12.✓\begin{aligned} \Delta &= 3(-5) - (-1)(3) + (-2)(0)\\ &= -15 + 3 + 0 = -12. \quad\checkmark \end{aligned}
    A different line, different minors, the same value. Agreement between two expansions is the best guard against a sign slip.
the determinantsigns 3−13−5 −2000 −1 +−+−++−+ − element −1, sign −, minor −12: (−1)(−1)(−12) = −12
Part (i), the only nonzero term along row 2: delete the row and column of a23=−1a_{23} = -1 to get the minor M23=−12M_{23} = -12⁠, and take the sign from position (2, 3) of the pattern.

Part (ii): no zeros, so expand along row 1

  1. Choose row 1

    No entry is 0, so no line is easier than another. We take row 1, 3,−4,53, -4, 5⁠, with signs +,−,++, -, +⁠.

    With no zeros, all three terms must be computed whichever line we use. Row 1 has the simplest signs to remember.

  2. The three minors along row 1

    M11=∣1−231∣=1⋅1−3(−2)=7M12=∣1−221∣=1⋅1−2(−2)=5M13=∣1123∣=1⋅3−2⋅1=1\begin{aligned} M_{11} &= \begin{vmatrix} 1 & -2 \\ 3 & 1 \end{vmatrix} = 1\cdot1 - 3(-2) = 7\\ M_{12} &= \begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = 1\cdot1 - 2(-2) = 5\\ M_{13} &= \begin{vmatrix} 1 & 1 \\ 2 & 3 \end{vmatrix} = 1\cdot3 - 2\cdot1 = 1 \end{aligned}
    Delete row 1 and, in turn, columns 1, 2, 3. For example M12M_{12} keeps columns 1 and 3 of rows 2 and 3: 1,−21, -2 and 2,12, 1⁠. In M11M_{11}⁠, subtracting 3(−2)=−63(-2) = -6 gives 1+6=71 + 6 = 7⁠; in M12M_{12}⁠, subtracting 2(−2)=−42(-2) = -4 gives 1+4=51 + 4 = 5⁠.

  3. Signed sum

    Δ=(+1)(3)(7)+(−1)(−4)(5)+(+1)(5)(1)=21+20+5=46.\begin{aligned} \Delta &= (+1)(3)(7) + (-1)(-4)(5)\\ &\quad + (+1)(5)(1)\\ &= 21 + 20 + 5 = 46. \end{aligned}
    The middle term has sign −- and element −4-4⁠; two negatives give ++⁠, so it adds 4×5=204\times5 = 20⁠.

  4. Confirm along column 1

    Column 1 is 3,1,23, 1, 2 with signs +,−,++, -, +⁠:

    M11=7 (from Step 2)M21=∣−4531∣=−4−15=−19M31=∣−451−2∣=8−5=3\begin{aligned} M_{11} &= 7 \ \text{(from Step 2)}\\ M_{21} &= \begin{vmatrix} -4 & 5 \\ 3 & 1 \end{vmatrix} = -4 - 15 = -19\\ M_{31} &= \begin{vmatrix} -4 & 5 \\ 1 & -2 \end{vmatrix} = 8 - 5 = 3 \end{aligned} Δ=3(7)−1(−19)+2(3)=21+19+6=46.✓\begin{aligned} \Delta &= 3(7) - 1(-19) + 2(3)\\ &= 21 + 19 + 6 = 46. \quad\checkmark \end{aligned}
    M21M_{21}⁠: delete row 2 and column 1, leaving −4,5-4, 5 and 3,13, 1⁠, so (−4)(1)−3⋅5=−19(-4)(1) - 3\cdot5 = -19⁠. M31M_{31}⁠: delete row 3 and column 1, leaving −4,5-4, 5 and 1,−21, -2⁠, so (−4)(−2)−1⋅5=3(-4)(-2) - 1\cdot5 = 3⁠.

Part (iii): a determinant that comes out 0

  1. Choose row 1

    Every row and every column has exactly one zero, and the zeros lie on the main diagonal. Row 1 is 0,1,20, 1, 2 with signs +,−,++, -, +⁠; its zero is the first element.

    Since every line has one zero, each expansion needs two minors. Row 1 is as good as any.

  2. The minors along row 1

    M11=∣0−330∣=0−3(−3)=9M12=∣−1−3−20∣=0−(−2)(−3)=−6M13=∣−10−23∣=−3−0=−3\begin{aligned} M_{11} &= \begin{vmatrix} 0 & -3 \\ 3 & 0 \end{vmatrix} = 0 - 3(-3) = 9\\ M_{12} &= \begin{vmatrix} -1 & -3 \\ -2 & 0 \end{vmatrix}\\ &= 0 - (-2)(-3) = -6\\ M_{13} &= \begin{vmatrix} -1 & 0 \\ -2 & 3 \end{vmatrix} = -3 - 0 = -3 \end{aligned}
    M12M_{12}⁠: rows 2 and 3 without column 2 are −1,−3-1, -3 and −2,0-2, 0⁠. The product (−2)(−3)=6(-2)(-3) = 6 is positive, and it is subtracted, giving −6-6⁠. M11M_{11} is multiplied by 0 below, but we show it for completeness.

