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Matrix Equation

Nonhomogeneous matrix equations of the form

 Ax=b
(1)

can be solved by taking the matrix inverse to obtain

 x=A^(-1)b.
(2)

This equation will have a nontrivial solution iff the determinant det(A)!=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×n matrix equation

 [a_(11) a_(12) ... a_(1n); a_(21) a_(22) ... a_(2n); | | ... |; a_(n1) a_(n2) ... a_(nn)][x_1; x_2; |; x_n]=[0; 0; |; 0]
(3)

to be solved for the x_is, consider the determinant

 |a_(11) a_(12) ... a_(1n); a_(21) a_(22) ... a_(2n); | | ... |; a_(n1) a_(n2) ... a_(nn)|.
(4)

Now multiply by x_1, which is equivalent to multiplying the first column (or any column) by x_1,

 x_1|a_(11) a_(12) ... a_(1n); a_(21) a_(22) ... a_(2n); | | ... |; a_(n1) a_(n2) ... a_(nn)|=|a_(11)x_1 a_(12) ... a_(1n); a_(21)x_1 a_(22) ... a_(2n); | | ... |; a_(n1)x_1 a_(n2) ... a_(nn)|.
(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|a_(11) a_(12) ... a_(1n); a_(21) a_(22) ... a_(2n); | | ... |; a_(n1) a_(n2) ... a_(nn)|=|a_(11)x_1+a_(12)x_2+...+a_(1n)x_n a_(12) ... a_(1n); a_(21)x_1+a_(22)x_2+...+a_(2n)x_n a_(22) ... a_(2n); | | ... |; a_(n1)x_1+a_(n2)x_2+...+a_(nn)x_n a_(n2) ... a_(nn)|.
(6)

But from the original matrix, each of the entries in the first columns is zero since

 a_(i1)x_1+a_(i2)x_2+...+a_(in)x_n=0,
(7)

so

 |0 a_(12) ... a_(1n); 0 a_(22) ... a_(2n); | | ... |; 0 a_(n2) ... a_(nn)|=0.
(8)

Therefore, if there is an x_1!=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_is 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

 Ax=b
(9)

is x_1=x+deltax_1, where deltax_1 is an error term. The first solution therefore gives

 Ax_1=A(x+deltax_1)=b+deltab
(10)
 Adeltax_1=deltab,
(11)

where deltab is found by solving (10)

 deltab=Ax_1-b.
(12)

Combining (11) and (12) then gives

 deltax_1=A^(-1)deltab=A^(-1)(Ax_1-b)=x_1-A^(-1)b.
(13)

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