How do you find the determinant of a 3x3 matrix?

Cofactor expansion along row 1: det([[a,b,c],[d,e,f],[g,h,i]]) = a(ei-fh) - b(di-fg) + c(dh-eg) Steps: 1. For element a (position 0,0): delete row 1 and col 1. Compute det of remaining 2x2: ei-fh. Multiply by a. 2. For element b (position 0,1): delete row 1 and col 2. Compute det: di-fg. Multiply by -b (sign changes). 3. For element c (position 0,2): delete row 1 and col 3. Compute det: dh-eg. Multiply by +c. 4. Sum: a(ei-fh) - b(di-fg) + c(dh-eg). Example: [[1,2,3],[4,5,6],[7,8,9]] = 1(5*9-6*8) - 2(4*9-6*7) + 3(4*8-5*7) = 1(-3) - 2(-6) + 3(-3) = -3 + 12 - 9 = 0

What is the Sarrus rule for 3x3 determinants?

Sarrus rule is a memory shortcut for 3x3 determinants only: 1. Write the 3x3 matrix and append columns 1 and 2 to the right 2. Multiply along three downward-right diagonals (positive terms) 3. Multiply along three upward-right diagonals (negative terms) 4. det = sum of positive terms - sum of negative terms For [[a,b,c],[d,e,f],[g,h,i]]: Positive: aei + bfg + cdh Negative: ceg + afh + bdi det = (aei + bfg + cdh) - (ceg + afh + bdi) Warning: Sarrus rule does NOT generalise to 4x4 or larger matrices. For those, use cofactor expansion or row reduction.

What does det = 0 mean for a 3x3 matrix?

A 3x3 determinant of 0 means the matrix is singular. Consequences: - No inverse matrix exists - The system Ax = b has either no solution or infinitely many - The three row vectors are linearly dependent (at least one row is a linear combination of the others) - The parallelepiped formed by the three row vectors has zero volume - The transformation collapses 3D space onto a plane, line, or point Example: [[1,2,3],[4,5,6],[7,8,9]] has det=0 because row 3 = row 1 + 2*(row 2 - row 1). The rows lie in the same plane. To check for singularity quickly: does any row equal a linear combination of the others? If yes, det = 0.

What is the cofactor sign pattern for a 3x3 matrix?

The cofactor Cij = (-1)^(i+j) * Mij where Mij is the (i,j) minor determinant. The sign pattern for a 3x3 is: | + - + | | - + - | | + - + | Position (0,0): sign = (-1)^0 = + Position (0,1): sign = (-1)^1 = - Position (0,2): sign = (-1)^2 = + Position (1,0): sign = (-1)^1 = - ... and so on, alternating. You can expand along ANY row or column -- the result is the same. Choosing a row/column with many zeros reduces arithmetic significantly. Example: expanding along a row with a zero entry means that entire cofactor term vanishes, saving one 2x2 determinant calculation.

How does a 3x3 determinant relate to volume?

The absolute value of the 3x3 determinant equals the volume of the parallelepiped formed by the three row vectors. If the rows are vectors v1, v2, v3, then: Volume = |det([v1, v2, v3])| Sign interpretation: - det > 0: right-handed orientation (standard) - det < 0: left-handed orientation (reflected) - det = 0: degenerate (flat parallelepiped, zero volume) For linear transformations: - A matrix with det=2 scales all volumes by a factor of 2 - det=-1 reflects and preserves volume - det=0 collapses 3D space to a lower dimension This is also why det(AB) = det(A)*det(B): composing two transformations multiplies their volume scale factors.

3x3 Matrix Determinant Calculator — Cofactor Expansion and Sarrus Rule

The 3x3 Matrix Determinant Calculator finds det([[a,b,c],[d,e,f],[g,h,i]]) using your choice of cofactor expansion or Sarrus rule. Enter your 3x3 matrix and see the complete step-by-step working: each 2x2 minor is extracted and computed, the cofactor sign is applied, and all three contributions are ...

ENTER 3×3 MATRIX

QUICK EXAMPLES

DETERMINANT

det = a·C₀₀ + b·C₀₁ + c·C₀₂

= 1·(-3) + 2·(6) + 3·(-3)

= -3 +12 -9

Invertible?

No (singular)

Sign

Zero

|det|

Volume scale

0×

Singular matrix — det = 0

No inverse exists. The rows are linearly dependent. The parallelepiped formed by the rows has zero volume.

COFACTOR EXPANSION — ROW 1

det = a·M₀₀ − b·M₀₁ + c·M₀₂

+ 1 × M₀0 (delete row 1, col 1)

1×

5

6

8

9

=-3→ cofactor =-3

Contribution: 1 × -3 = -3

− 2 × M₀1 (delete row 1, col 2)

2×

4

6

7

9

=-6→ cofactor =6

Contribution: 2 × 6 = 12

+ 3 × M₀2 (delete row 1, col 3)

3×

4

5

7

8

=-3→ cofactor =-3

Contribution: 3 × -3 = -9

det = -3 +12 -9 = 0

Created with❤️byeaglecalculator.com

HOW TO USE

1
Enter the nine values of your 3x3 matrix, row by row. The determinant is computed automatically — you do not need to press a button. Use Tab to move between cells quickly.
2
Choose your method: Cofactor Expansion (the standard algebraic method, works for any size) or Sarrus Rule (a 3x3-only diagonal shortcut). Both give the same result.
3
Cofactor Expansion shows each of the three 2x2 minors extracted from row 1, the cofactor (with sign), and the contribution to the final sum. Every arithmetic step is shown.
4
Sarrus Rule appends columns 1 and 2 to the right of the matrix, then sums three downward-right diagonal products and subtracts three upward-right diagonal products.
5
The result is shown in green (positive), red (negative), or grey (zero/singular). If det = 0, a warning explains the matrix has no inverse and its rows are linearly dependent.

WORKED EXAMPLE

Matrix A = [[1,2,3],[0,1,4],[5,6,0]]. Cofactor expansion along row 1: C00 = +|[[1,4],[6,0]]| = 1*0-4*6 = -24. Contribution: 1*(-24) = -24. C01 = -|[[0,4],[5,0]]| = -(0*0-4*5) = 20. Contribution: 2*20 = 40. C02 = +|[[0,1],[5,6]]| = 0*6-1*5 = -5. Contribution: 3*(-5) = -15. det = -24 + 40 + (-15) = 1. Invertible: yes (det=1, perfect for integer inverse).

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Last updated: April 29, 2026 · Formula verified by EagleCalculator team · Eagle-eyed accuracy for every calculation.