# NCERT Solutions for Ex 4.2 Class 12 Maths Chapter 4

Mathematics is one of the most important subjects in the school life of students, and even after they finish school, it is likely that they will need some knowledge of Mathematics in their professional lives. The Class 12 Mathematics curriculum includes the basics of Mathematics and its uses in everyday life. NCERT textbook solutions like the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 are a great resource for the students to learn Mathematics. These solutions provide a consistent learning experience for students. Therefore, Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2, so that students can score higher marks in their board examinations.

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## NCERT Solutions for Class 12 Maths Chapter 4 – Determinants

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## Access NCERT Solutions for Class 12 Maths Chapter 4 – Determinants

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### NCERT Solutions For Class 12 Math Chapter 4 Exercise 4.2

Mathematics requires a lot of practice so that students can improve their skills. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 are considered to be one of the best resources for students to score good marks in board examinations. This is because NCERT solutions cover all the topics that can be asked in board examinations. Moreover, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 are curated by experienced and certified professionals. The NCERT Solutions also help teachers to track the progress of students and measure the students’ level of understanding. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 help students understand the basic concepts of the subject, and they are perfect for students who find Mathematics difficult. Practising the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 helps students increase their pace of solving the problems of Mathematics, which is very essential for them to score well in the board examinations.

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### What are Determinants?

Determinants are mathematical properties that help to explain the relationship between two or more variables. They can be used to describe how a change in one variable affects the other variables. There are many types of determinants. Some determinants are symmetrical and others are not. Some determinants are positive and others are negative. Determinants also have different properties, such as Absolute Value, Signed Magnitude, and Traceability. Determinants play an important role in statistics and algebra. They can be used to solve problems, determine relationships, and predict outcomes. Determinants are a valuable tool for solving equations and systems of equations. Students can refer to the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 to have a better understanding of determinants.

### What are the Different Properties of Determinants?

Different properties of determinants include being associative, commutative, distributive, and sometimes idempotent.

The properties of determinants can be used to solve a variety of mathematics problems. A few examples are provided below:

• The determinant of a matrix can be used to determine the eigenvalues and eigenvectors of the matrix. The determinant of a linear system can be used to find the solutions to the system. The determinant of a triangular matrix can be used to find the rotation matrix that will transform the triangle into a square.

The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 help students in having a better understanding of the topic of Determinants.

### NCERT Solutions for Class 12 Maths

One of the best ways to improve the mathematical skills of a student is to go through the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2. These solutions help students to build strong fundamentals and learn how to apply those skills in real-world scenarios. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 are designed for students to improve their mathematical skills quickly and efficiently. Students can find the solutions complicated, but they are helpful resources and practice materials. Solving the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 can help the student practice the concepts and techniques of mathematics which are very essential to score better marks in any examinations. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 are helpful resources that help students build strong fundamentals.

### NCERT Solutions Class 12 Maths Chapter 4 – Other Exercises

There are many topics covered in Class 12, Chapter-4 Determinants, including the Introduction of Determinants, Properties of Determinants, Area of a Triangle, Minors and Cofactors, Adjoints and Inverses of Matrices, and Applications of Determinants and Matrices. Extramarks provides the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 so that students do not waste their time searching for answers and get authentic solutions without having to look anywhere else. Among the many tools provided by Extramarks are K12 study materials, live doubt-solving sessions, and so much more in order to help students resolve all their queries and doubts. As a result, students are able to concentrate on their goals, excel in their studies, and score highly in their board examinations.

Exercise 4.2 Class 12th is based on the various properties of determinants. Students can also practice the exemplar questions given before Exercise  4.2 Class 12th Maths to have a clear understanding of the concepts of the exercise.

Overall, there are six exercises in the chapter and students require rigorous practice to grasp the concepts.

### Properties of Determinants

There are six properties of determinants in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 that students should practice thoroughly.

1. If the rows and columns are interchanged, the value of a Determinant remains unchanged.
2. If any two rows (or columns) of a Determinant are interchanged, then the sign of the Determinant changes.
3. If any two rows (or columns) of a Determinant are identical (all corresponding elements are the same), then the value of the Determinant is zero.
4. If each element of a row (or a column) of a Determinant gets multiplied by a constant k, then its value is multiplied by k.
5. If some or all elements of a row or column of a Determinant are expressed as the sum of two (or more) terms, then the Determinant can be expressed as the sum of two (or more) Determinants.
6. If to each element of any row or column of a Determinant, the equimultiples of corresponding elements of the other row (or column) are added, then the value of the Determinant remains the same. This means the value of the Determinant remains the same if we apply the operation Ri → Ri + kRj or Ci → Ci + k Cj.

### Class 12 Determinants Exercise 4.2

A Determinant can also be explained as an associated number to every Square Matrix. This may be thought of as a function that associates each square matrix with a unique number (real or complex). For learning more about Determinants, students can download  NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 provided by Extramarks. For studying Determinants, students must know the basic concept of Matrices, which was the previous chapter of the NCERT curriculum. In Mathematics, a Determinant is said to be a scalar value that is a function of the entries of a square matrix. NCERT is like the building block for any student who has to appear for the board examinations.

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### Chapter 4 – Determinants of  Other Exercises

 Chapter 4 – Determinants Other Exercises Exercise 4.1 8 Questions & Solutions (3 Short Answers, 5 Long Answers) Exercise 4.3 5 Questions & Solutions (2 Short Answers, 3 Long Answers) Exercise 4.4 5 Questions & Solutions (2 Short Answers, 3 Long Answers) Exercise 4.5 18 Questions & Solutions (4 Short Answers, 14 Long Answers) Exercise 4.6 16 Questions & Solutions (3 Short Answers, 13 Long Answers)

Q.1 Using the property of determinants and without expanding in, prove that:

$\left|\begin{array}{l}\mathrm{x}\mathrm{a}\mathrm{x}+\mathrm{a}\\ \mathrm{y}\mathrm{b}\mathrm{y}+\mathrm{b}\\ \mathrm{z}\mathrm{c}\mathrm{z}+\mathrm{c}\end{array}\right|=0$

Ans

$\begin{array}{l}|\begin{array}{l}\mathrm{x}\mathrm{a}\mathrm{x}+\mathrm{a}\\ \mathrm{y}\mathrm{b}\mathrm{y}+\mathrm{b}\\ \mathrm{z}\mathrm{c}\mathrm{z}+\mathrm{c}\end{array}|=|\begin{array}{l}\mathrm{x}\mathrm{a}\mathrm{x}\\ \mathrm{y}\mathrm{b}\mathrm{y}\\ \mathrm{z}\mathrm{c}\mathrm{z}\end{array}|+|\begin{array}{l}\mathrm{x}\mathrm{a}\mathrm{a}\\ \mathrm{y}\mathrm{b}\mathrm{b}\\ \mathrm{z}\mathrm{c}\mathrm{c}\end{array}|\\ =0+0\left[\begin{array}{l}\mathrm{If}\mathrm{two}\mathrm{columns}\mathrm{of}\mathrm{a}\mathrm{determinant}\mathrm{is}\\ \mathrm{identical},\mathrm{then}\mathrm{their}\mathrm{value}\mathrm{is}\mathrm{zero}.\end{array}\right]\\ =0\end{array}$

