Question

solve the following equations using cramer's rule x+2y-z= -1, 3x+8y+2z= 28, 4x+9y-z= 14​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The answers are x = -2,  y = 3 and z = 5

Givan: x+2y-z= -1

           3x+8y+2z= 28

           4x+9y-z= 14​

To find: Solving using cramer's rules

Solution: Write the given equations in the form AX = B

⇒  $\left[\begin{array}{ccc}1&2&-1\\3&8&2\\4&9&-1\end{array}\right]$  × $\left[\begin{array}{c}x\\y\\z\end{array}\right]$ = $\left[\begin{array}{c}-1\\28\\14\end{array}\right]$  

⇒ D =  |A| = $\left|\begin{array}{ccc}1&2&-1\\3&8&2\\4&9&-1\end{array}\right|$

= 1(-8-18) - 2(-3-8) -1(27-32) = 1(-26) - 2 (-11) -1 (-5)

= -26 + 22 + 5 = 1

⇒ D = 1

Now find D$_{x}$, D$_{y}$ and D$_{z}$  

To find D$_{x}$ replace 1st row with X

D$_{x}$ = $\left|\begin{array}{ccc}-1&2&-1\\28&8&2\\14&9&-1\end{array}\right|$  = -1(-8 -18) - 2(-28 -28) -1(252 - 112)

=  -1(-26) -2 (-56) -1 (140)  = 26 + 112 -140 = - 2  

⇒ D$_{x}$ = -2

To find D$_{y}$ replace 2nd row with X

D$_{y}$ = $\left|\begin{array}{ccc}1&-1&-1\\3&28&2\\4&14&-1\end{array}\right|$  = 1(-28 -28) +1(-3 -8) -1 (42 - 112)

= - 56 - 11 + 70 = 3  

⇒ D$_{y}$ = 3

To find D$_{z}$ replace 3rd row with X

D$_{z}$ = $\left|\begin{array}{ccc}1&2&-1\\3&8&28\\4&9&14\end{array}\right|$ = 1 (112 - 252) - 2(42 - 112) -1(27 -32)

=  - 140 - 2(-70) +5 = - 140+140 + 5 = 5

⇒ D$_{z}$ = 5  

⇒  $x = \frac{D_{x} }{D }$            ⇒  $y = \frac{D_{y} }{D }$           ⇒ $z = \frac{D_{z} }{D}$  

⇒  $x = \frac{-2 }{1 }$             ⇒ $y = \frac{3}{1 }$              ⇒  $z= \frac{5 }{1 }$  

⇒  x = -2               ⇒  y = 3              ⇒ z = 5

Therefore,  x = -2,  y = 3 and z = 5

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