# Use Cramer's rule to find the value of xx + 2y - 3z = -222x - 6y +2z = 0-5x +2y + z = 0 Use cramer's rule to find the value of x, would appreciate any assist on this.

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Given a system of equations:

`a_1x+b_1y+c_1z=d_1`

`a_2x+b_2y+c_2z=d_2`

`a_3x+b_3y+c_3z=d_3`

Cramer's Rule states that if there is a solution, then the solution can be found by:

`x=|[d_1,b_1,c_1],[d_2,b_2,c_2],[d_3,b_3,c_3]|/|[a_1,b_1,c_1],[a_2,b_2,c_2],[a_3,b_3,c_3]|`

where the straight bars indicate the determinant of the matrix.

The y value is found by replacing the column of the coefficients on y in the coefficient matrix by the column from the RHS, and the z value by replacing the column of the coefficients of z by the column from the RHS.

So `x=|[-22,2,-3],[0,-6,2],[0,2,1]|/|[1,2,-3],[2,-6,2],[-5,2,1]|=220/44=5`

** One method for finding the determinant is

`|[a,b,c],[d,e,f],[g,h,i]|=(aei+bfg+cdh)-( g e c+hfa+idb)`

or you can use expansion by minors **

`y=|[1,-22,-3],[2,0,2],[-5,0,1]|/|[1,2,-3],[2,-6,2],[-5,2,1]|=264/44=6`

`z=|[1,2,-22],[2,-6,0],[-5,2,0]|/|[1,2,-3],[2,-6,2],[-5,2,1]|=572/44=13`

**Thus the solution (x,y,z)=(5,6,13)**

If there is no solution, then the determinant of the coefficient matrix will be zero.