When attempting to solve a problem of type $AX=B$ where $A$ is the matrix of coefficients, $X$ contains the variables and $B$ is the right hand side, it turned out the $A$ was singular. As a result I could not premultiply both side by $A^{-1}$. In such a case, how would I solve for the variables without having to use augmented matrices?
Here is the question:
$$\begin{pmatrix} 1&2&1&1&1 \\ 2&1&2&1&1 \\ 1&2&3&1&1 \\ 2&2&1&1&3 \\3&3&5&2&2 \end{pmatrix}\begin{pmatrix} u \\ v \\ w \\ x \\ y \end{pmatrix}=\begin{pmatrix} 6.3 \\ 6.7 \\ 7.7 \\ 9.8 \\10.9 \end{pmatrix}$$
Since $A$ is singular I cannot do $X=A^{-1}B$ which is the method I would typically use to solve for the unknowns. In later parts of the book I am studying they elude to a method with which one can use to solve for $X$, but do not further elaborate. Does anyone have an idea on how to solve this problem (or any problem in general where A is singular) without the use augmented matrices or is that the only way?
Best Answer
Even when the system of equations is singular, you can find a least-squares solution by solving the system $A^{T}Ax=A^{T}b$.