case study based questions on determinants

Class 12 Maths: Case Study of Chapter 4 Determinants PDF Download

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In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Mathematics Chapter 4 Determinants Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Maths Determinants  to know their preparation level.

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In CBSE Class 12 Maths Paper, There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Determinants Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 12 Mathematics  Chapter 4 Determinants

Case Study/Passage-Based Questions

(i) What is the award money for Honesty?

Answer: (b) Rs.300

(ii) What is the award money for Punctuality?

Answer: (d) Rs.500

(iii) What is the award money for Hard work?

Answer: (b) Rs.400

(iv) If a matrix P is both symmetric and skew-symmetric, then IPI is equal to

Answer: (c) 0

(v) If P and Q are two matrices such that PQ = Q and QP = P, then IQ 2 1 is equal to

Answer: (a) IQI

Hope the information shed above regarding Case Study and Passage Based Questions for Class 12 Maths Chapter 4 Determinants with Answers Pdf free download has been useful to an extent. If you have any other queries of CBSE Class 12 Mathematics Determinants Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible. By Team Study Rate

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Case Study on Determinants Class 12 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Determinants Class 12 Maths can use this page to download the PDF file. 

The case study questions on Determinants are based on the CBSE Class 12 Maths Syllabus, and therefore, referring to the Determinants case study questions enable students to gain the appropriate knowledge and prepare better for the Class 12 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Determinants Class 12 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Determinants, therefore, they prepared a set of solutions along with the case study questions on Determinants.

The case study on Determinants Class 12 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Determinants case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Determinants Case Study Questions on Class 12 Maths?

There are three major reasons why one should solve Determinants case study questions on Class 12 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 12 Maths students, therefore, it is important to solve Determinants Case study questions as it will help better prepare for the Class 12 board exam preparation.
• Develop Problem-Solving Skills: Class 12 Maths Determinants case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 12 students develop their problem-solving skills, which are essential for success in any profession rather than Class 12 board exam preparation.
• Understand Real-Life Applications: Several Determinants Class 12 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Determinants as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Determinants?

Students can choose their own way to answer Case Study on Determinants Class 12 Maths, however, we believe following these three steps would help a lot in answering Class 12 Maths Determinants Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Determinants questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Determinants Class 12 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 12 Determinants?

 A few essential things to know to solve Case Study Questions on Class 12 Determinants are -

• Basic Formulas of Determinants: One of the most important things to know to solve Case Study Questions on Class 12 Determinants is to learn about the basic formulas or revise them before solving the case-based questions on Determinants.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 12 Maths Determinants case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Determinants Class 12 Maths?

Use Selfstudys.com to find Case Study on Determinants Class 12 Maths. For ease, here is a step-wise procedure to download the Determinants Case Study for Class 12 Maths in PDF for free of cost.

Since you are already on this page, you can scroll to the top section of this page to get access to the Case Study on Determinants. To help others reach this page let them know these steps:

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• Once the website gets loaded, click on the navigation button

• Find CBSE from the given menu

• Click on Case Study

• Choose Class 12 
• Search Maths and then navigate to the Determinants Class 12 Maths Case Study

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Class 12 Maths Case Study Questions

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Class 12 Maths question paper will have 1-2 Case Study Questions. These questions will carry 5 MCQs and students will attempt any four of them. As all of these are only MCQs, it is easy to score good marks with a little practice. Class 12 Maths Case Study Questions are available on the myCBSEguide App and Student Dashboard .

Why Case Studies in CBSE Syllabus?

CBSE has introduced case study questions in the CBSE curriculum recently. The purpose was to make students ready to face real-life challenges with the knowledge acquired in their classrooms. It means, there was a need to connect theories with practicals. Whatsoever the students are learning, they must know how to apply it in their day-to-day life. That’s why CBSE is emphasizing case studies and competency-based education .

Case Study Questions in Maths

Let’s have a look over the class 12 Mathematics sample question paper issued by CBSE, New Delhi. Question numbers 17 and 18 are case study questions.

Focus on concepts

If you go through each MCQ there, you will find that the theme/case study is common but the questions are based on different concepts related to the theme. It means, that if you have done ample practice on the various concepts, you can solve all these MCQs in minutes.

Easy Questions with a Practical Approach

The difficulty level of the questions is average or say easy in some cases. On the other hand, you get four options to choose from. So, you get two levels of support to get full marks with very little effort.

Practice Questions Regularly

Most of the time we feel that it’s easy and neglect it. But in the end, we have to pay for this negligence. This may happen here too. Although it’s easy to score good marks on the case study questions if you don’t practice such questions, you may lose your marks. So, we suggest students should practice at least 30-40 such questions before writing the board exam.

12 Maths Case-Based Questions

We are giving you some examples of case study questions here. We have arranged hundreds of such questions chapter-wise on the myCBSEguide App. It is the complete guide for CBSE students. You can download the myCBSEguide App and get more case study questions there.

Case Study Question – 1

• A is a diagonal matrix
• A is a scalar matrix
• A is a zero matrix
• A is a square matrix
• If A and B are two matrices such that AB = B and BA = A, then B 2 is equal to

Case Study Question – 2

• 4(x 3  – 24x 2   + 144x)
• 4(x 3 – 34x 2   + 244x)
• x 3  – 24x 2   + 144x
• 4x 3  – 24x 2   + 144x
• Local maxima at x = c 1
• Local minima at x = c 1
• Neither maxima nor minima at x = c 1
• None of these

Case Study Questions Matrices -1

Answer Key:

Case Study Questions Matrices – 2

Read the case study carefully and answer any four out of the following questions: Once a mathematics teacher drew a triangle ABC on the blackboard. Now he asked Jose,” If I increase AB by 11 cm and decrease the side BC by 11 cm, then what type of triangle it would be?” Jose said, “It will become an equilateral triangle.”

Again teacher asked Suraj,” If I multiply the side AB by 4 then what will be the relation of this with side AC?” Suraj said it will be 10 cm more than the three times AC.

Find the sides of the triangle using the matrix method and  answer the following questions:

• (a) 3  ×  3

Case Study Questions Determinants – 01

DETERMINANTS:  A determinant is a square array of numbers (written within a pair of vertical lines) that represents a certain sum of products. We can solve a system of equations using determinants, but it becomes very tedious for large systems. We will only do 2 × 2 and 3 × 3 systems using determinants. Using the properties of determinants solve the problem given below and answer the questions that follow:

Three shopkeepers Ram Lal, Shyam Lal, and Ghansham are using polythene bags, handmade bags (prepared by prisoners), and newspaper envelopes as carrying bags. It is found that the shopkeepers Ram Lal, Shyam Lal, and Ghansham are using (20,30,40), (30,40,20), and (40,20,30) polythene bags, handmade bags, and newspapers envelopes respectively. The shopkeeper’s Ram Lal, Shyam Lal, and Ghansham spent ₹250, ₹270, and ₹200 on these carry bags respectively.

• (b) Shyam Lal
• (a) Ram Lal

Case Study Questions Determinants – 02

Case study questions application of derivatives.

