Finding determinant of $3times3$ matrix












3












$begingroup$


$$A = left(begin{matrix}
lambda - 1 & -1 & -1 \
1 & lambda - 3 & 1 \
-3 & 1 & lambda + 1 \
end{matrix}right)$$



$$det A = begin{vmatrix}Aend{vmatrix} =
(lambda - 1) begin{vmatrix}
lambda - 3 & 1 \
1 & lambda + 1 \
end{vmatrix} + 1begin{vmatrix}
1 & 1 \
-3 & lambda + 1 \
end{vmatrix} -
1begin{vmatrix}
1 & lambda - 3 \
-3 & 1 \
end{vmatrix}\
= ((lambda - 1)(lambda - 3)(lambda + 1)) - (1 + (lambda + 1) + 3) + (1 +3lambda - 9) \
= (lambda - 1)(lambda - 3)(lambda + 1) + (lambda + 1) + 2(lambda - 3)$$



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







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$endgroup$

















    3












    $begingroup$


    $$A = left(begin{matrix}
    lambda - 1 & -1 & -1 \
    1 & lambda - 3 & 1 \
    -3 & 1 & lambda + 1 \
    end{matrix}right)$$



    $$det A = begin{vmatrix}Aend{vmatrix} =
    (lambda - 1) begin{vmatrix}
    lambda - 3 & 1 \
    1 & lambda + 1 \
    end{vmatrix} + 1begin{vmatrix}
    1 & 1 \
    -3 & lambda + 1 \
    end{vmatrix} -
    1begin{vmatrix}
    1 & lambda - 3 \
    -3 & 1 \
    end{vmatrix}\
    = ((lambda - 1)(lambda - 3)(lambda + 1)) - (1 + (lambda + 1) + 3) + (1 +3lambda - 9) \
    = (lambda - 1)(lambda - 3)(lambda + 1) + (lambda + 1) + 2(lambda - 3)$$



    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







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      $$A = left(begin{matrix}
      lambda - 1 & -1 & -1 \
      1 & lambda - 3 & 1 \
      -3 & 1 & lambda + 1 \
      end{matrix}right)$$



      $$det A = begin{vmatrix}Aend{vmatrix} =
      (lambda - 1) begin{vmatrix}
      lambda - 3 & 1 \
      1 & lambda + 1 \
      end{vmatrix} + 1begin{vmatrix}
      1 & 1 \
      -3 & lambda + 1 \
      end{vmatrix} -
      1begin{vmatrix}
      1 & lambda - 3 \
      -3 & 1 \
      end{vmatrix}\
      = ((lambda - 1)(lambda - 3)(lambda + 1)) - (1 + (lambda + 1) + 3) + (1 +3lambda - 9) \
      = (lambda - 1)(lambda - 3)(lambda + 1) + (lambda + 1) + 2(lambda - 3)$$



      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







      share|cite|improve this question











      $endgroup$




      $$A = left(begin{matrix}
      lambda - 1 & -1 & -1 \
      1 & lambda - 3 & 1 \
      -3 & 1 & lambda + 1 \
      end{matrix}right)$$



      $$det A = begin{vmatrix}Aend{vmatrix} =
      (lambda - 1) begin{vmatrix}
      lambda - 3 & 1 \
      1 & lambda + 1 \
      end{vmatrix} + 1begin{vmatrix}
      1 & 1 \
      -3 & lambda + 1 \
      end{vmatrix} -
      1begin{vmatrix}
      1 & lambda - 3 \
      -3 & 1 \
      end{vmatrix}\
      = ((lambda - 1)(lambda - 3)(lambda + 1)) - (1 + (lambda + 1) + 3) + (1 +3lambda - 9) \
      = (lambda - 1)(lambda - 3)(lambda + 1) + (lambda + 1) + 2(lambda - 3)$$



      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      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






















          3 Answers
          3






          active

          oldest

          votes


















          3












          $begingroup$

          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.



          enter image description here



          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



          $$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
          =(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
          =(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
          =(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
          =(lambda-3)(lambda^2-1-3)+4\
          =(lambda-3)(lambda^2-4)+4\
          =(lambda-3)(lambda-2)(lambda+2)+4$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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



















          4












          $begingroup$

          $$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -4lambda +16$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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



















          1












          $begingroup$

          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?






          share|cite|improve this answer











          $endgroup$









          • 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











          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          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.



          enter image description here



          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



          $$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
          =(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
          =(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
          =(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
          =(lambda-3)(lambda^2-1-3)+4\
          =(lambda-3)(lambda^2-4)+4\
          =(lambda-3)(lambda-2)(lambda+2)+4$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          3












          $begingroup$

          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.



          enter image description here



          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



          $$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
          =(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
          =(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
          =(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
          =(lambda-3)(lambda^2-1-3)+4\
          =(lambda-3)(lambda^2-4)+4\
          =(lambda-3)(lambda-2)(lambda+2)+4$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          3












          3








          3





          $begingroup$

          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.



          enter image description here



          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



          $$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
          =(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
          =(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
          =(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
          =(lambda-3)(lambda^2-1-3)+4\
          =(lambda-3)(lambda^2-4)+4\
          =(lambda-3)(lambda-2)(lambda+2)+4$$






          share|cite|improve this answer











          $endgroup$



          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.



          enter image description here



          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



          $$(lambda-1)(lambda-3)(lambda+1)+3-1-3(lambda-3)+(lambda+1)-(lambda-1)\
          =(lambda-1)(lambda-3)(lambda+1)+2-3(lambda-3)+2\
          =(lambda-1)(lambda-3)(lambda+1)+4-3(lambda-3)\
          =(lambda-3)left[(lambda-1)(lambda+1)-3right]+4\
          =(lambda-3)(lambda^2-1-3)+4\
          =(lambda-3)(lambda^2-4)+4\
          =(lambda-3)(lambda-2)(lambda+2)+4$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 22:26









          Jean-Claude Arbaut

          14.7k63464




          14.7k63464










          answered Dec 5 '18 at 21:39









          nafhgoodnafhgood

          1,803422




          1,803422












          • $begingroup$
            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












          • $begingroup$
            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











          4












          $begingroup$

          $$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -4lambda +16$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          4












          $begingroup$

          $$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -4lambda +16$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          4












          4








          4





          $begingroup$

          $$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -4lambda +16$$






          share|cite|improve this answer











          $endgroup$



          $$=(lambda - 1)Big((lambda - 3)(lambda + 1)-1Big) + ((lambda + 1) + 3) -(1 +3lambda - 9) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) + (lambda + 4) -(3lambda - 8) $$



          $$=(lambda - 1)Big(lambda ^2 - 2lambda -4Big) - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -2lambda +4 - 2lambda +12 $$



          $$=lambda ^3 - 3lambda^2 -4lambda +16$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 21:23

























          answered Dec 5 '18 at 21:09









          greedoidgreedoid

          38.6k114797




          38.6k114797












          • $begingroup$
            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












          • $begingroup$
            @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











          1












          $begingroup$

          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?






          share|cite|improve this answer











          $endgroup$









          • 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
















          1












          $begingroup$

          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?






          share|cite|improve this answer











          $endgroup$









          • 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














          1












          1








          1





          $begingroup$

          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?






          share|cite|improve this answer











          $endgroup$



          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?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 21:19

























          answered Dec 5 '18 at 21:11









          gimusigimusi

          1




          1








          • 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




            $begingroup$
            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


















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