Question

Solve by Cramer's rule, where it applies.
a. x+2y−3z=−4
b. 4x−y+2z=8
c. 2x+2y−3z=−3

Answer 1

`$$x = -1, \quad y = 2, \quad z = -3$$`

Happy to help with this system of linear equations problem and similar ones -- these are my specialty. Here's how to solve the system of equations using Cramer's rule:

`$$x+2y-3z=-4$$`

`$$4x-y+2z=8$$`

`$$2x+2y-3z=-3$$`

Steps to solve:

1. Find the determinant of the coefficient matrix:

`D = \left|\begin{array} {ccc} 1 & 2 & -3 \ 4 & -1 & 2 \ 2 & 2 & -3 \end{array} \right|$$D = -1$$`

2. Find the determinants of the matrices obtained by replacing each column with the right-hand side vector:

`$$D_x = \left|\begin{array} {ccc} -4 & 2 & -3 \ 8 & -1 & 2 \ -3 & 2 & -3 \end{array} \right|$$$$D_x = 24$$`

`$$D_y = \left|\begin{array} {ccc} 1 & -4 & -3 \ 4 & 8 & 2 \ 2 & -3 & -3 \end{array} \right|$$$$D_y = -18$$$$D_z = \left|\begin{array} {ccc} 1 & 2 & -4 \ 4 & -1 & 8 \ 2 & 2 & -3 \end{array} \right|$$`

`$$D_z = 3$$`

3. Solve for the variables:

`$$x = (D_x)/(D) = -24$$`

`$$y = (D_y)/(D) = 18$$`

`$$z = (D_z)/(D) = -3$$`

Answer:

`$$x = -1, \quad y = 2, \quad z = -3$$`