• R(x) = -x 2  + 200x + 150000
• R(x) = x 2  – 200x – 140000
• R(x) = 200x 2  + x + 150000
• R(x) = -x 2  + 100 x + 100000
• R'(x) > 0
• R'(x) < 0
• R”(x) = 0
• (a) -x 2  + 200x + 150000
• (a) R'(x) = 0
• (c) 257, -63

Case Study Questions Vector Algebra

• tan−1⁡(5/12)
• tan−1⁡(12/3)
• (b) 130 m/s
• (a)  tan−1⁡(5/12)
• (b) 170 m/s

More Case Study Questions

These are only some samples. If you wish to get more case study questions for CBSE class 12 maths, install the myCBSEguide App. It has class 12 Maths chapter-wise case studies with solutions.

12 Maths Exam pattern

Question Paper Design of CBSE class 12 maths is as below. It clearly shows that 20% weightage will be given to HOTS questions. Whereas 55% of questions will be easy to solve.

• No. chapter-wise weightage. Care to be taken to cover all the chapters
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections

12 Maths Prescribed Books

• Mathematics Part I – Textbook for Class XII, NCERT Publication
• Mathematics Part II – Textbook for Class XII, NCERT Publication
• Mathematics Exemplar Problem for Class XII, Published by NCERT
• Mathematics Lab Manual class XII, published by NCERT

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Case Based Questions (MCQ)

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Question 3 - Case Based Questions (MCQ) - Chapter 4 Class 12 Determinants

Last updated at April 2, 2024 by Teachoo

Two schools Oxford and Navdeep want to award  their selected students on the values of sincerity,  truthfulness and helpfulness. Oxford wants  to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students  respectively with a total award money of ₹ 1600.  Navdeep wants to spend ₹ 2300 to award its 4, 1  and 3 students on the respective values (by giving  the same amount to the three values as before).  The total amount of the award for one prize on  each is ₹ 900.

(i) x + y + z =  _______.

(a) 800  , (b) 900 , (c) 1000  .

(ii) 4x + y + 3z =  _______.

(a) 1600  , (c) 900  .

The value of y is  _______.

(a) 200  , (c) 300  .

The value of 2x + 3y is  _______.

(a) 1000  , (c) 1200  .

y – x =  _______.

(a) 100  .

Question Two schools Oxford and Navdeep want to award their selected students on the values of sincerity, truthfulness and helpfulness. Oxford wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹ 1600. Navdeep wants to spend ₹ 2300 to award its 4, 1 and 3 students on the respective values (by giving the same amount to the three values as before). The total amount of the award for one prize on each is ₹ 900. Given that Value of award for sincerity = Rs x Value of award for truthfulness = Rs y Value of award for helpfulness = Rs z Given that Total amount of the award for one prize on each is ₹ 900 x + y + z = 900 Also, Oxford wants to award ₹ x each, ₹ y each and ₹ z each for the three respective values to 3, 2 and 1 students respectively with a total award money of ₹ 1600 3x + 2y + z = 1600 And, Navdeep wants to spend ₹ 2300 to award its 4, 1 and 3 students on the respective values. 4x + y + 3z = 2300 Question 1 (i) x + y + z = _______. (a) 800 (b) 900 (c) 1000 (d) 1200 From (1) x + y + z = 900 So, the correct answer is (B) Question 2 (ii) 4x + y + 3z = _______. (a) 1600 (b) 2300 (c) 900 (d) 1200 From (3) 4x + y + 3z = 2300 So, the correct answer is (B) Question 3 The value of y is _______. (a) 200 (b) 250 (c) 300 (d) 350 Now, our equations are x + y + z = 900 3x + 2y + z = 1600 4x + y + 3z = 2300 Writing equation as AX = B [■8(1&1&1@3&2&1@4&1&3)] [■8(𝑥@𝑦@𝑧)] = [■8(900@1600@2300)] Hence A = [■8(1&1&1@3&2&1@4&1&3)], X = [■8(𝑥@𝑦@𝑧)] & B = [■8(900@1600@2300)] Calculating |A| |A| = |■8(1&1&1@3&2&1@4&1&3)| = 1 |■8(2&1@1&3)| – 1 |■8(3&1@4&3)| + 1 |■8(3&2@4&1)| = 1(6 − 1) − 1 (9 − 4) + 1 (3 − 8) = 1 (5) − 1 (5) + 1 (–5) = 5 − 5 − 5 = −5 Thus, |A| ≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Calculating A-1 A-1 = 1/(|A|) adj (A) adj (A) = [■8(A_11&A_12&A_13@A_21&A_22&A_23@A_31&A_32&A_33 )]^′ = [■8(A_11&A_21&A_31@A_12&A_22&A_23@A_13&A_32&A_33 )] And, A = [■8(1&1&1@3&2&1@4&1&3)] M11 = [■8(2&1@1&3)] = 6 − 1 = 5 M12 = [■8(3&1@4&3)] = 9 − 4 = 5 M13 = [■8(3&2@4&1)] = 3 − 8 = −5 M21 = [■8(1&1@1&3)] = 3 − 1 = 2 M22 = [■8(1&1@4&3)] = 3 − 4 = −1 M23 = [■8(1&1@4&1)] = 1 − 4 = −3 M31 = [■8(1&1@2&1)] = 1 − 2 = −1 M32 = [■8(1&1@3&1)] = 1 − 3 = −2 M33 = [■8(1&1@3&2)] = 2 − 3 = −1 Now, A11 = (–1)1+1 . M11 = (–1)2 . (5) = 5 A12 = (–1)1+2 . M12 = (–1)3 . (5) = −5 A13 = (–1)1+3 . M13 = (–1)4 . (−5) = −5 A21 = (–1)2+1 . M21 = (–1)3 . (2) = −2 A22 = (–1)2+2 . M22 = (–1)4 . (−1) = −1 A23 = (–1)2+3 . M23 = (–1)5 . (−3) = 3 A31 = (–1)3+1 . M31 = (–1)4 . (−1) = −1 A32 = (–1)3+2 . M32 = (–1)5 . (−2) = 2 A33 = (–1)3+3 . M33 = (–1)6 . (−1) = −1 Thus, adj (A) =[■8(5&−2&−1@−5&−1&2@−5&3&−1)] Now, A-1 = 1/(|A|) adj A Putting values = 1/(−5) [■8(5&−2&−1@−5&−1&2@−5&3&−1)] = 1/5 [■8(−5&2&1@5&1&−2@5&−3&1)] Also, X = A-1 B Putting values [█(■8(𝑥@𝑦)@𝑧)] = 1/5 [■8(−5&2&1@5&1&−2@5&−3&1)][■8(900@1600@2300)] [█(■8(𝑥@𝑦)@𝑧)] = 1/5 [■8(−5(900)+2(1600)+1(2300)@5(900)+1(1600)+(−2)(2300)@5(900)+(−3)(1600)+1(2300) )] [█(■8(𝑥@𝑦)@𝑧)] = 1/5 [■8(−4500+3200+2300@4500+1600−4600@4500−4800+2300)] = 1/5 [█(■8(1000@1500)@2000)] [█(■8(𝑥@𝑦)@𝑧)] = [█(■8(200@300)@400)] Hence, x = 200, y = 300 & z = 400 Since y = 300 So, the correct answer is (C) Question 4 The value of 2x + 3y is _______. (a) 1000 (b) 1100 (c) 1200 (d) 1300 Since, x = 200, y = 300 & z = 400 Thus, 2x + 3y = 2(200) + 3(300) = 400 + 900 = 1300 So, the correct answer is (d) Question 5 y – x = _______. (a) 100 (b) 200 (c) 300 (d) 400 Since, x = 200, y = 300 & z = 400 Thus, y − x = 300 − 200 = 100 So, the correct answer is (A)

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12th Class Mathematics Determinants Question Bank

Done case based (mcqs) - determinants total questions - 15.

