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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 $\bf{x}$?

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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
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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.

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