# Determinants Chapter 4 Class 12 Maths NCERT Textbook PDF

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### Chapter 4: Determinants

#### 4.3 Properties of Determinants

In the previous section, we have learned how to expand the determinants. In this section, we will study some properties of determinants which simplifies their evaluation by obtaining a maximum number of zeros in a row or a column.

These properties are true for determinants of any order. However, we shall restrict ourselves up to determinants of order 3 only.

#### 4.6 Adjoint and Inverse of a Matrix

In the previous chapter, we studied the inverse of a matrix. In this section, we shall discuss the condition for the existence of the inverse of a matrix. To find the inverse of a matrix A, i.e., A–1 we shall first define the adjoint of a matrix.

#### 4.7 Applications of Determinants and Matrices

In this section, we shall discuss the application of determinants and matrices for solving the system of linear equations in two or three variables and for checking the consistency of the system of linear equations.

Consistent system A system of equations is said to be consistent if its solution (one or more) exists. Inconsistent system A system of equations is said to be inconsistent if its solution does not exist.

### NCERT Solutions Class 12 Maths Chapter 4 Determinants

Question 1:

Evaluate the determinants in Exercises 1 and 2. = 2(−1) − 4(−5) = − 2 + 20 = 18

Question 2:

Evaluate the determinants in Exercises 1 and 2.

(i) (ii) (i) = (cos θ)(cos θ) − (−sin θ)(sin θ) = cos2 θ+ sin2 θ = 1

(ii) = (x2 − x + 1)(x + 1) − (x − 1)(x + 1)

x3 − x2 + x + x2 − x + 1 − (x2 − 1)

x3 + 1 − x2 + 1

x3 − x2 + 2

Question 3:

If , then show that The given matrix is .

Question 4:

If , then show that The given matrix is .

It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.

From equations (i) and (ii), we have:

Hence, the given result is proved.

Question 5:

Evaluate the determinants

(i) (iii) (ii) (iv) (i) Let .

It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.

(ii) Let .

By expanding along the first row, we have:

(iii) Let By expanding along the first row, we have:

(iv) Let By expanding along the first column, we have: