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Lets say there are n homogeneous equations of order n.
$$a_{11}x_1 + a_{12}x_2 +\ldots +a_{1n}x_n=0$$ $$a_{21}x_1 + a_{22}x_2 +\ldots +a_{2n}x_n=0$$ $$\vdots $$ $$a_{n1}x_1 + a_{n2}x_2 +\ldots +a_{nn}x_n=0$$

and lets say that $2$ of these equations are linearly dependent(eq(i) and eq(j))

then when we try to get if the system has non-trivial solution or not we find the determinant of the coefficients. And if the determinant is zero we say that it has infinite non-trivial solution.

But in this case the determinant will be zero because $2$ rows are same [row $i$ and row $j$ will be same after some row operation]. But the system will not have any other solution apart from the trivial solution. Then how can we say that the determinant being zero says that system has infinite solutions?

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  • $\begingroup$ We have corrected the way you have written equations. You must know for a further question that formulas are between dollar signs and indices necessitate "underscore" character. $\endgroup$ – Jean Marie Sep 21 '17 at 7:34
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The determinant is only defined for square matrices.
That means that $m$ must equal $n$.
On the other hand, if there is only one solution, then after row operations there must be $n$ pivots. There are only $m-1$ non-zero rows after subtracting a multiple of eq(i) from eq(ii). So $n$ must be less than $m$.

You only have $n-1$ equations, because equation (ii) gives no new information. You can use one of the equations to write one variable in terms of the others, then eliminate that variable from the other equations. Now you have $n-2$ equations in $n-1$ variables. Repeat the process and you have $n-3$ equations in $n-2$ variables, and eventually one equation in two variables $b_ix_i+b_jx_j=0$. This has infinitely many solutions $x_i=cb_j,x_j=-cb_i$. Next, you have written all the other variables in terms of $x_i$ and $x_j$, so you have infinitely many solutions for all the $x_k$.

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  • $\begingroup$ I have changed the number of equations from m to n. After the row operations $i$th row and $j$th row might be the same and thus the determinant will solve to zero. But as the other equations are not somehow derivable from the $i$th equation, we cant really say that the solution is infinitely many $\endgroup$ – shrijit Sep 21 '17 at 8:32
  • $\begingroup$ Yeah, i understood that the number of solutions will be infinite if the number of equations are less than the number of unknowns. But I am not able to visualize that how these (n-1) equations intersect at infinitely many places but only share the point (0,0 ... 0). $\endgroup$ – shrijit Sep 22 '17 at 3:06
  • $\begingroup$ You might have three planes that share a line, but have 60 degrees between each other so they look like a six-point asterisk from end-on. $\endgroup$ – Michael Sep 22 '17 at 4:48
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Determinant is only defined for square matrices.

If the homogeneous linear system is written as a square matrix and the corresponding determinant is zero, then indeed we have infinitely many solutions.

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