Latest Maharashtra State Board (SSC & HSC) 2026-27 Syllabus Digest & Solutions Updated!
MHBoardSolutionsSSC & HSC Digest
Class 11 (FYJC / HSC)Mathematics & Statistics2026-27 Syllabus

Chapter 4 Determinants and Matrices Ex 4.3 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 4 Determinants and Matrices Ex 4.3. Step-by-step solved exercises, numerical problems, and digest answers.

8 Solved Questions• 8 Diagrams• 2469 words

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Question 1 Maharashtra Board Solution
Solve the following linear equations by using Cramer’s Rule. x+y + z = 6, x – y + z = 2,.x + 2y – z = 2 x + y – 2z = -10, 2x +y – 3z = -19, 4x + 6y + z = 2 x + z = 1, y + z = 1, x + y = 4 = 3, = -13 and = -11
Solution & Step-by-Step Answer:
Given equations are x + y + z = 6, x – y + z = 2, x + 2y – z = 2. D = 1 2 -1 = 1(1 -2) – 1(-1 – 1) + 1(2 + 1) = 1 (-1)-1 (-2)+ 1(3) = -1 + 2 + 3 = 4 ≠ 0

Dx= ≤ft|{array}{ccc}
6 & 1 & 1 \\
2 & -1 & 1 \\
2 & 2 & -1
{array}
= 6(1 – 2) – 1(-2 – 2) + 1(4 + 2)
= 6(-1) -1 (-4) + 1(6)
= -6 + 4 + 6
= 4

Dy= ≤ft|{array}{ccc}
1 & 6 & 1 \\
1 & 2 & 1 \\
1 & 2 & -1
{array}
= 1(-2 – 2) – 6(-l – 1) + 1(2 = 1 (- 4) – 6 (- 2) + 1(0)
= -4+12 + 0 = 8

Dz= ≤ft|{array}{ccc}
1 & 1 & 6 \\
1 & -1 & 2 \\
1 & 2 & 2
{array}
= l(-2 – 4) – 1(2 – 2) + 6(2 + 1)
= l(-6)-l(0) + 6(3)
= -6 + 0+18 = 12

By Cramer’s Rule,

∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.

ii. Given equations are x+y- 2z = -10,
2x + y – 3z = -19,
Ax + 6y + z = 2.
D = ≤ft|{array}{ccc}
1 & 1 & -2 \\
2 & 1 & -3 \\
4 & 6 & 1
{array}
= 1(1 + 18)- 1(2+ 12)-2(12-4)
= 1(19)-1(14)-2(8)
= 19-14-16 = -11 ≠ 0

Dx= ≤ft|{array}{ccc}
-10 & 1 & -2 \\
-19 & 1 & -3 \\
2 & 6 & 1
{array}
= -10(1 + 18) – 1(-19 + 6) – 2(- 114 – 2)
= -10(19)- 1(-13) -2(-l 16)
= -190+ 13 + 232 = 55

Dy= ≤ft|{array}{ccc}
1 & -10 & -2 \\
2 & -19 & -3 \\
4 & 2 & 1
{array}
= 1(-19 + 6) – (-10)(2 + 12) – 2(4 + 76) = 1(-13) + 10(14) – 2(80)
= -13 + 140-160 = -33

Dz= ≤ft|{array}{ccc}
1 & 1 & -10 \\
2 & 1 & -19 \\
4 & 6 & 2
{array}
= 1(2+ 114)-1(4+ 76)-10(12-4)
= 1(116)-1(80)-10(8)
= 116-80-80.
= -44
By Cramer’s Rule,
∴ x = -5, y = 3 and z = 4 are the solutions of the given equations.
[Note: The question has been modified]

iii. Given equations are
x + z = 1, i.e.,x + 0y + z = 1,
y + z = 1, i.e., 0x + y + z = 1,
x + y = 4, i.e., x + y + 0z = 4.
D = ≤ft|{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
{array}
= 1(0 – 1) – 0 + 1(0 – 1)
= 1(-1)+1(-1)
= -1-1 = -2 ≠ 0
Dx= ≤ft|{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
{array}
= 1(0 – 1) – 0 + 1(1 -4) = l(-l)+l(-3)
= -1 – 3
= -4

