Finding determinant of $3times3$ matrix
$begingroup$
$$A = left(begin{matrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{matrix}right)$$
$$det A = begin{vmatrix}Aend{vmatrix} =
(lambda - 1) begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix} + 1begin{vmatrix}
1 & 1 \
-3 & lambda + 1 \
end{vmatrix} -
1begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}\
= ((lambda - 1)(lambda - 3)(lambda + 1)) - (1 + (lambda + 1) + 3) + (1 +3lambda - 9) \
= (lambda - 1)(lambda - 3)(lambda + 1) + (lambda + 1) + 2(lambda - 3)$$
The solution says that it is $(lambda - 2)(lambda + 2)(lambda - 3)$. I feel like I am so close, but I don't get what I am supposed to do to get to the solution.
linear-algebra determinant
$endgroup$
add a comment |
$begingroup$
$$A = left(begin{matrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{matrix}right)$$
$$det A = begin{vmatrix}Aend{vmatrix} =
(lambda - 1) begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix} + 1begin{vmatrix}
1 & 1 \
-3 & lambda + 1 \
end{vmatrix} -
1begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}\
= ((lambda - 1)(lambda - 3)(lambda + 1)) - (1 + (lambda + 1) + 3) + (1 +3lambda - 9) \
= (lambda - 1)(lambda - 3)(lambda + 1) + (lambda + 1) + 2(lambda - 3)$$
The solution says that it is $(lambda - 2)(lambda + 2)(lambda - 3)$. I feel like I am so close, but I don't get what I am supposed to do to get to the solution.
linear-algebra determinant
$endgroup$
add a comment |
$begingroup$
$$A = left(begin{matrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{matrix}right)$$
$$det A = begin{vmatrix}Aend{vmatrix} =
(lambda - 1) begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix} + 1begin{vmatrix}
1 & 1 \
-3 & lambda + 1 \
end{vmatrix} -
1begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}\
= ((lambda - 1)(lambda - 3)(lambda + 1)) - (1 + (lambda + 1) + 3) + (1 +3lambda - 9) \
= (lambda - 1)(lambda - 3)(lambda + 1) + (lambda + 1) + 2(lambda - 3)$$
The solution says that it is $(lambda - 2)(lambda + 2)(lambda - 3)$. I feel like I am so close, but I don't get what I am supposed to do to get to the solution.
linear-algebra determinant
$endgroup$
$$A = left(begin{matrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{matrix}right)$$
$$det A = begin{vmatrix}Aend{vmatrix} =
(lambda - 1) begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix} + 1begin{vmatrix}
1 & 1 \
-3 & lambda + 1 \
end{vmatrix} -
1begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}\
= ((lambda - 1)(lambda - 3)(lambda + 1)) - (1 + (lambda + 1) + 3) + (1 +3lambda - 9) \
= (lambda - 1)(lambda - 3)(lambda + 1) + (lambda + 1) + 2(lambda - 3)$$
The solution says that it is $(lambda - 2)(lambda + 2)(lambda - 3)$. I feel like I am so close, but I don't get what I am supposed to do to get to the solution.
linear-algebra determinant
linear-algebra determinant
edited Dec 5 '18 at 22:37
Jean-Claude Arbaut
14.7k63464
14.7k63464
asked Dec 5 '18 at 21:06
Evan KimEvan Kim
998
998
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
What is wrong with this solution? Here we line the two copy of the same matrix side by side and draw blue and red lines and add all products of numbers on the blue lines and substract products of numbers on the red lines. The solution is off by 4.
