**I. Select the correct option from the given alternatives.**

**Question 1.**

**Question 2.**

(a) k = -3

(b) k = -1

(c) k = 1

(d) k = 3

**Answer:**

(b) k = -1

Hint:

**Question 3.Answer:**

(b) D is independent of φ

**Question 4.The value of a for which the system of equations x + (a + 1)y + (a + 2 z = 0, ax + (a + 1)y + (a + 2)z = 0 and x + y + z = 0 has a non zero solution is**(a) 0

(b) -1

(c) 1

(d) 2

**Answer:**

(b) -1

Hint:

The given system of equations will have a non-zero solution, if

Question 5.

Hint:

**Question 6.The system 3x – y + 4z = 3, x + 2y – 3z = -2 and 6x + 5y + λz = -3 has atleast one solution when**

(a) λ = -5

(b) λ = 5

(c) λ = 3

(d) λ = -13

**Answer:**

(a) λ = -5

Hint:

The given system of equations will have more than one solution if

**Question 7.**

(a) 2, -7

(b) -2, 7

(c) 2, 7

(d) -2, -7

**Answer:**

(c) 2, 7

Hint:

**Question 8.**

(a) x = 3, y = 1

(b) x = 1, y = 3

(c) x = 0, y = 3

(d) x = 0, y = 0

**Answer:**

(d) x = 0, y = 0

**Question 9.If A(0, 0), B(1, 3) and C(k, 0) are vertices of triangle ABC whose area is 3 sq. units, then the value of k is**

(a) 2

(b) -3

(c) 3 or -3

(d) -2 or 2

**Answer:**

(d) -2 or 2

**Question 10.Which of the following is correct?**

(a) Determinant is a square matrix

(b) Determinant is number associated to matrix

(c) Determinant is a number associated with a square matrix

(d) None of these

**Answer:**

(c) Determinant is a number associated with a square matrix

**II. Answer the following questions.**

**Question 1.Evaluate:**

= 2(4 – 0) + 5(10 – 9) + 7(0 – 18)

= 2(4) + 5(1) + 7(-18)

= 8 + 5 – 126

= -113

= 1(4 + 28) + 3(0 + 36) + 12(0 – 18)

= 1(32) + 3(36) + 12(-18)

= 32 + 108 – 216

= -76

**Question 2.**

**Solution:**

**Question 3.Evaluate:by using properties.Solution:**

**Question 4.Find the minors and cofactors of elements of the determinants.Solution:**

**Question 5.Find the values of x, if**

**Question 6.Solution:**

**Question 7.Without expanding the determinants, show thatSolution:**

**Question 8.Solution:**

**Question 9.Solve the following linear equations by Cramer’s Rule.(i) 2x – y + z = 1, x + 2y + 3z = 8, 3x + y – 4z = 1(iii) 2x + 3y + 3z = 5, x – 2y + z = -4, 3x – y – 2z = 3(iv) x + y + 2z = 7, 3x + 4y – 5z = 5, 2x – y + 3z = 12Solution:**

(i) Given equations are

2x – y + z = 1

x + 2y + 3z = 8

3x + y – 4z = 1

= 2(-8 – 3) – (-1)(-4 – 9) + 1(1 – 6)

= 2(-11) + 1(-13) + 1(-5)

= -22 – 13 – 5

= -40 ≠ 0

= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2)

= 1(-11) + 1(-35) + 1(6)

= -11 – 35 + 6

= -40

= 2(-32 – 3) -1(-4 – 9) + 1(1 – 24)

= 2(-35) – 1(-13) + 1(-23)

= -70 + 13 – 23

= -80

= 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.

= 2(18 – 0) -2(0 – 18) + 0

= 2(18) – 2(-18)

= 36 + 36

= 72 ≠ 0

= 3(18 – 0) – 2(15 – 24) + 0

= 3(18) – 2(-9)

= 54 + 18

= 72

= 2(15 – 24) – 3(0 – 18) + 0

= 2(-9) – 3(-18)

= -18 + 54

= 36

= 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

= 2(4 + 1) – 3(-2 – 3) + 3(-1 + 6)

= 2(5) – 3(-5) + 3(5)

= 10 + 15 + 15

= 40 ≠ 0

= 5(4 + 1) – 3(8 – 3) + 3(4 + 6)

= 5(5) – 3(5) + 3(10)

= 25 – 15 + 30

= 40

= 2(8 – 3) – 5(-2 – 3) + 3(3 + 12)

= 2(5) – 5(-5) + 3(15)

= 10 + 25 + 45

= 80

= 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

= 1(12 – 5) – (-1)(9 + 10) + 2(-3 – 8)