Dy= ≤ft|{array}{lll}
1 & 1 & 1 \\
0 & 1 & 1 \\
1 & 4 & 0
{array}
= 1(0 – 4) – 1(0 – 1) + 1(0 – 1)
= 1(-4) – 1(-1) + 1(-1)
= -4 + 1 – 1
= -4

Dz= ≤ft|{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 4
{array}
= 1(4 – 1) – 0 + 1(0 – 1)
= 1(3) + 1(-1)
= 3 – 1
= 2

By Cramer’s Rule,

∴ x = 2, y = 2 and z = -1 are the solutions of the given equations.

Let = p, = q, = r
∴ The given equations become
-2p – q – 3r = 3, i.e., 2p + q + 3r = -3,
2p-3q + r = -13,
2p – 3r = -11, i.e., 2p + 0q – 3r = -11.
D = ≤ft|{array}{ccc}
2 & 1 & 3 \\
2 & -3 & 1 \\
2 & 0 & -3
{array}
= 2(9 – 0) – 1(-6 – 2) + 3(0 + 6)
= 2(9) – 1(-8) + 3(6)
= 18 + 8 + 18
= 44 ≠ 0

DP= ≤ft|{array}{ccc}
-3 & 1 & 3 \\
-13 & -3 & 1 \\
-11 & 0 & -3
{array}
= -3(9 – 0) – 1(39 + 11) + 3(0 – 33)
= -3(9) – 1(50) + 3(-33)
= -27 – 50 – 99
= -176

Dq= ≤ft|{array}{ccc}
2 & -3 & 3 \\
2 & -13 & 1 \\
2 & -11 & -3
{array}
= 2(39 + 11)- (-3)(-6 – 2) + 3(-22 + 26)
= 2(50) + 3(-8) + 3(4)
= 100 – 24 + 12
= 88

Dr= ≤ft|{array}{ccc}
2 & 1 & -3 \\
2 & -3 & -13 \\
2 & 0 & -11
{array}
= 2(33 – 0) – 1(-22 + 26) – 3(0 + 6)
= 2(33) – 1(4) – 3(6)
= 66 – 4 – 18
= 44

By Cramer’s Rule,

∴ x = , y = z = 1 are the solutions of the given equations.

Question 2 Maharashtra Board Solution
The sum of three numbers is 15. If the second number is subtracted from the sum of first and third numbers, then we get 5. When the third number is subtracted from the sum of twice the first number and the second number, we get 4. Find the three numbers.
Solution & Step-by-Step Answer:
Let the three numbers be x, y and z. According to the given conditions, x + y + z = 15, x + z-y = 5, i.e., x – y + z = 5, 2x + y – z = 4. D = = 1(1 – 1) – 1(-1 – 2) + 1(1 + 2) = 1(0) – 1(-3) + 1(3) = 0 + 3 + 3 = 6 ≠ 0

Dx= ≤ft|{array}{ccc}
15 & 1 & 1 \\
5 & -1 & 1 \\
4 & 1 & -1
{array}
= 15(1 – 1) – 1(-5 – 4) + 1(5 + 4)
= 15(0) – 1(-9) + 1(9)
= 0 + 9 + 9
= 18

Dy= ≤ft|{array}{ccc}
1 & 15 & 1 \\
1 & 5 & 1 \\
2 & 4 & -1
{array}
= 1(-5 – 4) – 15(-1 – 2) + 1(4 – 10)
= 1(-9) – 15(-3) + 1(-6)
= -9 + 45 – 6 = 30

Dz= ≤ft|{array}{ccc}
1 & 1 & 15 \\
1 & -1 & 5 \\
2 & 1 & 4
{array}
= 1(-4 – 5) – 1(4 – 10) + 15(1 + 2)
= 1(-9) – 1(-6) + 15(3)
= -9 + 6 + 45
= 42
By Cramer’s Rule,
∴ The three numbers are 3, 5 and 7.