Q.2 Using the property of determinants and without expanding in, prove that:

$\left|\begin{array}{l}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\\ \mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\\ \mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\end{array}\right|=0$

Ans

$\begin{array}{l}|\begin{array}{l}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\\ \mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\\ \mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\end{array}|=|\begin{array}{l}\mathrm{a}-\mathrm{b}+\mathrm{b}-\mathrm{c}+\mathrm{c}-\mathrm{a}\mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\\ \mathrm{b}-\mathrm{c}+\mathrm{c}-\mathrm{a}+\mathrm{a}-\mathrm{b}\mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\\ \mathrm{c}-\mathrm{a}+\mathrm{a}-\mathrm{b}+\mathrm{b}-\mathrm{c}\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\end{array}|\\ \left[\mathrm{Apply}\mathrm{ }{\mathrm{C}}_{1}\to {\mathrm{C}}_{1}+{\mathrm{C}}_{2}+{\mathrm{C}}_{3}\right]\\ \mathrm{ }=|\begin{array}{l}0\mathrm{b}-\mathrm{c}\mathrm{c}-\mathrm{a}\\ 0\mathrm{c}-\mathrm{a}\mathrm{a}-\mathrm{b}\\ 0\mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\end{array}|\\ \mathrm{ }=0\\ \left[\mathrm{Since},\mathrm{}\mathrm{all}\mathrm{the}\mathrm{elements}\mathrm{of}\mathrm{a}\mathrm{column}\mathrm{are}\mathrm{zero}\mathrm{.}\right]\end{array}$

Q.3 Using the property of determinants and without expanding in, prove that:

$\left|\begin{array}{l}2765\\ 3875\\ 5986\end{array}\right|=0$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{l}2765\\ 3875\\ 5986\end{array}\right|\\ \mathrm{Apply}{\mathrm{C}}_{\mathrm{3}}\to {\mathrm{C}}_{3}+{\mathrm{C}}_{1}\\ \mathrm{ }=\left|\begin{array}{l}2763+\mathrm{2}\\ 3872+\mathrm{3}\\ 5981+\mathrm{5}\end{array}\right|\\ \mathrm{ }=\left|\begin{array}{l}2763\\ 3872\\ 5981\end{array}\right|+\left|\begin{array}{l}272\\ 383\\ 595\end{array}\right|\\ \mathrm{ }=\left|\begin{array}{l}2763\\ 3872\\ 5981\end{array}\right|+\mathrm{0}\left[\because \mathrm{\Delta }=0\mathrm{because}\mathrm{two}\mathrm{columns}\mathrm{are}\mathrm{identical}\mathrm{.}\right]\\ \mathrm{Apply}{\mathrm{C}}_{\mathrm{3}}\to \frac{1}{9}{\mathrm{C}}_{3}\\ \mathrm{ }=9\left|\begin{array}{l}277\\ 388\\ 599\end{array}\right|\\ \mathrm{ }=9×0 \left[\because \mathrm{\Delta }=0\mathrm{because}\mathrm{two}\mathrm{columns}\mathrm{are}\mathrm{identical}\mathrm{.}\right]\\ \mathrm{ }=0\end{array}$

Q.4 Using the property of determinants and without expanding in, prove that:

$\left|\begin{array}{l}1\mathrm{bc}\mathrm{a}\left(\mathrm{b}+\mathrm{c}\right)\\ 1\mathrm{ca}\mathrm{b}\left(\mathrm{c}+\mathrm{a}\right)\\ 1\mathrm{ab}\mathrm{c}\left(\mathrm{a}+\mathrm{b}\right)\end{array}\right|=0$

Ans

$\begin{array}{l}\mathrm{\Delta }=|\begin{array}{l}\mathrm{1}\mathrm{bc}\mathrm{a}\left(\mathrm{b}+\mathrm{c}\right)\\ \mathrm{1}\mathrm{ca}\mathrm{b}\left(\mathrm{c}+\mathrm{a}\right)\\ \mathrm{1}\mathrm{ab}\mathrm{c}\left(\mathrm{a}+\mathrm{b}\right)\end{array}|\\ \mathrm{ }=|\begin{array}{l}\mathrm{1}\mathrm{bc}\mathrm{ab}+\mathrm{ac}\\ 1\mathrm{ca}\mathrm{bc}+\mathrm{ba}\\ \mathrm{1}\mathrm{ab}\mathrm{ca}+\mathrm{cb}\end{array}|\\ \mathrm{ }=|\begin{array}{l}\mathrm{1}\mathrm{bc}\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\\ 1\mathrm{ca}\mathrm{bc}+\mathrm{ba}+\mathrm{ca}\\ \mathrm{1}\mathrm{ab}\mathrm{ca}+\mathrm{cb}+\mathrm{ab}\end{array}|\left[\mathrm{Apply}\mathrm{ }{\mathrm{C}}_{3}\to {\mathrm{C}}_{3}+{\mathrm{C}}_{2}\right]\\ \mathrm{ }=\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)|\begin{array}{l}1\mathrm{bc}\mathrm{1}\\ \mathrm{1}\mathrm{ca}\mathrm{1}\\ 1\mathrm{ab}\mathrm{1}\end{array}|\left[\mathrm{Taking}\mathrm{common}\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)\mathrm{from}{\mathrm{C}}_{\mathrm{3}}.\right]\\ \mathrm{ }=\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)×0\left[\mathrm{\Delta }=0\mathrm{because}\mathrm{two}\mathrm{columns}\mathrm{are}\mathrm{identical}\mathrm{.}\right]\\ \mathrm{ }=0\end{array}$

Q.5

$\left|\begin{array}{l}\mathrm{b}+\mathrm{cq}+\mathrm{ry}+\mathrm{z}\\ \mathrm{c}+\mathrm{ar}+\mathrm{pz}+\mathrm{x}\\ \mathrm{a}+\mathrm{bp}+\mathrm{qx}+\mathrm{y}\end{array}\right|=2\left|\begin{array}{l}\\ \\ \end{array}\right|$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{a}+\mathrm{b}\mathrm{p}+\mathrm{q}\mathrm{x}+\mathrm{y}\end{array}|\\ =|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{a}\mathrm{p}\mathrm{x}\end{array}|+|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}|\\ ={\mathrm{\Delta }}_{1}+{\mathrm{\Delta }}_{2}\left[\mathrm{Let}\right]\\ \mathrm{ }{\mathrm{\Delta }}_{1}=|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{a}\mathrm{p}\mathrm{x}\end{array}|\\ =|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{a}\mathrm{p}\mathrm{x}\end{array}| \left[{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{3}\right]\\ =|\begin{array}{l}\mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{a}\mathrm{p}\mathrm{x}\end{array}| \left[{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-{\mathrm{R}}_{2}\right]\\ =\left(-1\right)|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}| \left[{\mathrm{R}}_{1}↔{\mathrm{R}}_{3}\right]\\ = =|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\end{array}|\\ {\mathrm{\Delta }}_{2}=|\begin{array}{l}\mathrm{b}+\mathrm{c}\mathrm{q}+\mathrm{r}\mathrm{y}+\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}|\\ =|\begin{array}{l}\mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{c}+\mathrm{a}\mathrm{r}+\mathrm{p}\mathrm{z}+\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}|\left[{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-{\mathrm{R}}_{3}\right]\\ =|\begin{array}{l}\mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}|\left[{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{1}\right]{\left(-1\right)}^{2}|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\end{array}| \left[{\mathrm{R}}_{1}↔{\mathrm{R}}_{3}\right]\\ =|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{c}\mathrm{r}\mathrm{z}\\ \mathrm{b}\mathrm{q}\mathrm{y}\end{array}|\left[{\mathrm{R}}_{1}↔{\mathrm{R}}_{2}\right]\\ =|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\end{array}|\left[{\mathrm{R}}_{2}↔{\mathrm{R}}_{3}\right]\\ \therefore \mathrm{ }\mathrm{\Delta }={\mathrm{\Delta }}_{1}+{\mathrm{\Delta }}_{2}\\ =\mathrm{2}|\begin{array}{l}\mathrm{a}\mathrm{p}\mathrm{x}\\ \mathrm{b}\mathrm{q}\mathrm{y}\\ \mathrm{c}\mathrm{r}\mathrm{z}\end{array}|\end{array}$

