1.Algebra Booster

Matrices and Determinants 7.59
Here D = (p – 2)(q – 3),
D 1
= (p – 2)(4q – 15)
D 2
= 0, D 3
= (p – 2)
Thus, the system of equations has infinitely many
15
solutions if p = 2, q = 3, . 4
30. Since the given system of equations are consistent, so
fi
fi
fi
fi
fi
2 3 -3
( c+ 2) ( c+ 4) -( c+ 6) = 0
2
( c+ 2)
2
( c+ 4)
2
-( c+
6)
2 0 -3
( c + 2) -2 -(c + 6) = 0
2
( c+ 2) - 4( c+ 5)
2
-(c+
6)
2 0 -3
- 2 ( c+ 2) 1 -( c+ 6) = 0
2
( c+ 2) 2( c+ 5)
2
-( c+
6)
(C 2
Æ C 2
+ C 3
)
2 0 -1
( c + 2) 1 - 4 = 0
2
( c+ 2) 2( c+ 5) - 8( c+
4)
(C 3
Æ C 3
+ C 1
)
0 0 -1
( c – 6) 1 - 4 = 0
2
( c -12c - 60) 2( c+ 5) - 8( c+
4)
( c – 6) 1
= 0
2
( c -12c - 60) 2( c+
5)
fi 2(c + 5)(c – 6) – (c 2 – 12c – 60) = 0
fi 2(c 2 – c – 30) – (c 2 – 12c – 60) = 0
fi c 2 + 10c = 0
fi c(c + 10) = 0
fi c = 0, c = –10
Here, c = 0 is not the solution
So c = –10
Solving, we get,
1 1
x=- , y = .
2 3
31. The system of equations has a solution, if
Here,
D = 0 = D 1
= D 2
= D 3
.
1 2 1
D = 1 3 4 = 0
1 5 10
(C 1
Æ C 2
+ 2C 3
)
1 2 1
D1
= k 3 4 = ( k -1)( k - 2)
2
k 5 10
1 1 1
D2
= 1 k 4 = ( k -1)( k - 2)
1
2
k 10
1 2 1
D3
= 1 3 k = ( k -1)( k - 2)
1 5
2
k
Thus, the system of equations has a solutions, if k = 1
and k = 2
When k = 1, x = 1 + 5t, y = –3t, t Œ R
When k = 2, x = –1 + 5t, y = 1 –3t, t Œ R
32. (i) The system of equations have a unique solution if
2 –3 5
3 1 l π 0
1 –7 8
fi 2(8 + 7l) + 3(24 – l) + 5(–21 – 1) = 0
fi 16 + 14l + 72 – 3l – 110 π 0
fi 11l – 22 π 0
fi l π 2
(ii) The system of equations has infinitely many solutions,
if D = 0 = D 1
= D 2
= D 3
We have,
and
2 –3 5
D = 3 1 l = 0fi l = 2
1 –7 8
12 –3 5
D1
= m 1 l = 0fi m = 7
17 –7 8
2 12 5
D2
= 3 m l = 0
1 17 8
2 –3 12
D3
= 3 1 m = 0fi m = 7
1 –7 17
Thus, the system of equations has infinitely many
solutions, if l = 2, m = 7.
(iii) The given system of equations has no solution if
D = 0 and any one of D 1
, D 2
, D 3
is non-zero.
Thus,
2 –3 5
D = 3 1 l = 0fi l = 2
1 –7 8