How to obtain these values if the complete solution is known? [closed]












0














Can you help me to understand the right steps to obtain the asked values?








linear-algebra







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closed as off-topic by Shailesh, Saad, max_zorn, Key Flex, Trevor Gunn Dec 1 at 4:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Saad, max_zorn, Key Flex, Trevor Gunn

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Welcome to MSE! It'll be helpful if you provide some description of what you've tried so far, so that people can clearly see your effort. This will also allow them to figure out where you are stuck so that they can tailor their advice.
    – platty
    Dec 1 at 2:16










  • Why not plug in a few values for $beta$ and see what you learn?
    – Michael Burr
    Dec 1 at 3:25










  • thanks for the help thats might be is a good idea
    – green life
    Dec 1 at 9:42


















0














Can you help me to understand the right steps to obtain the asked values?








linear-algebra







share|cite|improve this question















closed as off-topic by Shailesh, Saad, max_zorn, Key Flex, Trevor Gunn Dec 1 at 4:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Saad, max_zorn, Key Flex, Trevor Gunn

If this question can be reworded to fit the rules in the help center, please edit the question.













  • Welcome to MSE! It'll be helpful if you provide some description of what you've tried so far, so that people can clearly see your effort. This will also allow them to figure out where you are stuck so that they can tailor their advice.
    – platty
    Dec 1 at 2:16










  • Why not plug in a few values for $beta$ and see what you learn?
    – Michael Burr
    Dec 1 at 3:25










  • thanks for the help thats might be is a good idea
    – green life
    Dec 1 at 9:42
















0












0








0


1





Can you help me to understand the right steps to obtain the asked values?








linear-algebra







share|cite|improve this question















Can you help me to understand the right steps to obtain the asked values?








linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Dec 1 at 5:03









Saad

19.7k92252




19.7k92252










asked Dec 1 at 0:33









green life

156




156




closed as off-topic by Shailesh, Saad, max_zorn, Key Flex, Trevor Gunn Dec 1 at 4:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Saad, max_zorn, Key Flex, Trevor Gunn

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Shailesh, Saad, max_zorn, Key Flex, Trevor Gunn Dec 1 at 4:22


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Shailesh, Saad, max_zorn, Key Flex, Trevor Gunn

If this question can be reworded to fit the rules in the help center, please edit the question.












  • Welcome to MSE! It'll be helpful if you provide some description of what you've tried so far, so that people can clearly see your effort. This will also allow them to figure out where you are stuck so that they can tailor their advice.
    – platty
    Dec 1 at 2:16










  • Why not plug in a few values for $beta$ and see what you learn?
    – Michael Burr
    Dec 1 at 3:25










  • thanks for the help thats might be is a good idea
    – green life
    Dec 1 at 9:42




















  • Welcome to MSE! It'll be helpful if you provide some description of what you've tried so far, so that people can clearly see your effort. This will also allow them to figure out where you are stuck so that they can tailor their advice.
    – platty
    Dec 1 at 2:16










  • Why not plug in a few values for $beta$ and see what you learn?
    – Michael Burr
    Dec 1 at 3:25










  • thanks for the help thats might be is a good idea
    – green life
    Dec 1 at 9:42


















Welcome to MSE! It'll be helpful if you provide some description of what you've tried so far, so that people can clearly see your effort. This will also allow them to figure out where you are stuck so that they can tailor their advice.
– platty
Dec 1 at 2:16




Welcome to MSE! It'll be helpful if you provide some description of what you've tried so far, so that people can clearly see your effort. This will also allow them to figure out where you are stuck so that they can tailor their advice.
– platty
Dec 1 at 2:16












Why not plug in a few values for $beta$ and see what you learn?
– Michael Burr
Dec 1 at 3:25




Why not plug in a few values for $beta$ and see what you learn?
– Michael Burr
Dec 1 at 3:25












thanks for the help thats might be is a good idea
– green life
Dec 1 at 9:42






thanks for the help thats might be is a good idea
– green life
Dec 1 at 9:42












1 Answer
1






active

oldest

votes


















1














I am unable to find the values of $alpha_1,alpha_2$ but I have a relation between them involving $beta$. All notations are same as defined in the question.



