Determining $x_2$ in the solution of the system












0












$begingroup$


Task:
Determine $x_2$ in the solution of the system$$
begin{bmatrix}4&a&0\6&b&2\9&c&3end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}1\2\0end{bmatrix}$$when$$
left|begin{matrix}4&a&0\6&b&2\9&c&3end{matrix}right|=4$$
using Cramer's rule.



Options to choose from:




  1. $-4$

  2. $4$

  3. $6$

  4. $-2$


My answer:
$$begin{bmatrix}x_1\x_2\x_3end{bmatrix}×4=begin{bmatrix}1\2\0end{bmatrix}$$



I got $x_2=dfrac24$ which is not an option to choose. How do I do?






linear-algebra systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean determinant of matrix is $4$?
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:35










  • $begingroup$
    The information that I get from the task is that the 3x3 matrix is equal to 4
    $endgroup$
    – anders
    Dec 20 '18 at 9:37










  • $begingroup$
    The vertical lines are just another notion for the determinant of a matrix.
    $endgroup$
    – Student7
    Dec 20 '18 at 10:22
















0












$begingroup$


Task:
Determine $x_2$ in the solution of the system$$
begin{bmatrix}4&a&0\6&b&2\9&c&3end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}1\2\0end{bmatrix}$$when$$
left|begin{matrix}4&a&0\6&b&2\9&c&3end{matrix}right|=4$$
using Cramer's rule.



Options to choose from:




  1. $-4$

  2. $4$

  3. $6$

  4. $-2$


My answer:
$$begin{bmatrix}x_1\x_2\x_3end{bmatrix}×4=begin{bmatrix}1\2\0end{bmatrix}$$



I got $x_2=dfrac24$ which is not an option to choose. How do I do?






linear-algebra systems-of-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you mean determinant of matrix is $4$?
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:35










  • $begingroup$
    The information that I get from the task is that the 3x3 matrix is equal to 4
    $endgroup$
    – anders
    Dec 20 '18 at 9:37










  • $begingroup$
    The vertical lines are just another notion for the determinant of a matrix.
    $endgroup$
    – Student7
    Dec 20 '18 at 10:22














0












0








0





$begingroup$


Task:
Determine $x_2$ in the solution of the system$$
begin{bmatrix}4&a&0\6&b&2\9&c&3end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}1\2\0end{bmatrix}$$when$$
left|begin{matrix}4&a&0\6&b&2\9&c&3end{matrix}right|=4$$
using Cramer's rule.



Options to choose from:




  1. $-4$

  2. $4$

  3. $6$

  4. $-2$


My answer:
$$begin{bmatrix}x_1\x_2\x_3end{bmatrix}×4=begin{bmatrix}1\2\0end{bmatrix}$$



I got $x_2=dfrac24$ which is not an option to choose. How do I do?






linear-algebra systems-of-equations







share|cite|improve this question











$endgroup$




Task:
Determine $x_2$ in the solution of the system$$
begin{bmatrix}4&a&0\6&b&2\9&c&3end{bmatrix}begin{bmatrix}x_1\x_2\x_3end{bmatrix}=begin{bmatrix}1\2\0end{bmatrix}$$when$$
left|begin{matrix}4&a&0\6&b&2\9&c&3end{matrix}right|=4$$
using Cramer's rule.



Options to choose from:




  1. $-4$

  2. $4$

  3. $6$

  4. $-2$


My answer:
$$begin{bmatrix}x_1\x_2\x_3end{bmatrix}×4=begin{bmatrix}1\2\0end{bmatrix}$$



I got $x_2=dfrac24$ which is not an option to choose. How do I do?






linear-algebra systems-of-equations




linear-algebra systems-of-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 10:01









Yadati Kiran

1,7911619




1,7911619










asked Dec 20 '18 at 9:30









andersanders

615




615












  • $begingroup$
    Do you mean determinant of matrix is $4$?
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:35










  • $begingroup$
    The information that I get from the task is that the 3x3 matrix is equal to 4
    $endgroup$
    – anders
    Dec 20 '18 at 9:37