Q.6 By using properties of determinants, show that:

$\left|\begin{array}{l} 0a-b\\ -a0-c\\ bc 0\end{array}\right|=\mathbf{0}$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{l} 0\mathrm{a}-\mathrm{b}\\ -\mathrm{a}\mathrm{0}-\mathrm{c}\\ \mathrm{ }\mathrm{bc} 0\end{array}\right|\\ \mathrm{ }=\frac{1}{\mathrm{c}}\left|\begin{array}{l} 0\mathrm{ac}-\mathrm{bc}\\ -\mathrm{a}\mathrm{0}-\mathrm{c}\\ \mathrm{ }\mathrm{bc} 0\end{array}\right|\left[{\mathrm{R}}_{1}\to {\mathrm{cR}}_{1}\right]\\ \mathrm{ }=\frac{1}{\mathrm{c}}\left|\begin{array}{l} \mathrm{ }\mathrm{abac} \mathrm{ }0\\ -\mathrm{a}\mathrm{0}-\mathrm{c}\\ \mathrm{ }\mathrm{bc} 0\end{array}\right|\left[{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-{\mathrm{bR}}_{2}\right]\\ \mathrm{ }=\frac{\mathrm{a}}{\mathrm{c}}\left|\begin{array}{l} \mathrm{ }\mathrm{bc} \mathrm{ }0\\ -\mathrm{a}\mathrm{0}-\mathrm{c}\\ \mathrm{ }\mathrm{bc} 0\end{array}\right|\left[\mathrm{Taking}\mathrm{a}\mathrm{common}\mathrm{from}{\mathrm{R}}_{\mathrm{1}}\mathrm{.}\right]\\ \mathrm{ }=\frac{\mathrm{a}}{\mathrm{c}}×0\left[\mathrm{Two}\mathrm{rows}\mathrm{are}\mathrm{identical},\mathrm{so}\mathrm{\Delta }=0.\right]\\ \mathrm{ }=0\end{array}$

Q.7 By using properties of determinants, show that:

$\left|\begin{array}{l}-{\mathrm{a}}^{2}\mathrm{ab}\mathrm{ac}\\ \mathrm{ }\mathbf{b}\mathrm{a}-{\mathrm{b}}^{2}\mathbf{b}\mathrm{c}\\ \mathrm{ }\mathrm{ca}\mathrm{cb}-{\mathrm{c}}^{2}\end{array}\right|=4{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}$

Ans

$\begin{array}{c}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\mathrm{\Delta }\\ \mathrm{ }=\left|\begin{array}{l}-{\mathrm{a}}^{\mathrm{2}}\mathrm{abac}\\ \mathrm{ }\mathrm{ba}-{\mathrm{b}}^{2}\mathrm{bc}\\ \mathrm{ }\mathrm{cacb}-{\mathrm{c}}^{2}\end{array}\right|\\ \mathrm{ }=\mathrm{abc}\left|\begin{array}{l}-\mathrm{a} \mathrm{ }\mathrm{a} \mathrm{ }\mathrm{a}\\ \mathrm{ }\mathrm{b}-\mathrm{b} \mathrm{ }\mathrm{b}\\ \mathrm{ }\mathrm{c} \mathrm{ }\mathrm{c}-\mathrm{c}\end{array}\right|\left[\begin{array}{l}\mathrm{Applying} \mathrm{ }{\mathrm{C}}_{1}\to \frac{1}{\mathrm{a}}{\mathrm{C}}_{1},{\mathrm{C}}_{2}\to \frac{1}{\mathrm{b}}{\mathrm{C}}_{2}\mathrm{ }\\ \mathrm{and}\mathrm{ }{\mathrm{C}}_{3}\to \frac{1}{\mathrm{c}}{\mathrm{C}}_{3}\end{array}\right]\\ \mathrm{ }=\mathrm{abc}\left|\begin{array}{l}-\mathrm{a} \mathrm{ }\mathrm{a} \mathrm{ }\mathrm{a}\\ \mathrm{ }\mathrm{b}-\mathrm{b} \mathrm{ }\mathrm{b}\\ \mathrm{ }\mathrm{c} \mathrm{ }\mathrm{c}-\mathrm{c}\end{array}\right|\\ \mathrm{ }=\mathrm{abc}\left|\begin{array}{l}0 \mathrm{ }\mathrm{a} \mathrm{ }\mathrm{a}\\ 2\mathrm{b}-\mathrm{b} \mathrm{ }\mathrm{b}\\ 0 \mathrm{c}-\mathrm{c}\end{array}\right|\left[\mathrm{Applying}{\mathrm{C}}_{\mathrm{1}}\to {\mathrm{C}}_{\mathrm{1}}+{\mathrm{C}}_{\mathrm{3}}\right]\\ \mathrm{ }=\mathrm{abc}×-2\mathrm{b}\left|\begin{array}{l}\mathrm{a} \mathrm{ }\mathrm{a}\\ \mathrm{c}-\mathrm{c}\end{array}\right|\\ \mathrm{ }=\mathrm{abc}×-2\mathrm{b}\left(-\mathrm{ac}-\mathrm{ac}\right)\\ \mathrm{ }=\mathrm{abc}×-2\mathrm{b}\left(-2\mathrm{ac}\right)\\ \mathrm{ }=4{\mathrm{a}}^{\mathrm{2}}{\mathrm{b}}^{\mathrm{2}}{\mathrm{c}}^{\mathrm{2}}=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\end{array}$

Q.8 By using properties of determinants, show that:

