Multivariable-Calculus · Unit 2 · Video 3 · Interactive Practice

One Row, Three Minors, One Minus Sign: The 3x3 Determinant as Volume

IKey Formulas

FormulaNameWhat it says
∣a1a2b1b2∣=a1b2−a2b1\begin{vmatrix} a_1 & a_2 \\ b_1 & b_2 \end{vmatrix} = a_1 b_2 - a_2 b_12×22 \times 2 determinantThe building block
det⁡(A⃗,B⃗,C⃗)=a1M1−a2M2+a3M3\det(\vec{A}, \vec{B}, \vec{C}) = a_1 M_1 - a_2 M_2 + a_3 M_3Expansion along the first rowSigns +,−,++, -, +
M2=∣b1b3c1c3∣M_2 = \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix}The middle minorDelete row 1 and column 2: columns 1 and 3 survive
volume=∣det⁡(A⃗,B⃗,C⃗)∣\text{volume} = \bigl| \det(\vec{A}, \vec{B}, \vec{C}) \bigr|Volume theoremBox with edges A⃗,B⃗,C⃗\vec{A}, \vec{B}, \vec{C}

Key Insight: MjM_j is whatever 2×22 \times 2 is left after deleting row 1 and column jj, so the whole definition is det⁡(A⃗,B⃗,C⃗)=a1∣b2b3c2c3∣−a2∣b1b3c1c3∣+a3∣b1b2c1c2∣\det(\vec{A}, \vec{B}, \vec{C}) = a_1 \begin{vmatrix} b_2 & b_3 \\ c_2 & c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1 & b_3 \\ c_1 & c_3 \end{vmatrix} + a_3 \begin{vmatrix} b_1 & b_2 \\ c_1 & c_2 \end{vmatrix} and the minus on the middle term is forced by the orientation of space — without it ∣det⁡∣|\det| is no longer a volume.

IIVisualization 1 — Delete a Row and a Column

Each first-row entry keeps exactly the 2×22 \times 2 left when its own row and column are deleted.

IIIVisualization 2 — Three Terms, Six Terms

Multiplying the three first-row terms out gives six monomials, each taking one entry from every row and every column.

IVVisualization 3 — The Determinant as Volume

The determinant of three vectors is the signed volume of the box they span.

💡 The proof that ∣det⁡∣|\det| really is the volume waits for the cross product; until then the arithmetic above is the whole of the evidence.

VQuiz Questions

Problem 1 · Expand Along the First Row

Given: ∣123045106∣\begin{vmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{vmatrix} — evaluate it by expanding along the first row.

✅ Correct! 1(24)−2(−5)+3(−4)=24+10−12=221(24) - 2(-5) + 3(-4) = 24 + 10 - 12 = 22.
❌ The middle sign. You added all three terms: 24+(−10)+(−12)=224 + (-10) + (-12) = 2. The middle term is subtracted, and −2(−5)=+10-2(-5) = +10.
❌ The middle minor. M2M_2 deletes column 2, so it is ∣0516∣=−5\begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} = -5, not ∣0410∣=−4\begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} = -4.
❌ Not quite. The three minors are M1=24M_1 = 24, M2=−5M_2 = -5, M3=−4M_3 = -4; now combine them with signs +,−,++, -, +.
Show solution

Delete row 1 and column jj for each first-row entry:

M1=∣4506∣=24,M2=∣0516∣=−5,M3=∣0410∣=−4M_1 = \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} = 24, \qquad M_2 = \begin{vmatrix} 0 & 5 \\ 1 & 6 \end{vmatrix} = -5, \qquad M_3 = \begin{vmatrix} 0 & 4 \\ 1 & 0 \end{vmatrix} = -4

Combine with signs +,−,++, -, +:

det⁡=1(24)−2(−5)+3(−4)=24+10−12=22\det = 1(24) - 2(-5) + 3(-4) = 24 + 10 - 12 = 22

Note M2M_2 keeps columns 1 and 3 — the entries 0,50, 5 and 1,61, 6 — because column 2 is the one deleted.

Problem 2 · The Middle Term

Given: ∣314205167∣\begin{vmatrix} 3 & 1 & 4 \\ 2 & 0 & 5 \\ 1 & 6 & 7 \end{vmatrix} — find the middle term −a2M2-a_2 M_2 of its first-row expansion.

✅ Correct! M2=∣2517∣=9M_2 = \begin{vmatrix} 2 & 5 \\ 1 & 7 \end{vmatrix} = 9 and a2=1a_2 = 1, so the term is −1(9)=−9-1(9) = -9.
❌ The minor is right, the sign is not. M2=9M_2 = 9, but the middle term of the expansion is −a2M2-a_2 M_2, not +a2M2+a_2 M_2.
❌ Wrong columns. ∣2016∣=12\begin{vmatrix} 2 & 0 \\ 1 & 6 \end{vmatrix} = 12 keeps columns 1 and 2. Deleting column 2 leaves columns 1 and 3: ∣2517∣\begin{vmatrix} 2 & 5 \\ 1 & 7 \end{vmatrix}.
❌ Not quite. Cover row 1 and column 2 of the array and read off what is left.
Show solution

The entry a2=1a_2 = 1 sits in row 1, column 2. Delete that row and that column; the four surviving entries are 2,52, 5 (row B⃗\vec{B}) and 1,71, 7 (row C⃗\vec{C}):

M2=∣2517∣=2(7)−1(5)=9M_2 = \begin{vmatrix} 2 & 5 \\ 1 & 7 \end{vmatrix} = 2(7) - 1(5) = 9

The middle term carries the minus sign:

−a2M2=−1(9)=−9-a_2 M_2 = -1(9) = -9

For the record, the whole determinant is 3(−30)−1(9)+4(12)=−90−9+48=−513(-30) - 1(9) + 4(12) = -90 - 9 + 48 = -51.

