Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(A)
I. Select the correct option from the given alternatives.
II. Answer the following questions.
(ii) ≤ft|{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
{array}
= 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
⇒ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0
⇒ 1(-12) – 2x(-15) + 4x(-3) = 0
⇒ -12 + 30x – 12x = 0
⇒ 18x = 12
⇒ x =
Dx = ≤ft|{array}{ccc}
1 & -1 & 1 \\
8 & 2 & 3 \\
1 & 1 & -4
{array}
= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2)
= 1(-11) + 1(-35) + 1(6)
= -11 – 35 + 6
= -40
Dy = ≤ft|{array}{ccc}
2 & 1 & 1 \\
1 & 8 & 3 \\
3 & 1 & -4
{array}
= 2(-32 – 3) -1(-4 – 9) + 1(1 – 24)
= 2(-35) – 1(-13) + 1(-23)
= -70 + 13 – 23
= -80
Dz = ≤ft|{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 8 \\
3 & 1 & 1
{array}
= 2(2 – 8) – (-1)(1 – 24) + 1(1 – 6)
= 2(-6) + 1(-23) + 1(-5)
= -12 – 23 – 5
= -40
By Cramer’s Rule,
∴ x = 1, y = 2 and z = 1 are the solutions of the given equations.
(ii) Let = p, = q, = r
∴ The given equations become
p + q =
i.e., 2p + 2q = 3
i.e., 2p + 2q + 0 = 3
q + r =
i.e., 6q + 6r = 5,
i.e., 0p + 6q + 6r = 5
r + p =
i.e., 3r + 3p = 4,
i.e., 3p + 0q + 3r = 4
D = ≤ft|{array}{lll}
2 & 2 & 0 \\
0 & 6 & 6 \\
3 & 0 & 3
{array}
= 2(18 – 0) -2(0 – 18) + 0
= 2(18) – 2(-18)
= 36 + 36
= 72 ≠ 0
Dp = ≤ft|{array}{lll}
3 & 2 & 0 \\
5 & 6 & 6 \\
4 & 0 & 3
{array}
= 3(18 – 0) – 2(15 – 24) + 0
= 3(18) – 2(-9)
= 54 + 18
= 72
Dq = ≤ft|{array}{lll}
2 & 3 & 0 \\
0 & 5 & 6 \\
3 & 4 & 3
{array}
= 2(15 – 24) – 3(0 – 18) + 0
= 2(-9) – 3(-18)
= -18 + 54
= 36
Dr = ≤ft|{array}{lll}
2 & 2 & 3 \\
0 & 6 & 5 \\
3 & 0 & 4
{array}
= 2(24 – 0) – 2(0 – 15) + 3(0 – 18)
= 2(24) – 2(-15) + 3(-18)
= 48 + 30 – 54
= 24
By Cramer’s Rule,
∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.
(iii) Given equations are
2x + 3y + 3z = 5
x – 2y + z = -4
3x – y – 2z = 3
D = ≤ft|{array}{ccc}
2 & 3 & 3 \\
1 & -2 & 1 \\
3 & -1 & -2
{array}
= 2(4 + 1) – 3(-2 – 3) + 3(-1 + 6)
= 2(5) – 3(-5) + 3(5)
= 10 + 15 + 15
= 40 ≠ 0
Dx = ≤ft|{array}{ccc}
5 & 3 & 3 \\
-4 & -2 & 1 \\
3 & -1 & -2
{array}
= 5(4 + 1) – 3(8 – 3) + 3(4 + 6)
= 5(5) – 3(5) + 3(10)
= 25 – 15 + 30
= 40
Dy = ≤ft|{array}{ccc}
2 & 5 & 3 \\
1 & -4 & 1 \\
3 & 3 & -2
{array}
= 2(8 – 3) – 5(-2 – 3) + 3(3 + 12)
= 2(5) – 5(-5) + 3(15)
= 10 + 25 + 45
= 80
Dz = ≤ft|{array}{ccc}
2 & 3 & 5 \\
1 & -2 & -4 \\
3 & -1 & 3
{array}
= 2(-6 – 4) – 3(3 + 12) + 5(-1 + 6)
= 2(-10) – 3(15) + 5(5)
= -20 -45 + 25
= -40
By Cramer’s Rule,
∴ x = 1, y = 2 and z = -1 are the solutions of the given equations.
(iv) Given equations are
x – y + 2z = 7
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 = and z = are the solutions of the given equations.
(ii) 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,
≤ft|{array}{rrr}
3 & 1 & -2 \\
k & 2 & -3 \\
2 & -1 & -3
{array}=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
(iii) Given equations are
(k – 2)x + (k – 1)y = 17
⇒ (k – 2)x + (k – 1)y – 17 = 0
(k – 1)x + (k – 2)y = 18
⇒ (k – 1)x + (k – 2)y – 18 = 0
x + y = 5
⇒ x + y – 5 = 0
Since these equations are consistent,
⇒ -1(-5k + 10 + 18) – 1(-5k + 5 + 18) + 1(k – 1 – k + 2) = 0
⇒ -1(-5k + 28) – 1(-5k + 23) + 1(1) = 0
⇒ 5k – 28 + 5k – 23 + 1 = 0
⇒ 10k – 50 = 0
⇒ k = 5
(ii) Here, P(x1, y1) = P(3, 6)
Q(x2, y2) = Q(-1, 3)
R(x3, y3) = R(2, -1)
A(ΔPQR) = sq.units
(iii) Here, L(x1, y1) = L(1, 1)
M(x2, y2) = M(-2, 2)
N(x3, y3) = N(5, 4)
Since area cannot be negative,
A(ΔLMN) = sq.units
(ii) Here, L(x1, y1) = L(3, -5), M(x2, y2) = M(-2, k), N(x3, y3) = N(1, 4)
A(ΔLMN) = 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
(ii) Here, P(x1, y1) = P(5, 1)
Q(x2, y2) = Q(1, -1)
R(x3, y3) = R(11, 4)
If A(ΔPQR) = 0, then the points P, Q, R are collinear.
∴ The points P, Q, R are collinear.
