0 votes 0 votes Let $A$ be a $n \times n$ square matrix whose all columns are independent. Is $Ax = b$ always solvable? Actually, I know that $Ax= b$ is solvable if $b$ is in the column space of $A$. However, I am not sure if it is solvable for all values of $b$. Linear Algebra linear-algebra + – Debargha Bhattacharj asked Jun 6, 2019 Debargha Bhattacharj 628 views answer comment Share Follow See all 12 Comments See all 12 12 Comments reply Show 9 previous comments Debargha Bhattacharj commented Jun 7, 2019 reply Follow Share Let $A = \begin{bmatrix} col_{1} & col_{2} & col_{3} \end{bmatrix}$ be any arbitrary $3 \times 3$ matrix. Here, $col_{1}, col_{2}, col_{3}$ are the column vectors of A, each of dimension $3 \times 1$. Also, let $x =\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}$. $ \therefore Ax = b \Rightarrow col_{1}.x_{1} + col_{2}.x_{2} + col_{3}.x_{3} = b$, i.e., a linear combination of the column vectors of $A$. Therefore, finding the solution of the equation $Ax = b$ basically reduces to finding a linear combination of column vectors of matrix $A$ that result in vector $b$. Such a combination can only be found if vector $b$ lies in the column space of matrix $A$, otherwise not. 0 votes 0 votes Sourajit25 commented Jun 7, 2019 reply Follow Share $\begin{bmatrix} 1 & 2 & 3\\ 4& 5 &6 \\ 7& 8 & 10 \end{bmatrix}$ $\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$ $=$ $\begin{bmatrix} 1\\ 0\\ 0 \end{bmatrix}$ This has a solution of $\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$ $=$ $\begin{bmatrix} -(2/3)\\ -(2/3)\\ 1 \end{bmatrix}$ 0 votes 0 votes srestha commented Jun 7, 2019 reply Follow Share yes right. @ankitgupta.1729 u r not correct here 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes Few Inferences first : If the Rank of the Augmented Matrix $[A:b]$ $=$ Rank $[A]$ =$N$(Order of the matrix A) then it is solvable and can be solved by reducing by using Gaussian Elimination Method, and it has Unique Solution. If the Rank of the Augmented Matrix $[A:b]$ $=$ Rank $[A]$$ < N$ then it is solvable and has Infinitely many solutions. If the Rank of the Augmented Matrix $[A:b]$ $\neq$ Rank $[A]$ then it is not solvable. Now as per your question, if A is a vector whose all columns are linearly independent then in that case it has a rank = $n$ and whatever may be $b$ the augmented matrix $[A:b]$ will always have a rank = $n$ hence it is always solvable and will have unique solution. For more details you can refer to the lectures by Gilbert Strang or his book on linear algebra. Sourajit25 answered Jun 7, 2019 Sourajit25 comment Share Follow See 1 comment See all 1 1 comment reply Debargha Bhattacharj commented Jun 7, 2019 reply Follow Share Thanks. I was doing Gaussian Elimination but did some mistake. 0 votes 0 votes Please log in or register to add a comment.