Question

Solve by using Cramer’s rule

3x-y+2z=8 ; x+y+z=2 ; 2x+y-z=-1​

Answer 1

$3x-y+2z=8 ; x+y+z=2 ; 2x+y-z=-1$  the value of $x,y,z$ is$1,-1,2.$

Step-by-step explanation:

Given:

$3x-y+2z=8 ; x+y+z=2 ; 2x+y-z=-1$

To find:

$x,y,z$ value is using Cramers' rule

Solution:

Write the system as a matrix equation $AX=b$

where,

$A=\left[\begin{array}{ccc}3&-1&2\\1&1&1\\2&1&-1\end{array}\right]$

$X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$

$b=\left[\begin{array}{ccc}8\\2\\-1\end{array}\right]$

Write down the main matrix and find its determinant,

$|A|=\left[\begin{array}{ccc}3&-1&2\\1&1&1\\2&1&-1\end{array}\right]$

$|A|=8(-1-1)+1(-2+1)+2(2+1)$

$|A|=-16-1+6$

$|A|=-11=$Δ

Replace the 1st column of the main matrix with the solution vector $b$ and its determinant,

Δ$_{1}$$=\left[\begin{array}{ccc}8&-1&2\\2&1&1\\-1&1&-1\end{array}\right]$

   $=8(-1-1)+1(2-1)+2(2+1)$

   $=-16-1+6$

Δ$_{1}$ $=-11$

Replace the 2nd column of the main matrix with the solution vector $b$ and its determinant,

Δ$_{2}$$=\left[\begin{array}{ccc}3&8&2\\1&2&1\\2&-1&-1\end{array}\right]$

   $=3(-2+1)-8(-1-2)+2(-1-4)$

   $=-3+24-10$

Δ$_{2}$ $=11$

Replace the 3rd column of the main matrix with the solution vector $b$ and its determinant,

Δ$_{3}$ $=\left[\begin{array}{ccc}3&-1&8\\1&1&2\\2&1&-1\end{array}\right]$

   $=3(-1-2)+1(-1-4)+8(1-2)$

   $=-9-5-8$

Δ$_{3}$$=-22$

Now,

$x=$ Δ$_{1}$/ Δ

$x=\frac{-11}{-11}$

$x=1$

$y=$ Δ$_{2}$/Δ

$y=\frac{11}{-11}$

$y=-1$

$z=$Δ$_{3}$/Δ

$z=\frac{-22}{-11}$

$z=2$

So the final value of $x,y,z$ is$1, -1, 2.$

   

Answer 2

Answer:

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Step-by-step explanation: