
            it uses a matrix factorization instead of multiplying by the inverse.  Even though your linear algebra textbook might write
            x=A^(-1)*b as the solution to the system A*x=b, your textbook author never means for you to compute the inverse.
         </p>
         <p>These next statements give the same answer, so what's the big deal?</p><pre class="codeinput">S = inv(A) ;
x = S*b
x = A\b
</pre><pre class="codeoutput">x =
    0.6294
    0.4942
   -0.1937
x =
    0.6294
    0.4942
   -0.19