Firstly, I will use row operation in the given system of equations, as follows,



$$begin{bmatrix}3 & 5 & -4\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1\b_2\-4end{bmatrix}$$
Clearly, $u=-1+4beta,v=2,w=3beta$



Applying $R_{12}(+1)$ we get,
$$begin{bmatrix}0 & 3 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1+b_2\b_2\-4end{bmatrix}$$
Now, $R_{1}(frac{1}{3})$,
$$begin{bmatrix}0 & 1 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2\-4end{bmatrix}$$
Now, $R_{21}(+2)$ and $R_{31}-1$ we get,
$$begin{bmatrix}0 & 1 & 0\ -3 & 0 & 4\ alpha_1&0 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2+frac{2}{3}(b_1+b_2)\-4-frac{b_1+b_2}{3}end{bmatrix}$$
Now by simple observation we get,
$$v=frac{b_1+b_2}{3}...(1)\-u+4w=frac{2b_1+5b_2}{3}...(2)\alpha_1u+alpha_2w=-4-frac{b_1+b_2}{3}...(3)$$
Recalling the solutions we get,
$$u=-1+4beta...(4)\v=2...(5)\w=3beta...(6)$$
(1) and (4) $implies b_1+b_2=6...(7)$



(2), (4) and (6) $implies 9=2b_1+5b_2...(8)$



(7) & (8) $implies b_1=7$ and $b_2=1$. Finally by (3) we get,
$$-alpha_1+4alpha_1beta+3alpha_2beta=-frac{19}{3}$$



That's all I got. If anything wrong please let me know.






share|cite|improve this answer





















  • thank you very much i appreciate your help
    – green life
    Dec 1 at 9:39










  • i will update you if anythings wrong but i need some time to check and try
    – green life
    Dec 1 at 9:52


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














I am unable to find the values of $alpha_1,alpha_2$ but I have a relation between them involving $beta$. All notations are same as defined in the question.



Firstly, I will use row operation in the given system of equations, as follows,



$$begin{bmatrix}3 & 5 & -4\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1\b_2\-4end{bmatrix}$$
Clearly, $u=-1+4beta,v=2,w=3beta$



Applying $R_{12}(+1)$ we get,
$$begin{bmatrix}0 & 3 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1+b_2\b_2\-4end{bmatrix}$$
Now, $R_{1}(frac{1}{3})$,
$$begin{bmatrix}0 & 1 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2\-4end{bmatrix}$$
Now, $R_{21}(+2)$ and $R_{31}-1$ we get,
$$begin{bmatrix}0 & 1 & 0\ -3 & 0 & 4\ alpha_1&0 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2+frac{2}{3}(b_1+b_2)\-4-frac{b_1+b_2}{3}end{bmatrix}$$
Now by simple observation we get,
$$v=frac{b_1+b_2}{3}...(1)\-u+4w=frac{2b_1+5b_2}{3}...(2)\alpha_1u+alpha_2w=-4-frac{b_1+b_2}{3}...(3)$$
Recalling the solutions we get,
$$u=-1+4beta...(4)\v=2...(5)\w=3beta...(6)$$
(1) and (4) $implies b_1+b_2=6...(7)$



(2), (4) and (6) $implies 9=2b_1+5b_2...(8)$



(7) & (8) $implies b_1=7$ and $b_2=1$. Finally by (3) we get,
$$-alpha_1+4alpha_1beta+3alpha_2beta=-frac{19}{3}$$



That's all I got. If anything wrong please let me know.






share|cite|improve this answer





















  • thank you very much i appreciate your help
    – green life
    Dec 1 at 9:39










  • i will update you if anythings wrong but i need some time to check and try
    – green life
    Dec 1 at 9:52
















1














I am unable to find the values of $alpha_1,alpha_2$ but I have a relation between them involving $beta$. All notations are same as defined in the question.



Firstly, I will use row operation in the given system of equations, as follows,



$$begin{bmatrix}3 & 5 & -4\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1\b_2\-4end{bmatrix}$$
Clearly, $u=-1+4beta,v=2,w=3beta$



Applying $R_{12}(+1)$ we get,
$$begin{bmatrix}0 & 3 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1+b_2\b_2\-4end{bmatrix}$$
Now, $R_{1}(frac{1}{3})$,
$$begin{bmatrix}0 & 1 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2\-4end{bmatrix}$$
Now, $R_{21}(+2)$ and $R_{31}-1$ we get,
$$begin{bmatrix}0 & 1 & 0\ -3 & 0 & 4\ alpha_1&0 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2+frac{2}{3}(b_1+b_2)\-4-frac{b_1+b_2}{3}end{bmatrix}$$
Now by simple observation we get,
$$v=frac{b_1+b_2}{3}...(1)\-u+4w=frac{2b_1+5b_2}{3}...(2)\alpha_1u+alpha_2w=-4-frac{b_1+b_2}{3}...(3)$$
Recalling the solutions we get,
$$u=-1+4beta...(4)\v=2...(5)\w=3beta...(6)$$
(1) and (4) $implies b_1+b_2=6...(7)$



(2), (4) and (6) $implies 9=2b_1+5b_2...(8)$



(7) & (8) $implies b_1=7$ and $b_2=1$. Finally by (3) we get,
$$-alpha_1+4alpha_1beta+3alpha_2beta=-frac{19}{3}$$



That's all I got. If anything wrong please let me know.






share|cite|improve this answer





















  • thank you very much i appreciate your help
    – green life
    Dec 1 at 9:39










  • i will update you if anythings wrong but i need some time to check and try
    – green life
    Dec 1 at 9:52














1












1








1






I am unable to find the values of $alpha_1,alpha_2$ but I have a relation between them involving $beta$. All notations are same as defined in the question.



