# One more determinant related question.

#### quikwerk

Alright, here is another question that I need some help with, I think I have an idea of how to do it but it never hurts to make sure, though

The question states:
solve the following equation.

$$\displaystyle \begin{vmatrix} 0 & 0 & -1\\ 3 & x & 0\\ 2 & 0 & 3 \end{vmatrix}=x^2$$

My idea is to either, expand by minors, or, to evaluate it by using the main diagonals(?) method.
Which way would be quickest and easiest to do?
Or am I headed in the wrong direction entirely?

#### dwsmith

MHF Hall of Honor
Alright, here is another question that I need some help with, I think I have an idea of how to do it but it never hurts to make sure, though

The question states:
solve the following equation.

$$\displaystyle \begin{vmatrix} 0 & 0 & -1\\ 3 & x & 0\\ 2 & 0 & 3 \end{vmatrix}=x^2$$

My idea is to either, expand by minors, or, to evaluate it by using the main diagonals(?) method.
Which way would be quickest and easiest to do?
Or am I headed in the wrong direction entirely?
$$\displaystyle \begin{vmatrix} 0 & 0 & -1\\ 3 & x_1 & 0\\ 2 & 0 & 3 \end{vmatrix}=2x_1=x^2\rightarrow x_1=\frac{x^2}{2}$$

Last edited:

#### Soroban

MHF Hall of Honor
Hello, quikwerk!

Solve: . $$\displaystyle \begin{vmatrix}0 & 0 & \text{-}1\\ 3 & x & 0\\ 2 & 0 & 3 \end{vmatrix}\;=\;x^2$$

My idea is to either, expand by minors, or, to evaluate it by using the main diagonals method.
Which way would be quickest and easiest to do?
I'd expand by minors.

I can already see all of this: . $$\displaystyle \begin{vmatrix}0&0&\text{-}1 \\ 3&x&0 \\ 2&0&3\end{vmatrix} \;=\;x^2$$

. . $$\displaystyle 0\begin{vmatrix}x&0\\0&3\end{vmatrix} \:-\: 0\begin{vmatrix}3&0\\2&3\end{vmatrix} \:-\: 1\begin{vmatrix}3&x\\2&0\end{vmatrix} \;\;=\;\;x^2 \qquad\Rightarrow\qquad 0 \;-\; 0 \;-\;1(0 - 2x) \;\;=\;\;x^2$$

Hence: .$$\displaystyle 2x \:=\:x^2 \quad\Rightarrow\quad x^2-2x \:=\:0 \quad\Rightarrow\quad x(x-2) \:=\:0$$

Therefore: . $$\displaystyle x \;=\;0,\:2$$

#### pickslides

MHF Helper
Take the determinant of the LHS and solve, should be pretty easy.

#### dwsmith

MHF Hall of Honor
Hello, quikwerk!

I'd expand by minors.

I can already see all of this: . $$\displaystyle \begin{vmatrix}0&0&\text{-}1 \\ 3&x&0 \\ 2&0&3\end{vmatrix} \;=\;x^2$$

. . $$\displaystyle 0\begin{vmatrix}x&0\\0&3\end{vmatrix} \:-\: 0\begin{vmatrix}3&0\\2&3\end{vmatrix} \:-\: 1\begin{vmatrix}3&x\\2&0\end{vmatrix} \;\;=\;\;x^2 \qquad\Rightarrow\qquad 0 \;-\; 0 \;-\;1(0 - 2x) \;\;=\;\;x^2$$

Hence: .$$\displaystyle 2x \:=\:x^2 \quad\Rightarrow\quad x^2-2x \:=\:0 \quad\Rightarrow\quad x(x-2) \:=\:0$$

Therefore: . $$\displaystyle x \;=\;0,\:2$$

Don't we want the determinant to be
$$\displaystyle x^2$$?