$\begin{array}{l}\left(\mathrm{i}\right)\left|\begin{array}{l}1\mathrm{a}{\mathrm{a}}^{2}\\ 1\mathrm{b}{\mathbf{b}}^{2}\\ 1\mathrm{c}{\mathrm{c}}^{2}\end{array}\right|=\left(\mathrm{a}–\mathrm{b}\right)\left(\mathrm{b}–\mathrm{c}\right)\left(\mathrm{c}–\mathrm{a}\right)\\ \left(\mathrm{ii}\right) \left|\begin{array}{l}111\\ \mathrm{a}\mathrm{b}\mathrm{c}\\ {\mathrm{a}}^{3}{\mathrm{b}}^{3}{\mathrm{c}}^{3}\end{array}\right|=\left(\mathrm{a}–\mathrm{b}\right)\left(\mathrm{b}–\mathrm{c}\right)\left(\mathrm{c}–\mathrm{a}\right)\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{l}1\mathrm{a}{\mathrm{a}}^{\mathrm{2}}\\ \mathrm{1}\mathrm{b}{\mathrm{b}}^{\mathrm{2}}\\ \mathrm{1}\mathrm{c}{\mathrm{c}}^{\mathrm{2}}\end{array}|\\ \mathrm{ }=|\begin{array}{l}0\mathrm{a}-\mathrm{b}{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^{2}\\ \mathrm{0}\mathrm{b}-\mathrm{c}{\mathrm{b}}^{\mathrm{2}}-{\mathrm{c}}^{2}\\ \mathrm{1}\mathrm{c}{\mathrm{c}}^{\mathrm{2}}\end{array}|\left[\mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\mathrm{ }\mathrm{and} {\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{3}}\right]\\ \mathrm{ }=\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)|\begin{array}{l}01\mathrm{a}+\mathrm{b}\\ 01\mathrm{b}+\mathrm{c}\\ 1\mathrm{c}{\mathrm{c}}^{\mathrm{2}}\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ \mathrm{ }=\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)|\begin{array}{l}1\mathrm{a}+\mathrm{b}\\ 1\mathrm{b}+\mathrm{c}\end{array}|\\ \mathrm{ }=\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}-\mathrm{a}-\mathrm{b}\right)\\ \mathrm{ }=\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{c}-\mathrm{a}\right)=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \left(\mathrm{ii}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{l}11\mathrm{1}\\ \mathrm{a}\mathrm{b}\mathrm{c}\\ {\mathrm{a}}^{3}{\mathrm{b}}^{3}{\mathrm{c}}^{\mathrm{3}}\end{array}|\\ =|\begin{array}{l}0\mathrm{0}\mathrm{1}\\ \mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}\\ {\mathrm{a}}^{\mathrm{3}}-{\mathrm{b}}^{3}{\mathrm{b}}^{\mathrm{3}}-{\mathrm{c}}^{3}{\mathrm{c}}^{\mathrm{3}}\end{array}| \mathrm{ }\\ \left[\mathrm{Applying}{\mathrm{C}}_{\mathrm{1}}\to {\mathrm{C}}_{\mathrm{1}}-{\mathrm{C}}_{\mathrm{2}}\mathrm{ }\mathrm{and} {\mathrm{C}}_{\mathrm{2}}\to {\mathrm{C}}_{\mathrm{2}}-{\mathrm{C}}_{\mathrm{3}}\right]\\ =|\begin{array}{l}0\mathrm{0}\mathrm{1}\\ \mathrm{a}-\mathrm{b}\mathrm{b}-\mathrm{c}\mathrm{c}\\ \left(\mathrm{a}-\mathrm{b}\right)\left({\mathrm{a}}^{2}+\mathrm{ab}+{\mathrm{b}}^{2}\right)\left(\mathrm{b}-\mathrm{c}\right)\left({\mathrm{b}}^{2}+\mathrm{bc}+{\mathrm{c}}^{2}\right){\mathrm{c}}^{\mathrm{3}}\end{array}|\mathrm{ }\\ =\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)|\begin{array}{l}\mathrm{0}\mathrm{0}\mathrm{1}\\ 11\mathrm{c}\\ \left({\mathrm{a}}^{2}+\mathrm{ab}+{\mathrm{b}}^{2}\right)\left({\mathrm{b}}^{2}+\mathrm{bc}+{\mathrm{c}}^{2}\right){\mathrm{c}}^{\mathrm{3}}\end{array}|\\ =\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)|\begin{array}{l}11\\ \left({\mathrm{a}}^{2}+\mathrm{ab}+{\mathrm{b}}^{2}\right)\left({\mathrm{b}}^{2}+\mathrm{bc}+{\mathrm{c}}^{2}\right)\end{array}|\mathrm{ }\left[\mathrm{Expending}\mathrm{along}{\mathrm{R}}_{\mathrm{1}}\right]\\ =\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left\{\left({\mathrm{b}}^{2}+\mathrm{bc}+{\mathrm{c}}^{2}\right)-\left({\mathrm{a}}^{2}+\mathrm{ab}+{\mathrm{b}}^{2}\right)\right\}\\ =\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left({\mathrm{b}}^{2}+\mathrm{bc}+{\mathrm{c}}^{2}-{\mathrm{a}}^{2}-\mathrm{ab}-{\mathrm{b}}^{2}\right)\\ =\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{bc}+{\mathrm{c}}^{2}-{\mathrm{a}}^{2}-\mathrm{ab}\right)\\ =\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left\{{\mathrm{c}}^{2}-{\mathrm{a}}^{2}+\mathrm{bc}-\mathrm{ab}\right\}\\ =\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left\{\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{c}+\mathrm{a}\right)-\mathrm{b}\left(\mathrm{c}-\mathrm{a}\right)\right\}\\ =\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{c}-\mathrm{a}\right)\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\end{array}$

Q.9 By using properties of determinants, show that:

