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 solutionSolution:Here `x+3y-2z=-1`
`2x+5y+z=2`
`2x+6y-az=b`
| `|D|` | = | | `1` | `3` | `-2` | | | `2` | `5` | `1` | | | `2` | `6` | `-a` | |
|
`=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) 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 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 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`
