The Determinant of a 3 × 3 Matrix

Expanding along the first row

A 3 × 3 determinant is built from 2 × 2 determinants, which are already known: for (a b; c d) the determinant is ad − bc.

Take M = (2 1 3; 0 4 5; 1 0 6). To expand along row 1, go through the entries of row 1 one at a time. For each one, delete the row and the column it sits in. Four entries are left, making a 2 × 2 matrix, and its determinant is called the minor of that entry. Each entry of row 1 multiplies its own minor.

The three minors

Delete row 1 and column 1, through the 2. The entries left are 4, 5, 0 and 6, so the minor of 2 is the determinant of (4 5; 0 6), which is 4 × 6 − 5 × 0 = 24.

Row 1 and column 1 are struck out through the 2. The four entries left in gold, 4, 5, 0 and 6, make the minor of the 2, which is 24.

Delete row 1 and column 2, through the 1. What is left is (0 5; 1 6), and its determinant is 0 × 6 − 5 × 1 = −5.

Struck out through the 1, the four entries left are 0, 5, 1 and 6, from columns 1 and 3. Their determinant, −5, is the minor of the 1.

Delete row 1 and column 3, through the 3. What is left is (0 4; 1 0), and its determinant is 0 × 0 − 4 × 1 = −4.

Struck out through the 3, the four entries left are 0, 4, 1 and 0. Their determinant, −4, is the minor of the 3.

Signs and the sum

Multiply each entry of row 1 by its minor, then combine the three products with the signs plus, minus, plus:

det M = 2 × 24 − 1 × (−5) + 3 × (−4).

The first product is 48. The second is −5, and subtracting it adds 5. The third is −12. So det M = 48 + 5 − 12 = 41.

Two different minus signs meet in the middle term: the minus that belongs to the place in row 1, and the minus of the minor −5 itself. Each has to be kept.

Where the signs come from

Every place in a 3 × 3 matrix has a sign, and the signs alternate like a chessboard, starting with plus in the top left. Row 1 reads plus, minus, plus; row 2 reads minus, plus, minus; row 3 reads plus, minus, plus. The place in row i and column j has a plus when i + j is even and a minus when it is odd.

The sign belongs to the place, not to the number in it. The 1 in row 1, column 2 is positive, but its place carries a minus, so its product is subtracted.

The sign of each place. Expanding along row 1 uses the top row: plus, minus, plus.

Any row or column gives the same answer

A determinant can be expanded along any row or any column, using the signs of those places, and the answer is always the same. Column 1 of M holds 2, 0 and 1, with the signs plus, minus, plus. The 0 makes its whole term 0, so only two minors are needed. The minor of the 1 in row 3 is the determinant of (1 3; 4 5), which is 1 × 5 − 3 × 4 = −7. So det M = 2 × 24 − 0 + 1 × (−7) = 48 − 7 = 41.

Row 2 holds 0, 4 and 5, with the signs minus, plus, minus. The minor of the 4 is the determinant of (2 3; 1 6), which is 12 − 3 = 9, and the minor of the 5 is the determinant of (2 1; 1 0), which is 0 − 1 = −1. The 0 contributes nothing, so det M = 4 × 9 − 5 × (−1) = 36 + 5 = 41 again.

So choose the row or column with the most zeros: every 0 is a minor that never needs working out.

Column 1 of M holds a 0, so expanding down it needs only two minors, and it gives 2 × 24 + 1 × (−7) = 41.

Zeros and triangles

If row 1 is 3, 0, 0, two of the three terms vanish. For (3 0 0; 1 2 4; 5 1 3), the determinant is 3 times the minor of the 3, which is 2 × 3 − 4 × 1 = 2, so the determinant is 3 × 2 = 6.

A matrix with only zeros below its main diagonal, such as (2 1 3; 0 4 5; 0 0 6), has determinant 2 × 4 × 6 = 48, the product of its diagonal. Expanding down column 1 shows it: only the 2 survives, and its minor is the determinant of (4 5; 0 6), which is 24, so the determinant is 2 × 24 = 48.

What the number means

For a 2 × 2 matrix the determinant is the factor by which the matrix scales area. A 3 × 3 matrix moves the points of space, and its determinant is the factor by which it scales volume: M sends a cube of volume 1 to a slanted box of volume 41.

