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Is there a way to determine the nature of a system of equations without solving it? For example, given the system \begin{cases} 2x + y - 4z = 6 \\[4px] y - 2z = 2 \\[4px] 4x + 3y - 10z = -3 \end{cases} Can I tell that this system is independent without solving it?

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Not really. You can see whether the three equations are independent by computing the determinant $$ \det\begin{bmatrix} 2 & 1 & -4 \\ 0 & 1 & -2 \\ 4 & 3 & -10 \end{bmatrix} $$ but, unfortunately, it is $0$. Besides, computing a determinant with Laplace expansion is a very expensive computation.

Much simpler is going with Gaussian elimination: \begin{align} \begin{bmatrix} 2 & 1 & -4 & 6\\ 0 & 1 & -2 & 2\\ 4 & 3 & -10 & -3 \end{bmatrix} &\to \begin{bmatrix} 2 & 1 & -4 & 6\\ 0 & 1 & -2 & 2\\ 0 & 1 & -2 & -15 \end{bmatrix} && R_3\gets R_3-2R_1 \\[6px] &\to \begin{bmatrix} 2 & 1 & -4 & 6\\ 0 & 1 & -2 & 2\\ 0 & 0 & 0 & -17 \end{bmatrix} && R_3\gets R_3-R_2 \end{align} The last column is a pivot column, so the system is inconsistent.


The system would be solvable if $-10$ is changed into $14$. In this case we could go backwards \begin{align} \begin{bmatrix} 2 & 1 & -4 & 6\\ 0 & 1 & -2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix} &\to \begin{bmatrix} 2 & 0 & -2 & 4\\ 0 & 1 & -2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix} && R_1\gets R_1-R_2 \\[6px] &\to \begin{bmatrix} 1 & 0 & -1 & 2\\ 0 & 1 & -2 & 2\\ 0 & 0 & 0 & 0 \end{bmatrix} && R_1\gets\tfrac{1}{2}R_1 \end{align} The third unknown is free, and the solutions are \begin{cases} x=2+h \\[4px] y=2+2h \\[4px] z=h \end{cases} with arbitrary $h$.

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Yes just form matrix of coefficient of equation row- wise and check its determinant. If it's determinant is non-zero then system is independent otherwise dependent

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Let$$S=\begin{bmatrix}2 & 1 & -4 & 6\\ 0 & 1 & -2 & 2\\ 4 & 3 & -10 & -3\end{bmatrix}\text{ and }A=\begin{bmatrix}2 & 1 & -4 \\ 0 & 1 & -2 \\ 4 & 3 & -10 \end{bmatrix}.$$Since $\operatorname{rank}A<3$ and $\operatorname{rank}S=3$, the system is inconsistent. If $\operatorname{rank}A=3$, the system would be independent. And if $\operatorname{rank}A=\operatorname{rank}S<3$, the system would be dependent.

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