$\left|\begin{array}{l}\mathrm{x}{\mathrm{x}}^{2}\mathrm{yz}\\ \mathrm{y}{\mathrm{y}}^{2}\mathrm{zx}\\ \mathrm{z}{\mathrm{z}}^{2}\mathrm{xy}\end{array}\right|=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{z}-\mathrm{x}\right)\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{l}\mathrm{x}{\mathrm{x}}^{2}\mathrm{yz}\\ \mathrm{y}{\mathrm{y}}^{2}\mathrm{zx}\\ \mathrm{z}{\mathrm{z}}^{2}\mathrm{xy}\end{array}|\\ \mathrm{ }=|\begin{array}{l}\mathrm{x}-\mathrm{y}{\mathrm{x}}^{\mathrm{2}}-{\mathrm{y}}^{2}\mathrm{yz}-\mathrm{zx}\\ \mathrm{y}-\mathrm{z}{\mathrm{y}}^{\mathrm{2}}-{\mathrm{z}}^{2}\mathrm{zx}-\mathrm{xy}\\ \mathrm{z}{\mathrm{z}}^{2}\mathrm{xy}\end{array}|\\ \left[\mathrm{Applying}\mathrm{ }{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}},\mathrm{ }{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{3}}\right]\\ \mathrm{ }=|\begin{array}{l}\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{x}+\mathrm{y}\right)-\mathrm{z}\left(\mathrm{x}-\mathrm{y}\right)\\ \left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{y}+\mathrm{z}\right)-\mathrm{x}\left(\mathrm{y}-\mathrm{z}\right)\\ \mathrm{z} {\mathrm{z}}^{\mathrm{2}} \mathrm{xy}\end{array}|\\ \mathrm{ }=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)|\begin{array}{l}1\left(\mathrm{x}+\mathrm{y}\right)-\mathrm{z}\\ 1\left(\mathrm{y}+\mathrm{z}\right)-\mathrm{x}\\ \mathrm{ }\mathrm{z} {\mathrm{z}}^{\mathrm{2}}\mathrm{ } \mathrm{xy}\end{array}| \mathrm{ }\left[\begin{array}{l}\mathrm{Taking}\mathrm{common}\left(\mathrm{x}-\mathrm{y}\right)\mathrm{ }\\ \mathrm{and}\mathrm{ }\left(\mathrm{y}-\mathrm{z}\right)\mathrm{ }\mathrm{from}{\mathrm{R}}_{\mathrm{1}}\mathrm{and}{\mathrm{R}}_{\mathrm{2}}\\ \mathrm{respectively}\end{array}\right]\\ \mathrm{ }=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)|\begin{array}{l}0\left(\mathrm{x}-\mathrm{z}\right)\left(\mathrm{x}-\mathrm{z}\right)\\ 1\left(\mathrm{y}+\mathrm{z}\right)-\mathrm{x}\\ \mathrm{ }\mathrm{z} \mathrm{ }{\mathrm{z}}^{\mathrm{2}} \mathrm{xy}\end{array}|\\ \left[\mathrm{Applying}\mathrm{ }{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\right]\\ \mathrm{ }=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{z}-\mathrm{x}\right)|\begin{array}{l}0-1-1\\ 1\left(\mathrm{y}+\mathrm{z}\right)-\mathrm{x}\\ \mathrm{ }\mathrm{z} \mathrm{ } \mathrm{ }{\mathrm{z}}^{\mathrm{2}} \mathrm{xy}\end{array}|\left[\begin{array}{l}\mathrm{Taking}\mathrm{common}\\ \left(\mathrm{z}-\mathrm{x}\right)\mathrm{ }\mathrm{from}{\mathrm{R}}_{\mathrm{1}}\end{array}\right]\\ \mathrm{ }=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{z}-\mathrm{x}\right)|\begin{array}{l}00-1\\ 1\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)-\mathrm{x}\\ \mathrm{ }\mathrm{z}\mathrm{ } {\mathrm{z}}^{2}-\mathrm{xy} \mathrm{xy}\end{array}|\left[\begin{array}{l}\mathrm{Applying} \\ {\mathrm{C}}_{2}\to {\mathrm{C}}_{2}-{\mathrm{C}}_{3}\end{array}\right]\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \mathrm{ }=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{z}-\mathrm{x}\right)\left\{\left(-1\right)|\begin{array}{l}1\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\\ \mathrm{z} {\mathrm{z}}^{2}-\mathrm{xy}\end{array}|\right\}\\ \mathrm{ }=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{z}-\mathrm{x}\right)×\left(-1\right)\left\{\mathrm{ }{\mathrm{z}}^{2}-\mathrm{xy}-\left(\mathrm{z}\right)\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\right\}\\ \mathrm{ }=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{z}-\mathrm{x}\right)×\left(-1\right)\left\{\mathrm{ }{\mathrm{z}}^{2}-\mathrm{xy}-\mathrm{zx}-\mathrm{zy}-{\mathrm{z}}^{2}\right\}\\ \mathrm{ }=\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{z}-\mathrm{x}\right)\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right)\\ \mathrm{ }=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\end{array}\mathrm{a}$

Q.10 By using properties of determinants, show that:

$\begin{array}{l}\left(\mathrm{i}\right)\left|\begin{array}{l}\mathrm{x}+42\mathrm{x}2\mathrm{x}\\ 2\mathrm{x}\mathrm{x}+42\mathrm{x}\\ 2\mathrm{x}2\mathrm{x}\mathrm{x}+4\end{array}\right|=\left(5\mathrm{x}+4\right){\left(4-\mathrm{x}\right)}^{2}\\ \end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\mathrm{ }\left|\begin{array}{l}\mathrm{y}+\mathrm{k}\mathrm{y}\mathrm{y}\\ \mathrm{y}\mathrm{y}+4\mathrm{y}\\ \mathrm{y}\mathrm{y}\mathrm{y}+4\end{array}\right|={\mathrm{k}}^{2}\left(3\mathrm{y}+\mathrm{k}\right)\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{l}\mathrm{x}+42\mathrm{x}2\mathrm{x}\\ 2\mathrm{x}\mathrm{x}+42\mathrm{x}\\ 2\mathrm{x}2\mathrm{x}\mathrm{x}+\mathrm{4}\end{array}|\\ =|\begin{array}{l}5\mathrm{x}+42\mathrm{x}2\mathrm{x}\\ 5\mathrm{x}+\mathrm{5}\mathrm{x}+42\mathrm{x}\\ 5\mathrm{x}+\mathrm{5}2\mathrm{x}\mathrm{x}+\mathrm{4}\end{array}|\mathrm{ }\left[\mathrm{Apply}\mathrm{ }{\mathrm{C}}_{1}\to {\mathrm{C}}_{1}+{\mathrm{C}}_{2}+{\mathrm{C}}_{3}\right]\\ =\left(5\mathrm{x}+4\right)|\begin{array}{l}12\mathrm{x}2\mathrm{x}\\ 1\mathrm{x}+42\mathrm{x}\\ 12\mathrm{x}\mathrm{x}+\mathrm{4}\end{array}|\\ =\left(5\mathrm{x}+4\right)|\begin{array}{l}12\mathrm{x}2\mathrm{x}\\ 0\mathrm{4}-\mathrm{x}\mathrm{0}\\ 0\mathrm{0}\mathrm{4}-\mathrm{x}\end{array}| \mathrm{ }\left[\begin{array}{l}\mathrm{Apply}\\ {\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{1},{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{1}\end{array}\right]\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =\left(5\mathrm{x}+4\right)|\begin{array}{l}4-\mathrm{x}0\\ 04-\mathrm{x}\end{array}|\\ =\left(5\mathrm{x}+4\right){\left(4-\mathrm{x}\right)}^{2}\\ \mathrm{ }=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{L}.\mathrm{H}.\mathrm{S}.=|\begin{array}{l}\mathrm{y}+\mathrm{k}\mathrm{y}\mathrm{y}\\ \mathrm{y}\mathrm{y}+\mathrm{k}\mathrm{y}\\ \mathrm{y}\mathrm{y}\mathrm{y}+\mathrm{k}\end{array}|\\ \mathrm{ }=|\begin{array}{l}3\mathrm{y}+\mathrm{k}\mathrm{y} \mathrm{ }\mathrm{y}\\ 3\mathrm{y}+\mathrm{k}\mathrm{y}+\mathrm{k} \mathrm{y}\\ 3\mathrm{y}+\mathrm{k}\mathrm{y}\mathrm{y}+\mathrm{k}\end{array}|\mathrm{ }\left[\mathrm{Apply}\mathrm{ }{\mathrm{C}}_{1}\to {\mathrm{C}}_{1}+{\mathrm{C}}_{2}+{\mathrm{C}}_{3}\right]\\ \mathrm{ }=\left(3\mathrm{y}+\mathrm{k}\right)|\begin{array}{l}1\mathrm{y}\mathrm{ } \mathrm{ }\mathrm{y}\\ 1\mathrm{y}+\mathrm{k} \mathrm{y}\\ 1\mathrm{y}\mathrm{y}+\mathrm{k}\end{array}|\\ \left[\mathrm{Apply}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}},\mathrm{ }{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}}\right]\\ \mathrm{ }=\left(3\mathrm{y}+\mathrm{k}\right)|\begin{array}{l}1\mathrm{y}\mathrm{y}\\ 0\mathrm{k}\mathrm{0}\\ 00\mathrm{k}\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \mathrm{ }=\left(3\mathrm{y}+\mathrm{k}\right)|\begin{array}{l}\mathrm{k}0\\ 0\mathrm{k}\end{array}|\\ \mathrm{ }=\left(3\mathrm{y}+\mathrm{k}\right)×{\mathrm{k}}^{2}\\ \mathrm{ }={\mathrm{k}}^{\mathrm{2}}\left(3\mathrm{y}+\mathrm{k}\right)\end{array}$

