Show that following determinant is divisible by $\lambda^2$ and find the other factor. Show that $\begin{vmatrix}
a^2+\lambda &ab &ac \\ 
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$ is divisible by $\lambda^2$ and find the other factor.
My attempt is as follows:-
$$R_1\rightarrow R_1+R_2+R_3$$
$$\begin{vmatrix}
a(a+b+c)+\lambda &b(a+b+c)+\lambda &c(a+b+c)+\lambda \\ 
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=0$$
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\ 
-\lambda & \lambda & bc \\
0 & -\lambda & c^2+\lambda
\end{vmatrix}=0$$
Taking $\lambda^2$ common
$$\lambda^2\begin{vmatrix}
1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\ 
-1 & 1 & bc \\
0 & -1 & c^2+\lambda
\end{vmatrix}=0
$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&c-b &c(a+b+c)+\lambda \\ 
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0
$$
$$R_1\rightarrow R_1-R_3$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&2c-b &ca+bc \\ 
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
$$R_1\rightarrow R_1-R_2$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
2b-a&c-b &ca \\ 
-b & c & bc \\
0 & -c & c^2+\lambda
\end{vmatrix}=0$$
Now expanding it
$$\dfrac{\lambda^2}{bc}\left(c(2b^2c-abc+abc)+(c^2+\lambda)(2bc-ac+bc-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+(c^2+\lambda)(3bc-ac-b^2)\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(2b^2c^2+3bc^3-ac^3-b^2c^2+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(b^2c^2+3bc^3-ac^3+3bc\lambda-\lambda ac-\lambda b^2\right)=0$$
$$\dfrac{\lambda^2}{bc}\left(c^2(b^2+3bc-ac\right)+\lambda(3bc-ac-b^2)=0$$
So another factor seems to be $\dfrac{1}{bc}\left(c^2(b^2+3bc-ac)+\lambda\left(3bc-ac-b^2\right)\right)$
But actual answer is $a^2+b^2+c^2+\lambda$.
I tried to find my mistake, but everything seems correct. What am I missing here? Please help me in this.
 A: I did not go all the way, but the first (maybe the only) mistake is in the $C_1\rightarrow C_1-\dfrac{a}{b}C_2$ step. The second row will be $$ab-\frac ab(b^2+\lambda)=ab-ab-\frac ab\lambda=-\frac ab\lambda\ne-\lambda$$
A: Let me give you a much simpler solution, using the general fact that if $B$ is a square matrix of rank $1$, then $\det({\rm Id}_n+B) = 1+{\rm tr}(B)$. Let $A$ be the matrix you want to compute the determinant of. Then if $v = [a ~ b ~c]^\top$, we have that $A = \lambda{\rm Id}_3 + vv^\top$. This means that $$\begin{align}\det(A) &= \det(\lambda{\rm Id}_3+vv^\top) = \det\left(\lambda\left({\rm Id}_3 + \frac{1}{\lambda}vv^\top\right)\right) \\ &= \lambda^3 \det\left({\rm Id}_3 + \frac{1}{\lambda}vv^\top\right) = \lambda^3\left(1+ {\rm tr}\left(\frac{1}{\lambda}vv^\top\right)\right) \\ &= \lambda^3\left(1+ \frac{\|v\|^2}{\lambda}\right) = \lambda^3 + \lambda^2\|v\|^2 \\ &= \lambda^2(\lambda + \|v\|^2).\end{align}$$
A: Finally solved it.
