Cramer's Rule
Consider a system of n linear equations for n unknowns, represented in matrix multiplication form as follows:
Ax = b
where the n by n matrix A has a nonzero determinant, and the vector x = (x_1 ... x_n) is the column vector of the variables.
Then the theorem states that in this case the system has a unique solution, whose individual values for the unknowns are given by:
x_i = {det(A_i)}{det(A)} , i = 1... n .
where A_i is the matrix formed by replacing the ith column of A by the column vector b .
The column vecxtor b is: [-2, 5, 8].
We have A = |4 -2 9|, and det(A) = 4{4*2 - 1*(-3)} - (-2)*{3*2 - 1*1} + 9*{3*(-3) - 1*4}
|3 4 1| = 4*11 + 2*5 + 9*(-13)
|1 -3 2| = 44 + 10 - 117 = -63
i.e. det(A) = -63
And Ax = |-2 -2 9|, and det(Ax) = -2{4*2 - 1*(-3)} - (-2)*{5*2 - 8*1} + 9*{5*(-3) - 8*4}
|5 4 1| = -2*11 + 2*2 + 9*(-47)
|8 -3 2| = -22 + 4 - 423 = -441
i.e. det(Ax) = -441
And Ay = |4 -2 9|, and det(Ay) = 4{5*2 - 1*8} - (-2)*{3*2 - 1*1} + 9*{3*8 - 1*5}
|3 5 1| = 4*2 + 2*5 + 9*(19)
|1 8 2| = 8 + 10 + 171 = 189
i.e. det(Ay) = 189
And Az = |4 -2 -2|, and det(Az) = 4{4*8 - 5*(-3)} - (-2)*{3*8 - 5*1} + (-2)*{3*(-3) - 1*4}
|3 4 5| = 4*47 + 2*19 - 2*(-13)
|1 -3 8| = 188 + 38 + 26 = 252
i.e. det(Az) = 252
Using Cramer's Rule,
x = det(Ax)/det(A) = -441/-63 = 7
y = det(Ay)/det(A) = 189/-63 = -3
z = det(Az)/det(A) = 252/-63 = -4
Answer: the solution is :- x = 7, y = -3, z = -4