One more determinant related question.

Mar 2010
49
0
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
Mar 2010
3,093
582
Florida
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
May 2006
12,028
6,341
Lexington, MA (USA)
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
Sep 2008
5,237
1,625
Melbourne
Take the determinant of the LHS and solve, should be pretty easy.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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
\(\displaystyle x^2\)?