A) x - y = 50, 2x - y = 550 done clear

B) x - y = 50, 2x + y = 550 done clear

C) x + y = 50, 2x + y = 550 done clear

D) x + y = 50, 2x + y = 550. done clear

question_answer 2) Which of the following matrix equation is represented by the given information?

question_answer 3) The value of x (length of rectangular field) is

A) 150 m done clear

B) 400 m done clear

C) 200 m done clear

D) 320 m done clear

question_answer 4) The value of y (breadth of rectangular field) is

B) 200 m done clear

C) 430m done clear

D) 350m done clear

question_answer 5) How much is the area of rectangular field?

A) 60000 Sq.m. done clear

B) 30000 Sq.m. done clear

C) 30000 m done clear

D) 3000 m. done clear

A) 700 done clear

B) 7,000 done clear

C) 6,125 done clear

D) 7875 done clear

question_answer 7) What is the total amount of money (in Rs) collected by schools CVS and KVS?

A) 14,000 done clear

B) 15,725 done clear

C) 21,000 done clear

D) 13,125 done clear

question_answer 8) What is the total amount of money collected by all three schools DPS, CVS and KVS

A) Rs 15,775 done clear

B) Rs 14,000 done clear

C) Rs 21,000 done clear

D) Rs 17,125 done clear

question_answer 9) If the number of handmade fans and plates are interchanged for all the schools, then what is the total money collected by all schools?

A) Rs 18,000 done clear

B) Rs 6,750 done clear

C) Rs 5,000 done clear

D) Rs 21,250. done clear

question_answer 10) How many articles (in total) are sold by three schools?

A) 230 done clear

B) 130 done clear

C) 430 done clear

D) 330 done clear

A) \[5x-4y=40,\text{ }5x-8y=-80\] done clear

B) \[5x-4y=40,\text{ }5x-8y=80\] done clear

C) \[5x-4y=40,\text{ }5x+8y=-80\] done clear

D) \[5x+4y=40,\text{ }5x-8y=-80\] done clear

question_answer 12) Which of the following matrix equations represent the information given above?

question_answer 13) The number of children who were given some money by Seema, is

A) 30 done clear

B) 40 done clear

C) 23 done clear

D) 32 done clear

question_answer 14) How much amount is given to each child by Seema?

A) Rs 32 done clear

B) Rs 30 done clear

C) Rs 62 done clear

D) Rs 26 done clear

question_answer 15) How much amount Seema spends in distributing the money to all the students of the orphanage?

A) Rs 609 done clear

B) Rs 960 done clear

C) Rs 906 done clear

D) Rs 690 done clear

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3.4: Applications of the Determinant

• Last updated
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• Page ID 20006

• Ken Kuttler
• Brigham Young University via Lyryx
• Use determinants to determine whether a matrix has an inverse, and evaluate the inverse using cofactors.
• Apply Cramer’s Rule to solve a \(2\times 2\) or a \(3\times 3\) linear system.
• Given data points, find an appropriate interpolating polynomial and use it to estimate points.

A Formula for the Inverse

The determinant of a matrix also provides a way to find the inverse of a matrix. Recall the definition of the inverse of a matrix in Definition 2.6.1 . We say that \(A^{-1}\) , an \(n \times n\) matrix, is the inverse of \(A\) , also \(n \times n\) , if \(AA^{-1} = I\) and \(A^{-1}A=I\) .

We now define a new matrix called the cofactor matrix of \(A\) . The cofactor matrix of \(A\) is the matrix whose \(ij^{th}\) entry is the \(ij^{th}\) cofactor of \(A\) . The formal definition is as follows.

Definition \(\PageIndex{1}\): The Cofactor Matrix

Let \(A=\left[ a_{ij}\right]\) be an \(n\times n\) matrix. Then the cofactor matrix of \(A\) , denoted \(\mathrm{cof}\left( A\right)\) , is defined by \(\mathrm{cof}\left( A\right) =\left[ \mathrm{cof}\left(A\right)_{ij}\right]\) where \(\mathrm{cof}\left(A\right)_{ij}\) is the \(ij^{th}\) cofactor of \(A\) .

Note that \(\mathrm{cof}\left(A\right)_{ij}\) denotes the \(ij^{th}\) entry of the cofactor matrix.

We will use the cofactor matrix to create a formula for the inverse of \(A\) . First, we define the adjugate of \(A\) to be the transpose of the cofactor matrix. We can also call this matrix the classical adjoint of \(A\) , and we denote it by \(adj \left(A\right)\) .

In the specific case where \(A\) is a \(2 \times 2\) matrix given by \[A = \left[ \begin{array}{rr} a & b \\ c & d \end{array} \right]\nonumber \] then \({adj}\left(A\right)\) is given by \[{adj}\left(A\right) = \left[ \begin{array}{rr} d & -b \\ -c & a \end{array} \right]\nonumber \]

In general, \({adj}\left(A\right)\) can always be found by taking the transpose of the cofactor matrix of \(A\) . The following theorem provides a formula for \(A^{-1}\) using the determinant and adjugate of \(A\) .

Theorem \(\PageIndex{1}\): The Inverse and the Determinant

Let \(A\) be an \(n\times n\) matrix. Then \[A \; {adj}\left(A\right) = {adj}\left(A\right)A = {\det \left(A\right)} I\nonumber \]

Moreover \(A\) is invertible if and only if \(\det \left(A\right) \neq 0\) . In this case we have: \[A^{-1} = \frac{1}{\det \left(A\right)} {adj}\left(A\right)\nonumber \]

Notice that the first formula holds for any \(n \times n\) matrix \(A\) , and in the case \(A\) is invertible we actually have a formula for \(A^{-1}\) .

Consider the following example.

Example \(\PageIndex{1}\): Find Inverse Using the Determinant

Find the inverse of the matrix \[A=\left[ \begin{array}{rrr} 1 & 2 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 1 \end{array} \right]\nonumber \] using the formula in Theorem \(\PageIndex{1}\) .

According to Theorem \(\PageIndex{1}\) , \[A^{-1} = \frac{1}{\det \left(A\right)} {adj}\left(A\right)\nonumber \]

First we will find the determinant of this matrix. Using Theorems 3.2.1 , 3.2.2 , and 3.2.4 , we can first simplify the matrix through row operations. First, add \(-3\) times the first row to the second row. Then add \(-1\) times the first row to the third row to obtain \[B = \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 0 & -6 & -8 \\ 0 & 0 & -2 \end{array} \right]\nonumber \] By Theorem 3.2.4 , \(\det \left(A\right) = \det \left(B\right)\) . By Theorem 3.1.2 , \(\det \left(B\right) = 1 \times -6 \times -2 = 12\) . Hence, \(\det \left(A\right) = 12\) .

