Example of Finding the Inverse of a 2x2 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 =		1	1
		-1	3

Formula for calculating the inverse matrix:

A^-1 = 1 / det A *		A₁₁	A₂₁
		A₁₂	A₂₂

A₁₁ ... A₂₂ 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 =		1	1
		-1	3

= 1 * 3 - 1 * ( -1) = 3 + 1 = 4

Determinant A is not zero. It is possible to calculate inverse matrix.

Let's calculate numbers (algebraic additions) A₁₁ ... A₂₂

	1	1
	-1	3

Row number 1
Column number 1

Row 1 and column 1
have been deleted

A₁₁

( -1) ^{1 + 1}

= 3

	1	1
	-1	3

Row number 1
Column number 2

Row 1 and column 2
have been deleted

A₁₂

( -1) ^{1 + 2}

-1

= 1

	1	1
	-1	3

Row number 2
Column number 1

Row 2 and column 1
have been deleted

A₂₁

( -1) ^{2 + 1}

= -1

	1	1
	-1	3

Row number 2
Column number 2

Row 2 and column 2
have been deleted

A₂₂

( -1) ^{2 + 2}

= 1

Result:

A^-1 = 1 / det A *		A₁₁	A₂₁
		A₁₂	A₂₂

A^-1 = 1 / 4 *		3	-1
		1	1

A^-1 =		3/4	-1/4
		1/4	1/4

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.

	3	-1
	1	1

	1	1
	-1	3

	b₁₁	b₁₂
	b₂₁	b₂₂

b₁₁ = 3 * 1 + ( -1) * ( -1) = 3 + 1 = 4

	3	-1
	1	1

	1	1
	-1	3

	4	b₁₂
	b₂₁	b₂₂

b₁₂ = 3 * 1 + ( -1) * 3 = 3 - 3 = 0

	3	-1
	1	1

	1	1
	-1	3

	4	0
	b₂₁	b₂₂

b₂₁ = 1 * 1 + 1 * ( -1) = 1 - 1 = 0

	3	-1
	1	1

	1	1
	-1	3

	4	0
	0	b₂₂

b₂₂ = 1 * 1 + 1 * 3 = 1 + 3 = 4

	3	-1
	1	1

	1	1
	-1	3

	4	0
	0	4

It is necessary to multiply the result by 1/4

1/4 *		4	0
		0	4

	1	0
	0	1

= E

Thus, the found matrix A^-1 is inverse for the given matrix A.

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