  3. Signed sum

    Δ=(+1)(0)(9)+(−1)(1)(−6)+(+1)(2)(−3)=0+6−6=0.\begin{aligned} \Delta &= (+1)(0)(9) + (-1)(1)(-6)\\ &\quad + (+1)(2)(-3)\\ &= 0 + 6 - 6 = 0. \end{aligned}
    The two nonzero terms are equal and opposite, +6+6 and −6-6⁠, so they cancel exactly. A determinant can be 0 even when no row or column is all zeros.

  4. Confirm along column 1, and see why it must be 0

    Column 1 is 0,−1,−20, -1, -2 with signs +,−,++, -, +⁠. Its minors are M11=9M_{11} = 9 (Step 2) and

    M21=∣1230∣=0−3⋅2=−6M31=∣120−3∣=−3−0=−3\begin{aligned} M_{21} &= \begin{vmatrix} 1 & 2 \\ 3 & 0 \end{vmatrix} = 0 - 3\cdot2 = -6\\ M_{31} &= \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix} = -3 - 0 = -3 \end{aligned} Δ=0−(−1)(−6)+(−2)(−3)=−6+6=0.✓\begin{aligned} \Delta &= 0 - (-1)(-6) + (-2)(-3)\\ &= -6 + 6 = 0. \quad\checkmark \end{aligned}

    The matrix has a special shape: each entry below the diagonal is the negative of its mirror image above it, and the diagonal is 0. With letters,

    ∣0ab−a0c−b−c0∣=0−a∣−ac−b0∣+b∣−a0−b−c∣=−a(bc)+b(ac)=0.\begin{aligned} &\begin{vmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{vmatrix}\\ &= 0 - a\begin{vmatrix} -a & c \\ -b & 0 \end{vmatrix} + b\begin{vmatrix} -a & 0 \\ -b & -c \end{vmatrix}\\ &= -a(bc) + b(ac) = 0. \end{aligned}

    Here a=1a = 1⁠, b=2b = 2⁠, c=−3c = -3⁠.

    The letter calculation expands along row 1 exactly as in Step 3. The first minor is (−a)(0)−(−b)(c)=bc(-a)(0) - (-b)(c) = bc and the second is (−a)(−c)−(−b)(0)=ac(-a)(-c) - (-b)(0) = ac⁠. The two terms −abc-abc and +abc+abc always cancel. So every determinant of this shape is 0, whatever a,b,ca, b, c are. (Matrices of this shape are the skew symmetric matrices of Chapter 3, Section 3.6.)
    Basics: transpose and identity

Part (iv): one zero in column 1

  1. Choose column 1

    Column 1 is 2,0,32, 0, 3⁠, with signs +,−,++, -, +⁠. Rows 2 and 3 and column 3 also contain one zero each; column 1 has the advantage that its nonzero elements are positive.

    No line has two zeros, so every choice needs two minors. Expanding down a column works exactly like along a row, using the signs of that column.

  2. The minors down column 1

    M11=∣2−1−50∣=0−(−5)(−1)=−5M21=∣−1−2−50∣=0−(−5)(−2)=−10M31=∣−1−22−1∣=1−(−4)=5\begin{aligned} M_{11} &= \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix}\\ &= 0 - (-5)(-1) = -5\\ M_{21} &= \begin{vmatrix} -1 & -2 \\ -5 & 0 \end{vmatrix}\\ &= 0 - (-5)(-2) = -10\\ M_{31} &= \begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix}\\ &= 1 - (-4) = 5 \end{aligned}
    M11M_{11}⁠: delete row 1 and column 1, leaving 2,−12, -1 and −5,0-5, 0⁠; (−5)(−1)=5(-5)(-1) = 5 is subtracted. M31M_{31}⁠: delete row 3 and column 1, leaving −1,−2-1, -2 and 2,−12, -1⁠; (−1)(−1)=1(-1)(-1) = 1 and 2(−2)=−42(-2) = -4⁠, so 1−(−4)=51 - (-4) = 5⁠.

  3. Signed sum

    Δ=(+1)(2)(−5)+(−1)(0)(−10)+(+1)(3)(5)=−10+0+15=5.\begin{aligned} \Delta &= (+1)(2)(-5) + (-1)(0)(-10)\\ &\quad + (+1)(3)(5)\\ &= -10 + 0 + 15 = 5. \end{aligned}
    The zero element removes the middle term. The other two are 2×(−5)=−102\times(-5) = -10 and 3×5=153\times5 = 15⁠.