Q.11 By using properties of determinants, show that:

$\begin{array}{l}\left(\mathrm{i}\right)\left|\begin{array}{l}\mathrm{a}-\mathrm{b}-\mathrm{c}2\mathrm{a} \mathrm{ }2\mathrm{a}\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a} \mathrm{ } \mathrm{ }2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}\right|={\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}^{3}\\ \left(\mathrm{ii}\right)\left|\begin{array}{l}\mathrm{x}+\mathrm{y}+2\mathrm{z}\mathrm{x} \mathrm{ } \mathrm{ }\mathrm{y}\\ \mathrm{z} \mathrm{ }\mathrm{y}+\mathrm{z}+2\mathrm{x} \mathrm{ }\mathrm{y}\\ \mathrm{z} \mathrm{ } \mathrm{x}\mathrm{ } \mathrm{z}+\mathrm{x}+2\mathrm{y}\end{array}\right|=2{\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)}^{3}\end{array}$

Ans

$\begin{array}{l}\left(\mathrm{i}\right)|\begin{array}{l}\mathrm{a}-\mathrm{b}-\mathrm{c}2\mathrm{a} 2\mathrm{a}\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a} \mathrm{ }2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}|\mathrm{=}{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}^{\mathrm{3}}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=|\begin{array}{l}\mathrm{a}-\mathrm{b}-\mathrm{c}2\mathrm{a} 2\mathrm{a}\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a} 2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}|\\ =|\begin{array}{l}\mathrm{a}+\mathrm{b}+\mathrm{c}\mathrm{a}+\mathrm{b}+\mathrm{c}\mathrm{a}+\mathrm{b}+\mathrm{c}\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a} 2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}|\left[\begin{array}{l}\mathrm{Apply}\\ {\mathrm{R}}_{1}\to {\mathrm{R}}_{1}+{\mathrm{R}}_{2}+{\mathrm{R}}_{3}\end{array}\right]\\ =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)|\begin{array}{l}111\\ 2\mathrm{b}\mathrm{b}-\mathrm{c}-\mathrm{a}\mathrm{ }2\mathrm{b}\\ 2\mathrm{c}2\mathrm{c}\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}|\\ =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)|\begin{array}{l}1 \mathrm{ }00\\ 2\mathrm{b}-\mathrm{b}-\mathrm{c}-\mathrm{a} 0\\ 2\mathrm{c}0-\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}| \left[\begin{array}{l}{\mathrm{C}}_{2}\to {\mathrm{C}}_{2}-{\mathrm{C}}_{1},\mathrm{ }\\ {\mathrm{C}}_{3}\to {\mathrm{C}}_{3}-{\mathrm{C}}_{1}\end{array}\right]\\ =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)|\begin{array}{l}-\mathrm{b}-\mathrm{c}-\mathrm{a}\mathrm{ }\mathrm{0}\\ 0 -\mathrm{c}-\mathrm{a}-\mathrm{b}\end{array}|\\ =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right){\left(-\mathrm{b}-\mathrm{c}-\mathrm{a}\right)}^{2}\\ ={\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}^{3}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{l}\mathrm{x}+\mathrm{y}+2\mathrm{z}\mathrm{x} \mathrm{y}\\ \mathrm{z} \mathrm{ }\mathrm{y}+\mathrm{z}+2\mathrm{x} \mathrm{y}\\ \mathrm{z} \mathrm{ } \mathrm{x} \mathrm{ }\mathrm{z}+\mathrm{x}+2\mathrm{y}\end{array}|\\ =|\begin{array}{l}2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\mathrm{x}\mathrm{y}\\ 2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\mathrm{y}+\mathrm{z}+2\mathrm{x} \mathrm{y}\\ 2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\mathrm{x}\mathrm{ } \mathrm{z}+\mathrm{x}+2\mathrm{y}\end{array}| \left[{\mathrm{C}}_{1}\to {\mathrm{C}}_{1}+{\mathrm{C}}_{2}+{\mathrm{C}}_{3}\right]\\ =\left(2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\right)|\begin{array}{l}1\mathrm{x}\mathrm{y}\\ 1\mathrm{y}+\mathrm{z}+2\mathrm{x} \mathrm{y}\\ 1\mathrm{x}\mathrm{ } \mathrm{z}+\mathrm{x}+2\mathrm{y}\end{array}|\\ =\left(2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\right)|\begin{array}{l}1\mathrm{x}\mathrm{y}\\ 0\mathrm{y}+\mathrm{z}+\mathrm{x} \mathrm{0}\\ 00\mathrm{z}+\mathrm{x}+\mathrm{y}\end{array}|\left[\begin{array}{l}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{1}\\ {\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{1}\end{array}\right]\\ ={\left(2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\right)}^{3}|\begin{array}{l}1\mathrm{x}\mathrm{y}\\ 01 0\\ 001\end{array}|\\ \mathrm{Expanding} \mathrm{along} \mathrm{ }{\mathrm{C}}_{1},\mathrm{we}\mathrm{get}\\ ={\left(2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\right)}^{3}|\begin{array}{l}10\\ 01\end{array}|\\ ={\left(2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\right)}^{3}\left(1-0\right)\\ ={\left(2\mathrm{x}+2\mathrm{y}+2\mathrm{z}\right)}^{3}=\mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{ }\end{array}$

Q.12 By using properties of determinants, show that:

