Title 4. DETERMINANTS.pdf ✅

TOPIC-1 Determinants, Minors & Co-factors Concepts Covered:  Determinant value of a matrix,  Co-factor and Minor of a matrix  Inverse of matrix using Adjoint method,  Area of triangle with the help of determinant Revision Notes Determinants, Minors & Co-factors (a) Determinant: A unique number (real or complex) can be associated to every square matrix A = [aij] of order m. This number is called the determinant of the square matrix A, where aij = (i, j) th element of A. For instance, if A = a b c d       then, determinant of matrix A is written as |A| = a b c d = det (A) and its value is given by ad – bc. [Board 2023] (b)Minors: Minor of an element aij of a determinant (or a determinant corresponding to matrix A) is the determinant obtained by deleting its i th row and j th column in which aij lies. Minor of aij is denoted by Mij. Hence, we can get 9 minors corresponding to the 9 elements of a third order (i.e., 3 × 3) determinant. (c)Co-factors: Cofactor of an element aij, denoted by Aij, is defined by Aij = (–1)(i + j) Mij, where Mij is minor of aij. Sometimes Cij is used in place of Aij to denote the co- factor of element aij. 1. ADJOINT OF A SQUARE MATRIX: Let A = [aij] be a square matrix. Also, assume B = [Aij], where Aij is the cofactor of the elements aij in matrix A. Then the transpose BT of matrix B is called the adjoint of matrix A and it is denoted by “adj (A)”. [Board 2023] To find adjoint of a 2 × 2 matrix: Follow this, A = a b c d       or adj A = d b c a − −       . For example, consider a square matrix of order 3 as A = 1 2 3 234 2 0 5         , then in order to find the adjoint matrix A, we find a matrix B (formed by the co-factors of elements of matrix A as mentioned above in the definition) i.e., B = 15 2 6 10 1 4 1 2 1 − − − − − −         . Hence, adj A = BT = 15 10 1 2 1 2 6 4 1 − − − − − −         2. SINGULAR MATRIX AND NON-SINGULAR MATRIX: (a) Singular matrix: A square matrix A is said to be singular if |A| = 0 i.e., its determinant is zero. [Board 2023] e.g. A = 1 2 3 4 5 12 1 1 3 = 1(15 – 12) – 2(12 – 12) + 3(4 – 5) = 3 – 0 – 3 = 0 \ A is singular matrix. (b)Non-singular matrix: A square matrix A is said to be non-singular if |A| ≠ 0 . [Board 2023] e.g. A = 0 1 1 1 0 1 1 1 0 = 0 (0 – 1) – 1(0 – 1) + 1(1 – 0) = 0 + 1 + 1 = 2 ≠ 0 \ A is non-singular matrix. l A square matrix A is invertible if and only if A is non- singular. Learning Objectives After going through this chapter, the students will be able to:  Learn to find determinant of a matrix of order 2×2 or 3×3 , minor and cofactor  Learn to calculate the adjoint and inverse of a square matrix  Learn to evaluate area of a triangle with given coordinates using determinants  Learn to determine the consistency, inconsistency and number of solutions of system of linear equations  Learn to apply the properties of matrices and determinants in solving the system of linear equations in two or three variables LIST OF TOPICS Topic-1: Determinants, Minors & Co-factors Topic-2: Solutions of System of Linear Equa- tions 4 DETERMINANTS CHAPTER


Determinants |Adj P| = ad – bc |Adj P| = |P| = 9 Short Answer Type Questions-I (2 marks each) 1. If A and B are square matrices of order 3 such that |A| = –1, |B| = 3, then find the value of |2AB|. R & U [Foreign 2017] Sol. |2AB| = 23 × |A| × |B| 1 = 8 × (–1) × 3 = –24 1 [Marking Scheme Foreign, 2017] 2. If A is a square matrix of order 3 such that A2 = 2A, then find the value of |A|. R & U [SQP 2020-21] Sol. A2 = 2A Þ |AA| = |2A| Þ |A| |A| = 8|A| ( |AB| = |A| |B| and |2A| = 23 |A| ) 1⁄2 Þ |A| (|A| – 8) = 0 1 Þ |A| = 0 or 8 1⁄2 [Marking Scheme SQP, 2020-21] Commonly Made Error Students make mistakes in applying the property |kA|= kn|A|. Instead they take it as |kA|= k|A|. Practice more problems involving properties of determinants. Answering Tip 3. If the area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq units, then find the value of k. E [OEB] Sol.Given area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq units. We have, 1 2 3 0 1 3 0 1 0 1 − k = ±9 Þ −3 0 1 3 0 1 0 1 k = ±18 Þ –3(0 – k) – 0 + 1(3k – 0) = ±18 1 Þ 3k + 3k = ±18 Þ 6k = ±18 Þ k = ±3 1 Short Answer Type Questions-II (3 marks each) 1. Using determinants, find the area of DPQR with vertices P(3, 1), Q(9, 3) and R(5, 7). Also, find the equation of line PQ using determinants. R & U [Delhi Set-2, 2023-24] Sol. Ar.(D PQR) = 1 1 2 2 3 3 1 1 1 2 1 x y x y x y = 311 1 931 2 571 = ( ) ( ) ( ) 1 3 3 7 1 9 5 1 63 15 2 −− −+ − = 1 12 4 48 2 − −+ = 32 16 sq. units 2 = Equation of the line PQ 1 311 931 x y = 0 x(1 – 3) –y(3 – 9) + 1(9 – 9) = 0 –2x+6y = 0 x –3y = 0 [Marking Scheme Delhi Set-2, 2023-24] Commonly Made Error To find equation of line students generally forget to put determinant equals to zero. Keep remember to put determinant equals to zero while solving questions related to equation of line. Answering Tip 2. Show that the determinant x x x sin cos sin cos θ θ θ θ − − 1 1 is independent of q. A [Delhi Set-3, 2023-24] Sol. x x x sin cos sin cos θ θ θ θ − − 1 1 = x x x x − − 1 − 1 1 sin sin cos θ θ θ + − − cos sin cos θ θ θ x 1 = x (–x2 –1) – sin q (–x sin q – cos q) + cos q (– sin q + x cos q) = – x3 – x + x sin2 q + sinq cos q – sin q cos q + x cos2 q = – x3 – x + x (sin2 q + cos2 q) = – x3 – x + x = – x3 It is independent of q. Hence Proved. [Marking Scheme, Delhi Set-3, 2023-24] Long Answer Type Questions (5 marks each) 1. Find the adjoint of the matrix A = − − − − −         1 2 2 2 1 2 2 2 1 and hence show that A(adj A) = |A|I3 . E [OEB] Sol. We have, A = − − − − −         1 2 2 2 1 2 2 2 1 Let Cij be the cofactor of the element aij of |A|. Now, cofactors of |A| are