# Determinants Class 12 Mathematics Important Questions

Please refer to Determinants Class 12 Mathematics Important Questions with solutions provided below. These questions and answers have been provided for Class 12 Mathematics based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 12 Mathematics for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 12.

## Class 12 Mathematics Important Questions Determinants

Question.

⇒ x(–x2 – 1) –sin θ(–xsin θ – cos θ) + cos θ (–sin θ + xcos θ) = 8
⇒ –x3 – x + xsin2θ + sin θ cos θ – sin θ cos θ + xcos2θ = 8
⇒ –x3 – x + x(sin2θ + cos2θ) = 8
⇒ –x3 – x + x = 8 ⇒ x3 + 8 = 0
⇒ (x + 2) (x2 – 2x + 4) = 0 ⇒ x + 2 = 0
[∵  x2 – 2x + 4 > 0 ∀ x]
⇒ x = –2

Question.

Question. What is the value of the determinant

Question. Find the co-factor of a12 in the following :

Question. In the interval π/2 < x < π, find the value of x for which the matrix

Question. What positive value of x makes the following pair of determinants equal ?

⇒ 2x2 – 15 = 32 – 15 ⇒ x = 4                           [ x > 0]

Question. Find the value of x from

Question. Evaluate :

Question.

Question.

⇒ 2x2 – 40 = 18 + 14
⇒ 2x2 = 72 ⇒ x2 = 36
⇒ x = ± 6

Question.

⇒ 12x + 14 = 32 – 42
⇒ 12x = – 10 – 14 = – 24
⇒ x = – 2.

Question. Write the value of the determinant

Question.

⇒ Δ = 2[8(86) – 9 (75)] – 7[3(86) – 5(75)] + 65[3(9) – 5(8)]
= 2(688 – 675) – 7(258 – 375) + 65(27 – 40)
= 2(13) – 7(– 117) + 65(–13)
= 26 + 819 – 845 = 0

Question. Using properties of determinants, prove the following :

Question. Show that Δ = Δ1, where

Question. Using properties of determinants, prove the following :

Question. Using properties of determinants, prove the following :

50. Using properties of determinants, prove that

= (a – b) (b – c) [(b2 + bc + c2) – (a2 + ab + b2)]
= (a – b) (b – c) [(c2 – a2) + b(c – a)]
= (a – b) (b – c) (c – a) (c + a + b)
= (a – b) (b – c) (c – a) (c + a + b) = R.H.S.

Question. Using properties of determinants, prove the following :

Question. Using properties of determinants, prove that

Question. Using properties of determinants, prove that

Question. Using properties of determinants, solve the following for x.

Question. Using properties of determinants, solve the following for x :

Question. Using properties of determinants, prove the following :

Question.. Using properties of determinants, show that

Question. Prove, using the properties of determinants

Question.

and I is the identify matrix of order 2, then show that A2 = 4A – 3I. Hence find A–1.

Question. A trust invested some money in two type of bonds. The first bond pays 10% interest and second bond pays 12% interest. The trust received ₹ 2,800 as interest. However, if trust had interchanged money in bonds they would have got ₹ 100 less as interest. Using matrix method, find the amount invested by the trust. Interest received on this amount will be given to Helpage India as donation. Which value is reflected in this question?
Answer. Let x be invested in the first bond and y be invested in the second bond.
According to question,

If the rate of interest had been interchanged, then the total interest earned is ₹ 100 less than the previous interest. i.e., ₹ 2700.

Question. A coaching institute of English (Subject) conducts classes in two batches I and II and fees for rich and poor children are different. In batch I, it has 20 poor and 5 rich children and total monthly collection is ₹ 9,000, whereas in batch II, it has 5 poor and 25 rich children and total monthly collection is ₹ 26,000. Using matrix method, find monthly fees paid by each child of two types. What values the coaching institute is inculcating in the society?

Let the monthly fees paid by poor and rich children be ` x and ` y, respectively.
For batch I :
20x + 5y = 9000                                 …(i)
For batch II :
5x + 25y = 26000                               …(ii)
The system of equations (i) and (ii) can be written as
AX = B

⇒ x = 200, y = 1000
Hence, the monthly fees paid by each poor child is ₹ 200 and the monthly fees paid by each rich child is ₹ 1000.
By offering discount to the poor children, the coaching institute offers an unbiased chance for the development and enhancement of the weaker section of our society.