Now, we need to find \({adj} \left(A\right)\) . To do so, first we will find the cofactor matrix of \(A\) . This is given by \[\mathrm{cof}\left( A\right) = \left[ \begin{array}{rrr} -2 & -2 & 6 \\ 4 & -2 & 0 \\ 2 & 8 & -6 \end{array} \right]\nonumber \] Here, the \(ij^{th}\) entry is the \(ij^{th}\) cofactor of the original matrix \(A\) which you can verify. Therefore, from Theorem \(\PageIndex{1}\) , the inverse of \(A\) is given by \[A^{-1} = \frac{1}{12}\left[ \begin{array}{rrr} -2 & -2 & 6 \\ 4 & -2 & 0 \\ 2 & 8 & -6 \end{array} \right] ^{T}= \left[ \begin{array}{rrr} -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \\ -\frac{1}{6} & -\frac{1}{6} & \frac{2}{3} \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{array} \right]\nonumber \]

Remember that we can always verify our answer for \(A^{-1}\) . Compute the product \(AA^{-1}\) and \(A^{-1}A\) and make sure each product is equal to \(I\) .

Compute \(A^{-1}A\) as follows \[A^{-1}A = \left[ \begin{array}{rrr} -\frac{1}{6} & \frac{1}{3} & \frac{1}{6} \\ -\frac{1}{6} & -\frac{1}{6} & \frac{2}{3} \\ \frac{1}{2} & 0 & -\frac{1}{2} \end{array} \right] \left[ \begin{array}{rrr} 1 & 2 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 1 \end{array} \right] = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] = I\nonumber \] You can verify that \(AA^{-1} = I\) and hence our answer is correct.

We will look at another example of how to use this formula to find \(A^{-1}\) .

Example \(\PageIndex{2}\): Find the Inverse From a Formula

Find the inverse of the matrix \[A=\left[ \begin{array}{rrr} \frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{6} & \frac{1}{3} & - \frac{1}{2} \\ -\frac{5}{6} & \frac{2}{3} & - \frac{1}{2} \end{array} \right]\nonumber \] using the formula given in Theorem \(\PageIndex{1}\) .

First we need to find \(\det \left(A\right)\) . This step is left as an exercise and you should verify that \(\det \left(A\right) = \frac{1}{6}.\) The inverse is therefore equal to \[A^{-1} = \frac{1}{(1/6)}\; {adj} \left(A\right) = 6\; {adj} \left(A\right)\nonumber \]

We continue to calculate as follows. Here we show the \(2 \times 2\) determinants needed to find the cofactors. \[A^{-1} = 6\left[ \begin{array}{rrr} \left| \begin{array}{rr} \frac{1}{3} & -\frac{1}{2} \\ \frac{2}{3} & -\frac{1}{2} \end{array} \right| & -\left| \begin{array}{rr} -\frac{1}{6} & -\frac{1}{2} \\ -\frac{5}{6} & -\frac{1}{2} \end{array} \right| & \left| \begin{array}{rr} -\frac{1}{6} & \frac{1}{3} \\ -\frac{5}{6} & \frac{2}{3} \end{array} \right| \\ -\left| \begin{array}{rr} 0 & \frac{1}{2} \\ \frac{2}{3} & -\frac{1}{2} \end{array} \right| & \left| \begin{array}{rr} \frac{1}{2} & \frac{1}{2} \\ -\frac{5}{6} & -\frac{1}{2} \end{array} \right| & -\left| \begin{array}{rr} \frac{1}{2} & 0 \\ -\frac{5}{6} & \frac{2}{3} \end{array} \right| \\ \left| \begin{array}{rr} 0 & \frac{1}{2} \\ \frac{1}{3} & -\frac{1}{2} \end{array} \right| & -\left| \begin{array}{rr} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{6} & -\frac{1}{2} \end{array} \right| & \left| \begin{array}{rr} \frac{1}{2} & 0 \\ -\frac{1}{6} & \frac{1}{3} \end{array} \right| \end{array} \right] ^{T}\nonumber \]

Expanding all the \(2\times 2\) determinants, this yields \[A^{-1} = 6\left[ \begin{array}{rrr} \frac{1}{6} & \frac{1}{3} & \frac{1}{6} \\ \frac{1}{3} & \frac{1}{6} & -\frac{1}{3} \\ -\frac{1}{6} & \frac{1}{6} & \frac{1}{6} \end{array} \right] ^{T}= \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 1 & -2 & 1 \end{array} \right]\nonumber \]

Again, you can always check your work by multiplying \(A^{-1}A\) and \(AA^{-1}\) and ensuring these products equal \(I\) . \[A^{-1}A = \left[ \begin{array}{rrr} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 1 & -2 & 1 \end{array} \right] \left[ \begin{array}{rrr} \frac{1}{2} & 0 & \frac{1}{2} \\ -\frac{1}{6} & \frac{1}{3} & - \frac{1}{2} \\ -\frac{5}{6} & \frac{2}{3} & - \frac{1}{2} \end{array} \right] = \left[ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]\nonumber \] This tells us that our calculation for \(A^{-1}\) is correct. It is left to the reader to verify that \(AA^{-1} = I\) .

The verification step is very important, as it is a simple way to check your work! If you multiply \(A^{-1}A\) and \(AA^{-1}\) and these products are not both equal to \(I\) , be sure to go back and double check each step. One common error is to forget to take the transpose of the cofactor matrix, so be sure to complete this step.

We will now prove Theorem \(\PageIndex{1}\) .

(of Theorem \(\PageIndex{1}\) ) Recall that the \((i,j)\) -entry of \({adj}(A)\) is equal to \(\mathrm{cof}(A)_{ji}\) . Thus the \((i,j)\) -entry of \(B=A\cdot {adj}(A)\) is : \[B_{ij}=\sum_{k=1}^n a_{ik} {adj} (A)_{kj}= \sum_{k=1}^n a_{ik} \mathrm{cof} (A)_{jk}\nonumber \] By the cofactor expansion theorem, we see that this expression for \(B_{ij}\) is equal to the determinant of the matrix obtained from \(A\) by replacing its \(j\) th row by \(a_{i1}, a_{i2}, \dots a_{in}\) — i.e., its \(i\) th row.

If \(i=j\) then this matrix is \(A\) itself and therefore \(B_{ii}=\det A\) . If on the other hand \(i\neq j\) , then this matrix has its \(i\) th row equal to its \(j\) th row, and therefore \(B_{ij}=0\) in his case. Thus we obtain: \[A \; {adj}\left(A\right) = {\det \left(A\right)} I\nonumber \] Similarly we can verify that: \[{adj}\left(A\right)A = {\det \left(A\right)} I\nonumber \] And this proves the first part of the theorem.

Further if \(A\) is invertible, then by Theorem 3.2.5 we have: \[1 = \det \left( I \right) = \det \left( A A^{-1} \right) = \det \left( A \right) \det \left( A^{-1} \right)\nonumber \] and thus \(\det \left( A \right) \neq 0\) . Equivalently, if \(\det \left( A \right) = 0\) , then \(A\) is not invertible.

