Calculate the determinant of this matrix

[Nemanjavuk69]

I have the following matrix
[math]\begin{matrix} 1 & 0 & -1\\ 2 & 3 & -1\\ 0 & 6 & 0 \end{matrix}[/math]
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
[math]\begin{matrix} 1-\lambda & 0 & -1\\ 2 & 3-\lambda & -1\\ 0 & 6 & -\lambda \end{matrix}[/math]
However, when doing the calculations I always end up with [imath]-\lambda^3+4\lambda^2-3\lambda-6[/imath] and that is not true, the actual equation should be [imath]-\lambda^3+4\lambda^2-9\lambda-6[/imath]

I am row expanding from the first column, so [math]a_{11}, a_{21}, a_{31}[/math]
I can therefor calculate the determinant as followed...
[math](1-\lambda) \cdot \begin{vmatrix} 3-\lambda & -1\\ 6 & -\lambda \end{vmatrix} -2 \cdot \begin{vmatrix} 0 & -1\\ 6 & -\lambda \end{vmatrix}[/math][math](1-\lambda) \cdot (3-\lambda) \cdot (-\lambda) + 6 - 2 \cdot 6[/math][math](1-\lambda) \cdot (-3\lambda+\lambda^2) + 6[/math][math]-3\lambda + \lambda^2 + 3\lambda^2 - \lambda^3 - 6[/math][math]-\lambda^3 + 4\lambda^2 - 3\lambda - 6[/math]
However, as stated before, the result is not right. Instead of [imath]-3\lambda[/imath] it should be [imath]-9\lambda[/imath]. Where am I doing something wrong? Any help is gladely appreciated.

 

[Dr.Peterson]

I have the following matrix
[math]\begin{matrix} 1 & 0 & -1\\ 2 & 3 & -1\\ 0 & 6 & 0 \end{matrix}[/math]
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
[math]\begin{matrix} 1-\lambda & 0 & -1\\ 2 & 3-\lambda & -1\\ 0 & 6 & -\lambda \end{matrix}[/math]
However, when doing the calculations I always end up with [imath]-\lambda^3+4\lambda^2-3\lambda-6[/imath] and that is not true, the actual equation should be [imath]-\lambda^3+4\lambda^2-9\lambda-6[/imath]

I am row expanding from the first column, so [math]a_{11}, a_{21}, a_{31}[/math]
I can therefor calculate the determinant as followed...
[math](1-\lambda) \cdot \begin{vmatrix} 3-\lambda & -1\\ 6 & -\lambda \end{vmatrix} -2 \cdot \begin{vmatrix} 0 & -1\\ 6 & -\lambda \end{vmatrix}[/math][math](1-\lambda) \cdot (3-\lambda) \cdot (-\lambda) + 6 - 2 \cdot 6[/math][math](1-\lambda) \cdot (-3\lambda+\lambda^2) + 6[/math][math]-3\lambda + \lambda^2 + 3\lambda^2 - \lambda^3 - 6[/math][math]-\lambda^3 + 4\lambda^2 - 3\lambda - 6[/math]
However, as stated before, the result is not right. Instead of [imath]-3\lambda[/imath] it should be [imath]-9\lambda[/imath]. Where am I doing something wrong? Any help is gladely appreciated.

Your
[math](1-\lambda) \cdot (3-\lambda) \cdot (-\lambda) + 6 - 2 \cdot 6[/math][math](1-\lambda) \cdot (-3\lambda+\lambda^2) - 6[/math][math]-3\lambda + \lambda^2 + 3\lambda^2 - \lambda^3 - 6[/math][math]-\lambda^3 + 4\lambda^2 - 3\lambda - 6[/math]should be
[math](1-\lambda) \cdot {\color{Red} (}(3-\lambda) \cdot (-\lambda) + 6{\color{Red} )} - 2 \cdot 6[/math][math](1-\lambda) \cdot(-3\lambda+\lambda^2{\color{Red} +6}) {\color{Red} - 12}[/math][math]-3\lambda + \lambda^2 {\color{Red} +6} + 3\lambda^2 - \lambda^3 {\color{Red} - 6\lambda - 12}[/math][math]-\lambda^3 + 4\lambda^2 - {\color{Red} 9}\lambda - 6[/math]

 

[Otis]

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
[imath]\;[/imath]

 

[Steven G]

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!

 

[Dr.Peterson]

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
[math]\begin{matrix} 1-\lambda & 0 & -1\\ 2 & 3-\lambda & -1\\ 0 & 6 & -\lambda \end{matrix}[/math]

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.

 

[Nemanjavuk69]

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!