Find the value of h,k for which the system of equations has a Unique solution calculator

SolutionHelp

$7x-7y+6z=-4,-8x+7y+3z=6,-37x+35y+hz=k$

$x+y+z=6,x+2y+3z=10,x+2y+hz=k$

$2x+y+z=5,3x+5y+2z=15,Ax+y+4z=8$

$x+2y=3,5x+ky=-7$

$2x+3y=5,4x+ky=10$

$3x+5y=0,kx+10y=0$

Example

1. Find the value of h,k for which the system of equations x+3y-2z=-1,2x+5y+z=2,2x+6y-az=b has No solution

Solution:
Here

$x+3y-2z=-1$

$2x+5y+z=2$

$2x+6y-az=b$

$|D|$

$1$	$3$	$-2$
$2$	$5$	$1$
$2$	$6$	$-a$

$1$

	$5$	$1$
	$6$	$-a$

$-3$

	$2$	$1$
	$2$	$-a$

$-2$

	$2$	$5$
	$2$	$6$

$=1 xx (5 × (-a) - 1 × 6) -3 xx (2 × (-a) - 1 × 2) -2 xx (2 × 6 - 5 × 2)$

$=1 xx (-5a -6) -3 xx (-2a -2) -2 xx (12 -10)$

$=1 xx (-5a-6) -3 xx (-2a-2) -2 xx (2)$

$= -5a-6 +6a+6 -4$

$=a-4$

$->(1)$

$|D_1|$

$-1$	$3$	$-2$
$2$	$5$	$1$
$b$	$6$	$-a$

$-1$

	$5$	$1$
	$6$	$-a$

$-3$

	$2$	$1$
	$b$	$-a$

$-2$

	$2$	$5$
	$b$	$6$

$=(-1) xx (5 × (-a) - 1 × 6) -3 xx (2 × (-a) - 1 × b) -2 xx (2 × 6 - 5 × b)$

$=(-1) xx (-5a -6) -3 xx (-2a -b) -2 xx (12 -5b)$

$=(-1) xx (-5a-6) -3 xx (-2a-b) -2 xx (-5b+12)$

$= 5a+6 +6a+3b +10b-24$

$=11a+13b-18$

$->(2)$

$|D_2|$

$1$	$-1$	$-2$
$2$	$2$	$1$
$2$	$b$	$-a$

$1$

	$2$	$1$
	$b$	$-a$

$+1$

	$2$	$1$
	$2$	$-a$

$-2$

	$2$	$2$
	$2$	$b$

$=1 xx (2 × (-a) - 1 × b) +1 xx (2 × (-a) - 1 × 2) -2 xx (2 × b - 2 × 2)$

$=1 xx (-2a -b) +1 xx (-2a -2) -2 xx (2b -4)$

$=1 xx (-2a-b) +1 xx (-2a-2) -2 xx (2b-4)$

$= -2a-b -2a-2 -4b+8$

$=-4a-5b+6$

$->(3)$

$|D_3|$

$1$	$3$	$-1$
$2$	$5$	$2$
$2$	$6$	$b$

$1$

	$5$	$2$
	$6$	$b$

$-3$

	$2$	$2$
	$2$	$b$

$-1$

	$2$	$5$
	$2$	$6$

$=1 xx (5 × b - 2 × 6) -3 xx (2 × b - 2 × 2) -1 xx (2 × 6 - 5 × 2)$

$=1 xx (5b -12) -3 xx (2b -4) -1 xx (12 -10)$

$=1 xx (5b-12) -3 xx (2b-4) -1 xx (2)$

$= 5b-12 -6b+12 -2$

$=-b-2$

$->(4)$

From

$(1)$ , we get

$=>a-4=0$

$=>a=4$

substitute

$a=4$ in equation

$(2)$ , we get

$=>13b+11*4-18=0$

$=>13b=-11*4+18$

$=>13b=-44+18$

$=>13b=-26$

$=>b=(-26)/13$

$=>b=-2$

substitute

$a=4$ in equation

$(3)$ , we get

$=>-5b-4*4+6=0$

$=>-5b=4*4-6$

$=>-5b=16-6$

$=>-5b=10$

$=>b=10/(-5)$

$=>b=-2$

From

$(4)$ , we get

$=>-b-2=0$

$=>-b=2$

$=>b=-2$

The system has unique solutions if

$D!=0$ , so

$a!=4$

The system has infinite solution if

$D=D_1=D_2=D_3=0$ , so

$a=4$ and

$b=-2$

System has no solution if

$D=0$ and at least one of

$D_1,D_2,D_3$ is nonzero, so

$a=4$ and

$b!=-2$