Question. Two schools A and B decided to award prizes to their students for three values, team spirit, truthfulness and tolerance at the rate of ₹ x, ₹ y and ₹ z per student respectively. School A, decided to award a total of ₹ 1,100 for the three values to 3, 1 and 2 students respectively while school B decided to award ₹ 1,400 for the three values to 1, 2 and 3 students respectively. If one prize for all the three values together amount to ₹ 600 then (i) Represent the above situation by a matrix equation after forming linear equations. (ii) Is it possible to solve the system of equations so obtained using matrices? (iii) Which value you prefer to be rewarded most and why?
Answer. 118. (i) Given, value of prize for team spirit = ₹ x
Value of prize for truthfulness = ₹ y
Value of prize for tolerance = ₹ z
Linear equation for School A is 3x + y + 2 z = 1100
Linear equation for School B is x + 2 y + 3z = 1400
Linear equation for Prize is x + y + z = 600
The corresponding matrix equation is PX = Q

= 3·(2 – 3) – 1·(1 – 3) + 2·(1 – 2)
= – 3 + 2 – 2 = – 3 ≠ 0
Thus, P–1 exists. So, system of equations has unique
solution and it is given by X = P–1Q
Now, cofactors of elements of P are
A11 = –1, A12 = 2, A13 = –1,
A21 = 1, A22 = 1, A23 = –2,
A31 = –1, A32 = –7, A33 = 5

Question.

divisible by (x + y + z), and hence find the quotient.

⇒ Δ = – (x + y + z) (x3 + y3 + z3 – 3xyz)
[(x – y) (y – x) – (x – z) (y – z)]
⇒ Δ = –(x + y + z) (x3 + y3 + z3 –3xyz)
(xy + yz + zx – x2 –y2 – z2)
⇒ Δ = (x + y + z) (x3 + y3 + z3 – 3xyz)
(x2 + y2 + z2 – xy – yz – zx)
Hence, Δ is divisible by (x + y + z) and quotient is
(x3 + y3 + z3 – 3xyz)(x2 + y2 + z2 – xy – yz – zx)

Question. Using properties of determinants, prove the following :

Since each element in third column of determinant is the sum of two elements, therefore determinant can be expressed as the sum of two determinants given by

= (1 + pxyz) (y – x) (z – x) [1(z + x – y – x) – 0 + 0]
= (1 + pxyz) (y – x) (z – x) (z – y)
= (1 + pxyz) (x – y) (y – z) (z – x) = R.H.S.

Question.

Now, A11 = cos α, A12 = – sin α, A13 = 0,
A21 = sin α, A22 = cos α, A23 = 0,
A31 = 0, A32 = 0, A33 = 1

= cos α (cos α – 0) + sin α (sin α – 0) + 0 = 1 …(iii)
From (i), (ii) and (iii), we get

Question. Using elementary transformations, find the inverse of the matrix

and use it to solve the following system of linear equations:
8x + 4y + 3z = 19; 2x + y + z = 5;
x + 2y + 2z = 7

Question. A shopkeeper has 3 varieties of pens ‘A’, ‘B’ and ‘C’. Meenu purchased 1 pen of each variety for a total of ₹ 21. Jeevan purchased 4 pens of ‘A’ variety, 3 pens of ‘B’ variety and 2 pens of ‘C’ variety for ₹ 60. While Shikha purchased 6 pens of ‘A’ variety, 2 pens of ‘B’ variety and 3 pens of ‘C ’ variety for ₹ 70. Using matrix method, find cost of each variety of pen.
Answer. Let one pen of variety ‘A’ costs ₹ x, one pen of variety ‘B’ costs ₹ y and one pen of variety ‘C’ costs ₹ z.
According to question,
x + y + z = 21             (For Meenu)
4x + 3y + 2z = 60       (For Jeevan)
6x + 2y + 3z = 70       (For Shikha)
The given system of equations can be written as
AX = B

∴ Cost of 1 pen of variety ‘A’ = ₹ 5
Cost of 1 pen of variety ‘B’ = ₹ 8
Cost of 1 pen of variety ‘C’ = ₹ 8