Firstly, I will use row operation in the given system of equations, as follows,



$$begin{bmatrix}3 & 5 & -4\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1\b_2\-4end{bmatrix}$$
Clearly, $u=-1+4beta,v=2,w=3beta$



Applying $R_{12}(+1)$ we get,
$$begin{bmatrix}0 & 3 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1+b_2\b_2\-4end{bmatrix}$$
Now, $R_{1}(frac{1}{3})$,
$$begin{bmatrix}0 & 1 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2\-4end{bmatrix}$$
Now, $R_{21}(+2)$ and $R_{31}-1$ we get,
$$begin{bmatrix}0 & 1 & 0\ -3 & 0 & 4\ alpha_1&0 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2+frac{2}{3}(b_1+b_2)\-4-frac{b_1+b_2}{3}end{bmatrix}$$
Now by simple observation we get,
$$v=frac{b_1+b_2}{3}...(1)\-u+4w=frac{2b_1+5b_2}{3}...(2)\alpha_1u+alpha_2w=-4-frac{b_1+b_2}{3}...(3)$$
Recalling the solutions we get,
$$u=-1+4beta...(4)\v=2...(5)\w=3beta...(6)$$
(1) and (4) $implies b_1+b_2=6...(7)$



(2), (4) and (6) $implies 9=2b_1+5b_2...(8)$



(7) & (8) $implies b_1=7$ and $b_2=1$. Finally by (3) we get,
$$-alpha_1+4alpha_1beta+3alpha_2beta=-frac{19}{3}$$



That's all I got. If anything wrong please let me know.






share|cite|improve this answer












I am unable to find the values of $alpha_1,alpha_2$ but I have a relation between them involving $beta$. All notations are same as defined in the question.



Firstly, I will use row operation in the given system of equations, as follows,



$$begin{bmatrix}3 & 5 & -4\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1\b_2\-4end{bmatrix}$$
Clearly, $u=-1+4beta,v=2,w=3beta$



Applying $R_{12}(+1)$ we get,
$$begin{bmatrix}0 & 3 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}b_1+b_2\b_2\-4end{bmatrix}$$
Now, $R_{1}(frac{1}{3})$,
$$begin{bmatrix}0 & 1 & 0\ -3 & -2 & 4\ alpha_1&1 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2\-4end{bmatrix}$$
Now, $R_{21}(+2)$ and $R_{31}-1$ we get,
$$begin{bmatrix}0 & 1 & 0\ -3 & 0 & 4\ alpha_1&0 & alpha_2end{bmatrix}begin{bmatrix}u\v\wend{bmatrix}=begin{bmatrix}frac{b_1+b_2}{3}\b_2+frac{2}{3}(b_1+b_2)\-4-frac{b_1+b_2}{3}end{bmatrix}$$
Now by simple observation we get,
$$v=frac{b_1+b_2}{3}...(1)\-u+4w=frac{2b_1+5b_2}{3}...(2)\alpha_1u+alpha_2w=-4-frac{b_1+b_2}{3}...(3)$$
Recalling the solutions we get,
$$u=-1+4beta...(4)\v=2...(5)\w=3beta...(6)$$
(1) and (4) $implies b_1+b_2=6...(7)$



(2), (4) and (6) $implies 9=2b_1+5b_2...(8)$



(7) & (8) $implies b_1=7$ and $b_2=1$. Finally by (3) we get,
$$-alpha_1+4alpha_1beta+3alpha_2beta=-frac{19}{3}$$



That's all I got. If anything wrong please let me know.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 1 at 4:21









Sujit Bhattacharyya

945317




945317












  • thank you very much i appreciate your help
    – green life
    Dec 1 at 9:39










  • i will update you if anythings wrong but i need some time to check and try
    – green life
    Dec 1 at 9:52


















  • thank you very much i appreciate your help
    – green life
    Dec 1 at 9:39










  • i will update you if anythings wrong but i need some time to check and try
    – green life
    Dec 1 at 9:52
















thank you very much i appreciate your help
– green life
Dec 1 at 9:39




thank you very much i appreciate your help
– green life
Dec 1 at 9:39












i will update you if anythings wrong but i need some time to check and try
– green life
Dec 1 at 9:52




i will update you if anythings wrong but i need some time to check and try
– green life
Dec 1 at 9:52



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