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Show that the following two matrices are inverses of each other.

A = [ 1 4 −1 −3 ] , B = [ −3 −4 1 1 ]
A B = [ 1 4 −1 −3 ] [ −3 −4 1 1 ] = [ 1 ( −3 ) + 4 ( 1 ) 1 ( −4 ) + 4 ( 1 ) −1 ( −3 ) + −3 ( 1 ) −1 ( −4 ) + −3 ( 1 ) ] = [ 1 0 0 1 ] B A = [ −3 −4 1 1 ] [ 1 4 −1 −3 ] = [ −3 ( 1 ) + −4 ( −1 ) −3 ( 4 ) + −4 ( −3 ) 1 ( 1 ) + 1 ( −1 ) 1 ( 4 ) + 1 ( −3 ) ] = [ 1 0 0 1 ]
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Finding the multiplicative inverse using matrix multiplication

We can now determine whether two matrices are inverses, but how would we find the inverse of a given matrix? Since we know that the product of a matrix and its inverse is the identity matrix, we can find the inverse of a matrix by setting up an equation using matrix multiplication .

Finding the multiplicative inverse using matrix multiplication

Use matrix multiplication to find the inverse of the given matrix.

A = [ 1 −2 2 −3 ]

For this method, we multiply A by a matrix containing unknown constants and set it equal to the identity.

[ 1 −2 2 −3 ]     [ a b c d ] = [ 1 0 0 1 ]

Find the product of the two matrices on the left side of the equal sign.

[ 1 −2 2 −3 ]     [ a b c d ] = [ 1 a −2 c 1 b −2 d 2 a −3 c 2 b −3 d ]

Next, set up a system of equations with the entry in row 1, column 1 of the new matrix equal to the first entry of the identity, 1. Set the entry in row 2, column 1 of the new matrix equal to the corresponding entry of the identity, which is 0.

1 a −2 c = 1      R 1 2 a −3 c = 0      R 2

Using row operations, multiply and add as follows: ( −2 ) R 1 + R 2 R 2 . Add the equations, and solve for c .

1 a 2 c = 1 0 + 1 c = 2 c = 2

Back-substitute to solve for a .

a −2 ( −2 ) = 1 a + 4 = 1 a = −3

Write another system of equations setting the entry in row 1, column 2 of the new matrix equal to the corresponding entry of the identity, 0. Set the entry in row 2, column 2 equal to the corresponding entry of the identity.

1 b −2 d = 0 R 1 2 b −3 d = 1 R 2

Using row operations, multiply and add as follows: ( −2 ) R 1 + R 2 = R 2 . Add the two equations and solve for d .

1 b −2 d = 0 0 + 1 d = 1 d = 1

Once more, back-substitute and solve for b .

b −2 ( 1 ) = 0 b −2 = 0 b = 2
A −1 = [ −3 2 −2 1 ]
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Finding the multiplicative inverse by augmenting with the identity

Another way to find the multiplicative inverse is by augmenting with the identity. When matrix A is transformed into I , the augmented matrix I transforms into A −1 .

For example, given

A = [ 2 1 5 3 ]

augment A with the identity

[ 2 1 5 3   |   1 0 0 1 ]

Perform row operations    with the goal of turning A into the identity.

  1. Switch row 1 and row 2.
    [ 5 3 2 1   |   0 1 1 0 ]
  2. Multiply row 2 by −2 and add to row 1.
    [ 1 1 2 1   |   −2 1 1 0 ]
  3. Multiply row 1 by −2 and add to row 2.
    [ 1 1 0 −1   |   −2 1 5 −2 ]
  4. Add row 2 to row 1.
    [ 1 0 0 −1   |   3 −1 5 −2 ]
  5. Multiply row 2 by −1.
    [ 1 0 0 1   |   3 −1 −5 2 ]

The matrix we have found is A −1 .

A −1 = [ 3 −1 −5 2 ]

Finding the multiplicative inverse of 2×2 matrices using a formula

When we need to find the multiplicative inverse of a 2 × 2 matrix, we can use a special formula instead of using matrix multiplication or augmenting with the identity.

If A is a 2 × 2 matrix, such as

A = [ a b c d ]

the multiplicative inverse of A is given by the formula

A −1 = 1 a d b c [ d b c a ]

where a d b c 0. If a d b c = 0 , then A has no inverse.

Using the formula to find the multiplicative inverse of matrix A

Use the formula to find the multiplicative inverse of

A = [ 1 −2 2 −3 ]

Using the formula, we have

A −1 = 1 ( 1 ) ( −3 ) ( −2 ) ( 2 ) [ −3 2 −2 1 ] = 1 −3 + 4 [ −3 2 −2 1 ] = [ −3 2 −2 1 ]
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Questions & Answers

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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
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mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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4
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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Practice Key Terms 2

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Source:  OpenStax, College algebra. OpenStax CNX. Feb 06, 2015 Download for free at https://legacy.cnx.org/content/col11759/1.3
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