Find value of a for which system has non-trivial solution. $$ \begin{cases} (5-a)x +4y + 2z = 0 \\ 4x +(5-a)y + 2z = 0 \\ 2x + 2y + (2-a)z = 0 \end{cases} $$ I know that non-trivial solutions have determinant equal to zero. We can solve it by using putting determinants equal to zero, make upper/lower triangle and then multiply principal diagonal elements.
But I want to solve this by using row-echelon form. How can I solve it?
My Attempt: $$\pmatrix{(5-a) & 4 & 2 \\ 0 & -(a-9)(a-1)&a(a-1)\\0&0&-(a-6)(a-1)}$$
I convert it into row echelon form. Then I am multiplying Principal Diagonal elements and putting them to zero knowing that it will give me the value of a. But I am getting wrong values.i.e a=4 , 13
I don't know how you got your RREF, but here's I'd do it.
$$\begin{align}\pmatrix{5-a & 4 & 2 \\ 4 & 5-a & 2 \\ 2 & 2 & 2-a}&\longrightarrow\pmatrix{1 & 1 & 1-\frac a2 \\ 0 & 1-a & -2+2a \\ 0 & -1+a & 2-(5-a)(1-\frac a2)}&\matrix{\small {R_1\leftrightarrow R_3 \\ R_1 \to \frac 12R_1 \\ R_2 \to R_2 - 4R_1 \\ R_3\to R_3-(5-a)R_1}} \\ &\longrightarrow \pmatrix{1 & 1 & 1-\frac a2 \\ 0 & 1-a & -2+2a \\ 0 & 0 & 2a-(5-a)(1-\frac a2)}&\matrix{\small {R_3\to R_3+R_2}}\end{align}$$