Cramer's Rule Method Examples

1. Solve linear equations x+y=2 and 2x+3y=4 using Cramer's Rule Method

Solution:
The equations can be expressed as

x+y-2=0 $x+y-2=0$

2x+3y-4=0 $2x+3y-4=0$

Use Cramer's Rule to find the values of x, y, z.

xDx=-yDy=1D $(x)/D_x=(-y)/D_y=(1)/D$

Dx $D_x$

	1 $1$	-2 $-2$
	3 $3$	-4 $-4$

=1×(-4)-(-2)×3 $=1 × (-4) - (-2) × 3$

=-4+6 $=-4 +6$

=2 $=2$

Dy $D_y$

	1 $1$	-2 $-2$
	2 $2$	-4 $-4$

=1×(-4)-(-2)×2 $=1 × (-4) - (-2) × 2$

=-4+4 $=-4 +4$

=0 $=0$

D $D$

	1 $1$	1 $1$
	2 $2$	3 $3$

=1×3-1×2 $=1 × 3 - 1 × 2$

=3-2 $=3 -2$

=1 $=1$

xDx=-yDy=1D $(x)/D_x=(-y)/D_y=(1)/D$

$:.(x)/2=(-y)/0=(1)/1$

$:.(x)/2=(1)/1,(-y)/0=(1)/1$

$:.x=(2)/(1),y=(0)/(1)$

$:.x=2,y=0$

2. Solve linear equations 2x+7y-11=0 and 3x-y-5=0 using Cramer's Rule Method

Solution:
The equations can be expressed as

$2x+7y-11=0$

$3x-y-5=0$

Use Cramer's Rule to find the values of x, y, z.

$(x)/D_x=(-y)/D_y=(1)/D$

$D_x$

	$7$	$-11$
	$-1$	$-5$

$=7 × (-5) - (-11) × (-1)$

$=-35 -11$

$=-46$

$D_y$

	$2$	$-11$
	$3$	$-5$

$=2 × (-5) - (-11) × 3$

$=-10 +33$

$=23$

$D$

	$2$	$7$
	$3$	$-1$

$=2 × (-1) - 7 × 3$

$=-2 -21$

$=-23$

$(x)/D_x=(-y)/D_y=(1)/D$

$:.(x)/-46=(-y)/23=(1)/-23$

$:.(x)/-46=(1)/-23,(-y)/23=(1)/-23$

$:.x=(-46)/(-23),y=(-23)/(-23)$

$:.x=2,y=1$

3. Solve linear equations 3x-y=3 and 7x+2y=20 using Cramer's Rule Method

Solution:
The equations can be expressed as

$3x-y-3=0$

$7x+2y-20=0$

Use Cramer's Rule to find the values of x, y, z.

$(x)/D_x=(-y)/D_y=(1)/D$

$D_x$

	$-1$	$-3$
	$2$	$-20$

$=-1 × (-20) - (-3) × 2$

$=20 +6$

$=26$

$D_y$

	$3$	$-3$
	$7$	$-20$

$=3 × (-20) - (-3) × 7$

$=-60 +21$

$=-39$

$D$

	$3$	$-1$
	$7$	$2$

$=3 × 2 - (-1) × 7$

$=6 +7$

$=13$

$(x)/D_x=(-y)/D_y=(1)/D$

$:.(x)/26=(-y)/-39=(1)/13$

$:.(x)/26=(1)/13,(-y)/-39=(1)/13$

$:.x=(26)/(13),y=(39)/(13)$

$:.x=2,y=3$

4. Solve linear equations 2x-y=11 and 5x+4y=1 using Cramer's Rule Method

Solution:
The equations can be expressed as

$2x-y-11=0$

$5x+4y-1=0$

Use Cramer's Rule to find the values of x, y, z.

$(x)/D_x=(-y)/D_y=(1)/D$

$D_x$

	$-1$	$-11$
	$4$	$-1$

$=-1 × (-1) - (-11) × 4$

$=1 +44$

$=45$

$D_y$

	$2$	$-11$
	$5$	$-1$

$=2 × (-1) - (-11) × 5$

$=-2 +55$

$=53$

$D$

	$2$	$-1$
	$5$	$4$

$=2 × 4 - (-1) × 5$

$=8 +5$

$=13$

$(x)/D_x=(-y)/D_y=(1)/D$

$:.(x)/45=(-y)/53=(1)/13$

$:.(x)/45=(1)/13,(-y)/53=(1)/13$

$:.x=(45)/(13),y=(-53)/(13)$

$:.x=45/13,y=-53/13$

This material is intended as a summary. Use your textbook for detail explanation.
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1. Examples