Question

Apply Cramer’s rule to solve the following system of equations:

2x x x 4 1 + 2 + 3 =
x x 2x 2 1 − 2 + 3 =
3x 2x x 0 1 − 2 − 3 =

Answer 1

Initial system is : 2x + y + z = 4

                             x - y + 2z = 2

                             3x - 2y - z = 0

$$x = \frac{det_x}{det} \quad y = \frac{det_y}{det} \quad z = \frac{det_z}{det}$$

$$det \begin{pmatrix} 2 & 1 & 1 \\ 1 & -1 & 2 \\ 3 & -2 & -1 \end{pmatrix} = 2 \begin{vmatrix} -1 & 2 \\ -2 & -1 \end{vmatrix} - 1\begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} + 1\begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix}$$

$$2( -1*-1 - 2 * -2) - 1(1*-1 - 2*3) + 1(1*-2 - (-1)*3) = 10 + 7 + 1 = 18$$

$$det_x = det \begin{pmatrix} 4 & 1 & 1 \\ 2 & -1 & 2 \\ 0 & -2 & -1 \end{pmatrix} = 4 \begin{vmatrix} -1 & 2 \\ -2 & -1 \end{vmatrix} - 1\begin{vmatrix} 2 & 2 \\ 0 & -1 \end{vmatrix} + 1\begin{vmatrix} 2 & -1 \\ 0 & -2 \end{vmatrix}$$

$$4(-1 *-1 - 2*-2) - 1(2*-1 - 2*0) + 1(2*-2 - (-1)*0) = 20 + 2 - 4 = 18$$

$$det_y = det \begin{pmatrix} 2 & 4 & 1 \\ 1 & 2 & 2 \\ 3 & 0 & -1 \end{pmatrix} = 2 \begin{vmatrix} 2 & 2 \\ 0 & -1 \end{vmatrix} - 4\begin{vmatrix} 1 & 2 \\ 3 & -1 \end{vmatrix} + 1\begin{vmatrix} 1 & 2 \\ 3 & 0 \end{vmatrix}$$

$$2(2*-1 - 2*0) - 4(1*-1 - 2*3) + 1(1*0 - 2*3) = -4 + 28 - 6 = 18$$

$$det_z = det \begin{pmatrix} 2 & 1 & 4 \\ 1 & -1 & 2 \\ 3 & -2 & 0 \end{pmatrix} = 2 \begin{vmatrix} -1 & 2 \\ -2 & 0 \end{vmatrix} - 1\begin{vmatrix} 1 & 2 \\ 3 & 0 \end{vmatrix} + 4\begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix}$$

$$2(-1*0 - 2*-2) - 1(1*0 - 2*3) + 4(1*-2 - (-1)*3) = 8 + 6 + 4 = 18$$

$$x = \frac{det_x}{det} \quad y = \frac{det_y}{det} \quad z = \frac{det_z}{det}$$

$$x = \frac{18}{18} \quad y = \frac{18}{18} = 1 \quad z= \frac{18}{18} = 1$$