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Matrix Equation

Nonhomogeneous matrix equations of the form

{\hbox{\sf A}}{\bf x}={\bf b}
\end{displaymath} (1)

can be solved by taking the Matrix Inverse to obtain
{\bf x}={\hbox{\sf A}}^{-1}{\bf b}.
\end{displaymath} (2)

This equation will have a nontrivial solution Iff the Determinant $\mathop{\rm det}({\hbox{\sf A}})\not=0$. In general, more numerically stable techniques of solving the equation include Gaussian Elimination, LU Decomposition, or the Square Root Method.

For a homogeneous $n\times n$ Matrix equation

a_{11} & a_{12} & \cdots & a_{1n}\cr
=\left[{\matrix{0\cr 0\cr \vdots\cr 0\cr}}\right]
\end{displaymath} (3)

to be solved for the $x_i$s, consider the Determinant
a_{11} & a_{12} & \cdots & a_{1n}\cr
...& \vdots\cr
a_{n1} & a_{n2} & \cdots & a_{nn}\cr}\right\vert.
\end{displaymath} (4)

Now multiply by $x_1$, which is equivalent to multiplying the first column (or any column) by $x_1$,
a_{11} & a_{12} & \cdots & a_{1n}\cr
a_{n1}x_1 & a_{n2} & \cdots & a_{nn}\cr}\right\vert.
\end{displaymath} (5)

The value of the Determinant is unchanged if multiples of columns are added to other columns. So add $x_2$ times column 2, ..., and $x_n$ times column $n$ to the first column to obtain

x_1\left\vert\matrix{ a_{11} & a_{12} & \cdots & a_{1n}\cr a...
..._2+\ldots+a_{nn}x_n & a_{n2} & \cdots & a_{nn}\cr}\right\vert.
\end{displaymath} (6)

But from the original Matrix, each of the entries in the first columns is zero since
\end{displaymath} (7)

0 & a_{12} & \cdots & a_{1n}\cr
0 & a_{...
... & \vdots\cr
0 & a_{n2} & \cdots & a_{nn}\cr}\right\vert = 0.
\end{displaymath} (8)

Therefore, if there is an $x_1\not = 0$ which is a solution, the Determinant is zero. This is also true for $x_2$, ..., $x_n$, so the original homogeneous system has a nontrivial solution for all $x_i$s only if the Determinant is 0. This approach is the basis for Cramer's Rule.

Given a numerical solution to a matrix equation, the solution can be iteratively improved using the following technique. Assume that the numerically obtained solution to

{\hbox{\sf A}}{\bf x}={\bf b}
\end{displaymath} (9)

is ${\bf x}_1={\bf x}+\delta{\bf x}_1$, where $\delta {\bf x}_1$ is an error term. The first solution therefore gives
{\hbox{\sf A}}{\bf x}_1={\hbox{\sf A}}({\bf x}+\delta{\bf x}_1) = {\bf b}+\delta{\bf b}
\end{displaymath} (10)

{\hbox{\sf A}}\delta{\bf x}_1=\delta {\bf b},
\end{displaymath} (11)

where $\delta{\bf b}$ is found by solving (10)
\delta{\bf b} = {\hbox{\sf A}}{\bf x}_1-{\bf b}.
\end{displaymath} (12)

Combining (11) and (12) then gives
\delta {\bf x}_1={\hbox{\sf A}}^{-1}\delta{\bf b}={\hbox{\sf...
... A}}{\bf x}_1-{\bf b}) = {\bf x}_1-{\hbox{\sf A}}^{-1}{\bf b}.
\end{displaymath} (13)

See also Cramer's Rule, Gaussian Elimination, LU Decomposition, Matrix, Matrix Addition, Matrix Inverse, Matrix Multiplication, Normal Equation, Square Root Method

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© 1996-9 Eric W. Weisstein