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I am trying to learn linear algebra on my own and am going through some old exams I found. I'm not sure how to solve this one. My initial thought was to find the inverse of $A$ and multiply $b$ with it, but apparently since it's not a square matrix I can't take the inverse of $A$. I also considered trying to combine rows to simplify and find some variables that way that I could use to solve the rest of the system but wasn't successful with that. I'd appreciate some guidance from the community.

Solve the linear system $Ax=b$ where

$$A= \begin{bmatrix} 3 & -4 & -3 & 1 & 2 & 1 \\ 0 & 2 & 1 & -1 & -3 & -1 \\ 0 & 0 & 0 & 0 & 3 & 1 \\ \end{bmatrix} $$

and

$$b= \begin{bmatrix} -1 \\ 4 \\ 6 \\ \end{bmatrix} $$ and write your solution in vector form.

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$$A= \begin{bmatrix} 3 & -4 & -3 & 1 & 2 & 1 \\ 0 & 2 & 1 & -1 & -3 & -1 \\ 0 & 0 & 0 & 0 & 3 & 1 \\ \end{bmatrix} $$

and

$$b= \begin{bmatrix} -1 \\ 4 \\ 6 \\ \end{bmatrix} $$

The matrix $A$ is already in row echelon form. The pivot columns are the first column, second column, and the fifth columns. They are the columns that contain leading non-zero entries of each row.

Assign parameter $r$,$s$, and $t$ to the third, fourth, and sixth columns respectively.

$$x_3=r$$

$$x_4=s$$ $$x_6=t$$

The last row tells us that $3x_5+t=6$. Hence, we have

$$x_5=\frac{6-t}{3}$$

Try to express $x_1$ and $x_2$ in $r,s$, and $t$ too.

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You have only three equations and six unknowns, as the $x$ vector must be six long to be able to multiply it by $A$. The rows of $A$ are linearly independent, so there will be a three dimensional space of solutions. You will be able to choose three variables at whim, then calculate the other three. The last row insists that one of the variables you choose must be $x_5$ or $x_6$ and one of the free variables must be the other.

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You're right that you can't solve this system of equations by finding an inverse of $A$. No such matrix exists because $A$ isn't square! In general, you can solve this problem by performing Gaussian elimination on the augmented matrix $\left(A \mid b\right)$ and as long as there are no equations of the form $c = 0$ in the reduced row-echelon form of $(A\mid b)$, where $c$ is itself nonzero, you will have a solution, or perhaps infinitely many.

Consider the augmented matrix $(A \mid b)$ whose first $6$ columns are the columns of $A$ and whose seventh column is $b$. Add row $3$ to row $2$ and replace row $2$ with the result. Proceed until $(A \mid b)$ is in reduced row-echelon form. That should be enough to get you started.

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It is clear that in that equation there are more variables than equations, so the system is indetermined.

However, you may use a least squares aproximation to solve it in the sense of minimizing the 2-norm error. Mathematically this means:

$\bar{x}=argmin_x ||Ax-b||_2$

The answer to this optimización problem is $\bar{x}=A^{\dagger}b$ where $A^{\dagger}$ is th so called Moore-Penrose pseudoinverse.

This ovbiously is not the only solución to the equations, however, it gives you an aproximation that might be useful that minimizes the 2-norm error.

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