1. Determine whether the system of linear equations 2x+2y+z=5,x-y+z=1,3x+y+2z=4 is consistent
Solution:
Here `2x+2y+z=5`
`x-y+z=1`
`3x+y+2z=4`
| `|D|` | = | | `2` | `2` | `1` | | | `1` | `-1` | `1` | | | `3` | `1` | `2` | |
|
`=2 xx (-1 × 2 - 1 × 1) -2 xx (1 × 2 - 1 × 3) +1 xx (1 × 1 - (-1) × 3)`
`=2 xx (-2 -1) -2 xx (2 -3) +1 xx (1 +3)`
`=2 xx (-3) -2 xx (-1) +1 xx (4)`
`= -6 +2 +4`
`=0`
| `|D_1|` | = | | `5` | `2` | `1` | | | `1` | `-1` | `1` | | | `4` | `1` | `2` | |
|
`=5 xx (-1 × 2 - 1 × 1) -2 xx (1 × 2 - 1 × 4) +1 xx (1 × 1 - (-1) × 4)`
`=5 xx (-2 -1) -2 xx (2 -4) +1 xx (1 +4)`
`=5 xx (-3) -2 xx (-2) +1 xx (5)`
`= -15 +4 +5`
`=-6`
| `|D_2|` | = | | `2` | `5` | `1` | | | `1` | `1` | `1` | | | `3` | `4` | `2` | |
|
`=2 xx (1 × 2 - 1 × 4) -5 xx (1 × 2 - 1 × 3) +1 xx (1 × 4 - 1 × 3)`
`=2 xx (2 -4) -5 xx (2 -3) +1 xx (4 -3)`
`=2 xx (-2) -5 xx (-1) +1 xx (1)`
`= -4 +5 +1`
`=2`
| `|D_3|` | = | | `2` | `2` | `5` | | | `1` | `-1` | `1` | | | `3` | `1` | `4` | |
|
`=2 xx (-1 × 4 - 1 × 1) -2 xx (1 × 4 - 1 × 3) +5 xx (1 × 1 - (-1) × 3)`
`=2 xx (-4 -1) -2 xx (4 -3) +5 xx (1 +3)`
`=2 xx (-5) -2 xx (1) +5 xx (4)`
`= -10 -2 +20`
`=8`
Here `D=0,D_1=-6,D_2=2,D_3=8`
According to cramer's rule if D=0 at least any one D1,D2 and D3 is non-zero then the system of equation is inconsistent.
Hence, given system has no solution
2. Determine whether the system of linear equations 2x+5y=16,3x+y=11 is consistent
Solution:
Here `2x+5y=16`
`3x+y=11`
Comparing `2x+5y=16` with `a_1x+b_1y+c_1=0`
we get `a_1=2,b_1=5,c_1=-16`
Comparing `3x+y=11` with `a_2x+b_2y+c_2=0`
we get `a_2=3,b_2=1,c_2=-11`
`a_1/a_2=(2)/(3)`
`b_1/b_2=(5)/(1)=5`
`a_1/a_2!=b_1/b_2`
So the given system has a unique solution (System of equation is consistent)
This material is intended as a summary. Use your textbook for detail explanation.
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