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 is consistent

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)`

`|D_1|`

=

`6`   `1`   `1`   `10`   `2`   `3`   `k`   `2`   `h`

=

`6`   ×    `2`   `3`   `2`   `h`

`-1`   ×    `10`   `3`   `k`   `h`

`+1`   ×    `10`   `2`   `k`   `2`

`=6 xx (2 × h - 3 × 2) -1 xx (10 × h - 3 × k) +1 xx (10 × 2 - 2 × k)`

`=6 xx (2h -6) -1 xx (10h -3k) +1 xx (20 -2k)`

`=6 xx (2h-6) -1 xx (10h-3k) +1 xx (-2k+20)`

`= 12h-36 -10h+3k -2k+20`

`=2h+k-16` `->(2)`

`|D_2|`

=

`1`   `6`   `1`   `1`   `10`   `3`   `1`   `k`   `h`

=

`1`   ×    `10`   `3`   `k`   `h`

`-6`   ×    `1`   `3`   `1`   `h`

`+1`   ×    `1`   `10`   `1`   `k`

`=1 xx (10 × h - 3 × k) -6 xx (1 × h - 3 × 1) +1 xx (1 × k - 10 × 1)`

`=1 xx (10h -3k) -6 xx (h -3) +1 xx (k -10)`

`=1 xx (10h-3k) -6 xx (h-3) +1 xx (k-10)`

`= 10h-3k -6h+18 +k-10`

`=4h-2k+8` `->(3)`

`|D_3|`

=

`1`   `1`   `6`   `1`   `2`   `10`   `1`   `2`   `k`

=

`1`   ×    `2`   `10`   `2`   `k`

`-1`   ×    `1`   `10`   `1`   `k`

`+6`   ×    `1`   `2`   `1`   `2`

`=1 xx (2 × k - 10 × 2) -1 xx (1 × k - 10 × 1) +6 xx (1 × 2 - 2 × 1)`

`=1 xx (2k -20) -1 xx (k -10) +6 xx (2 -2)`

`=1 xx (2k-20) -1 xx (k-10) +6 xx (0)`

`= 2k-20 -k+10 +0`

`=k-10` `->(4)`


From `(1)`, we get

`=>h-3=0`

`=>h=3`


substitute `h=3` in equation `(2)`, we get

`=>k+2*3-16=0`

`=>k=-2*3+16`

`=>k=-6+16`

`=>k=10`


substitute `h=3` in equation `(3)`, we get

`=>-2k+4*3+8=0`

`=>-2k=-4*3-8`

`=>-2k=-12-8`

`=>-2k=-20`

`=>k=(-20)/(-2)`

`=>k=10`


From `(4)`, we get

`=>k-10=0`

`=>k=10`


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

The system has infinite solution if `D=D_1=D_2=D_3=0`, so `h=3` and `k=10`

System has no solution if `D=0` and at least one of `D_1,D_2,D_3` is nonzero, so `h=3` and `k!=10`

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2. Find the value of h,k for which the system of equations x+2y=3,5x+ky=-7 is consistent

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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