# Maharashtra Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 6 Determinants Miscellaneous Exercise 6 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 6 Determinants Miscellaneous Exercise 6

Question 1.
Evaluate:
(i) $$\left|\begin{array}{ccc} 2 & -5 & 7 \\ 5 & 2 & 1 \\ 9 & 0 & 2 \end{array}\right|$$
Solution:

= 2(4 – 0) + 5(10 – 9) + 7(0 – 18)
= 2(4) + 5(1) + 7(-18)
= 8 + 5 – 126
= -113

(ii) $$\left|\begin{array}{ccc} 1 & -3 & 12 \\ 0 & 2 & -4 \\ 9 & 7 & 2 \end{array}\right|$$
Solution:

= 1(4 + 28) + 3(0 + 36) + 12(0 – 18)
= 1(32) + 3(36) + 12(-18)
= 32 + 108 – 216
= -76

Question 2.
Find the value(s) of x, if
(i) $$\left|\begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & -5 \\ 1 & 2 x & 5 x^{2} \end{array}\right|=0$$
Solution:
$$\left|\begin{array}{ccc} 1 & 4 & 20 \\ 1 & -2 & -5 \\ 1 & 2 x & 5 x^{2} \end{array}\right|=0$$
∴ 1(-10x2 + 10x) – 4(5x2 + 5) + 20(2x + 2) = 0
∴ -10x2 + 10x – 20x2 – 20 + 40x + 40 = 0
∴ -30x2 + 50x + 20 = 0
∴ 3x2 – 5x – 2 = 0 ……[Dividing throughout by (-10)]
∴ 3x2 – 6x + x – 2 = 0
∴ 3x(x – 2) + 1(x – 2) = 0
∴ (x – 2) (3x + 1) = 0
∴ x – 2 = 0 or 3x + 1 = 0
∴ x = 2 or x = $$-\frac{1}{3}$$

(ii) $$\left|\begin{array}{ccc} 1 & 2 x & 4 x \\ 1 & 4 & 16 \\ 1 & 1 & 1 \end{array}\right|=0$$
Solution:
$$\left|\begin{array}{ccc} 1 & 2 x & 4 x \\ 1 & 4 & 16 \\ 1 & 1 & 1 \end{array}\right|=0$$
∴ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0
∴ 1(-12) – 2x(-15) + 4x(-3) = 0
∴ -12 + 30x – 12x = 0
∴ 18x = 12
∴ x = $$\frac{2}{3}$$

Question 3.
By using properties of determinants, prove that $$\left|\begin{array}{ccc} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \end{array}\right|=0$$.
Solution:

Question 4.
Without expanding the determinants, show that

Solution:

Question 5.
Solve the following linear equations by Cramer’s Rule.
(i) 2x – y + z = 1, x + 2y + 3z = 8, 3x + y – 4z = 1
Solution:
Given equations are
2x – y + z = 1
x + 2y + 3z = 8
3x + y – 4z = 1
D = $$\left|\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & -4 \end{array}\right|$$
= 2(-8 – 3) – (-1)(-4 – 9) + 1(1 – 6)
= 2(-11) + 1(-13) + 1(-5)
= -22 – 13 – 5
= -40 ≠ 0
Dx = $$\left|\begin{array}{ccc} 1 & -1 & 1 \\ 8 & 2 & 3 \\ 1 & 1 & -4 \end{array}\right|$$
= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2)
= 1(-11) + 1(-35) + 1(6)
= -11 – 35 + 6
= -40
Dy = $$\left|\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 8 & 3 \\ 3 & 1 & -4 \end{array}\right|$$
= 2(-32 – 3) – 1(-4 – 9) + 1(1 – 24)
= 2(-35) – 1(-13) + 1(-23)
= -70 + 13 – 23
= -80
Dz = $$\left|\begin{array}{ccc} 2 & -1 & 1 \\ 1 & 2 & 8 \\ 3 & 1 & 1 \end{array}\right|$$
= 2(2 – 8) – (-1)(1 – 24) + 1(1 – 6)
= 2(-6) + 1(-23) + 1(-5)
= -12 – 23 – 5
= -40
By Cramer’s Rule,
x = $$\frac{D_{x}}{D}=\frac{-40}{-40}$$ = 1
y = $$\frac{\mathrm{D}_{y}}{\mathrm{D}}=\frac{-80}{-40}$$ = 2
z = $$\frac{\mathrm{D}_{z}}{\mathrm{D}}=\frac{-40}{-40}$$ = 1
∴ x = 1, y = 2 and z = 1 are the solutions of the given equations.