$\left|\begin{array}{l}1\mathrm{x}{\mathrm{x}}^{2}\\ {\mathrm{x}}^{2}1\mathrm{x}\\ \mathrm{x}{\mathrm{x}}^{2}1\end{array}\right|={\left(1-{\mathrm{x}}^{3}\right)}^{2}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{l}\mathrm{1}\mathrm{x}{\mathrm{x}}^{\mathrm{2}}\\ {\mathrm{x}}^{2}1\mathrm{x}\\ \mathrm{x}{\mathrm{x}}^{2}\mathrm{1}\end{array}|\\ =|\begin{array}{l}\mathrm{1}+\mathrm{x}+{\mathrm{x}}^{2}\mathrm{1}+\mathrm{x}+{\mathrm{x}}^{2}\mathrm{1}+\mathrm{x}+{\mathrm{x}}^{2}\\ {\mathrm{x}}^{2}1\mathrm{x}\\ \mathrm{x}{\mathrm{x}}^{2}\mathrm{1}\end{array}|\left[{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}+{\mathrm{R}}_{2}+{\mathrm{R}}_{3}\right]\\ =\left(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^{2}\right)|\begin{array}{l}111\\ {\mathrm{x}}^{2}\mathrm{1}\mathrm{x}\\ \mathrm{x}{\mathrm{x}}^{2}\mathrm{1}\end{array}|\\ =\left(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^{2}\right)|\begin{array}{l}1\mathrm{ } 00\\ {\mathrm{x}}^{2}\mathrm{1}-{\mathrm{x}}^{2}\mathrm{x}-{\mathrm{x}}^{\mathrm{2}}\\ \mathrm{x}{\mathrm{x}}^{\mathrm{2}}-\mathrm{x}\mathrm{1}-\mathrm{x}\end{array}|\left[\begin{array}{l}{\mathrm{C}}_{2}\to {\mathrm{C}}_{2}-{\mathrm{C}}_{1},\\ {\mathrm{C}}_{3}\to {\mathrm{C}}_{3}-{\mathrm{C}}_{1}\end{array}\right]\\ =\left(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^{2}\right)|\begin{array}{l}1 \mathrm{ } 00\\ {\mathrm{x}}^{2}\left(\mathrm{1}-\mathrm{x}\right)\left(\mathrm{1}+\mathrm{x}\right)\mathrm{x}\left(1-\mathrm{x}\right)\\ \mathrm{x}\mathrm{x}\left(\mathrm{x}-1\right)\left(\mathrm{1}-\mathrm{x}\right)\end{array}|\\ =\left(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^{2}\right)\left(1-\mathrm{x}\right)\left(1-\mathrm{x}\right)|\begin{array}{l}1 \mathrm{ } 00\\ {\mathrm{x}}^{2}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{x}\\ \mathrm{x}-\mathrm{x}1\end{array}|\\ \left[\begin{array}{l}\mathrm{Taking}\mathrm{}\left(1-\mathrm{x}\right) \mathrm{common}\mathrm{from}\\ {\mathrm{C}}_{\mathrm{2}}\mathrm{and}{\mathrm{C}}_{\mathrm{3}}\mathrm{.}\end{array}\right]\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =\left(\mathrm{1}+\mathrm{x}+{\mathrm{x}}^{2}\right)\left(1-\mathrm{x}\right)\left(1-\mathrm{x}\right)\left(1+\mathrm{x}+{\mathrm{x}}^{2}\right)\\ =\left\{\left(1-\mathrm{x}\right)\left(1+\mathrm{x}+{\mathrm{x}}^{2}\right)\right\}\left\{\left(1-\mathrm{x}\right)\left(1+\mathrm{x}+{\mathrm{x}}^{2}\right)\right\}\\ =\left(1-{\mathrm{x}}^{3}\right)\left(1-{\mathrm{x}}^{3}\right)\\ ={\left(1–{\mathrm{x}}^{\mathrm{3}}\right)}^{\mathrm{2}}=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \mathrm{Hence}\mathrm{Proved}\mathrm{.}\end{array}$

Q.13 By using properties of determinants, show that:

$\left|\begin{array}{l}1+{\mathrm{a}}^{2}-{\mathrm{b}}^{2}2\mathrm{ab}-2\mathrm{b}\\ 2\mathrm{ab}1-{\mathrm{a}}^{2}+{\mathrm{b}}^{2}2\mathrm{a}\\ 2\mathrm{b}-2\mathrm{a}1-{\mathrm{a}}^{2}-{\mathrm{b}}^{2}\end{array}\right|={\left(1+{\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)}^{3}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=|\begin{array}{l}\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^{\mathrm{2}}\mathrm{ } \mathrm{ }2\mathrm{ab} \mathrm{ }-2\mathrm{b}\\ 2\mathrm{ab}\mathrm{1}-{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}} \mathrm{ }2\mathrm{a}\\ 2\mathrm{b}-2\mathrm{a}\mathrm{1}-{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^{\mathrm{2}}\end{array}|\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{bR}}_{\mathrm{3}},{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{aR}}_{\mathrm{3}}\\ \mathrm{ }=|\begin{array}{ccc}\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}& \mathrm{0}& -\mathrm{b}\left(1+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}\right)\\ \mathrm{0}& \mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}& \mathrm{ }\mathrm{a}\left(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}\right)\\ 2\mathrm{b}& -2\mathrm{a}& \mathrm{1}-{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^{\mathrm{2}}\end{array}|\\ \mathrm{ }={\left(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}\right)}^{2}|\begin{array}{ccc}\mathrm{1}& \mathrm{0}& -\mathrm{b}\\ \mathrm{0}& \mathrm{1}& \mathrm{ }\mathrm{a}\\ 2\mathrm{b}& -2\mathrm{a}& \mathrm{1}-{\mathrm{a}}^{\mathrm{2}}-{\mathrm{b}}^{\mathrm{2}}\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \mathrm{ }={\left(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}\right)}^{2}\left[1|\begin{array}{l} 1\mathrm{a}\\ -2\mathrm{a}1-{\mathrm{a}}^{2}-{\mathrm{b}}^{2}\end{array}|+\left(-\mathrm{b}\right)|\begin{array}{l}01\\ 2\mathrm{b}-1\mathrm{a}\end{array}|\right]\\ \mathrm{ }={\left(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}\right)}^{2}\left(1-{\mathrm{a}}^{2}-{\mathrm{b}}^{2}+2{\mathrm{a}}^{2}-\mathrm{b}×-2\mathrm{b}\right)\\ \mathrm{ }={\left(\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}\right)}^{2}\left(1-{\mathrm{a}}^{2}-{\mathrm{b}}^{2}+2{\mathrm{a}}^{2}+2{\mathrm{b}}^{2}\right)\\ \mathrm{ }={\left(1+{\mathrm{a}}^{2}+{\mathrm{b}}^{2}\right)}^{\mathrm{3}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence},\mathrm{ }\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.14 By using properties of determinants, show that:

