Inverse of a 3×3 Matrix Calculator

Published on: October 5 2025

This Inverse of a 3×3 Matrix Calculator helps you find the inverse of a 3×3 matrix and shows each step clearly. It works by finding the determinant, computing the cofactors and adjugate matrix, and then multiplying by the reciprocal of the determinant. This makes it useful for checking answers, understanding how a matrix inverse is found, and practising linear algebra step by step.

Step-by-step method

  1. Identify the entries a through i in the 3×3 matrix.
  2. Find the determinant by breaking it into the a-part, b-part, and c-part.
  3. Find each 2×2 minor to build the minor matrix M.
  4. Apply the cofactor sign pattern S to M to get the cofactor matrix C.
  5. Transpose C to get Cᵀ, which is adj( A ).
  6. Multiply adj( A ) by 1 / det( A ) and simplify.

Determinant formula

For
A =
a
b
c
d
e
f
g
h
i
det( A ) = a(ei − fh) − b(di − fg) + c(dh − eg)

Inverse formula

A−1 =
1
det( A )
·
adj( A )
where adj( A ) = CT
C = S applied to M
M = minor matrix
S = cofactor sign pattern =
+
−
+
−
+
−
+
−
+
Use this only if det( A ) 0

Example 1: Take the matrix below.

A =
1
2
3
0
1
4
5
6
0

Step 1 - Identify the entries a through i in the 3×3 matrix.

In this problem: For a 3×3 matrix, label the entries from left to right and top to bottom as a through i.

a = 1, b = 2, c = 3
d = 0, e = 1, f = 4
g = 5, h = 6, i = 0

Step 2A - Find the a-part of the determinant.

In this problem: The a-part is a(ei − fh).

a(ei − fh) = ( 1 )×(( 1 )×( 0 ) − ( 4 )×( 6 )) = -24

Step 2B - Find the b-part of the determinant.

In this problem: The b-part is b(di − fg). This part is subtracted in the determinant formula.

b(di − fg) = ( 2 )×(( 0 )×( 0 ) − ( 4 )×( 5 )) = -40

Step 2C - Find the c-part of the determinant.

In this problem: The c-part is c(dh − eg).

c(dh − eg) = ( 3 )×(( 0 )×( 6 ) − ( 1 )×( 5 )) = -15

Step 2D - Combine the determinant parts.

In this problem: The determinant is 1. Since 1 0, the matrix has an inverse.

det( A ) = ( -24 ) − ( -40 ) + ( -15 ) = 1,
1 0

Step 3A - Find M11.

In this problem: Remove row 1 and column 1, then take the determinant of the remaining 2×2 matrix.

M11 = det
1
4
6
0
= ( 1 )×( 0 ) − ( 4 )×( 6 ) = -24

Step 3B - Find M12.

In this problem: Remove row 1 and column 2, then take the determinant of the remaining 2×2 matrix.

M12 = det
0
4
5
0
= ( 0 )×( 0 ) − ( 4 )×( 5 ) = -20

Step 3C - Find M13.

In this problem: Remove row 1 and column 3, then take the determinant of the remaining 2×2 matrix.

M13 = det
0
1
5
6
= ( 0 )×( 6 ) − ( 1 )×( 5 ) = -5

Step 3D - Find M21.

In this problem: Remove row 2 and column 1, then take the determinant of the remaining 2×2 matrix.

M21 = det
2
3
6
0
= ( 2 )×( 0 ) − ( 3 )×( 6 ) = -18

Step 3E - Find M22.

In this problem: Remove row 2 and column 2, then take the determinant of the remaining 2×2 matrix.

M22 = det
1
3
5
0
= ( 1 )×( 0 ) − ( 3 )×( 5 ) = -15

Step 3F - Find M23.

In this problem: Remove row 2 and column 3, then take the determinant of the remaining 2×2 matrix.

M23 = det
1
2
5
6
= ( 1 )×( 6 ) − ( 2 )×( 5 ) = -4

Step 3G - Find M31.

In this problem: Remove row 3 and column 1, then take the determinant of the remaining 2×2 matrix.

M31 = det
2
3
1
4
= ( 2 )×( 4 ) − ( 3 )×( 1 ) = 5

Step 3H - Find M32.

In this problem: Remove row 3 and column 2, then take the determinant of the remaining 2×2 matrix.

M32 = det
1
3
0
4
= ( 1 )×( 4 ) − ( 3 )×( 0 ) = 4

Step 3I - Find M33.

In this problem: Remove row 3 and column 3, then take the determinant of the remaining 2×2 matrix.

M33 = det
1
2
0
1
= ( 1 )×( 1 ) − ( 2 )×( 0 ) = 1

Step 4 - Apply the cofactor sign pattern S to M to get the cofactor matrix C.

In this problem: M is the minor matrix and S is the cofactor sign pattern. Apply S to M to get the cofactor matrix C.

M =
-24
-20
-5
-18
-15
-4
5
4
1
S =
+
−
+
−
+
−
+
−
+
C = S applied to M =
-24
20
-5
18
-15
4
5
-4
1

Step 5 - Transpose C to get Cᵀ, which is adj( A ).