Question 3 Maharashtra Board Solution
Examine the consistency of the following equations. i. 2x – y + 3 = 0, 3x + y – 2 = 0, 11x + 2y – 3 = 0 ii. 2x + 3y – 4 = 0, x + 2y = 3, 3x + 4y + 5 = 0 iii. x + 2y – 3 = 0,7x + 4y – 11 = 0,2x + 4y – 6 = 0
Solution & Step-by-Step Answer:
i. Given equations are 2x – y + 3 = 0, 3x + y – 2 = 0, 11x + 2y – 3 = 0. D = = 2(-3 + 4) – (-l)(-9 + 22) + 3(6-11) = 2(1)+1(13)+ 3(-5) = 2 + 13-15 = 0 ∴ The given equations are consistent.

ii. Given equations are 2x + 3y – 4 = 0,
x + 2y = 3, i.e., x + 2y – 3 = 0,
3x + 4y + 5 = 0.
≤ft|{array}{ccc}
2 & 3 & -4 \\
1 & 2 & -3 \\
3 & 4 & 5
{array}
= 2(10 + 12) – 3(5 + 9) – 4(4 – 6)
= 2 (22) – 3(14) – 4(-2)
= 44 – 42 + 8
= 10 ≠ 0
∴ The given equations are not consistent.

iii. Given equations are x + 2y – 3 =
7x + 4y – 11 =0,
2x + 4y – 6 = 0.
≤ft|{array}{ccc}
1 & 2 & -3 \\
7 & 4 & -11 \\
2 & 4 & -6
{array}
= 1(-24 + 44) – 2(-42 + 22) – 3(28 – 8)
= 1(20) – 2(-20) – 3(20)
= 20 + 40 – 60
= 0
∴ The given equations are consistent.

Question 4 Maharashtra Board Solution
Find k, if the following equations are consistent. i. 2x + 3y-2 = 0,2x + 4y-k = 0,x-2j + 3k = 0 ii. kx + 3,y +1 = 0, x + 2y+1 = 0, x + y = 0
Solution & Step-by-Step Answer:
i. Given equations are 2x + 3y – 2 = 0, 2x + 4y – k = 0, x – 2y + 3k = 0. Since these equations are consistent, = 0 ∴ 2(12k – 2k) – 3(6k + k) – 2(- 4 – 4) = 0 ∴ 2(10k) – 3(7k) – 2(- 8) = 0 ∴ 20k – 21k + 16 = 0 ∴ k = 16

Given equations are are
kx + 3y + 1 = 0,
x + 2y +1=0,
x + y = 0, i.e., x + y + 0 = 0.
Since these equations are consistent,
≤ft|{array}{lll}
k & 3 & 1 \\
1 & 2 & 1 \\
1 & 1 & 0
{array} = 0
∴ k(0 – 1) – 3(0 – 1) + 1(1 – 2) = 0
∴ k(-1) – 3(-1) + 1(-1) = 0
∴ -k + 3 – 1 = 0
∴ k = 2.

Question 5 Maharashtra Board Solution
Find the area of triangle whose vertices are i. A (5,8), B (5,0), C (1,0) ii. P(3/2, 1), Q(4,2), R(4, -1/2) iii. M (0, 5), N (- 2, 3), T (1, – 4)
Solution & Step-by-Step Answer:
i. Here, A(x1, y1) ≡ A(5, 8), B(x2, y2) = B(5, 0), C(x3, y3) = C(1,0) Area of a triangle = A(ΔABC) = = [5(0 – 0) – 8(5 – 1) + 1(0 – 0)] = [0 – 8(4) + 0] = (-32) = -16 Since area cannot be negative, A(ΔABC) = 16 sq. units

ii. Here, P(x1, y1) ≡ P(3/2, 1), Q(x2, y2) ≡ Q(4, 2), R(x3, y3) ≡ R(4,- )