Finally if \(\det \left( A \right) \neq 0\) , then the above formula shows that \(A\) is invertible and that: \[A^{-1} = \frac{1}{\det \left(A\right)} {adj}\left(A\right)\nonumber \]

This completes the proof.

This method for finding the inverse of \(A\) is useful in many contexts. In particular, it is useful with complicated matrices where the entries are functions, rather than numbers.

Example \(\PageIndex{3}\): Inverse for Non-Constant Matrix

Suppose \[A\left( t\right) =\left[ \begin{array}{ccc} e^{t} & 0 & 0 \\ 0 & \cos t & \sin t \\ 0 & -\sin t & \cos t \end{array} \right]\nonumber \] Show that \(A\left( t\right) ^{-1}\) exists and then find it.

First note \(\det \left( A\left( t\right) \right) = e^{t}(\cos^2 t + \sin^2 t) = e^{t}\neq 0\) so \(A\left( t\right) ^{-1}\) exists.

The cofactor matrix is \[C\left( t\right) =\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & -e^{t}\sin t & e^{t}\cos t \end{array} \right]\nonumber \] and so the inverse is \[\frac{1}{e^{t}}\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & -e^{t}\sin t & e^{t}\cos t \end{array} \right] ^{T}= \left[ \begin{array}{ccc} e^{-t} & 0 & 0 \\ 0 & \cos t & -\sin t \\ 0 & \sin t & \cos t \end{array} \right]\nonumber \]

Cramer’s Rule

Another context in which the formula given in Theorem \(\PageIndex{1}\) is important is Cramer’s Rule . Recall that we can represent a system of linear equations in the form \(AX=B\) , where the solutions to this system are given by \(X\) . Cramer’s Rule gives a formula for the solutions \(X\) in the special case that \(A\) is a square invertible matrix. Note this rule does not apply if you have a system of equations in which there is a different number of equations than variables (in other words, when \(A\) is not square), or when \(A\) is not invertible.

Suppose we have a system of equations given by \(AX=B\) , and we want to find solutions \(X\) which satisfy this system. Then recall that if \(A^{-1}\) exists, \[\begin{aligned} AX&=B \\ A^{-1}\left(AX\right)&=A^{-1}B \\ \left(A^{-1}A\right)X&=A^{-1}B \\ IX&=A^{-1}B\\ X &= A^{-1}B\end{aligned}\] Hence, the solutions \(X\) to the system are given by \(X=A^{-1}B\) . Since we assume that \(A^{-1}\) exists, we can use the formula for \(A^{-1}\) given above. Substituting this formula into the equation for \(X\) , we have \[X=A^{-1}B=\frac{1}{\det \left( A\right) }{adj}\left( A\right)B\nonumber \] Let \(x_i\) be the \(i^{th}\) entry of \(X\) and \(b_j\) be the \(j^{th}\) entry of \(B\) . Then this equation becomes \[x_i = \sum_{j=1}^{n}\left[ a_{ij}\right]^{-1}b_{j}=\sum_{j=1}^{n}\frac{1} {\det \left( A\right) } {adj}\left( A\right) _{ij}b_{j}\nonumber \] where \({adj}\left(A\right)_{ij}\) is the \(ij^{th}\) entry of \({adj}\left(A\right)\) .

By the formula for the expansion of a determinant along a column, \[x_{i}=\frac{1}{\det \left( A\right) }\det \left[ \begin{array}{ccccc} \ast & \cdots & b_{1} & \cdots & \ast \\ \vdots & & \vdots & & \vdots \\ \ast & \cdots & b_{n} & \cdots & \ast \end{array} \right]\nonumber \] where here the \(i^{th}\) column of \(A\) is replaced with the column vector \(\left[ b_{1}\cdots \cdot ,b_{n}\right] ^{T}\) . The determinant of this modified matrix is taken and divided by \(\det \left( A\right)\) . This formula is known as Cramer’s rule.

We formally define this method now.

Procedure \(\PageIndex{1}\): Using Cramer’s Rule

Suppose \(A\) is an \(n\times n\) invertible matrix and we wish to solve the system \(AX=B\) for \(X =\left[ x_{1},\cdots ,x_{n}\right] ^{T}.\) Then Cramer’s rule says \[x_{i}= \frac{\det \left(A_{i}\right)}{\det \left(A\right)}\nonumber \] where \(A_{i}\) is the matrix obtained by replacing the \(i^{th}\) column of \(A\) with the column matrix \[B = \left[ \begin{array}{c} b_1 \\ \vdots \\ b_n \end{array} \right]\nonumber \]

We illustrate this procedure in the following example.

Example \(\PageIndex{4}\): Using Cramer's Rule

Find \(x,y,z\) if \[\left[ \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 1 \\ 2 & -3 & 2 \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right]\nonumber \]

We will use method outlined in Procedure \(\PageIndex{1}\) to find the values for \(x,y,z\) which give the solution to this system. Let \[B = \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right]\nonumber\]

In order to find \(x\) , we calculate \[x = \frac{\det \left(A_{1}\right)}{\det \left(A\right)}\nonumber \] where \(A_1\) is the matrix obtained from replacing the first column of \(A\) with \(B\) .

Hence, \(A_1\) is given by \[A_1 = \left[ \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 3 & -3 & 2 \end{array} \right]\nonumber \]

Therefore, \[x= \frac{\det \left(A_{1}\right)}{\det \left(A\right)} = \frac{\left| \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 3 & -3 & 2 \end{array} \right| }{\left| \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 1 \\ 2 & -3 & 2 \end{array} \right| }=\frac{1}{2}\nonumber \]

Similarly, to find \(y\) we construct \(A_2\) by replacing the second column of \(A\) with \(B\) . Hence, \(A_2\) is given by \[A_2 = \left[ \begin{array}{rrr} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 3 & 2 \end{array} \right]\nonumber \]

Therefore, \[y=\frac{\det \left(A_{2}\right)}{\det \left(A\right)} = \frac{\left| \begin{array}{rrr} 1 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 3 & 2 \end{array} \right| }{\left| \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 1 \\ 2 & -3 & 2 \end{array} \right| }=-\frac{1}{7}\nonumber \]

Similarly, \(A_3\) is constructed by replacing the third column of \(A\) with \(B\) . Then, \(A_3\) is given by \[A_3 = \left[ \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 2 \\ 2 & -3 & 3 \end{array} \right]\nonumber \]

Therefore, \(z\) is calculated as follows.

\[z= \frac{\det \left(A_{3}\right)}{\det \left(A\right)} = \frac{\left| \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 2 \\ 2 & -3 & 3 \end{array} \right| }{\left| \begin{array}{rrr} 1 & 2 & 1 \\ 3 & 2 & 1 \\ 2 & -3 & 2 \end{array} \right| }=\frac{11}{14}\nonumber \]

Cramer’s Rule gives you another tool to consider when solving a system of linear equations.

We can also use Cramer’s Rule for systems of non linear equations. Consider the following system where the matrix \(A\) has functions rather than numbers for entries.

