4.2 Cofactor Expansion in Three Dimensions

The last section defined $a d - b c$ for a $2 times 2$ matrix and read it three ways: as an area, as an orientation, and as a test for invertibility. All three readings survive in higher dimensions, but the formula does not — there is no two-term expression for a $3 times 3$ determinant. What there is instead is a rule that reduces the new case to the old one, applied three times.

Take

$M equals open parenthesis 2, 1, 0; 1, 3, 1; 0, 1, 2 close parenthesis$

The minor of an entry is the determinant of what remains when you delete that entry's row and column — a $2 times 2$ determinant, which you can already do. Deleting the first row and first column of $M$ leaves $open parenthesis 3, 1; 1, 2 close parenthesis$ , whose determinant is $3 (2) - 1 (1) = 5$ . That $5$ is the minor of the entry $2$ in the corner. The cofactor of an entry is its minor with a sign attached, $(- 1)^{i + j}$ for the entry in row $i$ and column $j$ — so the cofactor of that corner $2$ is $plus 5$ , and the word in this section's title just means "expansion in terms of those signed minors".

Cofactor expansion says: walk along any one row or column, multiply each entry by its minor, and add the results with alternating signs. Along the first row of $M$ the entries are $2$ , $1$ and $0$ , so

$det M = 2 \cdot 5 - 1 \cdot 2 + 0 \cdot 1 = 8$

where the three minors come from deleting the first row together with each column in turn.

5

The first minor ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

Delete the first row and first column of $M$ and take the determinant of the $2 times 2$ block left behind: $3 (2) - 1 (1) = 5$ . Every cofactor expansion is built from determinants this small, which is what makes the rule usable by hand.

2

The second minor ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

Delete the first row and the second column and the remaining block is $open parenthesis 1, 1; 0, 2 close parenthesis$ , of determinant $2$ . This is the term that carries the minus sign, which is a property of its position — second along the row — and has nothing to do with its entries.

8

The whole determinant ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The full $3 times 3$ determinant, computed by a computer algebra system: $8$ , matching the expansion $2 (5) - 1 (2) + 0 (1)$ done by hand above.

The signs, and choosing where to walk

The alternating signs follow a fixed checkerboard, starting with a plus in the top-left corner:

$(\begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix})$

The sign attached to a term depends only on where the entry sits, never on the entry itself — a common early error is to let a negative entry flip a sign twice.

Because every row and every column gives the same answer, you get to choose, and the choice is worth making. Expanding $M$ along its bottom row, whose entries are $0$ , $1$ and $2$ with signs $+$ , $-$ , $+$ :

$det M = 0 \cdot 1 - 1 \cdot 2 + 2 \cdot 5 = 8$

the same $8$ — and, as it happens, through the same three products $10$ , $minus 2$ , $0$ in a different order, since both walks use all three of the same minors. What changed is the work: the leading $0$ multiplied its minor away, so that minor never had to be computed at all. Hence the practical rule: expand along the row or column with the most zeros.

Watching it vanish

Fix the top-left $2 times 2$ block and put the bottom-right entry on a slider:

$M sub g equals open parenthesis 2, 1, 0; 1, 3, 1; 0, 1, g close parenthesis$

Expanding along the bottom row again gives $0 \cdot 1 - 1 \cdot 2 + g \cdot 5$ , so $det M_{g} = 5 g - 2$ : a straight line in $g$ , hitting zero at $g equals 0.4$ and recovering the value $8$ at $g equals 2$ .

- 2 + 5 \cdot g

The whole family at once ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The determinant computed symbolically in $g$ . It is linear, because $g$ appears in exactly one entry and a cofactor expansion uses each entry at most once — determinants are linear in any single row or column, though emphatically not in the matrix as a whole.

A semi-transparent patch of a plane floating in space, spanned by a red arrow to (2, 1, 0) and a blue arrow to (1, 3, 1) that both start at the origin. A green arrow rises from the origin to (0, 1, g), where a slider sets g, and a dashed amber segment joins the green arrow's tip to a black dot at (0, 1, 0.4) lying in the parallelogram's plane. Dragging the slider down to 0.4 closes that gap, laying the green arrow flat into the parallelogram; there the determinant of the three columns is zero, and above or below it the determinant is the signed amount 5g - 2. — A patch of the plane spanned by the first two columns, $open parenthesis 2, 1, 0 close parenthesis$ in red and $open parenthesis 1, 3, 1 close parenthesis$ in blue. The green arrow is the third column $open parenthesis 0, 1, g close parenthesis$ , with $g$ on a slider, and the dashed amber segment measures the gap between its tip and the black dot at $open parenthesis 0, 1, 0.4 close parenthesis$ — where that column would have to sit to lie flat in the same plane.

Explore the figure

At the default $g equals 2$ the green arrow stands well clear of the blue-and-red sheet, and $det M_{g} = 5 (2) - 2 = 8$ , the determinant you computed by hand.
Drag $g$ down and watch two things at once: the dashed gap shrinking, and the value $5 g minus 2$ heading for zero. They arrive together — the geometry and the arithmetic are reporting the same event.
At $g equals 0.4$ the gap closes and the green arrow lies in the sheet. The three columns are now dependent: as the algebra confirms, $(0, 1, 0.4) = - 0.2 (2, 1, 0) + 0.4 (1, 3, 1)$ .
Keep going to $g = - 1$ . The green arrow has swung out below the sheet and $det M_{g} = 5 (- 1) - 2 = - 7$ . As in the last section, passing through dependence is the only way for the sign to change. (Avoid stopping at exactly $g equals 0$ to judge this: there the third column is $open parenthesis 0, 1, 0 close parenthesis$ , which lies along the $y$ -axis and is hard to pick out against it.)

Example 1 Expand along the cheapest line

Compute the determinant of $N$ with rows $open parenthesis 4, 0, 0 close parenthesis$ , $open parenthesis 7, 2, 1 close parenthesis$ and $open parenthesis minus 3, 5, 3 close parenthesis$ , choosing an expansion that minimizes the arithmetic. Then say what the answer implies about the columns of $N$ .

The first row has two zeros, so expand along it and only one term survives:

$det N = 4 \cdot det (\begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix}) - 0 + 0 = 4 (2 \cdot 3 - 1 \cdot 5) = 4 (1) = 4$

One $2 times 2$ determinant instead of three. Expanding along the first column instead would have cost all three minors, since its entries $4$ , $7$ , $minus 3$ are all nonzero — the answer would still be $4$ , but at triple the work.

Since $det N equals 4 not equal to 0$ , the Invertible Matrix Theorem applies in full: the columns of $N$ are linearly independent, they span $ℝ^{3}$ , $N$ is invertible, and $N 𝐱 = 𝐛$ has exactly one solution for every $𝐛$ . Note that the size of the determinant played no part in that conclusion — only that it was not zero. What the value $4$ does tell you is the subject of the next section.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Its chapter sequence is aligned to Interactive Linear Algebra (Margalit & Rabinoff, Georgia Tech, GNU FDL); this work is original, copies nothing from it, and is not affiliated with or endorsed by its authors. License: CC-BY-NC-SA-4.0.