# Non-Trivial Solution and Row -Echelon Form

Find value of a for which system has non-trivial solution.
{(5−a)x+4y+2z=04x+(5−a)y+2z=02x+2y+(2−a)z=0
\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:
((5−a)420−(a−9)(a−1)a(a−1)00−(a−6)(a−1))\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

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Have you already gotten it in RREF?
– Bye_World
2 days ago

Yes. I tried but I found the wrong value of a.
– Knowledge 32
2 days ago

1

Edit into your question your attempt and we’ll see where you went wrong. Note on mathjax: to get (1234)⟶(5678)\pmatrix{1 & 2 \\ 3 &4} \longrightarrow \pmatrix{5 & 6 \\ 7 & 8} you type $$\pmatrix{1 & 2 \\ 3 &4} \longrightarrow \pmatrix{5 & 6 \\ 7 & 8}$$.
– Bye_World
2 days ago

@Bye_World I edited the question.
– Knowledge 32
2 days ago

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