Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I would like to solve a system of linear equations $\bf{Ax=b}$ where the matrix $\bf{A}$ is lower triangular. Now if $\det(\bf{A}) = 0$, does this imply that I cannot solve for $\bf{x}$? Under which conditions can I solve for $\bf{x}$?

share|improve this question
1  
In general, $\det(A) = 0$ implies there is no inverse matrix. So, we can not solve by multiplying both sides by the inverse to get $$x = A^{-1} b.$$ But, it doesn't mean it can not be solved. –  Graphth Nov 8 '12 at 17:58

1 Answer 1

up vote 1 down vote accepted

No, it doesn't. The most you can say from the fact that $\det A = 0$ is that either there are no solutions, or there are infinitely many solutions (assuming that the matrix is over an infinite field such as $\mathbb{R}$). For the conditions under which you can solve for $\mathbf{x}$, you can look at the Kronecker-Capelli theorem and the Fredholm's theorem.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.