# Why square matrix with zero determinant have non trivial solution

While I was solving problem to determine if given set of vectors are linearly independent or not, the solution given was that the coefficient matrix A of the homogeneous linear system of those vectors

k1a1 + k2b1 + k3c1 = 0

k1a2 + k2b2 + k3c2 = 0

k1a3 + k2b3 + k3c3 = 0

was a 3x3 square matrix, and determinant(A) = 0

Thus it has a non trivial solution and hence the vectors are linearly dependent.

I am unaware about how if A is nxn square matrix and if determinant(A) = 0, then the system is having a non trivial solution.

One way to find the determinant is to bring the matrix in to row echelon form by row operations (which do not alter the determinant) and multiply the diagonal entries. If the determinant is $0$, one diagonal entry must be zero, and you can work your way back up the triangular matrix to obtain a solution with at least one "degree of freedom", i.e. , there is a nontrivial solution.
So, for one vector be linearly indenpendent all constants should be = 0 Thus, if we applly the rule of "sarrus" we obtain : $\operatorname{det}(A)= -(k_1*k_2*k_3)-(k_1*k_2*k_3)-(k_1*k_2*k_3)+(k_1*k_2*k_3)+(k_1*k_2*k_3)+(k_1*k_2*k_3)$
$\operatorname {det}(A)= -3(k_1*k_2*k_3)+3(k_1*k_2*k_3)$
$\operatorname {det}(A)= 0$
However, if we use method of elimination of Gauss-Jordan, we found $a_1= -k_2/k_1(b_1)-k_3/k_1(c_1)$ and $a_2, b_2, c_2, a_3, b_3, c_3 = 0$