Solving Systems
In[104]:=
Clear[a,b,x]
m={{1, 5}, {2, 1}};
MatrixForm[m]
Out[104]=
1 5 2 1
In[105]:=
v={a,b}
Out[105]=
{a, b}
In[106]:=
x=LinearSolve[m, v]
Out[106]=
5 (2 a - b) 2 a - b
{a - -----------, -------}
9 9
Note that x is a solution to mx=v if and only if v is a linear combination of the columns of m (see Manipulation with Matrices above). Let's use some of our tools to check this out. (See Basic Manipulations above.)
Column1 of m is:
In[107]:=
column1=Transpose[m][[1]]
Out[107]=
{1, 2}
Column2 of m is:
In[108]:=
column2=Transpose[m] [[2]]
Out[108]=
{5, 1}
Here is our linear combination of the columns of m. Is it v?
In[109]:=
x[[1]]column1+x[[2]]column2
v
Out[109]=
5 (2 a - b) 2 a - b
{a, 2 (a - -----------) + -------}
9 9
Out[110]=
{a, b}
Maybe this will help.
In[111]:=
Simplify[x[[1]]column1+x[[2]]column2]
v
Out[111]=
{a, b}
Out[112]=
{a, b}