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