The chapter on Determinants in the NCERT Class 12 Maths textbook (Part 1) introduces the concept of determinants, their properties, and their applications. Here’s a summary:
1. Determinant:
- A determinant is a scalar value associated with a square matrix.
- It is denoted by |A| or det(A) for a square matrix A.
- Determinants have important applications in solving systems of linear equations and finding the inverse of a matrix.
2. Calculating Determinants:
For a 2×2 matrix:
|A| = |a b| = ad – bc
|c d|
- For a 3×3 matrix: There are a couple of methods, including expansion along a row or column using minors and cofactors.
3. Minors and Cofactors:
- Minor: The minor M<sub>ij</sub> of an element a<sub>ij</sub> is the determinant of the submatrix obtained by deleting the i-th row and j-th column.
- Cofactor: The cofactor C<sub>ij</sub> of an element a<sub>ij</sub> is given by C<sub>ij</sub> = (-1)<sup>i+j</sup> M<sub>ij</sub>.
4. Properties of Determinants:
- The determinant of a matrix and its transpose are equal (|A| = |Aᵀ|).
- If any two rows or columns of a determinant are interchanged, the sign of the determinant changes.
- If any row or column of a determinant is multiplied by a scalar k, the determinant is also multiplied by k.
- If any two rows or columns of a determinant are identical or proportional, the determinant is zero.
- If each element of a row or column is expressed as the sum of two or more terms, the determinant can be expressed as the sum of two or more determinants.
- The value of a determinant remains unchanged if any row or column is added to or subtracted from another row or column multiplied by a constant.
5. Applications of Determinants:
- Solving systems of linear equations: Cramer’s rule provides a method to solve systems of linear equations using determinants.
- Finding the area of a triangle: The area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is given by:
Area = (1/2) |x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)|
Exercise 4.1
Evaluate the following determinants in Exercise 1 and 2.
1.
Ans :
2(-1) – 4(-5)
= -2 + 20
= 18
2. (i)
(ii)
Ans :
(i)
(cosθ)(cosθ) – (-sinθ) (sinθ)=
cos 2 θ + sin 2 θ
= 1
(ii) ( x 2 − x + 1)( x + 1) − ( x − 1)( x + 1)
= x 3 − x 2 + x + x 2 − x + 1 − ( x 2 − 1)
= x 3 + 1 − x 2 + 1
= x 3 − x 2 + 2
3. If A
then show that |2A| = 4|A|
Ans :
Given: A =
then 2A = 2 x
Hence, proved.
4. If A =
then show that 3|A| = 27|A|
Ans :
Given: A =
then 3A = 3
Hence Proved.
5. Evaluate the determinants:
(i)(ii)(iii)(iv)
Ans :
(i) Given:
It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation. =
(ii) Given:
(iii) Given:
= 0 + 6 – 6 = 0
(iv) Given:
= -10 + 15 = 5
6. If A =
find |A|
Ans :
Expanding along first row, =
7. Find the value of x if: (i)
(ii)
Ans :
(i)
⇒ 2 x 1 – 5 x 4
= 2x * x – 6 x 4
⇒ 2 – 20
= 2x 2 – 24
⇒ 2x 2 = 6
⇒ x 2 = 3
⇒ x = ± √3
(ii)
⇒ 2 x 5 – 4 x 3
= x * 5 – 2x – 3
⇒10 – 12
= 5x – 6x
⇒ – 2 = -x
⇒ x = 2
8. If
then x is equal to:
(A) 6 (B) ± 6 (C) – 6 (D) 0
Ans :
⇒x * x – 18 x 2
= 6 x 6 – 18 x 2
⇒x 2 – 36
= 36 – 36
⇒x 2 – 36
= 0
⇒x = ± 6
Therefore, (B) is correct.
