Page 153 - 35Linear Algebra

7.5 Inverse Matrix 153

At this point, we know M −1 assuming we didn’t goof up. However, row reduction is a
lengthy and involved process with lots of room for arithmetic errors, so we should check
our answer, by confirming that MM −1 = I (or if you prefer M −1 M = I):
     
−1 2 −3 −5 4 −3 1 0 0
MM −1 =  2 1 0   10 −7 6  =  0 1 0 
4 −2 5 8 −6 5 0 0 1

The product of the two matrices is indeed the identity matrix, so we’re done.

Reading homework: problem 5

7.5.3 Linear Systems and Inverses

If M −1 exists and is known, then we can immediately solve linear systems
associated to M.
Example 94 Consider the linear system:

−x +2y −3z = 1
2x + y = 2
4x −2y +5z = 0
 
1
2
The associated matrix equation is MX =   , where M is the same as in the
0
previous section, so the system above is equivalent to the matrix equation

    −1        
x −1 2 −3 1 −5 4 −3 1 3
y = 2 1 0 2 = 10 −7 6 2 = −4 .
           
z 4 −2 5 0 8 −6 5 0 −4
   
x 3
That is, the system is equivalent to the equation   =  −4 , and it is easy to
y

z −4
see what the solution(s) to this equation are.
In summary, when M −1 exists

Mx = v ⇔ x = M −1 v .

Reading homework: problem 5

153