(ii) $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-2$$, $$\frac{1}{x}-\frac{2}{y}+\frac{1}{z}=3$$, $$\frac{2}{x}-\frac{1}{y}+\frac{3}{z}=-1$$
Solution:
Let $$\frac{1}{x}$$ = p, $$\frac{1}{y}$$ = q, $$\frac{1}{z}$$ = r
The given equations become
p + q + r = -2
p – 2q + r = 3
2p – q + 3r = -1
D = $$\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 2 & -1 & 3 \end{array}\right|$$
= 1(-6 + 1) – 1(3 – 2) + 1(-1 + 4)
= -5 – 1 + 3
= -3
Dp = $$\left|\begin{array}{rrr} -2 & 1 & 1 \\ 3 & -2 & 1 \\ -1 & -1 & 3 \end{array}\right|$$
= -2(-6 + 1) – 1(9 + 1) + 1(-3 – 2)
= 10 – 10 – 5
= -5
Dq = $$\left|\begin{array}{ccc} 1 & -2 & 1 \\ 1 & 3 & 1 \\ 2 & -1 & 3 \end{array}\right|$$
= 1(9 + 1) + 2(3 – 2) + 1(-1 – 6)
= 10 + 2 – 7
= 5
Dr = $$\left|\begin{array}{rrr} 1 & 1 & -2 \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{array}\right|$$
= 1(2 + 3) – 1(-1 – 6) – 2(-1 + 4)
= 5 + 7 – 6
= 6
By Cramer’s Rule,
p = $$\frac{\mathrm{D}_{\mathrm{p}}}{\mathrm{D}}=\frac{-5}{-3}=\frac{5}{3}$$
q = $$\frac{\mathrm{D}_{y}}{\mathrm{D}}=\frac{-80}{-40}$$ = 2
r = $$\frac{D_{2}}{D}=\frac{-40}{-40}$$ = 1
∴ x = $$\frac{3}{5}$$, y = $$\frac{-3}{5}$$, z = $$\frac{-1}{2}$$ 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 = $$\left|\begin{array}{ccc} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{array}\right|$$
= 1(12 – 5) – (-1)(9 + 10) + 2(-3 – 8)
= 1(7) + 1(19) + 2(-11)
= 7 + 19 – 22
= 4 ≠ 0
Dx = $$\left|\begin{array}{ccc} 7 & -1 & 2 \\ 5 & 4 & -5 \\ 12 & -1 & 3 \end{array}\right|$$
= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48)
= 7(7)+ 1(75) +2(-53)
= 49 + 75 – 106
= 18
Dy = $$\left|\begin{array}{ccc} 1 & 7 & 2 \\ 3 & 5 & -5 \\ 2 & 12 & 3 \end{array}\right|$$
= 1(15 + 60) – 7(9 + 10) + 2(36 – 10)
= 1(75) – 7(19) + 2(26)
= 75 – 133 + 52
= -6
Dz = $$\left|\begin{array}{ccc} 1 & -1 & 7 \\ 3 & 4 & 5 \\ 2 & -1 & 12 \end{array}\right|$$
= 1(48 + 5) – (-1)(36 – 10) + 7(-3 – 8)
= 1(53)+ 1(26) + 7(-11)
= 53 + 26 – 77
= 2
By Cramer’s Rule,
x = $$\frac{\mathrm{D}_{x}}{\mathrm{D}}=\frac{18}{4}=\frac{9}{2}$$
y = $$\frac{D_{y}}{D}=\frac{-6}{4}=\frac{-3}{2}$$
z = $$\frac{D_{z}}{D}=\frac{2}{4}=\frac{1}{2}$$
∴ x = $$\frac{9}{2}$$, y = $$\frac{-3}{2}$$ and z = $$\frac{1}{2}$$ are the solutions of the given equations.

