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   a x = b    ( 1 a ) a x = ( 1 a ) b ( a −1     ) a x = ( a −1 ) b [ ( a −1 ) a ] x = ( a −1 ) b              1 x = ( a −1 ) b                x = ( a −1 ) b

The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable.

We will investigate this idea in detail, but it is helpful to begin with a 2 × 2 system and then move on to a 3 × 3 system.

Solving a system of equations using the inverse of a matrix

Given a system of equations, write the coefficient matrix A , the variable matrix X , and the constant matrix B . Then

A X = B

Multiply both sides by the inverse of A to obtain the solution.

( A −1 ) A X = ( A −1 ) B [ ( A −1 ) A ] X = ( A −1 ) B I X = ( A −1 ) B X = ( A −1 ) B

If the coefficient matrix does not have an inverse, does that mean the system has no solution?

No, if the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.

Solving a 2 × 2 system using the inverse of a matrix

Solve the given system of equations using the inverse of a matrix.

3 x + 8 y = 5 4 x + 11 y = 7

Write the system in terms of a coefficient matrix, a variable matrix, and a constant matrix.

A = [ 3 8 4 11 ] , X = [ x y ] , B = [ 5 7 ]

Then

[ 3 8 4 11 ]     [ x y ] = [ 5 7 ]

First, we need to calculate A −1 . Using the formula to calculate the inverse of a 2 by 2 matrix, we have:

A −1 = 1 a d b c [ d b c a ]        = 1 3 ( 11 ) −8 ( 4 ) [ 11 −8 −4 3 ]        = 1 1 [ 11 −8 −4 3 ]

So,

A −1 = [ 11 −8 −4 3 ]

Now we are ready to solve. Multiply both sides of the equation by A −1 .

( A −1 ) A X = ( A −1 ) B [ 11 −8 −4 3 ]     [ 3 8 4 11 ]     [ x y ] = [ 11 −8 −4 3 ]     [ 5 7 ] [ 1 0 0 1 ]     [ x y ] = [ 11 ( 5 ) + ( −8 ) 7 −4 ( 5 ) + 3 ( 7 ) ] [ x y ] = [ −1 1 ]

The solution is ( −1 , 1 ) .

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Can we solve for X by finding the product B A −1 ?

No, recall that matrix multiplication is not commutative, so A −1 B B A −1 . Consider our steps for solving the matrix equation.

( A −1 ) A X = ( A −1 ) B [ ( A −1 ) A ] X = ( A −1 ) B I X = ( A −1 ) B X = ( A −1 ) B

Notice in the first step we multiplied both sides of the equation by A −1 , but the A −1 was to the left of A on the left side and to the left of B on the right side. Because matrix multiplication is not commutative, order matters.

Solving a 3 × 3 system using the inverse of a matrix

Solve the following system using the inverse of a matrix.

5 x + 15 y + 56 z = 35 −4 x −11 y −41 z = −26 x −3 y −11 z = −7

Write the equation A X = B .

[ 5 15 56 −4 −11 −41 −1 −3 −11 ]     [ x y z ] = [ 35 −26 −7 ]

First, we will find the inverse of A by augmenting with the identity.

[ 5 15 56 −4 −11 −41 −1 −3 −11 | 1 0 0 0 1 0 0 0 1 ]

Multiply row 1 by 1 5 .

[ 1 3 56 5 −4 −11 −41 −1 −3 −11 | 1 5 0 0 0 1 0 0 0 1 ]

Multiply row 1 by 4 and add to row 2.

[ 1 3 56 5 0 1 19 5 −1 −3 −11 | 1 5 0 0 4 5 1 0 0 0 1 ]

Add row 1 to row 3.

[ 1 3 56 5 0 1 19 5 0 0 1 5 | 1 5 0 0 4 5 1 0 1 5 0 1 ]

Multiply row 2 by −3 and add to row 1.

[ 1 0 1 5 0 1 19 5 0 0 1 5 | 11 5 −3 0 4 5 1 0 1 5 0 1 ]

Multiply row 3 by 5.

[ 1 0 1 5 0 1 19 5 0 0 1 | 11 5 −3 0 4 5 1 0 1 0 5 ]

Multiply row 3 by 1 5 and add to row 1.

[ 1 0 0 0 1 19 5 0 0 1 | −2 −3 1 4 5 1 0 1 0 5 ]

Multiply row 3 by 19 5 and add to row 2.

[ 1 0 0 0 1 0 0 0 1 | −2 −3 1 −3 1 −19 1 0 5 ]

So,

A −1 = [ −2 −3 1 −3 1 −19 1 0 5 ]

Multiply both sides of the equation by A −1 . We want A −1 A X = A −1 B :

[ −2 −3 1 −3 1 −19 1 0 5 ]     [ 5 15 56 −4 −11 −41 −1 −3 −11 ]     [ x y z ] = [ −2 −3 1 −3 1 −19 1 0 5 ]     [ 35 −26 −7 ]

Thus,

A −1 B = [ −70 + 78 −7 −105 −26 + 133 35 + 0 −35 ] = [ 1 2 0 ]

The solution is ( 1 , 2 , 0 ) .

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Questions & Answers

12, 17, 22.... 25th term
Alexandra Reply
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Akash
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Abhi
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
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I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
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Commplementary angles
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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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