1. Find the value of h,k for which the system of equations x+y+z=6,x+2y+3z=10,x+2y+hz=k has a Unique solution

Solution:
Here `x+y+z=6`
`x+2y+3z=10`
`x+2y+hz=k`

`|D|`

=

`1`   `1`   `1`   `1`   `2`   `3`   `1`   `2`   `h`

=

`1`   ×    `2`   `3`   `2`   `h`

`-1`   ×    `1`   `3`   `1`   `h`

`+1`   ×    `1`   `2`   `1`   `2`

`=1 xx (2 × h - 3 × 2) -1 xx (1 × h - 3 × 1) +1 xx (1 × 2 - 2 × 1)`

`=1 xx (2h -6) -1 xx (h -3) +1 xx (2 -2)`

`=1 xx (2h-6) -1 xx (h-3) +1 xx (0)`

`= 2h-6 -h+3 +0`

`=h-3` `->(1)`


From `(1)`, we get

`=>h-3=0`

`=>h=3`


The system has unique solutions if `D!=0`, so `h!=3`

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2. Find the value of h,k for which the system of equations x+2y=3,5x+ky=-7 has a Unique solution

Solution:
Here `x+2y=3`
`5x+ky=-7`

Comparing `x+2y=3` with `a_1x+b_1y+c_1=0`

we get `a_1=1,b_1=2,c_1=-3`

Comparing `5x+ky=-7` with `a_2x+b_2y+c_2=0`

we get `a_2=5,b_2=k,c_2=7`

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

`(1)/(5)!=(2)/(k)`

`k!=10`


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

`(1)/(5)=(2)/(k)=(3)/(-7)`

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`

`(1)/(5)=(2)/(k)!=(3)/(-7)`

`(1)/(5)=(2)/(k)`

`k=10`

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