Question 6.
Find the value(s) of k, if the following equations are consistent.
(i) 3x + y – 2 = 0, kx + 2y – 3 = 0 and 2x – y = 3
Solution:
Given equations are
3x + y – 2 = 0
kx + 2y – 3 = 0
2x – y = 3 i.e. 2x – y – 3 = 0
Since, these equations are consistent.
∴ $$\left|\begin{array}{rrr} 3 & 1 & -2 \\ k & 2 & -3 \\ 2 & -1 & -3 \end{array}\right|=0$$
∴ 3(-6 – 3) – 1(-3k + 6) – 2(-k – 4) = 0
∴ 3(-9) – 1 (-3k + 6) – 2(-k – 4) = 0
∴ -27 + 3k – 6 + 2k + 8 = 0
∴ 5k – 25 = 0
∴ k = 5

(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.
∴ $$\left|\begin{array}{lll} \mathrm{k} & 3 & 4 \\ 1 & \mathrm{k} & 3 \\ 3 & 4 & 5 \end{array}\right|=0$$
∴ 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 = $$\frac{14}{5}$$

Question 7.
Find the area of triangles whose vertices are
(i) A(-1, 2), B(2, 4), C(0, 0)
Solution:
Here, A(x1, y1) ≡ A(-1, 2), B(x2, y2) ≡ B(2, 4), C(x3, y3) ≡ C(0, 0)
Area of a triangle = $$\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|$$
∴ A(∆ABC) = $$\frac{1}{2}\left|\begin{array}{ccc} -1 & 2 & 1 \\ 2 & 4 & 1 \\ 0 & 0 & 1 \end{array}\right|$$
= $$\frac{1}{2}$$ [-1(4 – 0) – 2(2 – 0) + 1(0 – 0)]
= $$\frac{1}{2}$$ (-4 – 4)
= $$\frac{1}{2}$$ (-8)
= -4
Since, area cannot be negative.
∴ A(∆ABC) = 4 sq.units

(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 = $$\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|$$
A(∆PQR) = $$\frac{1}{2}\left|\begin{array}{ccc} 3 & 6 & 1 \\ -1 & 3 & 1 \\ 2 & -1 & 1 \end{array}\right|$$
= $$\frac{1}{2}$$ [3(3 + 1) – 6(-1 – 2) + 1(1 – 6)]
= $$\frac{1}{2}$$ [3(4) – 6(-3) + 1(-5)]
= $$\frac{1}{2}$$ (12 + 18 – 5)
∴ A(∆PQR) = $$\frac{25}{2}$$ 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 = $$\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|$$
A(∆LMN) = $$\frac{1}{2}\left|\begin{array}{rrr} 1 & 1 & 1 \\ -2 & 2 & 1 \\ 5 & 4 & 1 \end{array}\right|$$
= $$\frac{1}{2}$$ [1(2 – 4) -1(-2 – 5) + 1(-8 – 10)]
= $$\frac{1}{2}$$ [1(-2) – 1(-7) + 1(-18)]
= $$\frac{1}{2}$$ (-2 + 7 – 18)
= $$\frac{-13}{2}$$
Since, area cannot be negative.
∴ A(∆LMN) = $$\frac{13}{2}$$ sq.units

Question 8.
Find the value of k,
(i) if the area of ∆PQR is 4 square units and vertices are P(k, 0), Q(4, 0), R(0, 2).
Solution:
Here, P(x1, y1) ≡ P(k, 0), Q(x2, y2) ≡ Q(4, 0), R(x3, y3) ≡ R(0, 2)
A(∆PQR) = 4 sq.units
Area of a triangle = $$\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|$$
∴ ±4 = $$\frac{1}{2}\left|\begin{array}{lll} k & 0 & 1 \\ 4 & 0 & 1 \\ 0 & 2 & 1 \end{array}\right|$$
∴ ±4 = $$\frac{1}{2}$$ [k(0 – 2) – 0 + 1(8 – 0)]
∴ ±8 = -2k + 8
∴ 8 = -2k + 8 or -8 = -2k + 8
∴ -2k = 0 or 2k = 16
∴ k = 0 or k = 8

(ii) if area of ∆LMN is $$\frac{33}{2}$$ 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) = $$\frac{33}{2}$$ sq.units
Area of a triangle = $$\frac{1}{2}\left|\begin{array}{lll} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right|$$
∴ $$\pm \frac{33}{2}=\frac{1}{2}\left|\begin{array}{ccc} 3 & -5 & 1 \\ -2 & \mathrm{k} & 1 \\ 1 & 4 & 1 \end{array}\right|$$
∴ ±$$\frac{33}{2}$$ = $$\frac{1}{2}$$ [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