  • $begingroup$
    The vertical lines are just another notion for the determinant of a matrix.
    $endgroup$
    – Student7
    Dec 20 '18 at 10:22


















  • $begingroup$
    Do you mean determinant of matrix is $4$?
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:35










  • $begingroup$
    The information that I get from the task is that the 3x3 matrix is equal to 4
    $endgroup$
    – anders
    Dec 20 '18 at 9:37










  • $begingroup$
    The vertical lines are just another notion for the determinant of a matrix.
    $endgroup$
    – Student7
    Dec 20 '18 at 10:22
















$begingroup$
Do you mean determinant of matrix is $4$?
$endgroup$
– Yadati Kiran
Dec 20 '18 at 9:35




$begingroup$
Do you mean determinant of matrix is $4$?
$endgroup$
– Yadati Kiran
Dec 20 '18 at 9:35












$begingroup$
The information that I get from the task is that the 3x3 matrix is equal to 4
$endgroup$
– anders
Dec 20 '18 at 9:37




$begingroup$
The information that I get from the task is that the 3x3 matrix is equal to 4
$endgroup$
– anders
Dec 20 '18 at 9:37












$begingroup$
The vertical lines are just another notion for the determinant of a matrix.
$endgroup$
– Student7
Dec 20 '18 at 10:22




$begingroup$
The vertical lines are just another notion for the determinant of a matrix.
$endgroup$
– Student7
Dec 20 '18 at 10:22










1 Answer
1






active

oldest

votes


















1












$begingroup$

$underline{text{Cramer's rule:}}$ $x_i=dfrac{Delta_i}{Delta }$ where $Delta_i$ is the determinant of matrix $A$ with its $i^{th}$ column replaced by vector $b$ and $Delta$ is the determinant of the matrix $A$.




$$x_2=dfrac{left|begin{matrix}4 &1 &0\6 &2 &2\9 &0 &3end{matrix}right|}{4}=dfrac{24}{4}=6$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    what is the second column in A?
    $endgroup$
    – anders
    Dec 20 '18 at 9:41










  • $begingroup$
    Its $(a: b: c)^T$
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:41












  • $begingroup$
    what does the delta mean?
    $endgroup$
    – anders
    Dec 20 '18 at 9:43










  • $begingroup$
    Delta means determinant of matrix concerned.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:44










  • $begingroup$
    sorry for being stupid but I don't understand how I should do
    $endgroup$
    – anders
    Dec 20 '18 at 9:45











Your Answer





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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

$underline{text{Cramer's rule:}}$ $x_i=dfrac{Delta_i}{Delta }$ where $Delta_i$ is the determinant of matrix $A$ with its $i^{th}$ column replaced by vector $b$ and $Delta$ is the determinant of the matrix $A$.




$$x_2=dfrac{left|begin{matrix}4 &1 &0\6 &2 &2\9 &0 &3end{matrix}right|}{4}=dfrac{24}{4}=6$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    what is the second column in A?
    $endgroup$
    – anders
    Dec 20 '18 at 9:41










  • $begingroup$
    Its $(a: b: c)^T$
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:41












  • $begingroup$
    what does the delta mean?
    $endgroup$
    – anders
    Dec 20 '18 at 9:43










  • $begingroup$
    Delta means determinant of matrix concerned.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:44










  • $begingroup$
    sorry for being stupid but I don't understand how I should do
    $endgroup$
    – anders
    Dec 20 '18 at 9:45
















1












$begingroup$

$underline{text{Cramer's rule:}}$ $x_i=dfrac{Delta_i}{Delta }$ where $Delta_i$ is the determinant of matrix $A$ with its $i^{th}$ column replaced by vector $b$ and $Delta$ is the determinant of the matrix $A$.




$$x_2=dfrac{left|begin{matrix}4 &1 &0\6 &2 &2\9 &0 &3end{matrix}right|}{4}=dfrac{24}{4}=6$$







share|cite|improve this answer











$endgroup$













  • $begingroup$
    what is the second column in A?
    $endgroup$
    – anders
    Dec 20 '18 at 9:41