The determinant is thus
$$(lambda-1)(lambda-3)(lambda+1)\+(-3)(1)(-1)\+(-1)(1)(1)\-(-1)(lambda-3)(-3)\-(-1)(1)(lambda+1)\-(lambda-1)(1)(1)$$
That is
$$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
=(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
=(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
=(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
=(lambda-3)(lambda^2-1-3)+4\
=(lambda-3)(lambda^2-4)+4\
=(lambda-3)(lambda-2)(lambda+2)+4$$
$endgroup$
$begingroup$
You are using the rule of Sarrus, and your solution is actually correct. But you should write this in LaTeX.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 21:57
$begingroup$
Really nice, maybe it is difficult make the lines with MathJax! I don't know how to obtain that :(
$endgroup$
– gimusi
Dec 5 '18 at 22:07
$begingroup$
Yes, very nice. This was very painful to keep track of, but I got the same answer and was able to follow along the math.
$endgroup$
– Evan Kim
Dec 5 '18 at 22:20
1
$begingroup$
I reworked your answer in LaTeX (except the matrices, made with Excel), can you check and confirm it's ok?
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:27
$begingroup$
Great! thank you soo much! The red line is not correct there should be one for (-1)(lambda-3)(-3) and there is none for (-1)(lambda-3)(-3).
$endgroup$
– nafhgood
Dec 6 '18 at 2:17
|
show 1 more comment
$begingroup$
$$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -4lambda +16$$
$endgroup$
$begingroup$
I am trying to find the characteristic polynomial of A. I just posted this part of the problem because I thought the characteristic polynomial was just finding the determinant. I think I did |lambda I - A| wrong...let me check
$endgroup$
– Evan Kim
Dec 5 '18 at 21:21
$begingroup$
@EvanKim Thanks for your kind inputs in your question post and here. Just to let you know, it's good to contain any questions on the initial question form. Regarding to your questions, the eigenvalues are given by the $lambda$s that nullify the determinant, simply meaning that you have to find the solutions to the equation $det(lambda I - A) = 0$. On the initial question, greedoid's answer follows exactly where you were mistaken.
$endgroup$
– Rebellos
Dec 5 '18 at 21:36
$begingroup$
so $λ^3−3λ^2−4λ+16$ is in fact $(λ−2)(λ+2)(λ−3)?$
$endgroup$
– Evan Kim
Dec 5 '18 at 21:47
$begingroup$
@EvanKim No, if only because the constant term of your expression after expansion is obviously $12$.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:07
add a comment |
$begingroup$
HINT
We can simplify a little bit summing the third to the first row
$$begin{vmatrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}= begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}$$
and using Laplace expansion in the first row to obtain
$$begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}
=(lambda -4) cdot begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix}
+lambdacdot begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}$$
Can you proceed from here?
$endgroup$
2
$begingroup$
Posts should be posted in final form. Comments should be limited to giving information relevant to the post. Discussions about votes are mostly off-topic. Excessive "+1 great posts" as a matter of fact can lead to suspensions.
$endgroup$
– quid♦
Dec 5 '18 at 21:52
2
$begingroup$
The text in the comment box explicitly says that these comments should be avoided and they are also discouraged in other ways. To do it sometimes will not cause issues, to do it all the time/often can lead to problems.
$endgroup$
– quid♦
Dec 5 '18 at 21:59
$begingroup$
Why the downvote for a correct answer?
$endgroup$
– gimusi
Dec 7 '18 at 13:02
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What is wrong with this solution? Here we line the two copy of the same matrix side by side and draw blue and red lines and add all products of numbers on the blue lines and substract products of numbers on the red lines. The solution is off by 4.
The determinant is thus
$$(lambda-1)(lambda-3)(lambda+1)\+(-3)(1)(-1)\+(-1)(1)(1)\-(-1)(lambda-3)(-3)\-(-1)(1)(lambda+1)\-(lambda-1)(1)(1)$$
That is
$$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
=(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
=(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
=(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
=(lambda-3)(lambda^2-1-3)+4\
=(lambda-3)(lambda^2-4)+4\
=(lambda-3)(lambda-2)(lambda+2)+4$$
$endgroup$
$begingroup$
You are using the rule of Sarrus, and your solution is actually correct. But you should write this in LaTeX.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 21:57
$begingroup$
Really nice, maybe it is difficult make the lines with MathJax! I don't know how to obtain that :(
$endgroup$
– gimusi
Dec 5 '18 at 22:07
$begingroup$
Yes, very nice. This was very painful to keep track of, but I got the same answer and was able to follow along the math.
