"Convert the following system of linear equations into an augmented matrix and then solve it"
$$x_1 + x_2 - x_3 - x_4 = 7$$ $$2x_1 - x_2 + 3x_3 + x_4 = 1$$ $$x_1 - 5x_2 + 9x_3 - x_4 = 3$$
My approach: So the augmented matrix, A, is
$$aug A = \begin{bmatrix}1&1&-1&-1&7\\2&-1&3&1&1\\1&-5&9&-1&3\end{bmatrix}$$
Then we perform elementary row operations to reduce the matrix into row echelon form:
$$B = \begin{bmatrix}1&1&-1&-1&7\\0&-3&5&3&-13\\0&-6&10&0&-4\end{bmatrix}$$
$$C = \begin{bmatrix}1&1&-1&-1&7\\0&-3&5&3&-13\\0&0&0&-6&22\end{bmatrix}$$
Matrix C is our system of linear equations in echelon form. From the last row of matrix C, $$-6x_4 =22$$ $$x_4 = \frac {-11}3$$
Now we let x_3 = t for t is any real number
Then we solve the two equations from the first two rows of matrix C,
$$x_2 = \frac 13(-2 - 5t)$$
$$x_1 = \frac 83t + 2\frac 23 $$
SO this should be the solutions to the equations. However, according to the answer key the solution is the empty set. I inputted the regular matrix into a determinant calculator and indeed the determinant was zero. However, I managed to get a solution to the system of linear equations. Can someone tell me where I went wrong?
To answer the more general question of when a matrix has no solution:
A system of equations can have one of three things: a unique solution, infinitely many solutions, and no solution.
Case One: unique solution
An augmented matrix has an unique solution when the equations are all consistent and the number of variables is equal to the number of rows. Simply put if the non-augmented matrix has a nonzero determinant, then it has a solution given by $\vec x = A^{-1}\vec b$.
Case Two: Infinitely many solutions
The number of rows is less than the number of variables. Think of it this way, we need one equation to solve for one unknown variable, two equations to solve for two variables, three equations to solve for three variables, and so on... The number of rows represents(at most) the number of independent equations we have so if it's less than the number of columns which represents the number of variables, we most likely have the case of infinite solutions. Why infinite solutions? We cannot nail down at least of the variables to a value so we let it take on any real value. Note: The non-augmented matrix must have determinant zero for this to be the case
Case Three: No Solutions
We have no solutions when the equations(represented by the rows) make no sense aka they are inconsistent. For example, when we put our augmented matrix in row reduced echelon form we have a non-zero value in the augmented column and zeroes every else. How can this be? It can't. A sum of all zeros is always zero.