When Does A Matrix Have Infinite Solutions
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When does this matrix have zero, one and infinite solution?
WEBIf a2 + 3a − 4 = 0 a 2 + 3 a − 4 = 0 and −3a + 3 = 0 − 3 a + 3 = 0 then you will have infinitely many solutions. If a2 + 3a − 4 = 0 a 2 + 3 a − 4 = 0 and −3a + 3 ≠ 0 − 3 a + 3 ≠ 0, then you will not have any solutions. The point is that a row (0 0 0 1) (0 0 0 1) will correspond to the equation 0x1 + 0x2 + 0x3 = 1 0 x 1 + 0 x
Math.stackexchange.comWhy does a row of zeroes in a matrix of linear systems mean …
WEB$\begingroup$ If you have a $4\times 3$ matrix, then you have three variables not two. So you should visualize this in terms of coordinates $(x,y,t)$ in 3D, not as coordinates $(x,y)$ in the plane. So you should visualize this in terms of coordinates $(x,y,t)$ in 3D, not as coordinates $(x,y)$ in the plane.
Math.stackexchange.comInverse matrices for linear equations with infinite solutions
WEBJan 21, 2020 · If the lines are parallel to each other and confounded, there is an infinite number of solutions. But for non-invertible matrices, consider the linear system $\begin{cases}2x + \frac{1}{3}y &= 7\\8x + \frac{4}{3}y &= 28\end{cases}$ We see that the lines are parallel to each other and confounded, so there is an infinite number of solutions.
Math.stackexchange.comWhy does a singular matrix imply that it does not have a solution?
WEBApr 22, 2015 · The formula has det(A) in the denominator of the unique solution values, where A is the coefficient matrix (only the first 3 columns of your augmented matrix). Clearly if det(A) is zero, then your solution can't exist. Hence, if the A is non singular, the solution can't exist. Your matrix is non singular, hence a real solution does not exist.
Math.stackexchange.comWhen does this matrix have no solultions, infinite solutions and 1
WEBNov 12, 2019 · So im supposed to decide for what h and k this matrix has no solultions, infinite solutions and a unique solution $$\left[ \begin{array}{cc|c} 1&h&1\\ 3&3&k\\ \end{array} \right]$$ By gaussian elimination i get
Math.stackexchange.comWhen does system of equations have infinitely many solutions?
WEBNov 6, 2015 · Notice that this also entails that all non-square matrices have either 0 or infinitely many solutions. But notice that in general you actually don't have to invert a matrix to find its solutions. Row reduction will bring the matrix in a form where it is easy to read off a) a particular solution (if any such exists) and b) a basis of the null space.
Math.stackexchange.comlinear algebra - Matrices and unique and infinite solutions
WEBJan 25, 2016 · And as you may know a matrix with as many rows as variables may have an unique solution or infinite numbers of solutions (or even no solution at all). Example1: (0 0 1 0 0 1)x = (42 α β) of course has no solution because the first row must become zero. Example2: (0 0 1 0 0 1)x = (0 α β) has the same solutions as. (1 0 0 1)x = (α β)
Math.stackexchange.comlinear algebra - Does a system with infinite solutions have linearly
WEBFeb 10, 2022 · Since Ax = b has infinitely many solutions, it must have at least two distinct solutions X1 and X2. Therefore it is linearly dependent as X1=/=X2. (This is because for a function to be linearly independent, X1=X2==Xn=0) I saw that other solutions used some form of summation to prove it but this is the only one that makes sense to me.
Math.stackexchange.comConditions for infinite solutions - Mathematics Stack Exchange
WEBDec 6, 2019 · 1)unique solution. 2) infinite solutions. now I understand that for a system AX = B A X = B to have unique solutions , the coefficient matrix should be non singular because only then X =A−1B X = A − 1 B will exist . However here’s the part that’s confusing. If we solve the coefficient matrix and equate it to zero we get the value of
Math.stackexchange.comlinear algebra - Does the existence of a free variable mean there …
WEBNov 3, 2019 · The question is: Is the matrix consistent with a unique solution, inconsistent, or consistent with an infinite solution? row reducing gives: $$\begin{bmatrix} 1&0&\frac32\\ 0&1&0\\ 0&0&0 \end{bmatrix}$$ the answer given is: consistent, unique solution. However, I thought if there is a free variable that indicates infinite solutions?
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