Question 669520
I'll try to "draw" the matrices. Hope you see it.
The determinant, D,  is
....... 2   -1    0
....... 3    1   -1
....... 0    1    1
Which evaluates to 7. Hope you know how to do this. Here's the arithmetic
(1)  (2*1*1 + (-1)*(-1)*0 + 0*3*1) - (0*1*0 + 2*1*(-1) + 3*(-1)*1) or
(2)  (2 + 0 + 0) - (0 -2 -3) or
(3) 2 + 5 = 7 or
(4) D = 7
To find x, replace the first column of D with the vector, V, of the right side of your system of equations, namely,
(5) V = (4 10 3)
The value of x is given by the determinate of that matrix, X, divided by D, where the determinant of X is
......  4   -1    0
...... 10    1   -1
......  3    1    1
Which evalaluates to 21. Use the same arithmetic as in (1) to (3) to get
(6) X = 21
Then the value of x is given by
(7) x = X/D or
(8) x = 21/7 or
(9) x = 3
Now do the same for y and z, except for y we replace the second column of D with V and the third column of D with V to get z.
You should get
(10) y = 2 and
(11) z = 1
Now place x, y, and z into to your original three equations to check the answers and get
Is (2*3 - 2 = 4)?
Is (6 - 2 = 4)?
Is (4 = 4)? Yes
Is (3*3 + 2 - 1 = 10)?
Is (9 + 1 = 10)?
Is (10 = 10) ? Yes
Is (2 + 1 = 3)?
Is (3 = 3)? Yes
Answer: Using Cramer's rule we calculate the values of x,y,z to be (3,2,1).