This tool finds the determinant of a 3 by 3 matrix and, when the determinant is not zero, the inverse. Each entry can be an integer, a decimal, or a simple fraction such as 3/4. The arithmetic stays exact: every result is a reduced fraction.
The numbers are meaningful when a homework matrix is 3 by 3 and you need the volume scale, the adjugate, or a matrix that undoes A. The page writes the cofactor expansion, then multiplies A by the inverse and checks for the identity.
This page does not compute eigenvalues, and it does not row reduce a general linear system. For 2 by 2 products, use the 2 by 2 matrix calculator. For stretch factors of a 2 by 2 map, use the eigenvalue calculator.
To row reduce a matrix and solve a linear system with exact fractions, use the reduced row echelon form calculator.
What the determinant tells you
The determinant tells you the signed volume scale of the linear map. An inverse exists exactly when that number is not zero.
For a 3 by 3 matrix, cofactor expansion along the first row builds the determinant from three 2 by 2 minors. The sign pattern on that row is plus, minus, plus. Multiply each entry of the first row by its cofactor and add the three products.
The matrix of minors stores every 2 by 2 determinant, including the six that the first row expansion does not use for the determinant itself. The cofactor matrix applies the alternating signs. The adjugate is the transpose of the cofactor matrix, not the transpose of A. The inverse is 1 over the determinant times the adjugate, and this page writes each entry as a reduced fraction.
If the determinant is 0, the matrix is singular and has no inverse. The columns are linearly dependent, so the map collapses volume and no matrix undoes it. Rank below 3 is the same dependence, seen by elimination. Rank 3 agrees with a determinant that is not zero. The transpose is the same array with rows and columns swapped. It is included so you can compare it with the adjugate and see that those two arrays are different jobs.
Type 1/3 when you mean one third. The decimal 0.333 is 333/1000, which is not 1/3. Dropping a cofactor sign, transposing A instead of the cofactor matrix, and dividing before you transpose are the usual places a hand calculation goes wrong.
A fully worked example
Let A have rows (1, 2, 3), (0, 1, 4), and (5, 6, 0). That is the matrix already filled in above. The determinant is 1, so the inverse equals the adjugate. The inverse rows are (−24, 18, 5), (20, −15, −4), and (−5, 4, 1). The product with A is the identity, and the rank is 3. The steps below are written in the page, so you can read them with scripts turned off.
Cofactor expansion of det(A) along row 1.
Minor M₁₁ deletes row 1 and column 1, leaving the 2 by 2 entries 1, 4; 6, 0.
M₁₁ = (1)(0) − (4)(6) = 0 − 24 = −24.
The cofactor sign for row 1, column 1 is +. So C₁₁ = +(−24) = −24.
Minor M₁₂ deletes row 1 and column 2, leaving the 2 by 2 entries 0, 4; 5, 0.
M₁₂ = (0)(0) − (4)(5) = 0 − 20 = −20.
The cofactor sign for row 1, column 2 is −. So C₁₂ = −(−20) = 20.
Minor M₁₃ deletes row 1 and column 3, leaving the 2 by 2 entries 0, 1; 5, 6.
M₁₃ = (0)(6) − (1)(5) = 0 − 5 = −5.
The cofactor sign for row 1, column 3 is +. So C₁₃ = +(−5) = −5.
det(A) = a₁₁ C₁₁ + a₁₂ C₁₂ + a₁₃ C₁₃.
a₁₁ C₁₁ = (1)(−24) = −24.
a₁₂ C₁₂ = (2)(20) = 40.
a₁₃ C₁₃ = (3)(−5) = −15.
det(A) = −24 + 40 − 15 = 1.
The other six minors complete the matrix of minors.
Minor M₂₁ deletes row 2 and column 1, leaving the 2 by 2 entries 2, 3; 6, 0.
M₂₁ = (2)(0) − (3)(6) = 0 − 18 = −18.
The cofactor sign for row 2, column 1 is −. So C₂₁ = −(−18) = 18.
Minor M₂₂ deletes row 2 and column 2, leaving the 2 by 2 entries 1, 3; 5, 0.
M₂₂ = (1)(0) − (3)(5) = 0 − 15 = −15.
The cofactor sign for row 2, column 2 is +. So C₂₂ = +(−15) = −15.
Minor M₂₃ deletes row 2 and column 3, leaving the 2 by 2 entries 1, 2; 5, 6.
M₂₃ = (1)(6) − (2)(5) = 6 − 10 = −4.
The cofactor sign for row 2, column 3 is −. So C₂₃ = −(−4) = 4.
Minor M₃₁ deletes row 3 and column 1, leaving the 2 by 2 entries 2, 3; 1, 4.
M₃₁ = (2)(4) − (3)(1) = 8 − 3 = 5.
The cofactor sign for row 3, column 1 is +. So C₃₁ = +(5) = 5.
Minor M₃₂ deletes row 3 and column 2, leaving the 2 by 2 entries 1, 3; 0, 4.
M₃₂ = (1)(4) − (3)(0) = 4 − 0 = 4.
The cofactor sign for row 3, column 2 is −. So C₃₂ = −(4) = −4.
Minor M₃₃ deletes row 3 and column 3, leaving the 2 by 2 entries 1, 2; 0, 1.
M₃₃ = (1)(1) − (2)(0) = 1 − 0 = 1.
The cofactor sign for row 3, column 3 is +. So C₃₃ = +(1) = 1.
Matrix of minors
Row 1: −24, −20, −5
Row 2: −18, −15, −4
Row 3: 5, 4, 1
Cofactor matrix
Row 1: −24, 20, −5
Row 2: 18, −15, 4
Row 3: 5, −4, 1
Adjugate, the transpose of the cofactor matrix
Row 1: −24, 18, 5
Row 2: 20, −15, −4
Row 3: −5, 4, 1
Inverse = (1/det(A)) times the adjugate. Here det(A) = 1.
Dividing by 1 leaves the adjugate unchanged, so A inverse equals the adjugate.
Row 1: −24, 18, 5
Row 2: 20, −15, −4
Row 3: −5, 4, 1
Check: A times A inverse. Each entry is a row of A dotted with a column of the inverse.
Row 1, column 1: (1)(−24) + (2)(20) + (3)(−5) = 1.
Row 1, column 2: (1)(18) + (2)(−15) + (3)(4) = 0.
Row 1, column 3: (1)(5) + (2)(−4) + (3)(1) = 0.
Row 2, column 1: (0)(−24) + (1)(20) + (4)(−5) = 0.
Row 2, column 2: (0)(18) + (1)(−15) + (4)(4) = 1.
Row 2, column 3: (0)(5) + (1)(−4) + (4)(1) = 0.
Row 3, column 1: (5)(−24) + (6)(20) + (0)(−5) = 0.
Row 3, column 2: (5)(18) + (6)(−15) + (0)(4) = 0.
Row 3, column 3: (5)(5) + (6)(−4) + (0)(1) = 1.
Row 1: 1, 0, 0
Row 2: 0, 1, 0
Row 3: 0, 0, 1
The product is the identity.
Rank is 3.
The three columns are linearly independent.
Transpose of A
Row 1: 1, 0, 5
Row 2: 2, 1, 6
Row 3: 3, 4, 0
Related reading
Determinants and linear maps live on the linear algebra pillar and in the glossary entries for determinant, matrix, and linear map. A study order is the linear algebra first course pathway.
The shorter arithmetic is the 2 by 2 matrix calculator. Stretch factors of a 2 by 2 map are on the 2 by 2 eigenvalue calculator and in eigenvalues in plain language. A zero determinant and a zero eigenvalue are the same collapse, seen two ways.