In this problem: Transpose the cofactor matrix by switching rows and columns.

adj( A ) = CT =
-24
18
5
20
-15
-4
-5
4
1

Step 6 - Multiply adj( A ) by 1 / det( A ) and simplify.

In this problem: Multiply each entry of adj( A ) by the reciprocal of the determinant and simplify.

A−1 =
1
1
·
-24
18
5
20
-15
-4
-5
4
1
=
-24
18
5
20
-15
-4
-5
4
1

Final answer:

-24
18
5
20
-15
-4
-5
4
1

Example 2: Take the matrix below.

A =
2
0
1
1
1
0
3
2
1

Step 1 - Identify the entries a through i in the 3×3 matrix.

In this problem: For a 3×3 matrix, label the entries from left to right and top to bottom as a through i.

a = 2, b = 0, c = 1
d = 1, e = 1, f = 0
g = 3, h = 2, i = 1

Step 2A - Find the a-part of the determinant.

In this problem: The a-part is a(ei − fh).

a(ei − fh) = ( 2 )×(( 1 )×( 1 ) − ( 0 )×( 2 )) = 2

Step 2B - Find the b-part of the determinant.

In this problem: The b-part is b(di − fg). This part is subtracted in the determinant formula.

b(di − fg) = ( 0 )×(( 1 )×( 1 ) − ( 0 )×( 3 )) = 0

Step 2C - Find the c-part of the determinant.

In this problem: The c-part is c(dh − eg).

c(dh − eg) = ( 1 )×(( 1 )×( 2 ) − ( 1 )×( 3 )) = -1

Step 2D - Combine the determinant parts.

In this problem: The determinant is 1. Since 1 0, the matrix has an inverse.

det( A ) = ( 2 ) − ( 0 ) + ( -1 ) = 1,
1 0

Step 3A - Find M11.

In this problem: Remove row 1 and column 1, then take the determinant of the remaining 2×2 matrix.

M11 = det
1
0
2
1
= ( 1 )×( 1 ) − ( 0 )×( 2 ) = 1

Step 3B - Find M12.

In this problem: Remove row 1 and column 2, then take the determinant of the remaining 2×2 matrix.

M12 = det
1
0
3
1
= ( 1 )×( 1 ) − ( 0 )×( 3 ) = 1

Step 3C - Find M13.

In this problem: Remove row 1 and column 3, then take the determinant of the remaining 2×2 matrix.

M13 = det
1
1
3
2
= ( 1 )×( 2 ) − ( 1 )×( 3 ) = -1

Step 3D - Find M21.

In this problem: Remove row 2 and column 1, then take the determinant of the remaining 2×2 matrix.

M21 = det
0
1
2
1
= ( 0 )×( 1 ) − ( 1 )×( 2 ) = -2

Step 3E - Find M22.

In this problem: Remove row 2 and column 2, then take the determinant of the remaining 2×2 matrix.

M22 = det
2
1
3
1
= ( 2 )×( 1 ) − ( 1 )×( 3 ) = -1

Step 3F - Find M23.

In this problem: Remove row 2 and column 3, then take the determinant of the remaining 2×2 matrix.

M23 = det
2
0
3
2
= ( 2 )×( 2 ) − ( 0 )×( 3 ) = 4

Step 3G - Find M31.

In this problem: Remove row 3 and column 1, then take the determinant of the remaining 2×2 matrix.

M31 = det
0
1
1
0
= ( 0 )×( 0 ) − ( 1 )×( 1 ) = -1

Step 3H - Find M32.

In this problem: Remove row 3 and column 2, then take the determinant of the remaining 2×2 matrix.

M32 = det
2
1
1
0
= ( 2 )×( 0 ) − ( 1 )×( 1 ) = -1

Step 3I - Find M33.

In this problem: Remove row 3 and column 3, then take the determinant of the remaining 2×2 matrix.

M33 = det
2
0
1
1
= ( 2 )×( 1 ) − ( 0 )×( 1 ) = 2

Step 4 - Apply the cofactor sign pattern S to M to get the cofactor matrix C.

In this problem: M is the minor matrix and S is the cofactor sign pattern. Apply S to M to get the cofactor matrix C.

M =
1
1
-1
-2
-1
4
-1
-1
2
S =
+
−
+
−
+
−
+
−
+
C = S applied to M =
1
-1
-1
2
-1
-4
-1
1
2

Step 5 - Transpose C to get Cᵀ, which is adj( A ).

In this problem: Transpose the cofactor matrix by switching rows and columns.

adj( A ) = CT =
1
2
-1
-1
-1
1
-1
-4
2

Step 6 - Multiply adj( A ) by 1 / det( A ) and simplify.

In this problem: Multiply each entry of adj( A ) by the reciprocal of the determinant and simplify.

A−1 =
1
1
·
1
2
-1
-1
-1
1
-1
-4
2
=
1
2
-1
-1
-1
1
-1
-4
2

Final answer:

1
2
-1
-1
-1
1
-1
-4
2
Matrix A