# Cramer's Rule

In the equation a1x + b1y = c1 and a2x + b2y = c2, according to Crammer's rule,
a) Write D in determinant form.
b) Write Dx in determinant form.
c) Write Dy in determinant form.
d) What is necessary condition for obtaining unique solution?

Given,
a1x + b1y = c1
a2x + b2y = c2
1. D = $\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}$
2. Dx = $\begin{vmatrix} c_1 & b_1 \\ c_2 & b_2 \end{vmatrix}$
3. DDy = $\begin{vmatrix} a_1 & c_1 \\ a_2 & c_2 \end{vmatrix}$
4. The necessary condition for obtaining unique solution is that the coefficient determinant should be non-zero.
i.e. $\begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix}$ ≠ 0

Solve the following system of linear equations using Cramer's rule:

2x - y = 3 and y + 3x = 7

Given,
2x - y = 3
y + 3x = 7

Now, From above equation,
D = $\begin{vmatrix} 2 & -1 \\ 3 & 1 \end{vmatrix}$ = 2 + 3 = 5
Dx = $\begin{vmatrix} 3 & -1 \\ 7 & 1 \end{vmatrix}$ = 3 + 7 = 10
Dy = $\begin{vmatrix} 2 & 3 \\ 3 & 7 \end{vmatrix}$ = 14 - 9 = 5

Finally,
x = $\frac{D_x}{D}$ = $\frac{10}{5}$ = 2
y = $\frac{D_y}{D}$ = $\frac{5}{5}$ = 1

Hence, x = 2 and y = 1.

y = 2x and x - $\frac 32$y + 1 = 0

Given, y - 2x or, 2x - y = 0 --- (i) x - $\frac 32$y = -1 --- (ii)

Now, From above equation,
D = $\begin{vmatrix} 2 & -1 \\ 1 & \frac{-3}{2} \end{vmatrix}$ = -3 + 1 = -2
Dx = $\begin{vmatrix} 0 & -1 \\ 1 & \frac{-3}{2} \end{vmatrix}$ = 0 - 1 = 1
Dy = $\begin{vmatrix} 2 & 0 \\ 1 & -1 \end{vmatrix}$ = -2

Finally,
x = $\frac{D_x}{D}$ = $\frac{1}{2}$
y = $\frac{D_y}{D}$ = 1

Hence, x = $\frac{1}{2}$ and y = 1.

$\frac 4x$ + $\frac 5y$ = 28 and $\frac 7x$ + $\frac 3y$ = 67

Given,
$\frac 4x$ + $\frac 5y$ = 28
$\frac 7x$ + $\frac 3y$ = 67

Now, From above equation,
D = $\begin{vmatrix} 4 & 5 \\ 7 & 3 \end{vmatrix}$ = 12 - 35 = -23
Dx = $\begin{vmatrix} 28 & 5 \\ 67 & 3 \end{vmatrix}$ = 84 - 335 = -251
Dy = $\begin{vmatrix} 4 & 28 \\ 7 & 67 \end{vmatrix}$ = 268 - 196 = 72

Finally,
$\frac 1x$ = $\frac{D_x}{D}$ = $\frac{-251}{-23}$x = $\frac{23}{251}$
$\frac 1y$ = $\frac{D_y}{D}$ = $\frac{72}{-23}$y = $\frac{-23}{72}$

Hence, x = $\frac{23}{251}$ and y = $\frac{-23}{72}$

4(x - 1) + 5(y + 2) = 10 and 5(x - 1) - 3(y + 2) + 6 = 0

Given, 4(x - 1) + 5(y + 2) = 10 or, 4x - 4 + 5y + 10 = 10
4x + 5y = 4 --- (i) 5(x - 1) - 3(y + 2) + 6 = 0 or, 8x - 5 - 3y - 6 + 6 = 0 5x - 3y = 5 --- (ii)

Now, From above equation,
D = $\begin{vmatrix} 4 & 5 \\ 5 & -3 \end{vmatrix}$ = -12 - 25 = -37
Dx = $\begin{vmatrix} 4 & 5 \\ 5 & -3 \end{vmatrix}$ = -12 - 25 = -37
Dy = $\begin{vmatrix} 4 & 5 \\ 5 & 4 \end{vmatrix}$ = 0

Finally,
x = $\frac{D_x}{D}$ = 1
y = $\frac{D_y}{D}$ = $\frac{0}{-37}$ = 0

Hence, x = 1 and y = 0.

3xy - 10y = 6x and 5xy + 3x = 21y

Given, 3xy - 10y = 6x or, 6x + 10y = 3xy
$\frac 6y$ + $\frac{10}{x}$ = 3 (dividing both sides by xy) --- (i) 5xy + 3x = 21y or, 21y - 3x = 5xy $\frac{21}{x}$ - $\frac 3y$ = 5 (dividing both sides by xy) --- (ii)

Now, From above equation,
D = $\begin{vmatrix} 10 & 6 \\ 21 & -3 \end{vmatrix}$ = -30 - 126 = -156
Dx = $\begin{vmatrix} 3 & 6 \\ 5 & -3 \end{vmatrix}$ = -9 - 30 = -39
Dy = $\begin{vmatrix} 10 & 3 \\ 21 & 5 \end{vmatrix}$ = 50 - 63 = -13

Finally,
$\frac 1x$ = $\frac{D_x}{D}$ = $\frac{-39}{-156}$ = $\frac 14$x = 4
$\frac 1y$ = $\frac{D_y}{D}$ = $\frac{-13}{-156}$ = $\frac{1}{12}$y = 12

Hence, x = 4 and y = 12.

3y + 4x = 2xy and 18y - 4x = 5xy

Given, 3y + 4x = 2xy or, $\frac 3x$ + $\frac 4y$ = 2 (dividing both sides by xy) --- (i) 18y - 4x = 5xy or, $\frac{18}{x}$ - $\frac 4y$ = 5 (dividing both sides by xy) --- (ii)

Now, From above equation,
D = $\begin{vmatrix} 3 & 4 \\ 18 & -4 \end{vmatrix}$ = -12 - 72 = -84
Dx = $\begin{vmatrix} 2 & 4 \\ 5 & -4 \end{vmatrix}$ = -8 - 20 = -28
Dy = $\begin{vmatrix} 3 & 2 \\ 18 & 5 \end{vmatrix}$ = 15 - 36 = -21

Finally,
$\frac 1x$ = $\frac{D_x}{D}$ = $\frac{-28}{-84}$ = $\frac 13$x = 3
$\frac 1y$ = $\frac{D_y}{D}$ = $\frac{-21}{-84}$ = $\frac 14$y = 4

Hence, x = 3 and y = 4.