Using Cramer’s Rule

Example \(\PageIndex{5}\): Use Cramer's Rule for Non-Constant Matrix

Solve for \(z\) if \[\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & -e^{t}\sin t & e^{t}\cos t \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] =\left[ \begin{array}{c} 1 \\ t \\ {0.05in}t^{2} \end{array} \right]\nonumber \]

We are asked to find the value of \(z\) in the solution. We will solve using Cramer’s rule. Thus \[z={.05in} \frac{\left| \begin{array}{ccc} 1 & 0 & 1 \\ 0 & e^{t}\cos t & t \\ 0 & -e^{t}\sin t & t^{2} \end{array} \right| }{\left| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & e^{t}\cos t & e^{t}\sin t \\ 0 & -e^{t}\sin t & e^{t}\cos t \end{array} \right| }= t\left( \left( \cos t\right) t+\sin t\right) e^{-t}\nonumber \]

Polynomial Interpolation

In studying a set of data that relates variables \(x\) and \(y\) , it may be the case that we can use a polynomial to “fit” to the data. If such a polynomial can be established, it can be used to estimate values of \(x\) and \(y\) which have not been provided.

Example \(\PageIndex{6}\): Polynomial Interpolation

Given data points \((1,4), (2,9), (3,12)\) , find an interpolating polynomial \(p(x)\) of degree at most \(2\) and then estimate the value corresponding to \(x = \frac{1}{2}\) .

We want to find a polynomial given by \[p(x) = r_0 + r_1x_1 + r_2x_2^2\nonumber \] such that \(p(1)=4, p(2)=9\) and \(p(3)=12\) . To find this polynomial, substitute the known values in for \(x\) and solve for \(r_0, r_1\) , and \(r_2\) . \[\begin{aligned} p(1) &= r_0 + r_1 + r_2 = 4\\ p(2) &= r_0 + 2r_1 + 4r_2 = 9\\ p(3) &= r_0 + 3r_1 + 9r_2 = 12\end{aligned}\]

Writing the augmented matrix, we have \[\left[ \begin{array}{rrr|r} 1 & 1 & 1 & 4 \\ 1 & 2 & 4 & 9 \\ 1 & 3 & 9 & 12 \end{array} \right]\nonumber\]

After row operations, the resulting matrix is \[\left[ \begin{array}{rrr|r} 1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 8 \\ 0 & 0 & 1 & -1 \end{array} \right]\nonumber \]

Therefore the solution to the system is \(r_0 = -3, r_1 = 8, r_2 = -1\) and the required interpolating polynomial is \[p(x) = -3 + 8x - x^2\nonumber \]

To estimate the value for \(x = \frac{1}{2}\) , we calculate \(p(\frac{1}{2})\) : \[\begin{aligned} p(\frac{1}{2}) &= -3 + 8(\frac{1}{2}) - (\frac{1}{2})^2\\ &= -3 + 4 - \frac{1}{4} \\ &= \frac{3}{4}\end{aligned}\]

This procedure can be used for any number of data points, and any degree of polynomial. The steps are outlined below.

Procedure \(\PageIndex{2}\): Finding an Interpolation Polynomial

Suppose that values of \(x\) and corresponding values of \(y\) are given, such that the actual relationship between \(x\) and \(y\) is unknown. Then, values of \(y\) can be estimated using an interpolating polynomial \(p(x)\) . If given \(x_1, ..., x_n\) and the corresponding \(y_1, ..., y_n\) , the procedure to find \(p(x)\) is as follows:

• The desired polynomial \(p(x)\) is given by \[p(x) = r_0 + r_1 x + r_2 x^2 + ... + r_{n-1}x^{n-1}\nonumber \]
• \(p(x_i) = y_i\) for all \(i = 1, 2, ...,n\) so that \[\begin{array}{c} r_0 + r_1x_1 + r_2 x_1^2 + ... + r_{n-1}x_1^{n-1} = y_1 \\ r_0 + r_1x_2 + r_2 x_2^2 + ... + r_{n-1}x_2^{n-1} = y_2 \\ \vdots \\ r_0 + r_1x_n + r_2 x_n^2 + ... + r_{n-1}x_n^{n-1} = y_n \end{array}\nonumber \]
• Set up the augmented matrix of this system of equations \[\left[ \begin{array}{rrrrr|r} 1 & x_1 & x_1^2 & \cdots & x_1^{n-1} & y_1 \\ 1 & x_2 & x_2^2 & \cdots & x_2^{n-1} & y_2 \\ \vdots & \vdots & \vdots & &\vdots & \vdots \\ 1 & x_n & x_n^2 & \cdots & x_n^{n-1} & y_n \\ \end{array} \right]\nonumber \]
• Solving this system will result in a unique solution \(r_0, r_1, \cdots, r_{n-1}\) . Use these values to construct \(p(x)\) , and estimate the value of \(p(a)\) for any \(x=a\) .

This procedure motivates the following theorem.

Theorem \(\PageIndex{2}\): Polynomial Interpolation

Given \(n\) data points \((x_1, y_1), (x_2, y_2), \cdots, (x_n, y_n)\) with the \(x_i\) distinct, there is a unique polynomial \(p(x) = r_0 + r_1x + r_2x^2 + \cdots + r_{n-1}x^{n-1}\) such that \(p(x_i) = y_i\) for \(i=1,2,\cdots, n\) . The resulting polynomial \(p(x)\) is called the interpolating polynomial for the data points.

We conclude this section with another example.

Example \(\PageIndex{7}\): Polynomial Interpolation

Consider the data points \((0,1), (1,2), (3,22), (5,66)\) . Find an interpolating polynomial \(p(x)\) of degree at most three, and estimate the value of \(p(2)\) .

The desired polynomial \(p(x)\) is given by: \[p(x) = r_0 + r_1 x + r_2x^2 + r_3x^3\nonumber \]

Using the given points, the system of equations is \[\begin{aligned} p(0) &= r_0 = 1 \\ p(1) &= r_0 + r_1 + r_2 + r_3 = 2 \\ p(3) &= r_0 + 3r_1 + 9r_2 + 27r_3 = 22 \\ p(5) &= r_0 + 5r_1 + 25r_2 + 125r_3 = 66\end{aligned}\]

The augmented matrix is given by: \[\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 2 \\ 1 & 3 & 9 & 27 & 22 \\ 1 & 5 & 25 & 125 & 66 \end{array} \right]\nonumber\]

The resulting matrix is \[\left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & -2 \\ 0 & 0 & 1 & 0 & 3 \\ 0 & 0 & 0 & 1 & 0 \end{array} \right]\nonumber\]

Therefore, \(r_0 = 1, r_1 = -2, r_2 = 3, r_3 = 0\) and \(p(x) = 1 -2x + 3x^2\) . To estimate the value of \(p(2)\) , we compute \(p(2) = 1 -2(2) + 3(2^2) = 1 - 4 + 12 = 9\) .