= 1(7) + 1(19) + 2(-11)

= 7 + 19 – 22

= 4 ≠ 0

= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48)

= 7(7) + 1(75) + 2(-53)

= 49 + 75 – 106

= 18

= 1(15 + 60) – 7(9 + 10) + 2(36 – 10)

= 1(75) – 7(19) + 2(26)

= 75 – 133 + 52

= -6

= 1(48 + 5) – (-1)(36 – 10) + 7(-3 – 8)

= 1(53) + 1(26) + 7(-11)

= 53 + 26 – 77

= 2

By Cramer’s Rule,

**Question 10.Find the value of k, if the following equations are consistent.(i) (k + 1)x + (k – 1)y + (k – 1) = 0(k – 1)x + (k + 1)y + (k – 1) = 0(k – 1)x + (k – 1)y + (k + 1) = 0(ii) 3x + y – 2 = 0, kx + 2y – 3 = 0 and 2x – y = 3(iii) (k – 2)x + (k – 1)y = 17, (k – 1)x +(k – 2)y = 18 and x + y = 5Solution:**

(i) Given equations are

(k + 1)x + (k – 1)y + (k – 1) = 0

(k – 1)x + (k + 1)y + (k – 1) = 0

(k – 1)x + (k – 1)y + (k + 1) = 0

Since these equations are consistent,

⇒ 2(2k + 2 + 2k – 2) – 0 + (k – 1) (4 – 0) = 0

⇒ 2(4k) + (k – 1)4 = 0

⇒ 8k + 4k – 4 = 0

⇒ 12k – 4 = 0

(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,

⇒ 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

**Question 11.Find the area of triangle whose vertices are(i) A(-1, 2), B(2, 4), C(0, 0)(ii) P(3, 6), Q(-1, 3), R(2, -1)(iii) L(1, 1), M(-2, 2), N(5, 4)Solution:**

Since area cannot be negative,

A(ΔABC) = 4 sq. units

Since area cannot be negative,

A(ΔLMN) =13/2 sq. units

**Question 12.Find the value of k,(i) if the area of a triangle is 4 square units and vertices are P(k, 0), Q(4, 0), R(0, 2).(ii) if area of triangle is 33/2 square units and vertices are L(3, -5), M(-2, k), N(1, 4).Solution:**

A(ΔPQR) = 4 sq. units

**Question 13.Find the area of quadrilateral whose vertices are A(0, -4), B(4, 0), C(-4,0), D (0, 4).Solution:**

A(0, -4), B(4, 0), C(-4, 0), D(0, 4)

∴ A(ABDC) = A(ΔABC) + A(ΔBDC)

= 16 + 16

= 32 sq.units

**Question 14.An amount of ₹ 5000 is put into three investments at the rate of interest of 6%, 7%, and 8% per annum respectively. The total annual income is ₹ 350. If the combined income from the first two investments is ₹ 70 more than the income from the third, find the amount of each investment.Solution:**

Let the amount of each investment be ₹ x, ₹ y and ₹ z.

According to the given conditions,

x + y + z = 5000,

6% x + 7% y + 8% z = 350

∴ The amounts of investments are ₹ 1750, ₹ 1500, and ₹ 1750.

**Question 15.Show that the lines x – y = 6, 4x – 3y = 20 and 6x + 5y + 8 = 0 are concurrent. Also, find the point of concurrence.Solution:**

Given equations of the lines are

x – y = 6, i.e., x – y – 6 = 0 ……(i)

4x – 3y = 20, i.e., 4x – 3y – 20 = 0 …..(ii)

6x + 5y + 8 = 0 ……(iii)

The given lines will be concurrent, if

= 1(-24 + 100) – (-1) (32 + 120) – 6(20 + 18)

= 1(76) + 1(152) – 6(38)

= 76 + 152 – 228

= 0

∴ The given lines are concurrent.

To find the point of concurrence, solve any two equations.

Multiplying (i) by 5, we get

5x – 5y – 30 = 0 …….(iv)

Adding (iii) and (iv), we get

11x – 22 = 0

∴ x = 2

Substituting x = 2 in (i), we get

2 – y – 6 = 0

∴ y = -4

∴ The point of concurrence is (2, -4).

**Question 16.Show that the following points are collinear using determinants:(i) L(2, 5), M(5, 7), N(8, 9)(ii) P(5,1), Q(1, -1), R(11, 4)Solution:**

∴ The points L, M, N are collinear.

If A(ΔPQR) = 0, then the points P, Q, R are collinear.

∴ The points P, Q, R are collinear.