In-Class use of Matlab  3/7/2012

Is A invertible?  If so, what is its inverse?

>> A=[1 2 3 4; 5 6 7 8; 1 0 1 0; 0 1 0 1]

A =

     1     2     3     4
     5     6     7     8
     1     0     1     0
     0     1     0     1

>> I4=eye(4)

I4 =

     1     0     0     0
     0     1     0     0
     0     0     1     0
     0     0     0     1

>> A1=[A I4]

A1 =

     1     2     3     4     1     0     0     0
     5     6     7     8     0     1     0     0
     1     0     1     0     0     0     1     0
     0     1     0     1     0     0     0     1

>> A2=rref(A1)

A2 =

  Columns 1 through 7 

    1.0000         0         0   -1.0000         0   -0.5000    3.5000
         0    1.0000         0    1.0000         0         0         0
         0         0    1.0000    1.0000         0    0.5000   -2.5000
         0         0         0         0    1.0000   -1.0000    4.0000

  Column 8 

    3.0000
    1.0000
   -3.0000
    4.0000

This A was not invertible.

Let's ask the same question for another A.

>> A=[1 2 3 4; 5 -6 7 -8; 1 0 1 0; 0 -1 0 1]

A =

     1     2     3     4
     5    -6     7    -8
     1     0     1     0
     0    -1     0     1

>> A1=[A I4]

A1 =

     1     2     3     4     1     0     0     0
     5    -6     7    -8     0     1     0     0
     1     0     1     0     0     0     1     0
     0    -1     0     1     0     0     0     1

>> A2=rref(A1)

A2 =

  Columns 1 through 7 

    1.0000         0         0         0   -0.3500   -0.1500    2.1000
         0    1.0000         0         0    0.0500   -0.0500    0.2000
         0         0    1.0000         0    0.3500    0.1500   -1.1000
         0         0         0    1.0000    0.0500   -0.0500    0.2000

  Column 8 

    0.2000
   -0.6000
   -0.2000
    0.4000

This A was invertible.  Its inverse was the right-hand side of the matrix.  

Let's check to be sure the inverse works. that is: Ainv*A=A*Ainv=I4.

>> Ainv=A2(1:4,5:8)

Ainv =

   -0.3500   -0.1500    2.1000    0.2000
    0.0500   -0.0500    0.2000   -0.6000
    0.3500    0.1500   -1.1000   -0.2000
    0.0500   -0.0500    0.2000    0.4000

>> A*Ainv

ans =

    1.0000   -0.0000   -0.0000         0
   -0.0000    1.0000   -0.0000   -0.0000
         0         0    1.0000         0
         0         0         0    1.0000

>> Ainv*A

ans =

    1.0000   -0.0000    0.0000    0.0000
         0    1.0000         0    0.0000
         0    0.0000    1.0000   -0.0000
         0         0         0    1.0000

So that worked.

Now lets look at that 4x2 matrix of maximal rank and compute its left-inverse


>> B=[1 -1; 1 1; 1 -1; 1 1]

B =

     1    -1
     1     1
     1    -1
     1     1

>> B1=[B I4]

B1 =

     1    -1     1     0     0     0
     1     1     0     1     0     0
     1    -1     0     0     1     0
     1     1     0     0     0     1


>> B2=rref(B1)

B2 =

    1.0000         0         0         0    0.5000    0.5000
         0    1.0000         0         0   -0.5000    0.5000
         0         0    1.0000         0   -1.0000         0
         0         0         0    1.0000         0   -1.0000

We want to use the left four columns (C).

>> C=B2(1:4,3:6)

C =

         0         0    0.5000    0.5000
         0         0   -0.5000    0.5000
    1.0000         0   -1.0000         0
         0    1.0000         0   -1.0000

Now we want a matrix D:=[I2|X] where X is *any* 2x2 matrix.

>> I2=eye(2)

I2 =

     1     0
     0     1

>> X=rand(2,2)

X =

    0.9501    0.6068
    0.2311    0.4860

>> D=[I2 X]

D =

    1.0000         0    0.9501    0.6068
         0    1.0000    0.2311    0.4860

The left-inverse of B should be the product D*C.


>> Binv=D*C

Binv =

    0.9501    0.6068   -0.4501   -0.1068
    0.2311    0.4860   -0.7311    0.0140

Now we check that the left-inverse works.

>> Binv*B

ans =

     1     0
     0     1

It worked.  Now I said it would not be a right-inverse, because of something having to do with the rank of B being too small.
Was I correct?

>> B*Binv

ans =

    0.7190    0.1209    0.2810   -0.1209
    1.1813    1.0928   -1.1813   -0.0928
    0.7190    0.1209    0.2810   -0.1209
    1.1813    1.0928   -1.1813   -0.0928

Yes I was.

Now our true objective was to determine the B-coordinates of the vector b=[5; -7; 5; -7].

That is, we wanted to solve B*x=b. 
Multiplying:
Binv*B*x=Binv*b,
I2*x=Binv*b, and
x=Binv*b.

So we can use Binv to find the solution.


>> Binv*[5;-7;5;-7]

ans =

   -1.0000
   -6.0000

>> 

x1=-1 x2=-6 is the solution.