Question

solve the equation by creamers rule x+y=3,y+z=5 z+x=4​

Answer 1

Answer:

By creamer's rule, $x=1,\ y=2,\ z=3$

Step-by-step explanation:

For the equations

$x+y=3\\\\y+z=5\\\\z+x=4$,  write the coefficient matrix

$D=\left[\begin{array}{ccc}1&1&0\\0&1&1\\1&0&1\end{array}\right]$ ,the variable matrix $X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ and solution matrix$B=\left[\begin{array}{ccc}3\\5\\4\end{array}\right]$

consider the matrix $D_{1}$ with replacing the first column of $D$ by $B$ to find the value of $x$.

$D_{1}=\left[\begin{array}{ccc}3&1&0\\5&1&1\\4&0&1\end{array}\right]$,

$det\ D_{1} =3(1-0)-1(5-4)=3-1=2$

$D_{2} =\left[\begin{array}{ccc}1&3&0\\0&5&1\\1&4&1\end{array}\right]$

$det\ D_{2}= 1(5-4)-3(0-1)=1+3=4$

$D_{3}=\left[\begin{array}{ccc}1&1&3\\0&1&5\\1&0&4\end{array}\right]$

$det D_{3}= 1(4-0) -1(0-5)+3(0-1)=4+5-3=6$

and $det D=1(1-0)-1(0-1)=1+1=2$.

By creamer's rule,

$x=\frac{detD_{1} }{detD} , y=\frac{det D_{2} }{det D} , Z=\frac{det D_{3} }{det D}$

Therefore $x=1,\ y=2,\ z=3$.

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