Question. A school wants to award its students for the value of honesty, regularity and hard work with a total cash award of ₹ 6,000. Three times the award money for hard work added to that given for honesty amounts to ₹ 11,000. The award money given for honesty and hard work together is double the one given for regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, honesty, regularity and hard work, suggest one more value which the school must include for awards.
Answer. Let the award money for honesty = `x Let the award money for regularity =` y
Let the award money for hard work = ` z
According to question, we have
x + y + z = 6000
3z + x = 11000
x + z – 2y = 0
The system of equations can be written as AX = B

= 1(0 + 6) – 1(1 – 3) + 1(–2 – 0)
= 6 – (–2) – 2 = 6 ≠ 0
∴ A is invertible. So, the given system has a unique solution given by X = A–1B
Now, A11 = 6, A12 = 2, A13 = –2
A21 = –3, A22 = 0, A23 = 3
A31 = 3, A32 = –2, A33 = –1

⇒ x = 500, y = 2000, z = 3500
One more value which the school can include for awards is discipline.

Question. The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the numbers of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards.
Answer. According to question, we have
x + y + z = 12
2x +3(y + z) = 33 ⇒ 2x + 3y + 3z = 33
x + z = 2y ⇒ x – 2y + z = 0
The system of equations can be written as AX = B

∴ A–1 exists. So, system of equations has a unique solution and it is given by X = A–1B
Now, A11 = 9, A12 = 1, A13 = –7,
A21 = –3, A22 = 0, A23 = 3,
A31 = 0, A32 = –1, A33 = 1

The management of the colony can include the awards for those members of the colony who help for keeping the environment of the colony free from pollution.

Question. Two institutions decided to award their employees for the three values of resourcefulness, competence and determination in the form of prizes at the rate of ₹ x, ₹ y and ₹ z respectively per person. The first institution decided to award respectively 4, 3 and 2 employees with a total prize money of ₹ 37,000 and the second institution decided to award respectively 5, 3 and 4 employees with a total prize money of ₹ 47,000. If all the three prizes per person together amount to ₹ 12,000 then using matrix method find the value of x, y and z. What values are described in the question?
Answer. According to question, we have
x + y + z = 12000
4x + 3y + 2z = 37000
5x + 3y + 4z = 47000
The system of equations can be written as AX = B

=1(12 – 6) – 1(16 – 10) + 1(12 – 15)
= 6 – 6 – 3 = – 3 ≠ 0
∴ A–1 exists. So, system of equations has a unique solution and it is given by X = A–1B
Now, A11 = 6, A12 = –6, A13 = –3,
A21 = –1, A22 = –1, A23 = 2,
A31 = –1, A32 = 2, A33 = –1

⇒ x = 4000, y = 5000; z = 3000
The values described in this question are
resourcefulness, competence and determination.

Question. Two factories decided to award their employees for three values of (a) adaptable to new techniques, (b) careful and alert in difficult situations and (c) keeping calm in tense situations, at the rate of ₹ x, ₹ y and ₹ z per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of ₹ 29,000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of ₹ 30,500. It the three prizes per person together cost ₹ 9,500; then
(i) Represent the above situation by a matrix equation and form linear equations using matrix multiplication.
(ii) Solve these equations using matrices.
(iii) Which values are reflected in this question?

Answer. (i) According to question, we have
x + y + z = 9500
2x + 4y + 3z = 29000
5x + 2y + 3z = 30500
The system of equations can be written as AX = B

= 1(12 – 6) – 1(6 – 15) + 1(4 – 20)
= 6 + 9 – 16 = – 1 ≠ 0
∴ A–1 exists. So, system of equations has a unique solution and it is given by X = A1B
Now, A11 = 6, A12 = 9, A13 = –16,
A21 = –1, A22 = –2, A23 = 3,
A31 = –1, A32 = –1, A33 = 2

⇒ x = 2500, y = 3000, z = 4000.
(iii) The factories honours the most, those employees who are keeping calm in tense situations.