  4. Confirm along row 1

    Row 1 is 2,−1,−22, -1, -2 with signs +,−,++, -, +⁠. M11=−5M_{11} = -5 (Step 2), and

    M12=∣0−130∣=0−3(−1)=3M13=∣023−5∣=0−3⋅2=−6\begin{aligned} M_{12} &= \begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} = 0 - 3(-1) = 3\\ M_{13} &= \begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix} = 0 - 3\cdot2 = -6 \end{aligned} Δ=2(−5)−(−1)(3)+(−2)(−6)=−10+3+12=5.✓\begin{aligned} \Delta &= 2(-5) - (-1)(3) + (-2)(-6)\\ &= -10 + 3 + 12 = 5. \quad\checkmark \end{aligned}
    Two sign combinations to watch: the middle sign −- times the element −1-1 gives ++⁠, and the element −2-2 times the minor −6-6 gives +12+12⁠.
Result

(i) −12(ii) 46\text{(i)}\ -12 \qquad \text{(ii)}\ 46 (iii) 0(iv) 5\text{(iii)}\ 0 \qquad \text{(iv)}\ 5 (Textbook answer: (i) −12, (ii) 46, (iii) 0, (iv) 5.)

Check it with numbers

A third, independent route for part (ii): the six-term formula (1) of Section 4.2.3,

Δ=a11a22a33−a11a32a23−a12a21a33+a12a31a23+a13a21a32−a13a31a22.\begin{aligned} \Delta &= a_{11}a_{22}a_{33} - a_{11}a_{32}a_{23}\\ &\quad - a_{12}a_{21}a_{33} + a_{12}a_{31}a_{23}\\ &\quad + a_{13}a_{21}a_{32} - a_{13}a_{31}a_{22}. \end{aligned}

With the entries of (ii):

3⋅1⋅1−3⋅3⋅(−2)−(−4)⋅1⋅1+(−4)⋅2⋅(−2)+5⋅1⋅3−5⋅2⋅1=3+18+4+16+15−10=46✓\begin{aligned} &3\cdot1\cdot1 - 3\cdot3\cdot(-2)\\ &\quad - (-4)\cdot1\cdot1 + (-4)\cdot2\cdot(-2)\\ &\quad + 5\cdot1\cdot3 - 5\cdot2\cdot1\\ &= 3 + 18 + 4 + 16 + 15 - 10\\ &= 46 \quad\checkmark \end{aligned}

The same formula gives −12-12⁠, 00 and 55 for parts (i), (iii) and (iv) ✓ (checked in the site's verification script).

Common mistakes
  1. Wrong sign for row 2. Row 2 starts with −-⁠: the signs are −,+,−-, +, -⁠. Using +,−,++, -, + in part (i) gives +12+12⁠.
  2. Deleting the wrong row or column for a minor. Always delete the row and the column of the element you are multiplying.
  3. Double minus slips such as −(−4)(5)=−20-(-4)(5) = -20⁠. Combine the sign and the element first: (−1)(−4)=+4(-1)(-4) = +4⁠, then multiply by the minor.
  4. Thinking a 0 answer means an error in part (iii). Zero is a perfectly good value; Part (iii), Step 4 shows it is forced by the shape of the matrix.
What to take away
  1. Expand along the row or column with most zeros; write its sign pattern before anything else.
  2. Each term is sign ×\times element ×\times minor; a zero element removes its term.
  3. A second expansion along another line is a quick, reliable check.
  4. A determinant of the form with 0 on the diagonal and aji=−aija_{ji} = -a_{ij} (order 3) is always 0.

Every fact used, and where it comes from

FactUsed inSource
Expansion along any row or column gives the same valueEvery part, Steps 1–4Section 4.2.3 (Lesson: expanding a 3 × 3 determinant)
Expand along the line with most zeros; sign ±1\pm1 by parity of i+ji+jEvery part, Step 1Remarks (i), (ii), Section 4.2.3
∣A∣=a11a22−a21a12|A| = a_{11}a_{22} - a_{21}a_{12} for order 2Every part, Steps 2, 4Section 4.2.2
The name "minor" MijM_{ij}Every part, Step 2Section 4.4 (Lesson: minors and cofactors)
Six-term formula for a 3×33\times3 determinantCheckEquation (1), Section 4.2.3
Skew symmetric matrices (aji=−aija_{ji} = -a_{ij}⁠)Part (iii), Step 4Chapter 3, Section 3.6
Similar expansionsEvery partExample 3, Example 4
All minors and values on this pageEvery part, CheckRecomputed in the site's verification script
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