# One more determinant related question.

• May 14th 2010, 03:19 PM
quikwerk
One more determinant related question.
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?
• May 14th 2010, 03:47 PM
dwsmith
Quote:

Originally Posted by 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?

$\displaystyle \begin{vmatrix} 0 & 0 & -1\\ 3 & x_1 & 0\\ 2 & 0 & 3 \end{vmatrix}=2x_1=x^2\rightarrow x_1=\frac{x^2}{2}$
• May 14th 2010, 03:49 PM
Soroban
Hello, quikwerk!

Quote:

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

• May 14th 2010, 03:49 PM
pickslides
Take the determinant of the LHS and solve, should be pretty easy.
• May 14th 2010, 03:50 PM
dwsmith
Quote:

Originally Posted by Soroban
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
$\displaystyle x^2$?