Question. Using matrices, solve the following system of equations :
x + y – z = 3; 2x + 3y + z = 10;
3x – y – 7z = 1
x + y – z = 3
2x + 3y + z = 10
3x + y – 7z = 1
The system of equations can be written as AX = B

= 1(–21 + 1) – (–14 – 3) – 1(– 2 – 9)
= – 20 + 17 + 11 = 8 􀁺􀀃0
∴ A–1 exists. So, system of equation has a unique solution given by X = A–1B
Now, A11 = –20, A12 = 17, A13 = –11, A21 = 8,
A22 = –4, A23 = 4, A31 = 4, A32 = –3, A33 = 1

Question.

find A–1 and hence solve the system of equations :
x + 2y + z = 4, – x + y + z = 0,
x – 3y + z = 4

= 1(1 + 3) – 2(– 1 – 1) + 1(3 – 1) = 10 ≠0
∴ A–1 exists.
Now, A11 = 4, A12 = 2, A13 = 2, A21 = –5, A22 = 0,
A23 = 5, A31 = 1, A32 = –2, A33 = 3

Since A–1 exists, therefore, system of equations has a unique solution given by

Question. Determine the product

and use it to solve the system of equations :
x – y + z = 4; x – 2y – 2z = 9;
2x + y + 3z = 1

Now the given system of equations is
x – y + z = 4
x – 2y – 2z = 9
2x + y + 3z = 1
The system of equations can be written as AX = B

Question.

Hence solve the system of equations
x + 2y – 3z = – 4; 2x + 3y + 2z = 2;
3x – 3y – 4z = 11.

Question.

two square matrices, find AB and hence solve the system of equations
x – y = 3, 2x + 3y + 4z = 17; and y + 2z = 7.

Since A–1 exists, so system of equations has a unique solution given by X = A–1 P

Question.

find A–1 and hence solve the system of equations
2x – 3y + 5z = 11; 3x + 2y – 4z = – 5;
x + y – 2z = – 3

= 2(– 4 + 4) + 3(– 6 + 4) + 5(3 – 2)
= – 6 + 5 = – 1 ≠ 0
∴ A–1 exists.
Now, A11 = 0, A12 = 2, A13 = 1, A21 = –1, A22 = –9,
A23 = –5, A31 = 2, A32 = 23, A33 = 13

The given system of equations is
2x – 3y + 5z = 11
3x + 2y – 4z = – 5
x + y – 2z = – 3
The system of equations can be written as
AX = B

Question.

find A–1. Hence solve the following system of equations :
x + 2y + 5z = 10, x – y – z = – 2,
2x + 3y – z = – 11

The given system of equations is
x + 2y + 5z = 10
x – y – z = – 2
2x + 3y – z = – 11
The given system of equations can be written as
AX = B

Question. Using matrix method, solve the following system of equations :

Question. Using matrix method, solve the following system of equations:
x + 2y + z = 7, x + 3z = 11, 2x – 3y = 1.
x + 2y + z = 7
x + 3z = 11
2x – 3y = 1
The given system of equations can be written as
AX = B

=2(6 – 0) + 3(3 – 1) = 12 + 6 = 18 ≠ 0
∴ A–1 exists. So, system of equations has a unique
solution given by X = A–1B
Now, A11 = 9, A12 = 6, A13 = –3, A21 = –3, A22 = –2,
A23 = 7, A31 = 6, A32 = –2, A33 = –2

Question. Using matrices, solve the following system of equations :
4x + 3y + 2z = 60, x + 2y + 3z = 45,
6x + 2y + 3z = 70.
Answer. The given system of equations are
4x + 3y + 2z = 60,
x + 2y + 3z = 45,
6x + 2y + 3z = 70
The given system of equations can be written as
AX = B

= 4 (6 – 6) – 3 (3 – 18) + 2 (2 – 12)
= 0 + 45 – 20 = 25 ≠ 0
∴ A–1 exists. So system of equations has a unique
solution X = A–1B
Now,
A11 = 0, A12 =15, A13 = – 10, A21 = –5, A22 = 0,
A23 = 10, A31 = 5, A32 = – 10, A33 = 5,

Question.

find A–1 and hence solve the following system of equations :
3x – 4y + 2z = –1, 2x + 3y + 5z = 7 and x + z = 2

= 9 – 12 – 6 = – 9 ≠ 0
∴ A–1 exists. So, system of equations has a unique
solution X = A–1B
Now, A11 = 3, A12 = 3, A13 = –3, A21 = 4, A22 = 1,
A23 = –4, A31 = –26, A32 = –11, A33 = 17

The given system of equations is
3x – 4y + 2z = –1,
2x + 3y + 5z = 7
x + z = 2
The given system of equations can be written as
AX = B

Question.

find AB. Hence solve the system of equations :
x – 2y = 10, 2x + y + 3z = 8 and – 2y + z = 7.