A 3 × 3 matrix has an inverse exactly when its determinant is not 0. When one row is a multiple of another, the determinant is 0. For (1 2 3; 2 4 6; 1 0 1), where row 2 is twice row 1, expanding along row 1 gives 1 × (4 × 1 − 6 × 0) − 2 × (2 × 1 − 6 × 1) + 3 × (2 × 0 − 4 × 1) = 4 + 8 − 12 = 0. That matrix flattens space onto a plane, and nothing can undo it.

The usual mistakes

Adding all three products. 2 × 24 + 1 × (−5) + 3 × (−4) = 48 − 5 − 12 = 31, not 41: the middle place carries a minus.

Adding inside a minor. The minor of the 1 is 0 × 6 − 5 × 1 = −5; adding the products gives 0 + 5 = 5, which is not a determinant.

Pairing the entries down the columns of the 2 × 2. For (0 5; 1 6), 0 × 1 − 5 × 6 = −30 is not the minor; each product runs along a diagonal.

Forgetting the entry. The minor 24 is multiplied by the 2 it belongs to, giving 48.

Taking the sign from the entry. A positive entry in a minus place is still subtracted.

A field from its corners

In the application below, the coordinates of the three corners of a field go into the rows of a 3 × 3 determinant with a column of 1s. Half of that determinant, expanded along row 1, is the area of the field, and a determinant of 0 shows that a fence post stands in line with two of the corners.

Worked example: The Area of a Triangular Field from the Coordinates of Its Corners, and a Fence Post Tested Against Two of Them

Question On a map with a grid of 10 m squares, the corners of a triangular field are A(2, 2), B(8, 4) and C(4, 9). The area of a triangle with corners (x₁, y₁), (x₂, y₂) and (x₃, y₃) is half of the determinant x₁y₁1x₂y₂1x₃y₃1, taken without its sign. (a) Find the determinant for A, B and C by expanding along the first row, and hence the area of the field in square meters. (b) A fence post stands at D(14, 6). Find the determinant for A, B and D, and say what it shows about the three posts.

1.Put the corners in the rows: 221841491. Each entry of the first row multiplies the 2 × 2 determinant left when its row and column are deleted, and the signs go +, −, +.

The corners go in the rows of 221841491, and the expansion along the first row takes the signs +, −, +.
2.The three 2 × 2 determinants are 4191 = 4 − 9 = −5, 8141 = 8 − 4 = 4 and 8449 = 72 − 16 = 56.

The three 2 × 2 determinants are 4191 = −5, 8141 = 4 and 8449 = 56.
3.So the determinant is 2 × (−5) − 2 × 4 + 1 × 56 = −10 − 8 + 56 = 38.

The determinant is 2 × (−5) − 2 × 4 + 1 × 56 = 38.
4.(a) The area is 12 × 38 = 19 grid squares. Each square is 10 × 10 = 100 square meters, so the field is 19 × 100 = 1900 square meters.

(a) The field is half of 38, which is 19 grid squares, and each square is 100 square meters: 1900 square meters.
5.For A, B and D: 2218411461 = 2 × (4 − 6) − 2 × (8 − 14) + 1 × (48 − 56) = −4 + 12 − 8 = 0.

For A, B and D: 2218411461 = −4 + 12 − 8 = 0.
6.(b) A triangle ABD would have no area, so A, B and D lie on one straight line: a fence from A through B runs straight on to D. Check: from A to B is 6 across and 2 up, and from A to D is 12 across and 4 up, which is the same direction.

(b) The determinant is 0, so A, B and D make no triangle: they lie on one straight line.

Answer: (a) The determinant is 38, so the field has an area of 19 grid squares, which is 1900 square meters; (b) the determinant is 0, so A, B and D lie on one straight line

Common mistakes

Using a plus sign for every term of the expansion. The middle term takes a minus sign, from the pattern +, −, + along the first row; with all plus signs the determinant comes out as −10 + 8 + 56 = 54 and the area as 27 squares, which is wrong.
Giving the area as 19 square meters. The coordinates count grid squares 10 m wide, so each square unit of area on the map is 100 square meters on the ground.

More determinants and inverses problems, worked step by step →