$endgroup$
– Evan Kim
Dec 5 '18 at 22:20
1
$begingroup$
I reworked your answer in LaTeX (except the matrices, made with Excel), can you check and confirm it's ok?
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:27
$begingroup$
Great! thank you soo much! The red line is not correct there should be one for (-1)(lambda-3)(-3) and there is none for (-1)(lambda-3)(-3).
$endgroup$
– nafhgood
Dec 6 '18 at 2:17
|
show 1 more comment
$begingroup$
What is wrong with this solution? Here we line the two copy of the same matrix side by side and draw blue and red lines and add all products of numbers on the blue lines and substract products of numbers on the red lines. The solution is off by 4.
The determinant is thus
$$(lambda-1)(lambda-3)(lambda+1)\+(-3)(1)(-1)\+(-1)(1)(1)\-(-1)(lambda-3)(-3)\-(-1)(1)(lambda+1)\-(lambda-1)(1)(1)$$
That is
$$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
=(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
=(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
=(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
=(lambda-3)(lambda^2-1-3)+4\
=(lambda-3)(lambda^2-4)+4\
=(lambda-3)(lambda-2)(lambda+2)+4$$
$endgroup$
$begingroup$
You are using the rule of Sarrus, and your solution is actually correct. But you should write this in LaTeX.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 21:57
$begingroup$
Really nice, maybe it is difficult make the lines with MathJax! I don't know how to obtain that :(
$endgroup$
– gimusi
Dec 5 '18 at 22:07
$begingroup$
Yes, very nice. This was very painful to keep track of, but I got the same answer and was able to follow along the math.
$endgroup$
– Evan Kim
Dec 5 '18 at 22:20
1
$begingroup$
I reworked your answer in LaTeX (except the matrices, made with Excel), can you check and confirm it's ok?
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:27
$begingroup$
Great! thank you soo much! The red line is not correct there should be one for (-1)(lambda-3)(-3) and there is none for (-1)(lambda-3)(-3).
$endgroup$
– nafhgood
Dec 6 '18 at 2:17
|
show 1 more comment
$begingroup$
What is wrong with this solution? Here we line the two copy of the same matrix side by side and draw blue and red lines and add all products of numbers on the blue lines and substract products of numbers on the red lines. The solution is off by 4.
The determinant is thus
$$(lambda-1)(lambda-3)(lambda+1)\+(-3)(1)(-1)\+(-1)(1)(1)\-(-1)(lambda-3)(-3)\-(-1)(1)(lambda+1)\-(lambda-1)(1)(1)$$
That is
$$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
=(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
=(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
=(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
=(lambda-3)(lambda^2-1-3)+4\
=(lambda-3)(lambda^2-4)+4\
=(lambda-3)(lambda-2)(lambda+2)+4$$
$endgroup$
What is wrong with this solution? Here we line the two copy of the same matrix side by side and draw blue and red lines and add all products of numbers on the blue lines and substract products of numbers on the red lines. The solution is off by 4.
The determinant is thus
$$(lambda-1)(lambda-3)(lambda+1)\+(-3)(1)(-1)\+(-1)(1)(1)\-(-1)(lambda-3)(-3)\-(-1)(1)(lambda+1)\-(lambda-1)(1)(1)$$
That is
$$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
=(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
=(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
=(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
=(lambda-3)(lambda^2-1-3)+4\
=(lambda-3)(lambda^2-4)+4\
=(lambda-3)(lambda-2)(lambda+2)+4$$
edited Dec 5 '18 at 22:26
Jean-Claude Arbaut
14.7k63464
14.7k63464
answered Dec 5 '18 at 21:39
nafhgoodnafhgood
1,797422
1,797422
$begingroup$
You are using the rule of Sarrus, and your solution is actually correct. But you should write this in LaTeX.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 21:57
$begingroup$
Really nice, maybe it is difficult make the lines with MathJax! I don't know how to obtain that :(
$endgroup$
– gimusi
Dec 5 '18 at 22:07
$begingroup$
Yes, very nice. This was very painful to keep track of, but I got the same answer and was able to follow along the math.
