Show that following determinant is divisible by $\lambda^2$ and find the other factor.

Question

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.

Answer 1

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 $$\begin{align}\det(A) &= \det(\lambda{\rm Id}_3+vv^\top) = \det\left(\lambda\left({\rm Id}_3 + \frac{1}{\lambda}vv^\top\right)\right) \\ &= \lambda^3 \det\left({\rm Id}_3 + \frac{1}{\lambda}vv^\top\right) = \lambda^3\left(1+ {\rm tr}\left(\frac{1}{\lambda}vv^\top\right)\right) \\ &= \lambda^3\left(1+ \frac{\|v\|^2}{\lambda}\right) = \lambda^3 + \lambda^2\|v\|^2 \\ &= \lambda^2(\lambda + \|v\|^2).\end{align}$$

Answer 2

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 $$ab-\frac ab(b^2+\lambda)=ab-ab-\frac ab\lambda=-\frac ab\lambda\ne-\lambda$$

Answer 3

Alternatively: $$\begin{vmatrix} a^2+\lambda &ab &ac \\ ab & b^2+\lambda & bc \\ ac & bc & c^2+\lambda \end{vmatrix}= \begin{vmatrix} a^2 &ab &ac \\ ab & b^2+\lambda & bc \\ ac & bc & c^2+\lambda \end{vmatrix}+ \begin{vmatrix} \lambda&ab&ac \\ 0 & b^2+\lambda & bc \\ 0 & bc & c^2+\lambda \end{vmatrix}=\\ a^2\begin{vmatrix} 1 &b &c \\ b & b^2+\lambda & bc \\ c & bc & c^2+\lambda \end{vmatrix}+ \lambda^2(b^2+c^2+\lambda)=\\ a^2\left(\begin{vmatrix} 1 &b &c \\ b & b^2 & bc \\ c & bc & c^2+\lambda \end{vmatrix}+ \begin{vmatrix} 1 &0 &c \\ b &\lambda & bc \\ c & 0 & c^2+\lambda \end{vmatrix}\right)+\lambda^2(b^2+c^2+\lambda)=\\ a^2\left(b^2\begin{vmatrix} 1 &1 &c \\ 1 & 1& c \\ c & c & c^2+\lambda \end{vmatrix}+\lambda^2\right)+\lambda^2(b^2+c^2+\lambda)=\\ a^2\lambda^2+\lambda^2(b^2+c^2+\lambda)=\\ \lambda^2(a^2+b^2+c^2+\lambda).$$

Answer 4

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 $$\operatorname{tr}(A) = \operatorname{tr}(vv^T) = \operatorname{tr} (v^Tv) = \| v \|^2.$$

Answer 5

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

$$\lambda^2\begin{vmatrix} 1-\dfrac{a}{b}&1-\dfrac{b}{c} &c(a+b+c)+\lambda \\ -\dfrac{a}{b} & 1 & bc \\ 0 & -\dfrac{b}{c} & c^2+\lambda \end{vmatrix}=0 $$

$$\dfrac{\lambda^2}{bc}\begin{vmatrix} b-a&c-b &c(a+b+c)+\lambda \\ -a & c & bc \\ 0 & -b & c^2+\lambda \end{vmatrix}=0 $$

$$R_1\rightarrow R_1-(R_2+R_3)$$

$$\dfrac{\lambda^2}{bc}\begin{vmatrix} b&0 &ca \\ -a & c & bc \\ 0 & -b & c^2+\lambda \end{vmatrix}=0$$

$$\dfrac{\lambda^2}{bc}\left(b(c^3+c\lambda+b^2c)+a^2bc\right)=0$$

$$\lambda^2\left(c^2+\lambda+b^2+a^2\right)=0$$

Answer 6

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.