Question

Solve equations x+y+z=1,2x+2y+3z=6,x+4y+9z=3 by using cramers rule.
Answer is 7,-10,4

Answer 1

Answer:

Please see the attachment

Answer 2

x = 7

y = -10

z = 4

Step-by-step explanation:

The given system of equations,

$x+y+z=1\\2x+2y+3z=6\\x+4y+9z=3$

Using cramer's rule

$x=\dfrac{|A_x|}{|A|}\\ y=\dfrac{|A_y|}{|A|}\\ z=\dfrac{|A_z|}{|A|}$

where,

$A=\begin{vmatrix}1&1&1\\2&2&3\\1&4&9\end{vmatrix}$

$B=\begin{vmatrix}1\\6\\3\end{vmatrix}$

Now find determinant of |A|

$|A|=1(2\cdot9-3\cdot4)-1(2\cdot9-1\cdot3)+1(2\cdot4-1\cdot2)=-3$

• Now the determinant $A_x$ replace first column by B

$A_x=\begin{vmatrix}1&1&1\\6&2&3\\3&4&9\end{vmatrix}$

$|A_x|=1\cdot \:6-1\cdot \:45+1\cdot \:18=-21$

• Now the determinant $A_y$ replace second column by B in A

$A_y=\begin{vmatrix}1&1&1\\2&6&3\\1&3&9\end{vmatrix}$

$|A_y|=1\cdot \:45-1\cdot \:15+1\cdot \:0=30$

• Now the determinant $A_z$ replace third column by B in A

$A_z=\begin{vmatrix}1&1&1\\2&2&6\\1&4&3\end{vmatrix}$

$|A_z|=1\cdot \left(-18\right)-1\cdot \:0+1\cdot \:6=-12$

$\therefore x=\dfrac{|A_x|}{|A|}=\dfrac{-21}{-3}=7\\ \Rightarrow y=\dfrac{|A_y|}{|A|}=\dfrac{30}{-3}=-10\\ \Rightarrow z=\dfrac{|A_z|}{|A|}=\dfrac{-12}{-3}=4$

Hence, the values are 7, -10 and 4

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