A simpler way to solve this determinant equation-$2$

Question

Question Statement:- Solve the following determinant equation $$\begin{vmatrix} x & 2 & 3 \\ 6 & x+4 & 4 \\ 7 & 8 & x+8 \\ \end{vmatrix}=0$$

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My Solution:-

$$\begin{vmatrix} x & 2 & 3 \\ 6 & x+4 & 4 \\ 7 & 8 & x+8 \\ \end{vmatrix}= \begin{vmatrix} x+13 & x+14 & x+15 \\ 6 & x+4 & 4 \\ 7 & 8 & x+8 \\ \end{vmatrix}\tag{$R_1\rightarrow R_1+R_2+R_3$}$$ $$(x+13)\begin{vmatrix} 1 & 1 & 1 \\ 6 & x+4 & 4 \\ 7 & 8 & x+8 \\ \end{vmatrix}+ \begin{vmatrix} 0 & 1 & 2 \\ 6 & x+4 & 4 \\ 7 & 8 & x+8 \\ \end{vmatrix}\tag{1}$$ Now lets solve the two matrices obtained in the last step above seperately. $$(x+13)\begin{vmatrix} 1 & 1 & 1 \\ 6 & x+4 & 4 \\ 7 & 8 & x+8 \\ \end{vmatrix}$$ $$=(x+13)\begin{vmatrix} 0 & 0 & 1 \\ 2 & x & 4 \\ -(x+1) & -x & x+8 \\ \end{vmatrix}\hspace{3cm}(C_1\rightarrow C_1-C_3, C_2\rightarrow C_2-C_3)$$ $$=x(x+13)\begin{vmatrix} 0 & 0 & 1 \\ 2 & 1 & 4 \\ -(x+1) & -1 & x+8 \\ \end{vmatrix}=x(x+13)(x-1)$$ And the second matrix can be simplified as $$\begin{vmatrix} 0 & 1 & 0 \\ 6 & x+4 & -4-2x \\ 7 & 8 & x-8 \\ \end{vmatrix}\tag{$C_3\rightarrow C_3-C_2$}$$ $$=-1\times(6x-48+28+14x)=-20(x-1)$$ So, we get $$\begin{vmatrix} x & 2 & 3 \\ 6 & x+4 & 4 \\ 7 & 8 & x+8 \\ \end{vmatrix}=0\implies x(x+13)(x-1)-20(x-1)=0\\ \implies (x-1)(x^2+13x-20)=0\implies x=1,\dfrac{-13\pm\sqrt{249}}{2}$$

As can be seen, my solution is way long and it is no better than opening the determinant. So if anyone can provide me with a better, simpler and a more intuitive solution I will be very thankful.

Answer 1

Your solution is fine but here is a more simple way to solve this problem:

$$\begin{vmatrix} x & 2 & 3 \\ 6 & x+4 & 4 \\ 7 & 8 & x+8 \\ \end{vmatrix} $$ = $$-\begin{vmatrix}-\begin{pmatrix} 0 & 2 & 3 \\ 6 & 4 & 4 \\ 7 & 8 & 8 \\ \end{pmatrix} - x I)\end{vmatrix} = 0$$ where $I$ is the identity matrix.

It lookes similar to the equation for eigenvalue.

So $x$ is eigenvalue for matrix:$$A :=\begin{pmatrix} 0 & -2 & -3 \\ -6 & -4 & -4 \\ -7 & -8 & -8 \\ \end{pmatrix}$$

Because $trace(A) = -12$ and $det(A) = -20$ so the characteristic polynomial of form: $-x^3 -12 x^2 +B x -20$. Where $B$ is an real number.

How to find $B$?

Let $x = 2$ into $det(-A-xI)$,then $det(-A-xI) = -10 = -76+2B \implies B=33$.

So it becomes solve this polynomial:$-x^3 -12 x^2 +33 x -20=0$

EDIT: Another way to find B is to use $x = 1$ as subsitution, as Francesco suggested in comment. Since the row become dependent then the determinant of matrix becomes zero so $-1-12-20+B =0 \implies B = 33$

Answer 2

Your computation is correct.

Here is a different procedure where we use the rule of Sarrus: the determinant is $$x(x+4)(x+8)+56+144-21(x+4)-32x-12(x+8)\\ =x^3+12x^2-33x+20=(x-1)\cdot(x^2+13x-20).$$ Here you have to guess that $1$ is a root of the third degree polynomial and then divide it by $(x-1)$ in order to obtain the other factor $x^2+13x-20$.