CBSE 12th Standard Maths Subject Determinants Case Study Questions With Solutions 2021

QB365 Provides the updated CASE Study Questions for Class 12 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

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12th Standard CBSE

Final Semester - June 2015

Case Study Questions

Let  \(\begin{equation} A=\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right] \end{equation}\)  , and U 1 U 2 are first and second columns respectively of a 2 x 2 matrix U. Also, let the column matrices U I and U 2 satisfying  \(\begin{equation} A U_{1}=\left[\begin{array}{l} 1 \\ 0 \end{array}\right] \end{equation}\)  and  \(\begin{equation} A U_{2}=\left[\begin{array}{l} 2 \\ 3 \end{array}\right] \end{equation}\)   Based on the above information, answer the following questions (i) The matrix U 1 + U 2 is equal to

(ii) The value of IUI is

(iii) If  \(\begin{equation} X=\left[\begin{array}{ll} 3 & 2 \end{array}\right] U\left[\begin{array}{l} 3 \\ 2 \end{array}\right] \end{equation}\) ,then the value of IXI =

(iv) The minor of element at the position a 22 in U is

(v) If  \(\begin{equation} U=\left[a_{i j}\right]_{2 \times 2} \end{equation}\)  , then the value of a 11 A 11 + a 12 A l2  where A ij  denotes the cofactor of a ij  is

(ii) What is the award money for Punctuality?

(iii) What is the award money for Hard work?

(iv) If a matrix P is both symmetric and skew-symmetric, then IPI is equal to

(v) If P and Q are two matrices such that PQ = Q and QP = P, then IQ 2 1 is equal to

(ii) What is the cost of one handmade bag?

(iii) What is the cost of one newspaper envelope

(iv) Keeping in mind the social conditions, which shopkeeper is better?

(v) Keeping in mind the environmental conditions, which shopkeeper is better?

Minor of an element aij of a determinant is the determinant obtained by deleting its ith row and /h column in a ij lies and is denoted by M ij . Cofactor of an element a ij' denoted by A ij ,is defined by  \(\begin{equation} A_{i j}=(-1)^{i+j} M_{i j} \end{equation}\) , where M ij is minor of a ij. Also, the determinant of a square matrix A is the sum of the products of the elements of any row (or column with their corresponding cofactors. For example if  \(\begin{equation} A=\left[a_{i j}\right]_{3 \times 3}, \text { then }|A|=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \end{equation}\)  . Based on the above information, answer the following questions (i) Find the sum of the cofactors of all the elements of  \(\begin{equation} \left|\begin{array}{cc} 1 & -2 \\ 4 & 3 \end{array}\right| \end{equation}\)  

(ii) Find the minor of a 21 of  \(\begin{equation} \left|\begin{array}{ccc} 5 & 6 & -3 \\ -4 & 3 & 2 \\ -4 & -7 & 3 \end{array}\right| \end{equation}\)  

(iii) In the determinant  \(\begin{equation} \left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\)  find the value of a 32 ·A 32

(iv) If  \(\begin{equation} \Delta=\left|\begin{array}{lll} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array}\right| \end{equation}\) , find the value of a 32 ·A 32  .

(v) If  \(\begin{equation} \Delta=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right| \end{equation}\) , then find the value of \(\begin{equation} |\Delta| \end{equation}\) .

(ii) What is the cost of one pen and one bag?

(iii) What is the cost of one pen and one instrument box?

(iv) Which of the following is correct?

(v) From the matrix equation AB = AC, it can be concluded that B = C provided

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Determinants Class 12 Mathematics Important Questions

Students can read the important questions given below for Determinants Class 12 Mathematics. All Determinants Class 12 Notes and questions with solutions have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT and KVS. You should read all notes provided by us and Class 12 Mathematics Important Questions provided for all chapters to get better marks in examinations. Mathematics Question Bank Class 12 is available on our website for free download in PDF.

Important Questions of Determinants Class 12

Very Short Answer Type Questions

Question.  If A is an invertible matrix of order 3 and |A| = 5, then find |Adj A|. Answer.

Question.  If A is a non-singular matrix of order 3 and |Adj A| = |A| K , then write the value of K. Answer.

Question.  If A is a square matrix of order 3 and |3A| = K|A|, then write the value of K. Answer.

Question.  A is a square matrix of order 3 and |A| = 7. Write the value of |adj. A|. Answer. |adj. A|= |A| n–1 = |A| 3 – 1 = (7) 2 = 49

Question.  If |A| = 2, where A is a 2 × 2 matrix, find |adj A|. Answer. |adj A|= |A| n – 1 = (2) 2 –1 = 2

Question.  What is the value of the determinant

Answer. Expanding along R1 = – 2(12 – 16) = – 2(– 4) = 8

Question. Find the minor of the element of second row and third column (a 23 ) in the following determinant :

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

Question.  Write the adjoint of the following matrix :

Question.  A matrix A of order 3 × 3 is such that |A| = 4. Find the value of |2A|. Answer.  |2A| = 2n|A| = 23(4) = 32

Question.  If A is an invertible matrix of order 3 and |A| = 5, then find the value of |adj A|. Answer.  |adj A|= |A| n – 1 = (5) 3–1 = (5) 2 = 25

⇒ x2 – x = 6 – 4 ⇒ x 2 – x = 2 ⇒ x 2 – x – 2 = 0 ⇒ x 2 – 2x + x – 2 = 0 ⇒ x(x – 2) + 1(x – 2) = 0 ⇒ (x – 2) (x + 1) = 0 ⇒ x = 2 or x = – 1 ∴ The positive value of x is 2.

Question.  Let A be a square matrix of order 3 × 3. Write the value of |2A|, where |A| = 4. Answer.  |2A| = 2 3 |A| = 8(4) = 32

Question.  The value of the determinant of a matrix A of order 3 × 3 is 4. Find the value of |5A|. Answer. |A| = 4 |5A| = 5 3 |A| = 125 (4) = 500

Question.  If the determinant of a matrix A of order 3 × 3 is of value 4, write the value of |3A|. Answer. |A| = 4 ⇒ |3A| = 3 3 |A| = 27(4) = 108

Question.  If A is a square matrix of order 3 such that |adj A| = 64, find |A|. Answer.  |adj A| = 64 |A| 2 = 64 [                                    ∵ |adj A| = |A| n–1 , Here n= 3] ∴ |A| = ± 8

Question. If A is an invertible square matrix of order 3 and |A| = 5, then find the value of |adj A|. Answer.  |adjA|= |A| n–1 = (5) 3–1 = 5 2 = 25                                             [Here |A| = 5, n = 3

Answer. (x + 1) (x + 2) – (x – 1) (x – 3) = 12 + 1 ⇒ (x 2 + 2x + x + 2) – (x 2 – 3x – x + 3) = 13 ⇒ 3x + 2 + 4x – 3 = 13 ⇒7x – 1 = 13                ⇒ 7x = 14

Question. If A ij i s the cofactor of the element a ij  of the

Question. Find the inverse of the matrix

Question. Find the maximum value of

Question. If A is a square matrix of order 2 and |adj A| = 9, find |A|. Answer. |adj A|= 9 …(given) |A| n – 1 = 9, (order, adj, n = 2) |A| 2 – 1 = 9 |A| = 9

Question. If for any 2 × 2 square matrix A,

Question. If A is an invertible matrix of order 2 and det (A) = 4, then write the value of det (A –1 ). Answer.