  • $begingroup$
    Its $(a: b: c)^T$
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:41












  • $begingroup$
    what does the delta mean?
    $endgroup$
    – anders
    Dec 20 '18 at 9:43










  • $begingroup$
    Delta means determinant of matrix concerned.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:44










  • $begingroup$
    sorry for being stupid but I don't understand how I should do
    $endgroup$
    – anders
    Dec 20 '18 at 9:45














1












1








1





$begingroup$

$underline{text{Cramer's rule:}}$ $x_i=dfrac{Delta_i}{Delta }$ where $Delta_i$ is the determinant of matrix $A$ with its $i^{th}$ column replaced by vector $b$ and $Delta$ is the determinant of the matrix $A$.




$$x_2=dfrac{left|begin{matrix}4 &1 &0\6 &2 &2\9 &0 &3end{matrix}right|}{4}=dfrac{24}{4}=6$$







share|cite|improve this answer











$endgroup$



$underline{text{Cramer's rule:}}$ $x_i=dfrac{Delta_i}{Delta }$ where $Delta_i$ is the determinant of matrix $A$ with its $i^{th}$ column replaced by vector $b$ and $Delta$ is the determinant of the matrix $A$.




$$x_2=dfrac{left|begin{matrix}4 &1 &0\6 &2 &2\9 &0 &3end{matrix}right|}{4}=dfrac{24}{4}=6$$








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 10:08

























answered Dec 20 '18 at 9:38









Yadati KiranYadati Kiran

1,7911619




1,7911619












  • $begingroup$
    what is the second column in A?
    $endgroup$
    – anders
    Dec 20 '18 at 9:41










  • $begingroup$
    Its $(a: b: c)^T$
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:41












  • $begingroup$
    what does the delta mean?
    $endgroup$
    – anders
    Dec 20 '18 at 9:43










  • $begingroup$
    Delta means determinant of matrix concerned.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:44










  • $begingroup$
    sorry for being stupid but I don't understand how I should do
    $endgroup$
    – anders
    Dec 20 '18 at 9:45


















  • $begingroup$
    what is the second column in A?
    $endgroup$
    – anders
    Dec 20 '18 at 9:41










  • $begingroup$
    Its $(a: b: c)^T$
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:41












  • $begingroup$
    what does the delta mean?
    $endgroup$
    – anders
    Dec 20 '18 at 9:43










  • $begingroup$
    Delta means determinant of matrix concerned.
    $endgroup$
    – Yadati Kiran
    Dec 20 '18 at 9:44










  • $begingroup$
    sorry for being stupid but I don't understand how I should do
    $endgroup$
    – anders
    Dec 20 '18 at 9:45
















$begingroup$
what is the second column in A?
$endgroup$
– anders
Dec 20 '18 at 9:41




$begingroup$
what is the second column in A?
$endgroup$
– anders
Dec 20 '18 at 9:41












$begingroup$
Its $(a: b: c)^T$
$endgroup$
– Yadati Kiran
Dec 20 '18 at 9:41






$begingroup$
Its $(a: b: c)^T$
$endgroup$
– Yadati Kiran
Dec 20 '18 at 9:41














$begingroup$
what does the delta mean?
$endgroup$
– anders
Dec 20 '18 at 9:43




$begingroup$
what does the delta mean?
$endgroup$
– anders
Dec 20 '18 at 9:43












$begingroup$
Delta means determinant of matrix concerned.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 9:44




$begingroup$
Delta means determinant of matrix concerned.
$endgroup$
– Yadati Kiran
Dec 20 '18 at 9:44












$begingroup$
sorry for being stupid but I don't understand how I should do
$endgroup$
– anders
Dec 20 '18 at 9:45




$begingroup$
sorry for being stupid but I don't understand how I should do
$endgroup$
– anders
Dec 20 '18 at 9:45


















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