Since area cannot be negative
A(ΔPQR) = 25/8 sq. units

iii. Here, M(x1, y1) ≡ M(0, 5), N(x2, y2) ≡ N(-2, 3)
T(x3, y3) ≡ T(1, -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(ΔMNT) = ≤ft|{array}{ccc}
0 & 5 & 1 \\
-2 & 3 & 1 \\
1 & -4 & 1
{array}
= [ 0 – 5 (-2 -1) + 1 (8 – 3)]
= [-5 (-3) + 1(5)]
= (15 + 5)
= (20)
= 10 sq. units

Question 6 Maharashtra Board Solution
Find the area of quadrilateral whose vertices are A (- 3,1), B (- 2, – 2), C (1,4), D (3, – 1).
Solution & Step-by-Step Answer:
A(-3, 1), B(-2, -2), C(l, 4), D(3, -1) A(□ ABDC) = A(ΔABD) + A(ΔADC) Area of triangle = A(ΔABD) = = [-3 (-2 + 1) – 1(-2 – 3) + 1(2 + 6) = [-3(-1) – 1(-5) + 1(8)] = (3 + 5 + 8) = (16) ∴ A(ΔABD) = 8 sq. units A(ΔADC) = = [-3(-1-4) – 1(3 – 1) + 1(12 + 1)] = [-3(-5) – 1(2) + 1(13)] = [15 – 2 + 13] = (26) ∴ A(ΔADC) = 13 sq. units ∴ A(□ ABDC) = A(ΔABD) + A(ΔADC) = 8 + 13 = 21 sq. units

Question 7 Maharashtra Board Solution
Find the value of k, if the area of triangle whose vertices are P (k, 0), Q (2,2), R (4,3) is sq. units.
Solution & Step-by-Step Answer:
Here, P(x1, y1) ≡ P(k, 0), Q(x2, y2) ≡ Q(2, 2), R(x3, y3) ≡ R(4,3) ∴ A(ΔPQR) = sq. units Area if triangle = ∴ ∴ ± (-k -2) ∴ ± 3 = -k – 2 ∴ 3 = -k – 2 or -3 = -k – 2 ∴ k = -5 or k = 1
Question 8 Maharashtra Board Solution
Examine the collinearity of the following set of points: i. A (3, – 1), B (0, – 3), C (12, 5) ii. P (3, – 5), Q (6,1), R (4, 2) iii. L(0,1/2), M(2,-1), N(-4, 7/2)
Solution & Step-by-Step Answer:
i. Here, A(x1, y1) ≡ A(3, -1), B(x2, y2) ≡ B(0, -3), C(x3, y3) ≡ C(12, 5) If A(∆ABC) = 0, then the points A, B, C are collinear. ∴ A(∆ABC) = = [3(-3 – 5) – (-1) (0 – 12) + 1(0 + 36)] = [3(-8)+ 1(-12)+ 1(36)] = (-24 – 12 + 36) = 0 ∴ The points A, B, C are collinear.

ii. Here, P(x1, y1) ≡ P(3, -5), Q(x2, y2) ≡ Q(6, 1), R(x3, y3) ≡ R(4,2)
∴ If A(∆PQR) = 0, then the points P,Q, R are collinear
∴ A(∆PQR) = ≤ft|{array}{ccc}
3 & -5 & 1 \\
6 & 1 & 1 \\
4 & 2 & 1
{array}
= [3(1-2) – (-5)(6 – 4) + 1(12 – 4)]
= [3(-1) + 5(2) + 1(8)]
= (-3 + 10 + 8)= ≠ 0
∴ The points P, Q, R are non-collinear.

iii. Here, L(x1, y1) ≡ L(0,1/2), M(x2, y2) ≡ M(2, -1), N(x3, y3) ≡ N(-4, 7/2)
If A(∆LMN) = 0, then the points L, M, N are collinear.
∴ A(∆LMN) = ≤ft|{array}{ccc}
0 & & 1 \\
2 & -1 & 1 \\
-4 & & 1
{array}
= [0 – ] (2 + 4) + 1(7 – 4)]
= [ – (6) + 1(3)]
= (-3 + 3) = 0
∴ The points L, M, N are collinear.

Previous Chapter
Chapter 4 Determinants and Matrices Ex 4.2
Next Chapter
Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(A)