$endgroup$
– Evan Kim
Dec 5 '18 at 22:20
1
$begingroup$
I reworked your answer in LaTeX (except the matrices, made with Excel), can you check and confirm it's ok?
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:27
$begingroup$
Great! thank you soo much! The red line is not correct there should be one for (-1)(lambda-3)(-3) and there is none for (-1)(lambda-3)(-3).
$endgroup$
– nafhgood
Dec 6 '18 at 2:17
|
show 1 more comment
$begingroup$
You are using the rule of Sarrus, and your solution is actually correct. But you should write this in LaTeX.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 21:57
$begingroup$
Really nice, maybe it is difficult make the lines with MathJax! I don't know how to obtain that :(
$endgroup$
– gimusi
Dec 5 '18 at 22:07
$begingroup$
Yes, very nice. This was very painful to keep track of, but I got the same answer and was able to follow along the math.
$endgroup$
– Evan Kim
Dec 5 '18 at 22:20
1
$begingroup$
I reworked your answer in LaTeX (except the matrices, made with Excel), can you check and confirm it's ok?
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:27
$begingroup$
Great! thank you soo much! The red line is not correct there should be one for (-1)(lambda-3)(-3) and there is none for (-1)(lambda-3)(-3).
$endgroup$
– nafhgood
Dec 6 '18 at 2:17
$begingroup$
You are using the rule of Sarrus, and your solution is actually correct. But you should write this in LaTeX.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 21:57
$begingroup$
You are using the rule of Sarrus, and your solution is actually correct. But you should write this in LaTeX.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 21:57
$begingroup$
Really nice, maybe it is difficult make the lines with MathJax! I don't know how to obtain that :(
$endgroup$
– gimusi
Dec 5 '18 at 22:07
$begingroup$
Really nice, maybe it is difficult make the lines with MathJax! I don't know how to obtain that :(
$endgroup$
– gimusi
Dec 5 '18 at 22:07
$begingroup$
Yes, very nice. This was very painful to keep track of, but I got the same answer and was able to follow along the math.
$endgroup$
– Evan Kim
Dec 5 '18 at 22:20
$begingroup$
Yes, very nice. This was very painful to keep track of, but I got the same answer and was able to follow along the math.
$endgroup$
– Evan Kim
Dec 5 '18 at 22:20
1
1
$begingroup$
I reworked your answer in LaTeX (except the matrices, made with Excel), can you check and confirm it's ok?
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:27
$begingroup$
I reworked your answer in LaTeX (except the matrices, made with Excel), can you check and confirm it's ok?
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:27
$begingroup$
Great! thank you soo much! The red line is not correct there should be one for (-1)(lambda-3)(-3) and there is none for (-1)(lambda-3)(-3).
$endgroup$
– nafhgood
Dec 6 '18 at 2:17
$begingroup$
Great! thank you soo much! The red line is not correct there should be one for (-1)(lambda-3)(-3) and there is none for (-1)(lambda-3)(-3).
$endgroup$
– nafhgood
Dec 6 '18 at 2:17
|
show 1 more comment
$begingroup$
$$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -4lambda +16$$
$endgroup$
$begingroup$
I am trying to find the characteristic polynomial of A. I just posted this part of the problem because I thought the characteristic polynomial was just finding the determinant. I think I did |lambda I - A| wrong...let me check
$endgroup$
– Evan Kim
Dec 5 '18 at 21:21
$begingroup$
@EvanKim Thanks for your kind inputs in your question post and here. Just to let you know, it's good to contain any questions on the initial question form. Regarding to your questions, the eigenvalues are given by the $lambda$s that nullify the determinant, simply meaning that you have to find the solutions to the equation $det(lambda I - A) = 0$. On the initial question, greedoid's answer follows exactly where you were mistaken.
