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`

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2. Find the value of h,k for which the system of equations kx-y=2,6x-2y=3 has a No solution

Solution:
Here `kx-y=2`
`6x-2y=3`

Comparing `kx-y=2` with `a_1x+b_1y+c_1=0`

we get `a_1=k,b_1=-1,c_1=-2`

Comparing `6x-2y=3` with `a_2x+b_2y+c_2=0`

we get `a_2=6,b_2=-2,c_2=-3`

For a unique solution
`a_1/a_2!=b_1/b_2`

`(k)/(6)!=(-1)/(-2)`

`k!=3`


For infinite solutions
`a_1/a_2=b_1/b_2=c_1/c_2`

`(k)/(6)=(-1)/(-2)=(2)/(3)`

There is no such value of `k`, which will satisfy the equation


For no solutions
`a_1/a_2=b_1/b_2!=c_1/c_2`

`(k)/(6)=(-1)/(-2)!=(2)/(3)`

`(k)/(6)=(-1)/(-2)`

`k=3`

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