Example of Finding the Inverse of a 3x3 Matrix

This solution has been done by the calculator presented on the site.

It is necessary to calculate a matrix A-1, inverse to the given one:
A = 4 3 2
2 1 -1
3 3 2
Formula for calculating the inverse matrix:
A-1 = 1 / det A * A11 A21 A31
A12 A22 A32
A13 A23 A33
A11 ... A33  are numbers (algebraic additions) that will be calculated later.
It is impossible to divide by zero. Therefore, if the determinant of A is zero, then it is impossible to calculate inverse matrix.
Let's calculate the determinant A.
det A = 4 3 2 =
2 1 -1
3 3 2
The elements of row 3 multiplied by -1 are added to the corresponding elements of row 1.   more info
4 + 3 * ( -1) 3 + 3 * ( -1) 2 + 2 * ( -1)
2 1 -1
3 3 2
This elementary transformation does not change the value of the determinant.
= 1 0 0 =
2 1 -1
3 3 2
Expand the determinant along the row 1.   more info
1 0 0
2 1 -1
3 3 2
Row number 1
Column number 1
Element Row 1 and column 1
have been deleted
( -1) 1 + 1 * 1 *
1 -1
3 2
1 0 0
2 1 -1
3 3 2
Row number 1
Column number 2
Element Row 1 and column 2
have been deleted
( -1) 1 + 2 * 0 *
2 -1
3 2
1 0 0
2 1 -1
3 3 2
Row number 1
Column number 3
Element Row 1 and column 3
have been deleted
( -1) 1 + 3 * 0 *
2 1
3 3
Products are summed. If the element is zero than product is zero too.
= ( -1) 1 + 1 * 1 * 1 -1 =
3 2
= 1 -1 =
3 2
= 1 * 2 - ( -1) * 3 =
= 2 + 3 =
= 5
Determinant A is not zero. It is possible to calculate inverse matrix.
Let's calculate numbers (algebraic additions)   A11 ... A33
4 3 2
2 1 -1
3 3 2
Row number 1
Column number 1
Row 1 and column 1
have been deleted
A11 = ( -1) 1 + 1 *
1 -1 =
3 2
= 1 * 2 - ( -1) * 3 = 2 + 3 = 5
4 3 2
2 1 -1
3 3 2
Row number 1
Column number 2
Row 1 and column 2
have been deleted
A12 = ( -1) 1 + 2 *
2 -1 =
3 2
= - ( 2 * 2 - ( -1) * 3 ) = - (4 + 3) = -7
4 3 2
2 1 -1
3 3 2
Row number 1
Column number 3
Row 1 and column 3
have been deleted
A13 = ( -1) 1 + 3 *
2 1 =
3 3
= 2 * 3 - 1 * 3 = 6 - 3 = 3
4 3 2
2 1 -1
3 3 2
Row number 2
Column number 1
Row 2 and column 1
have been deleted
A21 = ( -1) 2 + 1 *
3 2 =
3 2
= - ( 3 * 2 - 2 * 3 ) = - (6 - 6) = 0
4 3 2
2 1 -1
3 3 2
Row number 2
Column number 2
Row 2 and column 2
have been deleted
A22 = ( -1) 2 + 2 *
4 2 =
3 2
= 4 * 2 - 2 * 3 = 8 - 6 = 2
4 3 2
2 1 -1
3 3 2
Row number 2
Column number 3
Row 2 and column 3
have been deleted
A23 = ( -1) 2 + 3 *
4 3 =
3 3
= - ( 4 * 3 - 3 * 3 ) = - (12 - 9) = -3
4 3 2
2 1 -1
3 3 2
Row number 3
Column number 1
Row 3 and column 1
have been deleted
A31 = ( -1) 3 + 1 *
3 2 =
1 -1
= 3 * ( -1) - 2 * 1 = -3 - 2 = -5
4 3 2
2 1 -1
3 3 2
Row number 3
Column number 2
Row 3 and column 2
have been deleted
A32 = ( -1) 3 + 2 *
4 2 =
2 -1
= - ( 4 * ( -1) - 2 * 2 ) = - (-4 - 4) = 8
4 3 2
2 1 -1
3 3 2
Row number 3
Column number 3
Row 3 and column 3
have been deleted
A33 = ( -1) 3 + 3 *
4 3 =
2 1
= 4 * 1 - 3 * 2 = 4 - 6 = -2
Result:
A-1 = 1 / det A * A11 A21 A31
A12 A22 A32
A13 A23 A33
A-1 = 1 / 5 * 5 0 -5
-7 2 8
3 -3 -2
A-1 = 1 0 -1
-7/5 2/5 8/5
3/5 -3/5 -2/5
It is necessary to check that   A-1 * A = E.
We will use the penultimate form of the inverse matrix A-1. This will allow us to count without fractions.
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
b11 b12 b13
b21 b22 b23
b31 b32 b33
b11 = 5 * 4 + 0 * 2 + ( -5) * 3 = 20 + 0 - 15 = 5
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
5 b12 b13
b21 b22 b23
b31 b32 b33
b12 = 5 * 3 + 0 * 1 + ( -5) * 3 = 15 + 0 - 15 = 0
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
5 0 b13
b21 b22 b23
b31 b32 b33
b13 = 5 * 2 + 0 * ( -1) + ( -5) * 2 = 10 + 0 - 10 = 0
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
5 0 0
b21 b22 b23
b31 b32 b33
b21 = -7 * 4 + 2 * 2 + 8 * 3 = -28 + 4 + 24 = 0
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
5 0 0
0 b22 b23
b31 b32 b33
b22 = -7 * 3 + 2 * 1 + 8 * 3 = -21 + 2 + 24 = 5
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
5 0 0
0 5 b23
b31 b32 b33
b23 = -7 * 2 + 2 * ( -1) + 8 * 2 = -14 - 2 + 16 = 0
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
5 0 0
0 5 0
b31 b32 b33
b31 = 3 * 4 + ( -3) * 2 + ( -2) * 3 = 12 - 6 - 6 = 0
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
5 0 0
0 5 0
0 b32 b33
b32 = 3 * 3 + ( -3) * 1 + ( -2) * 3 = 9 - 3 - 6 = 0
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
5 0 0
0 5 0
0 0 b33
b33 = 3 * 2 + ( -3) * ( -1) + ( -2) * 2 = 6 + 3 - 4 = 5
5 0 -5
-7 2 8
3 -3 -2
*
4 3 2
2 1 -1
3 3 2
=
5 0 0
0 5 0
0 0 5
It is necessary to multiply the result by 1/5
1/5 * 5 0 0
0 5 0
0 0 5
=
1 0 0
0 1 0
0 0 1
= E
Thus, the found matrix A-1 is inverse for the given matrix A.
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