Answer:
x = 2 , y = 0 , z = 3
Step-by-step explanation:
Cramer's rule is a rule through which we can find the solution of linear equation.
we have the three linear equations as
x+2y+3z=11
2x+y+2z=10
3x+2y+z=9
AX=B
A: coefficient matrix
X= unknown vectors(x,y,z)
D = values of the linear equation (11 , 10 , 9)
now we find the determinant of the given linear equation
determinant of the matrix will be
A = ![\left[\begin{array}{ccc}1&2&3\\2&1&2\\3&2&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C2%261%262%5C%5C3%262%261%5Cend%7Barray%7D%5Cright%5D)
= 1(1-4) - 2(2-6) + 3(4 - 3)
= 1(-3) - 2(-4) + 3(1)
= -3+8+3 = 8
also D
so the determinant is Non zero we can apply Cramer's rule
we will be replacing the first column of the coefficient matrix A with the values of D
by replacing the first column we will get the value of the variable 'x'
Dx = ![\left[\begin{array}{ccc}11&2&3\\10&1&2\\9&2&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%262%263%5C%5C10%261%262%5C%5C9%262%261%5Cend%7Barray%7D%5Cright%5D)
= 11(1-4) -2(10-18) + 3(20-9) = -33+16+33 = 16
x = 
= 
= 2
similarly
Dy = ![\left[\begin{array}{ccc}1&11&3\\2&10&2\\3&9&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2611%263%5C%5C2%2610%262%5C%5C3%269%261%5Cend%7Barray%7D%5Cright%5D)
= 1(10-18) -11(2-6) + 3(18 -30) = -8 +44 -36 = 0
y = 
= 0
Dz= ![\left[\begin{array}{ccc}1&2&11\\2&1&10\\3&2&9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%2611%5C%5C2%261%2610%5C%5C3%262%269%5Cend%7Barray%7D%5Cright%5D)
= 1(9 - 20) -2(18 - 30) + 11(4 -3) = -11 +24 +11 = 24
z = 
= 
so we have the solution as
x = 2 , y = 0 , z = 3
Therefore the solution for the given linear equations is (2,0,3).