Question. If A is a 3 × 3 invertible matrix, then what will be the value of k if det(A –1 ) = (det A) k Answer.

Short Answer Type Questions-I

Question. If A is a skew-symmetric matrix of order 3, then prove that det A = 0. (All India) Answer.  A is a skew-symmetric matrix A t = – A ⇒ |A t | = |– A| ⇒ |A t | = (– 1) n |A| ⇒ |A t | = (– 1) 3 |A|                    …( ∵ = 3) ⇒ |A| = – |A|                             ( ∵ |A t | = |A|) ⇒ |A| + |A| = 0 ⇒ 2|A| = 0                                  ∴ |A| = 0

Short Answer Type Questions-II

Answer. To find : f (2x) – f (x) 1 st method :

Question. Using properties of determinants, prove the following :

Expanding along C 1 , we have = – 2abc 2 . 1(– ab – ab) = – 2abc 2 (– 2ab) = 4a 2 b 2 c 2 = R.H.S.

Question. Using properties of determinants, solve for x :

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? Answer.

⇒ |A| = 3(2 – 3) – 1(1 – 3) + 2(1 – 2) = – 3 + 2 – 2 = – 3 ≠ 0 A –1 exists, so equations have a unique solution. Hence system of equations can be solved.

Question. Using properties of determinants, prove that

Answer. Given : CD – AB = 0 DC = AB ⇒ D = C –1 AB             …(i) |C| = 16 – 15 = 1 ≠ 0;

Long Answer Type Questions

Question. Using matrices, solve the following system of equations : x + y + z = 1;x – 2y + 3z = 2; x – 3y + 5z = 3 Answer. Given equations can be written as

Question. Using properties of determinants, show the following :

Answer. Part I :

Part II : Given system of equations can be written in matrix form

AB. Use this to solve the following system of equations : x – y = 3, 2x + 3y + 4z = 17,                     y + 2z = 7 Answer.  Part I :

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Class 12 Maths Chapter 4 Determinants MCQs

Class 12 Maths Chapter 4 Determinants MCQs are available here to help the students of Class 12 score good marks in the CBSE board exam 2022-2023. Here, we provided multiple objective type questions that will help the students to practise for the exams. These multiple-choice questions form a useful source to prepare for exams as they are given as per the new CBSE guidelines for the academic year 2022-23. All the topics and subtopics of Chapter 4 Determinants of NCERT textbook are covered here in MCQs.

Get access to Class 12 Maths MCQs for all the chapters here at BYJU’S.

MCQs for Chapter 4 Determinants Class 12

Students can access several MCQs on matrices of Class 12 that cover various topics such as fundamentals of matrix and matrix algebra, i.e. mathematical operations on matrices.

Also, check:

• Determinants for Class 12 Notes
• Important Questions for Class 12 Maths Chapter 4 Determinants

Students must practise more MCQs to understand how to apply the concepts to solve problems in th exam. As we know, the chapter 4 determinants of Class 12 contain several topics and subtopics. They include finding the determinants up to order three only with real entries. Various properties of determinants, minors, cofactors and applications of determinants in finding the triangle area are included.

Download PDF – Chapter 4 Determinants MCQs

Besides, finding the adjoint and inverse of a square matrix, consistency and inconsistency of system of linear equations and solution of linear equations in two or three variables using the inverse of a matrix are included. All these topics are very important from the examination point of view. So, practising MCQs given below will help you get thorough with the formulas and applications of determinants.

MCQs for Class 12 Maths Chapter 4 Determinants with Solutions

Correct option: (c) ±6

So, (2x)(x) – (5)(8) = (6)(3) – (-2)(7)

2x 2 – 40 = 18 + 14

2x 2 = 32 + 40

x = √36 = ±6

Correct option: (c) ±4

Given that matrix A is singular.

Thus, the determinant of A is 0.

k(2k) – 8(4) = 0

2k 2 – 32 = 0

3. If A is a square matrix of order 3 and |A| = 5, then the value of |2A′| is

Correct option: (d) 40

According to the property of transpose of a matrix,

(kA′) = kA′

Also, from the property of determinant of a matrix,

and |kA| = k n |A|, where n is the order of matrix A.

Thus, |2A′| = 2 3 |A| {since A is a square matrix of order 3}

4. The area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units. The value of k will be

Correct option: (b) 3

The formula of area of the triangle with vertices (x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ) is given by:

Thus, the area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is:

⇒ -3(0 – k) – 0+ 1(3k – 0) = 18

⇒ 3k + 3k = 18

5. Given that A is a square matrix of order 3 and |A| = -4, then |adj A| is equal to

Correct option: (d) 16

Given that A is a square matrix of order 3 and |A| = -4.

We know that |adj A| = |A| n−1 , where n is the order of matrix A.

So, |adj A| = (−4) 3-1 = (-4) 2 = 16

(c) λ ≠ – 2

(d) None of these

Correct option: (d) None of these

The inverse of a matrix exists if its determinant is not equal to 0.

⇒ |A| = 2 (6 – 5) – λ (0 – 5) + (-3) (0 – 2) ≠ 0

⇒ 2 + 5λ + 6 ≠ 0

⇒ 5λ + 8 ≠ 0

Therefore, A-1 exists if and only if λ ≠ -8/5.

= (3)(2) – (-1)(1)

14A -1 = 14[adj A/ |A|] = (14/7) adj A

8. Which of the following is correct?

(a) Determinant is a square matrix.

(b) Determinant is a number associated with a matrix.

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

Correct option: (c) Determinant is a number associated with a square matrix.

We know that we can calculate determinant values only for square matrices.

Therefore, the determinant is a number associated with a square matrix.

9. Given that A = [a ij ] is a square matrix of order 3×3 and |A| = -7, then the value of ∑ i=1 3 a i2 A i2 , where A ij denotes the cofactor of element a ij is

Correct option: (b) -7

Order of matrix A is 3×3.

Now, ∑ i=1 3 a i2 A i2 = a 12 A 12 + a 22 A 22 + a 32 A 32

10. If A is an invertible matrix of order 2, then det (A –1 ) is equal to

(a) det (A)

(b) 1/det (A)

Correct option: (b) 1/det (A)

Given that the A is an invertible matrix of order 2.

If the matrix is invertible, then its determinant is not equal to 0.

We know that,

AA -1 = I, where I is the identity matrix

Taking determinant on both sides,

|AA -1 | = |I|

|A| |A -1 | = 1

|A -1 | = 1/|A| {since A is non-singular, |A| ≠ 0}

det(A -1 ) = 1/det(A)

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CBSE Assertion Reason Questions for Class 12 Maths Determinants Free PDF

Mere Bacchon, you must practice the CBSE Assertion Reason Questions for Class 12 Maths Determinants  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Assertion Reason Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Assertion Reason Questions , just click ‘ Download PDF ’.

Checkout our Assertion Reason Questions for other chapters

• Chapter 2 Inverse Trigonometric Functions Assertion Reason Questions
• Chapter 3 Matrices Assertion Reason Questions
• Chapter 5 Continuity and Differentiability Assertion Reason Questions
• Chapter 6 Applications of Derivatives Assertion Reason Questions

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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