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Use the formula to find the inverse of matrix A . Verify your answer by augmenting with the identity matrix.

A = [ 1 −1 2 3 ]

A −1 = [ 3 5 1 5 2 5 1 5 ]

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Finding the inverse of the matrix, if it exists

Find the inverse, if it exists, of the given matrix.

A = [ 3 6 1 2 ]

We will use the method of augmenting with the identity.

[ 3 6 1 3 | 1 0 0 1 ]
  1. Switch row 1 and row 2.
    [ 1 3 3 6 | 0 1 1 0 ]
  2. Multiply row 1 by −3 and add it to row 2.
    [ 1 2 0 0 | 1 0 −3 1 ]
  3. There is nothing further we can do. The zeros in row 2 indicate that this matrix has no inverse.
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Finding the multiplicative inverse of 3×3 matrices

Unfortunately, we do not have a formula similar to the one for a 2 × 2 matrix to find the inverse of a 3 × 3 matrix. Instead, we will augment the original matrix with the identity matrix and use row operations    to obtain the inverse.

Given a 3 × 3 matrix

A = [ 2 3 1 3 3 1 2 4 1 ]

augment A with the identity matrix

A | I = [ 2 3 1 3 3 1 2 4 1    |    1 0 0 0 1 0 0 0 1 ]

To begin, we write the augmented matrix    with the identity on the right and A on the left. Performing elementary row operations    so that the identity matrix    appears on the left, we will obtain the inverse matrix on the right. We will find the inverse of this matrix in the next example.

Given a 3 × 3 matrix, find the inverse

  1. Write the original matrix augmented with the identity matrix on the right.
  2. Use elementary row operations so that the identity appears on the left.
  3. What is obtained on the right is the inverse of the original matrix.
  4. Use matrix multiplication to show that A A −1 = I and A −1 A = I .

Finding the inverse of a 3 × 3 matrix

Given the 3 × 3 matrix A , find the inverse.

A = [ 2 3 1 3 3 1 2 4 1 ]

Augment A with the identity matrix, and then begin row operations until the identity matrix replaces A . The matrix on the right will be the inverse of A .

[ 2 3 1 3 3 1 2 4 1 | 1 0 0 0 1 0 0 0 1 ] Interchange  R 2 and  R 1 [ 3 3 1 2 3 1 2 4 1 | 0 1 0 1 0 0 0 0 1 ]
R 2 + R 1 = R 1 [ 1 0 0 2 3 1 2 4 1 | −1 1 0 1 0 0 0 0 1 ]
R 2 + R 3 = R 3 [ 1 0 0 2 3 1 0 1 0 | −1 1 0 1 0 0 −1 0 1 ]
R 3   R 2 [ 1 0 0 0 1 0 2 3 1 | −1 1 0 −1 0 1 1 0 0 ]
−2 R 1 + R 3 = R 3 [ 1 0 0 0 1 0 0 3 1 | −1 1 0 −1 0 1 3 −2 0 ]
−3 R 2 + R 3 = R 3 [ 1 0 0 0 1 0 0 0 1 | −1 1 0 −1 0 1 6 −2 −3 ]

Thus,

A −1 = B = [ −1 1 0 −1 0 1 6 −2 −3 ]
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Find the inverse of the 3 × 3 matrix.

A = [ 2 −17 11 −1 11 −7 0 3 −2 ]

A −1 = [ 1 1 2 2 4 −3 3 6 −5 ]

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Solving a system of linear equations using the inverse of a matrix

Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: X is the matrix representing the variables of the system, and B is the matrix representing the constants. Using matrix multiplication , we may define a system of equations with the same number of equations as variables as

A X = B

To solve a system of linear equations using an inverse matrix , let A be the coefficient matrix    , let X be the variable matrix, and let B be the constant matrix. Thus, we want to solve a system A X = B . For example, look at the following system of equations.

a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2

From this system, the coefficient matrix is

A = [ a 1 b 1 a 2 b 2 ]

The variable matrix is

X = [ x y ]

And the constant matrix is

B = [ c 1 c 2 ]

Then A X = B looks like

[ a 1 b 1 a 2 b 2 ]     [ x y ] = [ c 1 c 2 ]

Recall the discussion earlier in this section regarding multiplying a real number by its inverse, ( 2 −1 ) 2 = ( 1 2 ) 2 = 1. To solve a single linear equation a x = b for x , we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of a . Thus,

Questions & Answers

explain and give four Example hyperbolic function
Lukman Reply
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
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
Pawel
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Ifeanyi
on number 2 question How did you got 2x +2
Ifeanyi
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
Pawel
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Harshika Reply
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Harshika
find the subring of gaussian integers?
Rofiqul
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Shirley Reply
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Abdullahi
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Mark
I need quadratic equation link to Alpa Beta
Abdullahi Reply
find the value of 2x=32
Felix Reply
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
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Mark
use the y -intercept and slope to sketch the graph of the equation y=6x
Only Reply
how do we prove the quadratic formular
Seidu Reply
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Darius
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Shirley Reply
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Seidu
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Opoku
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Tric Reply
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
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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
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
Simone Reply
x+2y-z=7
Sidiki
what is the coefficient of -4×
Mehri Reply
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
Alfred Reply
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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