Find x: Determinant Equation 2

The last term on the first line is missing a 2.

When you are finished, you can put your values back into the equation and check -- that's what I did to realize you were wrong. (I didn't expect it to be wrong, but ...)
 
Personally, I would have expanded the first column: \(\displaystyle \left|\begin{array}{ccc} x & 1 & 2 \\1 & x & 3 \\ 0 & 1 & 2 \end{array}\right|=\)\(\displaystyle x\left|\begin{array}{cc} x & 3 \\ 1 & 2\end{array}\right|\)\(\displaystyle - 1\left|\begin{array}{cc}1 & 2 \\ 1 & 2 \end{array}\right|= x(2x- 3)- 0= 2x^2- 3x= -4x\).

\(\displaystyle 2x^2- 3x+ 4x= 0\)
 
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Dr.Peterson said:
The last term on the first line is missing a 2.

When you are finished, you can put your values back into the equation and check -- that's what I did to realize you were wrong. (I didn't expect it to be wrong, but ...)
Click to expand...

At least I understand the problem. I made a typo while rushing through.
 
HallsofIvy said:
Personally, I would have expanded the first column: \(\displaystyle \left|\begin{array}{ccc} x & 1 & 2 \\1 & x & 3 \\ 0 & 1 & 2 \end{array}\right|=\)\(\displaystyle x\left|\begin{array}{cc} x & 3 \\ 1 & 2\end{array}\right|\)\(\displaystyle - 1\left|\begin{array}{cc}1 & 2 \\ 1 & 2 \end{array}\right|= x(2x- 3)- 0= 2x^2- 3x= -4x\).

\(\displaystyle 2x^2- 3x+ 4x= 0\)
Click to expand...

Thank you. I'll take it from here.
 
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