"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?

  • $\begingroup$ Made a few small edits, you can revert if that was not your intention. Seemed like you made typos in $B_{35}$ and $C_{34}$. Also looks like you are missing a negative sign in $x_2$ $\endgroup$ – Shubham Johri Jan 14 at 21:33
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    $\begingroup$ What is the determinant of a non-square matrix? $\endgroup$ – amd Jan 14 at 21:33
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    $\begingroup$ Answer keys have been known to be wrong. Have you double-checked that you’ve copied the problem correctly? $\endgroup$ – amd Jan 14 at 21:35
  • $\begingroup$ Your solution is correct and the system, as written, has infinitely many solutions. You get no solution if the coefficient of $x_4$ in the last equation is $5$ instead of $-1$. $\endgroup$ – egreg Jan 14 at 21:40
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    $\begingroup$ @ShubhamJohri I didn't check the final details, but the system indeed has infinitely many solutions contrary to what the answer key claims. Indeed the solution should have: $x_2=(2+5t)/3$, $x_1=(8-2t)/3$. $\endgroup$ – egreg Jan 14 at 21:52

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 columns 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.


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