Thanks to all for looking into this problem, special thanks to @Andrei for pointing out the exact mistake.
$$C_1\rightarrow C_1-\dfrac{a}{b}C_2$$
$$C_2\rightarrow C_2-\dfrac{b}{c}C_3$$
$$\begin{vmatrix}
\lambda-\dfrac{a\lambda}{b}&\lambda-\dfrac{b\lambda}{c} &c(a+b+c)+\lambda \\ 
-\dfrac{a\lambda}{b} & \lambda & bc \\
0 & -\dfrac{b\lambda}{c} & c^2+\lambda
\end{vmatrix}=0$$
Taking $\lambda^2$ common
$$\lambda^2\begin{vmatrix}
1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\ 
-\dfrac{a}{b} & 1 & bc \\
0 & -\dfrac{b}{c} & c^2+\lambda
\end{vmatrix}=0
$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b-a&c-b &c(a+b+c)+\lambda \\ 
-a & c & bc \\
0 & -b & c^2+\lambda
\end{vmatrix}=0
$$
$$R_1\rightarrow R_1-(R_2+R_3)$$
$$\dfrac{\lambda^2}{bc}\begin{vmatrix}
b&0 &ca \\ 
-a & c & bc \\
0 & -b & c^2+\lambda
\end{vmatrix}=0$$
$$\dfrac{\lambda^2}{bc}\left(b(c^3+c\lambda+b^2c)+a^2bc\right)=0$$
$$\lambda^2\left(c^2+\lambda+b^2+a^2\right)=0$$
A: I got a new way to solve this question and its just beautiful:
Multiply $R_1$ by $a$, $R_2$ by $b$, $R_3$ by $c$
$$\dfrac{1}{abc}\begin{vmatrix}
a^3+a\lambda&a^2b&a^2c\\
ab^2&b^3+b\lambda&b^2c\\
ac^2&bc^2&c^3+c\lambda
\end{vmatrix}$$
Taking a common from first column, b from second column, c from third column
$$\begin{vmatrix}
a^2+\lambda&a^2&a^2\\
b^2&b^2+\lambda&b^2\\
c^2&c^2&c^2+\lambda
\end{vmatrix}$$
Now its simple , just do $$C_1\rightarrow C_1-C_2, C_2\rightarrow C_2-C_3$$
$$\begin{vmatrix}
\lambda&0&a^2\\
-\lambda&\lambda&b^2\\
0&-\lambda&c^2+\lambda
\end{vmatrix}$$
Now it's simple to solve. Hope it will be useful to somebody.
A: Alternatively:
$$\begin{vmatrix}
a^2+\lambda &ab &ac \\ 
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}=
\begin{vmatrix}
a^2 &ab &ac \\ 
ab & b^2+\lambda & bc \\
ac & bc & c^2+\lambda
\end{vmatrix}+
\begin{vmatrix}
\lambda&ab&ac \\ 
0 & b^2+\lambda & bc \\
0 & bc & c^2+\lambda
\end{vmatrix}=\\
a^2\begin{vmatrix}
1 &b &c \\ 
b & b^2+\lambda & bc \\
c & bc & c^2+\lambda
\end{vmatrix}+
\lambda^2(b^2+c^2+\lambda)=\\
a^2\left(\begin{vmatrix}
1 &b &c \\ 
b & b^2 & bc \\
c & bc & c^2+\lambda
\end{vmatrix}+
\begin{vmatrix}
1 &0 &c \\ 
b &\lambda & bc \\
c & 0 & c^2+\lambda
\end{vmatrix}\right)+\lambda^2(b^2+c^2+\lambda)=\\
a^2\left(b^2\begin{vmatrix}
1 &1 &c \\ 
1 & 1& c \\
c & c & c^2+\lambda
\end{vmatrix}+\lambda^2\right)+\lambda^2(b^2+c^2+\lambda)=\\
a^2\lambda^2+\lambda^2(b^2+c^2+\lambda)=\\
\lambda^2(a^2+b^2+c^2+\lambda).$$
A: The determinant in your problem is equal to $p(-\lambda),$ where $p$ is the characteristic polynomial of $A = vv^T,$ where $ v = [a, \ b, \ c]^T.$ The $0$-eigenspace (i.e, the kernel of $A$) is the two dimensional hyperplane given by $v^Tx = 0,$ so $0$ is an eigenvalue with algebraic multiplicity at least two and the final eigenvalue is given by $$\operatorname{tr}(A) = \operatorname{tr}(vv^T) = \operatorname{tr} (v^Tv) = \| v \|^2.$$ 