$endgroup$
– Rebellos
Dec 5 '18 at 21:36
$begingroup$
so $λ^3−3λ^2−4λ+16$ is in fact $(λ−2)(λ+2)(λ−3)?$
$endgroup$
– Evan Kim
Dec 5 '18 at 21:47
$begingroup$
@EvanKim No, if only because the constant term of your expression after expansion is obviously $12$.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:07
add a comment |
$begingroup$
$$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -4lambda +16$$
$endgroup$
$begingroup$
I am trying to find the characteristic polynomial of A. I just posted this part of the problem because I thought the characteristic polynomial was just finding the determinant. I think I did |lambda I - A| wrong...let me check
$endgroup$
– Evan Kim
Dec 5 '18 at 21:21
$begingroup$
@EvanKim Thanks for your kind inputs in your question post and here. Just to let you know, it's good to contain any questions on the initial question form. Regarding to your questions, the eigenvalues are given by the $lambda$s that nullify the determinant, simply meaning that you have to find the solutions to the equation $det(lambda I - A) = 0$. On the initial question, greedoid's answer follows exactly where you were mistaken.
$endgroup$
– Rebellos
Dec 5 '18 at 21:36
$begingroup$
so $λ^3−3λ^2−4λ+16$ is in fact $(λ−2)(λ+2)(λ−3)?$
$endgroup$
– Evan Kim
Dec 5 '18 at 21:47
$begingroup$
@EvanKim No, if only because the constant term of your expression after expansion is obviously $12$.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:07
add a comment |
$begingroup$
$$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -4lambda +16$$
$endgroup$
$$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$
$$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$
$$=lambda ^3 - 3lambda^2 -4lambda +16$$
edited Dec 5 '18 at 21:23
answered Dec 5 '18 at 21:09
greedoidgreedoid
39.7k114798
39.7k114798
$begingroup$
I am trying to find the characteristic polynomial of A. I just posted this part of the problem because I thought the characteristic polynomial was just finding the determinant. I think I did |lambda I - A| wrong...let me check
$endgroup$
– Evan Kim
Dec 5 '18 at 21:21
$begingroup$
@EvanKim Thanks for your kind inputs in your question post and here. Just to let you know, it's good to contain any questions on the initial question form. Regarding to your questions, the eigenvalues are given by the $lambda$s that nullify the determinant, simply meaning that you have to find the solutions to the equation $det(lambda I - A) = 0$. On the initial question, greedoid's answer follows exactly where you were mistaken.
$endgroup$
– Rebellos
Dec 5 '18 at 21:36
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so $λ^3−3λ^2−4λ+16$ is in fact $(λ−2)(λ+2)(λ−3)?$
$endgroup$
– Evan Kim
Dec 5 '18 at 21:47
$begingroup$
@EvanKim No, if only because the constant term of your expression after expansion is obviously $12$.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:07
add a comment |
$begingroup$
I am trying to find the characteristic polynomial of A. I just posted this part of the problem because I thought the characteristic polynomial was just finding the determinant. I think I did |lambda I - A| wrong...let me check
$endgroup$
– Evan Kim
Dec 5 '18 at 21:21
$begingroup$
@EvanKim Thanks for your kind inputs in your question post and here. Just to let you know, it's good to contain any questions on the initial question form. Regarding to your questions, the eigenvalues are given by the $lambda$s that nullify the determinant, simply meaning that you have to find the solutions to the equation $det(lambda I - A) = 0$. On the initial question, greedoid's answer follows exactly where you were mistaken.
