Answer to Question #311229 in Linear Algebra for james

Question

D. Cramer’s Rule; 1. 6x+3y= 6; 5x-3y=3 2. 2x-3y=5; 4x+10y=4

Answer 1

1.

$\begin{cases} 6x+3y= 6 \\ 5x-3y=3 \end{cases}$

$\Delta = det\begin{vmatrix} 6 & 3 \\ 5 & -3 \end{vmatrix} = 6\cdot(-3) - 3 \cdot 5 = -33$

$\Delta_1 = det\begin{vmatrix} 6 & 3 \\ 3 & -3 \end{vmatrix} = 6 \cdot (-3) - 3 \cdot 3 = -27$

$\Delta_2 = det\begin{vmatrix} 6 & 6 \\ 5 & 3 \end{vmatrix} = 6 \cdot 3 - 6 \cdot 5 = -12$

$x = \frac{\Delta_1}{\Delta} = \frac{-27}{-33} =\frac{9}{11}$

$y = \frac{\Delta_2}{\Delta} = \frac{-12}{-33} =\frac{4}{11}$

2.

$\begin{cases} 2x-3y=5 \\ 4x+10y=4 \end{cases}$

$\Delta = det\begin{vmatrix} 2 & -3 \\ 4 & 10 \end{vmatrix} = 2\cdot 10 - (-3) \cdot 4 = 32$

$\Delta_1 = det\begin{vmatrix} 5 & -3 \\ 4 & 10 \end{vmatrix} = 5\cdot 10 - (-3) \cdot 4 = 62$

$\Delta_2 = det\begin{vmatrix} 2 & 5 \\ 4 & 4 \end{vmatrix} = 2\cdot 4 - 5 \cdot 4 = -12$

$x = \frac{\Delta_1}{\Delta} = \frac{62}{32} = \frac{31}{16}$

$y = \frac{\Delta_2}{\Delta} = \frac{-12}{32} = \frac{-3}{8}$