$\left|\begin{array}{l}1+{\mathrm{a}}^{2}\mathrm{ab}\mathrm{ab}\\ \mathrm{ab}{\mathrm{b}}^{2}+1\mathrm{bc}\\ \mathrm{ca}\mathrm{cb}{\mathrm{c}}^{2}+1\end{array}\right|=1+{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{l}\mathrm{1}+{\mathrm{a}}^{2}\mathrm{ab}\mathrm{ac}\\ \mathrm{ab}{\mathrm{b}}^{\mathrm{2}}+\mathrm{1}\mathrm{bc}\\ \mathrm{ca}\mathrm{cb}{\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\end{array}|\\ \mathrm{Applying} {\mathrm{R}}_{1}\to \frac{1}{\mathrm{a}}{\mathrm{R}}_{1},{\mathrm{R}}_{2}\to \frac{1}{\mathrm{b}}{\mathrm{R}}_{2},{\mathrm{R}}_{3}\to \frac{1}{\mathrm{c}}{\mathrm{R}}_{3}\\ =\mathrm{abc}|\begin{array}{l}\frac{\mathrm{1}}{\mathrm{a}}+\mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{a}\mathrm{b}+\frac{\mathrm{1}}{\mathrm{b}}\mathrm{c}\\ \mathrm{a}\mathrm{b}\mathrm{c}+\frac{\mathrm{1}}{\mathrm{c}}\end{array}|\\ \mathrm{Applying} {\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-{\mathrm{R}}_{2},{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{3}\\ =\mathrm{abc}|\begin{array}{l}\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\mathrm{0}\\ 0 \mathrm{ }\frac{\mathrm{1}}{\mathrm{b}}-\frac{1}{\mathrm{c}}\\ \mathrm{a} \mathrm{b} \mathrm{ }\mathrm{c}+\frac{\mathrm{1}}{\mathrm{c}}\end{array}|\\ \mathrm{Applying} {\mathrm{C}}_{1}\to {\mathrm{aC}}_{1},{\mathrm{C}}_{2}\to {\mathrm{bC}}_{2},{\mathrm{C}}_{3}\to {\mathrm{cC}}_{3}\\ =\frac{\mathrm{abc}}{\mathrm{abc}}|\begin{array}{l}1-1\mathrm{0}\\ \mathrm{0} 1-1\\ {\mathrm{a}}^{\mathrm{2}} \mathrm{ }{\mathrm{b}}^{\mathrm{2}} \mathrm{ }{\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\end{array}|\\ \mathrm{Applying} {\mathrm{C}}_{2}\to {\mathrm{C}}_{2}+{\mathrm{C}}_{1}\\ =|\begin{array}{l}10\mathrm{0}\\ \mathrm{0}\mathrm{ }1-1\\ {\mathrm{a}}^{2}{\mathrm{a}}^{2}+\mathrm{ }{\mathrm{b}}^{\mathrm{2}} \mathrm{ }{\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\end{array}|\\ \mathrm{Expanding}\mathrm{}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ =|\begin{array}{l}1-1\\ {\mathrm{a}}^{2}+\mathrm{ }{\mathrm{b}}^{2}{\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\end{array}|\\ =\left({\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\right)-\left(-1\right)\left({\mathrm{a}}^{2}+\mathrm{ }{\mathrm{b}}^{\mathrm{2}}\right)\\ =1+{\mathrm{c}}^{2}+{\mathrm{a}}^{2}+\mathrm{ }{\mathrm{b}}^{\mathrm{2}}\\ =\mathrm{1}+{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}+{\mathrm{c}}^{\mathrm{2}}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.15 Let A b a square matrix of order 3 x 3, then |kA| is equal to

(A) k|A| (B) k2|A| (C) k3|A| (D) 3k|A|

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{A}\mathrm{be}\mathrm{a}\mathrm{square}\mathrm{matrix}\mathrm{of}\mathrm{order}3×3,\\ \mathrm{then} \mathrm{A}=\left[\begin{array}{l}{\mathrm{a}}_{1}{\mathrm{a}}_{2}{\mathrm{a}}_{3}\\ {\mathrm{b}}_{1}{\mathrm{b}}_{2}{\mathrm{b}}_{3}\\ {\mathrm{c}}_{1}{\mathrm{c}}_{2}{\mathrm{c}}_{3}\end{array}\right]\\ \mathrm{and}\\ \mathrm{kA}=\left[\begin{array}{l}{\mathrm{ka}}_{1}{\mathrm{ka}}_{2}{\mathrm{ka}}_{3}\\ {\mathrm{kb}}_{1}{\mathrm{kb}}_{2}{\mathrm{kb}}_{3}\\ {\mathrm{kc}}_{1}{\mathrm{kc}}_{2}{\mathrm{kc}}_{3}\end{array}\right]\\ |\mathrm{kA}|=|\begin{array}{l}{\mathrm{ka}}_{1}{\mathrm{ka}}_{2}{\mathrm{ka}}_{3}\\ {\mathrm{kb}}_{1}{\mathrm{kb}}_{2}{\mathrm{kb}}_{3}\\ {\mathrm{kc}}_{1}{\mathrm{kc}}_{2}{\mathrm{kc}}_{3}\end{array}|\\ ={\mathrm{k}}^{3}|\begin{array}{l}{\mathrm{a}}_{1}{\mathrm{a}}_{2}{\mathrm{a}}_{3}\\ {\mathrm{b}}_{1}{\mathrm{b}}_{2}{\mathrm{b}}_{3}\\ {\mathrm{c}}_{1}{\mathrm{c}}_{2}{\mathrm{c}}_{3}\end{array}|\left[\mathrm{Taking}\mathrm{k}\mathrm{common}\mathrm{from}{\mathrm{R}}_{\mathrm{1}}, {\mathrm{R}}_{\mathrm{2}}\mathrm{and}{\mathrm{R}}_{\mathrm{3}}.\right]\\ ={\mathrm{k}}^{3}|\mathrm{A}|\\ \mathrm{Thus},\mathrm{option}\left(\mathrm{C}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.16 Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these.

Ans

To every square matrix A = [aij] of order n, we can associate a number (real or complex) called determinant of the square matrix A, where aij = (i,j)th element of A.
Therefore, option (C) is correct

## 1. Are the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 difficult?

No, with regular practice and proper guidance, students can easily solve the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 and learn the concepts to score well in the board examinations.

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Yes, students should practice all the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2, as having a clear understanding of the basics and a fast calculation speed are crucial to scoring well on the examination.

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Extramarks provides students with NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2. Along with this, it provides students with K12 study material for their boards and live classes with the experts to provide detailed solutions for all the questions the students.

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Yes, according to the changes in the admission pattern of Delhi University, the University is conducting an entrance examination which is entirely based on the content of NCERT. Also, there are other universities which follow the same pattern. So yes, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2 will help students in the preparation for certain entrance examinations.

## 6. Is the NCERT Exemplar book needed for the preparation of the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2?

Mathematics is a subject which requires a lot of practice, practising the examples given in the NCERT book and the NCERT Exemplar will definitely help them in building the concepts of the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2. The more students will practice, the better they will get.

## 7. Is it necessary to know Matrices before learning the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2?

Yes, it is important to have a knowledge of Chapter 3 Matrices before starting to learn

Chapter 4 Determinants, as the Determinant is defined as the real or complex number that can be associated with a Square Matrix.

## 8. Why is it important to practice the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2?

Practising the NCERT Solutions is the first and the foremost step for preparing for Mathematics, as they structure and build the basic concepts of the students. If they practice the NCERT Textbook solutions, they can easily solve any complicated problem that occurs in their in-school, competitive or board examinations. The Extramarks’ website provides students with reliable study material so that they can score better marks in their examinations.

Click on the given link to download the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.2.