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

=================

  

 

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

=================

1 Answer
1

=================

I don’t know how you got your RREF, but here’s I’d do it.

(5−a4245−a2222−a)⟶(111−a201−a−2+2a0−1+a2−(5−a)(1−a2))R1↔R3R1→12R1R2→R2−4R1R3→R3−(5−a)R1⟶(111−a201−a−2+2a002a−(5−a)(1−a2))R3→R3+R2\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}