What I am struggling is the following problem. I have a linear matrix equation $Ax=b$, where $A$ is a known $n \times n$ large sparse real matrix, $x$ and $b$ are known $n \times 1$ vectors. Now, one entry of $A$ has changed into $a$ and we denote by this matrix $A'$. Since we make $b$ unchanged, the updated $A$ will cause the $x$ in the original linear eqaution accordingly changed to $x'$ such that $A'x'=b$. My goal is to find this new $x'$. A naive way is to re-solve $A'x'=b$. But since $A'$ is slightly different from $Ax=b$, is there any incremental way to fast solve $x'$ in $A'x'=b$ by taking advantage of the original equation $Ax=b$? Thank you very much for any of your kind suggestion!

Hi, everyone.

What I am struggling is the following problem. I have a linear matrix equation $Ax=b$, where $A$ is a known $n \times n$ large sparse real matrix, $x$ and $b$ are known $n \times 1$ vectors. Now, one entry of $A$ has changed into $a$ and we denote by this matrix $A'$. Since we make $b$ unchanged, the updated $A$ will cause the $x$ in the original linear eqaution accordingly changed to $x'$ such that $A'x'=b$. My goal is to find this new $x'$. A naive way is to re-solve $A'x'=b$. But since $A'$ is slightly different from $Ax=b$, is there any incremental way to fast solve $x'$ in $A'x'=b$ by taking advantage of the original equation $Ax=b$? Thank you very much for any of your kind suggestion!

Hi, everyone.

What I am struggling is the following problem. I have a linear matrix equation $Ax=b$, where $A$ is a known $n \times n$ large sparse real matrix, $x$ and $b$ are known $n \times 1$ vectors. Now, one entry of $A$ has changed into $a$ and we denote by this matrix $A'$. Since we make $b$ unchanged, the updated $A$ will cause the $x$ in the original linear eqaution accordingly changed to $x'$ such that $A'x'=b$. My goal is to find this new $x'$. A naive way is to re-solve $A'x'=b$. But since $A'$ is slightly different from $Ax=b$, is there any incremental way to fast solve $x'$ in $A'x'=b$ by taking advantage of the original equation $Ax=b$? Thank you very much for any of your kind suggestion!

What I am struggling is the following problem. I have a linear matrix equation $Ax=b$, where $A$ is a known $n \times n$ large sparse real matrix, $x$ and $b$ are known $n \times 1$ vectors. Now, one entry of $A$ has changed into $a$ and we denote by this matrix $A'$. Since we make $b$ unchanged, the updated $A$ will cause the $x$ in the original linear eqaution accordingly changed to $x'$ such that $A'x'=b$. My goal is to find this new $x'$. A naive way is to re-solve $A'x'=b$. But since $A'$ is slightly different from $Ax=b$, is there any incremental way to fast solve $x'$ in $A'x'=b$ by taking advantage of the original equation $Ax=b$? Thank you very much for any of your kind suggestion!