Is it possible that homogeneous system may have only trivial solution if determinant is zero?

Question

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?

Answer 1

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 $x_k$.

Answer 2

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.