Example of Finding the Inverse of a 3x3 Matrix

This solution was made using 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
		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}

		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}

		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

Let's check the result

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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