$endgroup$
– Rebellos
Dec 5 '18 at 21:36
$begingroup$
so $λ^3−3λ^2−4λ+16$ is in fact $(λ−2)(λ+2)(λ−3)?$
$endgroup$
– Evan Kim
Dec 5 '18 at 21:47
$begingroup$
@EvanKim No, if only because the constant term of your expression after expansion is obviously $12$.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:07
$begingroup$
I am trying to find the characteristic polynomial of A. I just posted this part of the problem because I thought the characteristic polynomial was just finding the determinant. I think I did |lambda I - A| wrong...let me check
$endgroup$
– Evan Kim
Dec 5 '18 at 21:21
$begingroup$
I am trying to find the characteristic polynomial of A. I just posted this part of the problem because I thought the characteristic polynomial was just finding the determinant. I think I did |lambda I - A| wrong...let me check
$endgroup$
– Evan Kim
Dec 5 '18 at 21:21
$begingroup$
@EvanKim Thanks for your kind inputs in your question post and here. Just to let you know, it's good to contain any questions on the initial question form. Regarding to your questions, the eigenvalues are given by the $lambda$s that nullify the determinant, simply meaning that you have to find the solutions to the equation $det(lambda I - A) = 0$. On the initial question, greedoid's answer follows exactly where you were mistaken.
$endgroup$
– Rebellos
Dec 5 '18 at 21:36
$begingroup$
@EvanKim Thanks for your kind inputs in your question post and here. Just to let you know, it's good to contain any questions on the initial question form. Regarding to your questions, the eigenvalues are given by the $lambda$s that nullify the determinant, simply meaning that you have to find the solutions to the equation $det(lambda I - A) = 0$. On the initial question, greedoid's answer follows exactly where you were mistaken.
$endgroup$
– Rebellos
Dec 5 '18 at 21:36
$begingroup$
so $λ^3−3λ^2−4λ+16$ is in fact $(λ−2)(λ+2)(λ−3)?$
$endgroup$
– Evan Kim
Dec 5 '18 at 21:47
$begingroup$
so $λ^3−3λ^2−4λ+16$ is in fact $(λ−2)(λ+2)(λ−3)?$
$endgroup$
– Evan Kim
Dec 5 '18 at 21:47
$begingroup$
@EvanKim No, if only because the constant term of your expression after expansion is obviously $12$.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:07
$begingroup$
@EvanKim No, if only because the constant term of your expression after expansion is obviously $12$.
$endgroup$
– Jean-Claude Arbaut
Dec 5 '18 at 22:07
add a comment |
$begingroup$
HINT
We can simplify a little bit summing the third to the first row
$$begin{vmatrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}= begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}$$
and using Laplace expansion in the first row to obtain
$$begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}
=(lambda -4) cdot begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix}
+lambdacdot begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}$$
Can you proceed from here?
$endgroup$
2
$begingroup$
Posts should be posted in final form. Comments should be limited to giving information relevant to the post. Discussions about votes are mostly off-topic. Excessive "+1 great posts" as a matter of fact can lead to suspensions.
$endgroup$
– quid♦
Dec 5 '18 at 21:52
2
$begingroup$
The text in the comment box explicitly says that these comments should be avoided and they are also discouraged in other ways. To do it sometimes will not cause issues, to do it all the time/often can lead to problems.
$endgroup$
– quid♦
Dec 5 '18 at 21:59
$begingroup$
Why the downvote for a correct answer?
$endgroup$
– gimusi
Dec 7 '18 at 13:02
add a comment |
$begingroup$
HINT
We can simplify a little bit summing the third to the first row
$$begin{vmatrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}= begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}$$
and using Laplace expansion in the first row to obtain
$$begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}
=(lambda -4) cdot begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix}
+lambdacdot begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}$$
Can you proceed from here?
$endgroup$
2
$begingroup$
Posts should be posted in final form. Comments should be limited to giving information relevant to the post. Discussions about votes are mostly off-topic. Excessive "+1 great posts" as a matter of fact can lead to suspensions.
