Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 6 Determinants Miscellaneous Exercise 6 Questions and Answers.
Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6
(ii) ≤ft|{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
{array}
Solution:
= 1(4 + 28) + 3(0 + 36) + 12(0 – 18)
= 1(32) + 3(36) + 12(-18)
= 32 + 108 – 216
= -76
(ii) ≤ft|{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
{array}=0
Solution:
≤ft|{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
{array}=0
∴ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0
∴ 1(-12) – 2x(-15) + 4x(-3) = 0
∴ -12 + 30x – 12x = 0
∴ 18x = 12
∴ x =
(ii) , ,
Solution:
Let = p, = q, = r
The given equations become
p + q + r = -2
p – 2q + r = 3
2p – q + 3r = -1
D = ≤ft|{array}{ccc}
1 & 1 & 1 \\
1 & -2 & 1 \\
2 & -1 & 3
{array}
= 1(-6 + 1) – 1(3 – 2) + 1(-1 + 4)
= -5 – 1 + 3
= -3
Dp= ≤ft|{array}{rrr}
-2 & 1 & 1 \\
3 & -2 & 1 \\
-1 & -1 & 3
{array}
= -2(-6 + 1) – 1(9 + 1) + 1(-3 – 2)
= 10 – 10 – 5
= -5
Dq= ≤ft|{array}{ccc}
1 & -2 & 1 \\
1 & 3 & 1 \\
2 & -1 & 3
{array}
= 1(9 + 1) + 2(3 – 2) + 1(-1 – 6)
= 10 + 2 – 7
= 5
Dr= ≤ft|{array}{rrr}
1 & 1 & -2 \\
1 & -2 & 3 \\
2 & -1 & -1
{array}
= 1(2 + 3) – 1(-1 – 6) – 2(-1 + 4)
= 5 + 7 – 6
= 6
By Cramer’s Rule,
p =
q = = 2
r = = 1
∴ x = , y = , z = are the solutions of the given equations.
(iii) x – y + 2z = 7, 3x + 4y – 5z = 5, 2x – y + 3z = 12
Solution:
Given equations are
x – y + 2z = 1
3x + 4y – 5z = 5
2x – y + 3z = 12
D = ≤ft|{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
{array}
= 1(12 – 5) – (-1)(9 + 10) + 2(-3 – 8)
= 1(7) + 1(19) + 2(-11)
= 7 + 19 – 22
= 4 ≠ 0
Dx= ≤ft|{array}{ccc}
7 & -1 & 2 \\
5 & 4 & -5 \\
12 & -1 & 3
{array}
= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48)
= 7(7)+ 1(75) +2(-53)
= 49 + 75 – 106
= 18
Dy= ≤ft|{array}{ccc}
1 & 7 & 2 \\
3 & 5 & -5 \\
2 & 12 & 3
{array}
= 1(15 + 60) – 7(9 + 10) + 2(36 – 10)
= 1(75) – 7(19) + 2(26)
= 75 – 133 + 52
= -6
Dz= ≤ft|{array}{ccc}
1 & -1 & 7 \\
3 & 4 & 5 \\
2 & -1 & 12
{array}
= 1(48 + 5) – (-1)(36 – 10) + 7(-3 – 8)
= 1(53)+ 1(26) + 7(-11)
= 53 + 26 – 77
= 2
By Cramer’s Rule,
x =
y =
z =
∴ x = , y = and z = are the solutions of the given equations.
(ii) kx + 3y + 4 = 0, x + ky + 3 = 0, 3x + 4y + 5 = 0
Solution:
Given equations are
kx + 3y + 4 = 0
x + ky + 3 = 0
3x + 4y + 5 = 0
Since, these equations are consistent.
∴ ≤ft|{array}{lll}
k & 3 & 4 \\
1 & k & 3 \\
3 & 4 & 5
{array}=0
∴ k(5k – 12) – 3(5 – 9) + 4(4 – 3k) = 0
∴ 5k2– 12k + 12 + 16 – 12k = 0
∴ 5k2– 24k + 28 = 0
∴ 5k2– 10k – 14k + 28 = 0
∴ 5k(k – 2) – 14(k – 2) = 0
∴ (k – 2) (5k – 14) = 0
∴ k – 2 = 0 or 5k – 14 = 0
∴ k = 2 or k =
(ii) P(3, 6), Q(-1, 3), R(2, -1)
Solution:
Here, P(x1, y1) ≡ P(3, 6), Q(x2, y2) ≡ Q(-1, 3), R(x3, y3) ≡ R(2, -1)
Area of a triangle = ≤ft|{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
{array}
A(∆PQR) = ≤ft|{array}{ccc}
3 & 6 & 1 \\
-1 & 3 & 1 \\
2 & -1 & 1
{array}
= [3(3 + 1) – 6(-1 – 2) + 1(1 – 6)]
= [3(4) – 6(-3) + 1(-5)]
= (12 + 18 – 5)
∴ A(∆PQR) = sq.units
(iii) L(1, 1), M(-2, 2), N(5, 4)
Solution:
Here, L(x1, y1) ≡ L(1, 1), M(x2, y2) ≡ M(-2, 2), N(x3, y3) ≡ N(5, 4)
Area of a triangle = ≤ft|{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
{array}
A(∆LMN) = ≤ft|{array}{rrr}
1 & 1 & 1 \\
-2 & 2 & 1 \\
5 & 4 & 1
{array}
= [1(2 – 4) -1(-2 – 5) + 1(-8 – 10)]
= [1(-2) – 1(-7) + 1(-18)]
= (-2 + 7 – 18)
=
Since, area cannot be negative.
∴ A(∆LMN) = sq.units
(ii) if area of ∆LMN is square units and vertices are L(3, -5), M(-2, k), N(1, 4).
Solution:
Here, L(x1, y1) ≡ L(3, -5), M(x2, y2) ≡ M(-2, k), N(x3, y3) ≡ N(1, 4)
A(∆LMN) = sq.units
Area of a triangle = ≤ft|{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
{array}
∴ ± =≤ft|{array}{ccc}
3 & -5 & 1 \\
-2 & k & 1 \\
1 & 4 & 1
{array}
∴ ± = [3(k – 4) – (-5) (-2 – 1) + 1(-8 – k)]
∴ ±33 = 3k – 12 – 15 – 8 – k
∴ 33 = 2k – 35
∴ 2k – 35 = 33 or 2k – 35 = -33
∴ 2k = 68 or 2k = 2
∴ k = 34 or k = 1