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Inverse of a 3x3 Matrix

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Date: 04/25/2001 at 02:05:57
From: Sharon Wasson
Subject: 3x3 Inverse Matrices

In my Advanced Algebra class, the book does not show how to obtain an
inverse matrix for 3x3 matrices. All it says is that the process is
complicated and that students should use a computer or graphing
calculator. I can find the determinant, use Cramer's Rule, and solve
any other way - including the graphing calculator - but cannot find
anything on this concept.  I would at least like to see it even if I
don't or can't use it. Can you help me?
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```
Date: 04/25/2001 at 15:46:58
From: Doctor Rob
Subject: Re: 3x3 Inverse Matrices

Thanks for writing to Ask Dr. Math, Sharon.

I like to use this method to find the inverse of a 3x3 (or larger
square) matrix. Write the matrix, and to its right append an identity
matrix of the same size. You'll have a 3x6 matrix. Now operate on the
matrix until the 3x3 left half is the identity matrix, by using
elementary row operations. Then the 3x3 right half will be the inverse
of the original matrix.

Elementary row operations are these:

1. Multiply a row by a nonzero constant.
2. Exchange two rows.
3. Add a constant multiple of one row to another.

Example:  Find the inverse of

[-1  3 -3]
[ 0 -6  5].
[-5 -3  1]

Augment the matrix with a 3x3 identity matrix:

[-1  3 -3  1  0  0]
[ 0 -6  5  0  1  0].
[-5 -3  1  0  0  1]

Now create a +1 in the upper left-hand corner by multiplying the first
row by -1:

[ 1 -3  3 -1  0  0]
[ 0 -6  5  0  1  0].
[-5 -3  1  0  0  1]

Now create 0's below it by adding a multiple of row 1 to the other
rows (in this case 5 times the first row to the third row):

[1  -3   3 -1  0  0]
[0  -6   5  0  1  0].
[0 -18  16 -5  0  1]

Now the first column looks fine, so we create a +1 at the (2,2) entry
by multiplying the second row by -1/6:

[1  -3    3 -1    0  0]
[0   1 -5/6  0 -1/6  0].
[0 -18   16 -5    0  1]

Now create 0's above and below that 1 by adding multiples of the
second row (3 and 18 times it) to the others (first and third):

[1  0  1/2 -1 -1/2  0]
[0  1 -5/6  0 -1/6  0].
[0  0   1  -5   -3  1]

Now the second column is right, and we already have a +1 at the (3,3)
entry.  Now create 0's above that by adding multiples of row 3 (-1/2
and 5/6 times it) to the rows above it (first and second):

[1  0  0   3/2    1 -1/2]
[0  1  0 -25/6 -8/3  5/6].
[0  0  1    -5   -3    1]

The left half is now a 3x3 identity matrix, so we stop.  Then the
inverse of the original matrix is the right half, or

[  3/2    1 -1/2]
[-25/6 -8/3  5/6]
[   -5   -3    1]

You can check this by multiplying it times the original matrix, and
getting the identity.

The explanation of this is that each elementary row operation
corresponds to multiplying a 3x3 matrix on the left:

1.  [c 0 0]   [1 0 0]      [1 0 0]
[0 1 0],  [0 c 0], or  [0 1 0]
[0 0 1]   [0 0 1]      [0 0 c]

(Multiply c times 1st, 2nd, or 3rd row, respectively.)

2.  [0 1 0]   [0 0 1]      [1 0 0]
[1 0 0],  [0 1 0], or  [0 0 1]
[0 0 1]   [1 0 0]      [0 1 0]

(Exchange rows 1 & 2, 1 & 3, or 2 & 3, respectively.)

3.  [1 0 0]   [1 0 0]   [1 c 0]   [1 0 0]   [1 0 c]      [1 0 0]
[c 1 0],  [0 1 0],  [0 1 0],  [0 1 0],  [0 1 0], or  [0 1 c]
[0 0 1]   [c 0 1]   [0 0 1]   [0 c 1]   [0 0 1]      [0 0 1]

(Add c times row 1 to row 2; add c times row 1 to row 3;
add c times row 2 to row 1; add c times row 2 to row 3;
add c times row 3 to row 1; or add c times row 3 to row 2,
respectively.)

A*A^(-1) = I.

Now multiply on the left by carefully selected ones of these
elementary operation matrices, according to the above procedure:

(En*...*E2*E1*A)*A^(-1) = En*...*E2*E1*I.

When En*...*E2*E1*A = I, which is what the above procedure makes
happen in the left half, then you get

A^(-1) = I*A^(-1) = En*...*E2*E1*I,

and the right side of this equation is what ends up in the right half.
That's A^(-1).

Another approach is to let the 9 entries in the inverse matrix be
unknowns, multiply the two matrices together, and set the result equal
to the identity matrix. That will get you a set of nine linear
equations in nine unknowns which you can solve simultaneously.

For example,

[-1  3 -3] [r s t]   [1 0 0]
[ 0 -6  5]*[u v w] = [0 1 0]
[-5 -3  1] [x y z]   [0 0 1]

gives you the equations

-r           + 3*u             - 3*x             = 1,
-s            + 3*v             - 3*y       = 0,
-t             + 3*w             - 3*z = 0,
-6*u             + 5*x             = 0,
-6*v             + 5*y       = 1,
-6*w             + 5*z = 0,
-5*r           - 3*u               + x             = 0,
-5*s            - 3*v               + y       = 0,
-5*t             - 3*w               + z = 1.

You need to solve these for r, s, ..., and z.  You can rearrange
this system to see that it is actually three sets of simultaneous
linear equations, each in three unknowns:

-r + 3*u - 3*x = 1
-6*u + 5*x = 0
-5*r - 3*u   + x = 0

-s + 3*v - 3*y = 0
-6*v + 5*y = 1
-5*s - 3*v   + y = 0

-t + 3*w - 3*z = 0
-6*w + 5*z = 0
-5*t - 3*w   + z = 1

These three systems all have the same coefficients on the left side,
but different right sides. You can solve them using Cramer's rule, if
you like. That will give you r, s, ..., and z, and thus the inverse of

An explicit formula for the inverse can also be given.  If

[a11 a12 a13]
A = [a21 a22 a23]
[a31 a32 a33]

and its determinant is d, then its inverse A^(-1) is given by

[ |a22 a23|   |a12 a13|  |a12 a13|]
[ |a32 a33|  -|a32 a33|  |a22 a23|]
[                                 ]
[ |a21 a23|   |a11 a13|  |a11 a13|]
A^(-1) = [-|a31 a33|   |a31 a33| -|a21 a23|] / d
[                                 ]
[ |a21 a22|   |a11 a12|  |a11 a12|]
[ |a31 a32|  -|a31 a32|  |a21 a22|]

The entry in the (i,j) position is found by taking the determinant of
the submatrix of A obtained by deleting the jth row and ith column,
and multiplying by (-1)^(i+j). Then the entire matrix is divided by d.

It is a long exercise to verify that this works.

If you need more help, feel free to write again.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
College Linear Algebra

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