$endgroup$
– quid♦
Dec 5 '18 at 21:52
2
$begingroup$
The text in the comment box explicitly says that these comments should be avoided and they are also discouraged in other ways. To do it sometimes will not cause issues, to do it all the time/often can lead to problems.
$endgroup$
– quid♦
Dec 5 '18 at 21:59
$begingroup$
Why the downvote for a correct answer?
$endgroup$
– gimusi
Dec 7 '18 at 13:02
add a comment |
$begingroup$
HINT
We can simplify a little bit summing the third to the first row
$$begin{vmatrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}= begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}$$
and using Laplace expansion in the first row to obtain
$$begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}
=(lambda -4) cdot begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix}
+lambdacdot begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}$$
Can you proceed from here?
$endgroup$
HINT
We can simplify a little bit summing the third to the first row
$$begin{vmatrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}= begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}$$
and using Laplace expansion in the first row to obtain
$$begin{vmatrix}
lambda -4 & 0 & lambda \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{vmatrix}
=(lambda -4) cdot begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix}
+lambdacdot begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}$$
Can you proceed from here?
edited Dec 5 '18 at 21:19
answered Dec 5 '18 at 21:11
gimusigimusi
92.8k84494
92.8k84494
2
$begingroup$
Posts should be posted in final form. Comments should be limited to giving information relevant to the post. Discussions about votes are mostly off-topic. Excessive "+1 great posts" as a matter of fact can lead to suspensions.
$endgroup$
– quid♦
Dec 5 '18 at 21:52
2
$begingroup$
The text in the comment box explicitly says that these comments should be avoided and they are also discouraged in other ways. To do it sometimes will not cause issues, to do it all the time/often can lead to problems.
$endgroup$
– quid♦
Dec 5 '18 at 21:59
$begingroup$
Why the downvote for a correct answer?
$endgroup$
– gimusi
Dec 7 '18 at 13:02
add a comment |
2
$begingroup$
Posts should be posted in final form. Comments should be limited to giving information relevant to the post. Discussions about votes are mostly off-topic. Excessive "+1 great posts" as a matter of fact can lead to suspensions.
$endgroup$
– quid♦
Dec 5 '18 at 21:52
2
$begingroup$
The text in the comment box explicitly says that these comments should be avoided and they are also discouraged in other ways. To do it sometimes will not cause issues, to do it all the time/often can lead to problems.
$endgroup$
– quid♦
Dec 5 '18 at 21:59
$begingroup$
Why the downvote for a correct answer?
$endgroup$
– gimusi
Dec 7 '18 at 13:02
2
2
$begingroup$
Posts should be posted in final form. Comments should be limited to giving information relevant to the post. Discussions about votes are mostly off-topic. Excessive "+1 great posts" as a matter of fact can lead to suspensions.
$endgroup$
– quid♦
Dec 5 '18 at 21:52
$begingroup$
Posts should be posted in final form. Comments should be limited to giving information relevant to the post. Discussions about votes are mostly off-topic. Excessive "+1 great posts" as a matter of fact can lead to suspensions.
$endgroup$
– quid♦
Dec 5 '18 at 21:52
2
2
$begingroup$
The text in the comment box explicitly says that these comments should be avoided and they are also discouraged in other ways. To do it sometimes will not cause issues, to do it all the time/often can lead to problems.
$endgroup$
– quid♦
Dec 5 '18 at 21:59
$begingroup$
The text in the comment box explicitly says that these comments should be avoided and they are also discouraged in other ways. To do it sometimes will not cause issues, to do it all the time/often can lead to problems.
$endgroup$
– quid♦
Dec 5 '18 at 21:59
$begingroup$
Why the downvote for a correct answer?
$endgroup$
– gimusi
Dec 7 '18 at 13:02
$begingroup$
Why the downvote for a correct answer?
$endgroup$
– gimusi
Dec 7 '18 at 13:02
add a comment |
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