Question.

find A–1 and hence solve the system of equations
x – 2y + z = 0, – y + z = – 2, 2x – 3z = 10.

Now the given linear equations are
x – 2y + z = 0
– y + z = – 2
2x – 3z = 10
The system of equations can be written as AX = B

Question.

find A–1. Using A–1, solve the following system of equations:
2x – y + z = – 3, 3x – z = 0, 2x + 6y – 2 = 0

= 2(0 + 6) + 1(0 + 2) + 1(18 – 0) = 12 + 2 + 18 = 32 ≠ 0
∴ A–1 exists.
Now, A11 = 6, A12 = –2, A13 = 18, A21 = 6, A22 = –2,
A23 = –14, A31 = 1, A32 = 5, A33 = 3

Now the given equations are
2x – y + z = – 3,
3x – z = 0,
2x + 6y – 2 = 0
The given system of equations can be written as AX = B

Question.

Hence, solve the following system of equations:
3x + 2y + z = 6, 4x – y + 2z = 5, 7x + 3y – 3z = 7.

= 3(3 – 6) – 2(– 12 – 14) + 1(12 + 7)
= – 9 + 52 + 19 = 62 ≠ 0 ∴ A– 1exists
Now,
A11 = – 3, A12 = 26, A13 = 19, A21 = 9, A22 = – 16,
A23 = 5, A31 = 5, A32 = – 2, A33 = – 11

Now, the given equations are
3x + 2y + z = 6
4x – y + 2z = 5
7x + 3y – 3z = 7
Given equations can be written as
AX = B

Question.

find A–1. Hence solve the following system of equations :
8x – 4y + z = 5, 10x + 6z = 4, 8x + y + 6z = 5/2

= 8(0 – 6) + 4(60 – 48) + 1(10 – 0)
= – 48 + 48 + 10 = 10 ≠ 0
∴ A–1 exists.
Now, A11 = – 6, A12 = – 12, A13 = 10, A21 = 25,
A22 = 40, A23 = – 40, A31 = – 24, A32 = – 38, A33 = 40

Since A–1 exists. So, system of equations has a unique solution given by X = A–1B

Question. Using matrices, solve the following system of equations :
x + y + z = 6, x + 2z = 7, 3x + y + z = 12.
Answer. Given equations can be written as
AX = B ⇒ A–1B

Question. Using matrices, solve the following system of equations :
2x + y + z = 7, x – y – z = –4, 3x + 2y + z = 10.
Answer. Given equations can be written as AX = B

Question. Using matrices, solve the following system of equations :
x + y + z = 6, x – y + z = 2, 2x + y – z = 1.
Answer. The given equations can be written as
AX = B ⇒ X = A–1 B

Question. Using matrices, solve the following system of equations for x, y and z :
x + 2y – 3z = 6, 3x + 2y – 2z = 3, 2x – y + z = 2.
Answer. The given equations can be written as
AX = B ⇒ X = A–1 B

= 1(2 – 2) – 2(3 + 4) – 3(– 3 – 4)
= – 14 + 21 = 7 ≠ 0
∴ A–1 exists.
Now, A11 = 0, A12 = – 7, A13 = – 7, A21 = 1, A22 = 7,
A23 = 5, A31 = 2, A32 = – 7, A33 = – 4,

Question. Using matrices, solve the following system of equations :
9x – 5y – 11z = 12, x – 3y + z = 1
2x + 3y – 7z = 2.
9x – 5y – 11z = 12
x – 3y + z = 1
2x + 3y – 7z = 2
Given equations can be written as
AX = B

Since, A–1 exists, therefore, system of equations has unique solution given by X = A–1· B

Question. Using matrices, solve the following system of equations :
2x – y + z = 2, 3x – z = 2, x + 2y = 3.