Calculate the determinant of this matrix

Nemanjavuk69

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Mar 23, 2022
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I have the following matrix
101231060\begin{matrix} 1 & 0 & -1\\ 2 & 3 & -1\\ 0 & 6 & 0 \end{matrix}
Following the text book linear algebra and its applications 6 global edition by David C. Lay on chapter 5.2 we can calculate the determinant by writing the characteristic equation like this
1λ0123λ106λ\begin{matrix} 1-\lambda & 0 & -1\\ 2 & 3-\lambda & -1\\ 0 & 6 & -\lambda \end{matrix}
However, when doing the calculations I always end up with λ3+4λ23λ6-\lambda^3+4\lambda^2-3\lambda-6 and that is not true, the actual equation should be λ3+4λ29λ6-\lambda^3+4\lambda^2-9\lambda-6

I am row expanding from the first column, so a11,a21,a31a_{11}, a_{21}, a_{31}
I can therefor calculate the determinant as followed...
(1λ)3λ16λ2016λ(1-\lambda) \cdot \begin{vmatrix} 3-\lambda & -1\\ 6 & -\lambda \end{vmatrix} -2 \cdot \begin{vmatrix} 0 & -1\\ 6 & -\lambda \end{vmatrix}(1λ)(3λ)(λ)+626(1-\lambda) \cdot (3-\lambda) \cdot (-\lambda) + 6 - 2 \cdot 6(1λ)(3λ+λ2)+6(1-\lambda) \cdot (-3\lambda+\lambda^2) + 63λ+λ2+3λ2λ36-3\lambda + \lambda^2 + 3\lambda^2 - \lambda^3 - 6λ3+4λ23λ6-\lambda^3 + 4\lambda^2 - 3\lambda - 6
However, as stated before, the result is not right. Instead of 3λ-3\lambda it should be 9λ-9\lambda. Where am I doing something wrong? Any help is gladely appreciated.
 
I have the following matrix
101231060\begin{matrix} 1 & 0 & -1\\ 2 & 3 & -1\\ 0 & 6 & 0 \end{matrix}
Following the text book linear algebra and its applications 6 global edition by David C. Lay on chapter 5.2 we can calculate the determinant by writing the characteristic equation like this
1λ0123λ106λ\begin{matrix} 1-\lambda & 0 & -1\\ 2 & 3-\lambda & -1\\ 0 & 6 & -\lambda \end{matrix}
However, when doing the calculations I always end up with λ3+4λ23λ6-\lambda^3+4\lambda^2-3\lambda-6 and that is not true, the actual equation should be λ3+4λ29λ6-\lambda^3+4\lambda^2-9\lambda-6

I am row expanding from the first column, so a11,a21,a31a_{11}, a_{21}, a_{31}
I can therefor calculate the determinant as followed...
(1λ)3λ16λ2016λ(1-\lambda) \cdot \begin{vmatrix} 3-\lambda & -1\\ 6 & -\lambda \end{vmatrix} -2 \cdot \begin{vmatrix} 0 & -1\\ 6 & -\lambda \end{vmatrix}(1λ)(3λ)(λ)+626(1-\lambda) \cdot (3-\lambda) \cdot (-\lambda) + 6 - 2 \cdot 6(1λ)(3λ+λ2)+6(1-\lambda) \cdot (-3\lambda+\lambda^2) + 63λ+λ2+3λ2λ36-3\lambda + \lambda^2 + 3\lambda^2 - \lambda^3 - 6λ3+4λ23λ6-\lambda^3 + 4\lambda^2 - 3\lambda - 6
However, as stated before, the result is not right. Instead of 3λ-3\lambda it should be 9λ-9\lambda. Where am I doing something wrong? Any help is gladely appreciated.
Your
(1λ)(3λ)(λ)+626(1-\lambda) \cdot (3-\lambda) \cdot (-\lambda) + 6 - 2 \cdot 6(1λ)(3λ+λ2)6(1-\lambda) \cdot (-3\lambda+\lambda^2) - 63λ+λ2+3λ2λ36-3\lambda + \lambda^2 + 3\lambda^2 - \lambda^3 - 6λ3+4λ23λ6-\lambda^3 + 4\lambda^2 - 3\lambda - 6should be
(1λ)((3λ)(λ)+6)26(1-\lambda) \cdot {\color{Red} (}(3-\lambda) \cdot (-\lambda) + 6{\color{Red} )} - 2 \cdot 6(1λ)(3λ+λ2+6)12(1-\lambda) \cdot(-3\lambda+\lambda^2{\color{Red} +6}) {\color{Red} - 12}3λ+λ2+6+3λ2λ36λ12-3\lambda + \lambda^2 {\color{Red} +6} + 3\lambda^2 - \lambda^3 {\color{Red} - 6\lambda - 12}λ3+4λ29λ6-\lambda^3 + 4\lambda^2 - {\color{Red} 9}\lambda - 6
 
I can [therefore] calculate the determinant as [follows]
...
(1−λ)⋅(3−λ)⋅(−λ)+6−2⋅6
Hi Nemanjavuk69. I'm getting a different polynomial. Is your work missing grouping symbols shown below? :)

(1 − λ)⋅((3 − λ)⋅(−λ) + 6) − 12
  \;
 
What are you talking about!
This 3x3 matrix that you said you want the determinant of only has real numbers as its entry. Therefore the determinant IS a real number and NOT a polynomial in λ.

Please please state the correct question!
 
What are you talking about!
This 3x3 matrix that you said you want the determinant of only has real numbers as its entry. Therefore the determinant IS a real number and NOT a polynomial in λ.

Please please state the correct question!
I chose not to point this out initially, in order to focus on the main issue; but now I will:
Following the text book linear algebra and its applications 6 global edition by David C. Lay on chapter 5.2 we can calculate the determinant by writing the characteristic equation like this
1λ0123λ106λ\begin{matrix} 1-\lambda & 0 & -1\\ 2 & 3-\lambda & -1\\ 0 & 6 & -\lambda \end{matrix}
I believe what was intended (or should have been!) is

we can calculate the characteristic equation by writing the determinant like this:​

I hope that makes the error in the statement of the problem clearer.
 
dear @Dr.Peterson and @Otis thank you for pointing out the missing parenthesis, that is what I was missing. Thank you both for helping me out. Have a wonderful day/ night forward!
 
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