Problem 3 · Volume of the Box

Given: A⃗=⟨1,2,1⟩\vec{A} = \langle 1, 2, 1 \rangle, B⃗=⟨2,0,4⟩\vec{B} = \langle 2, 0, 4 \rangle, C⃗=⟨0,3,1⟩\vec{C} = \langle 0, 3, 1 \rangle — find the determinant and the volume of the parallelepiped with these three edges.

What is det⁡(A⃗,B⃗,C⃗)\det(\vec{A}, \vec{B}, \vec{C})?

What is the volume of the box?

✅ Correct! The determinant is −10-10, so the volume is ∣−10∣=10|-10| = 10 and the triple A⃗,B⃗,C⃗\vec{A}, \vec{B}, \vec{C} is left-handed.
❌ Check the expansion. M1=−12M_1 = -12, M2=2M_2 = 2, M3=6M_3 = 6; combine them as 1(M1)−2(M2)+1(M3)1(M_1) - 2(M_2) + 1(M_3).
❌ That is the volume, not the determinant. The expansion gives 1(−12)−2(2)+1(6)=−101(-12) - 2(2) + 1(6) = -10. Only the volume takes the absolute value; the determinant keeps its sign, because the sign is what records the orientation.
❌ That is the answer without the middle minus sign. 1(−12)+2(2)+1(6)=−21(-12) + 2(2) + 1(6) = -2 adds the middle term instead of subtracting it. The determinant is −10-10, so the volume is 1010.
❌ A volume is never negative. The determinant is −10-10; the volume is its absolute value, and the sign only records orientation.
❌ Not quite. Once the determinant is known, the volume is ∣det⁡∣|\det| — nothing more to compute.
Show solution

The three minors, deleting row 1 and column jj in turn:

M1=∣0431∣=−12,M2=∣2401∣=2,M3=∣2003∣=6M_1 = \begin{vmatrix} 0 & 4 \\ 3 & 1 \end{vmatrix} = -12, \qquad M_2 = \begin{vmatrix} 2 & 4 \\ 0 & 1 \end{vmatrix} = 2, \qquad M_3 = \begin{vmatrix} 2 & 0 \\ 0 & 3 \end{vmatrix} = 6

Expand along the first row:

det⁡(A⃗,B⃗,C⃗)=1(−12)−2(2)+1(6)=−12−4+6=−10\det(\vec{A}, \vec{B}, \vec{C}) = 1(-12) - 2(2) + 1(6) = -12 - 4 + 6 = -10

The determinant is a signed quantity, so the volume is

volume=∣−10∣=10\text{volume} = \bigl| -10 \bigr| = 10

Dropping the middle minus sign would give −12+4+6=−2-12 + 4 + 6 = -2, whose absolute value 22 is not the volume of anything here.

Problem 4 · How Much the Minus Sign Is Worth

Given: a student expands a 3×33 \times 3 determinant along the first row but writes ++ on the middle term and gets 3030. The correct value is 66. Find the middle term −a2M2-a_2 M_2 of the correct expansion.

✅ Correct! The two answers differ by 2a2M2=242 a_2 M_2 = 24, so a2M2=12a_2 M_2 = 12 and the correct middle term is −12-12.
❌ That is a2M2a_2 M_2, not the term. Subtracting the two expansions does give a2M2=12a_2 M_2 = 12, but the middle term of the correct expansion is −a2M2-a_2 M_2, so it is −12-12.
❌ That is the whole gap, not the term. 30−6=2430 - 6 = 24 counts the middle term twice — once for dropping −a2M2-a_2M_2 and once for adding +a2M2+a_2M_2.
❌ Not quite. Write both expansions with the same M1,M2,M3M_1, M_2, M_3 and subtract one from the other.
Show solution

Both computations use the same three minors. Write them out:

correct=a1M1−a2M2+a3M3=6\text{correct} = a_1 M_1 - a_2 M_2 + a_3 M_3 = 6 student=a1M1+a2M2+a3M3=30\text{student} = a_1 M_1 + a_2 M_2 + a_3 M_3 = 30

Subtracting the first from the second cancels a1M1a_1 M_1 and a3M3a_3 M_3:

2a2M2=30−6=24⟹a2M2=122 a_2 M_2 = 30 - 6 = 24 \quad \Longrightarrow \quad a_2 M_2 = 12

The middle term of the correct expansion is therefore

−a2M2=−12-a_2 M_2 = -12

Check: a1M1+a3M3=6+12=18a_1 M_1 + a_3 M_3 = 6 + 12 = 18, and indeed 18−12=618 - 12 = 6 while 18+12=3018 + 12 = 30.

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