PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1QE3SLNu1ADZUg34K4p5T-y7gU0G_V_P2/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true Matrices and Determinants Matrices, Algebra of matrices, Types of matrices, Determinants & matrices of order two & three. Properties of determinants, Eva lua tion of determinants, Area of triangles using determinants. Adjoint & evaluation of inverse of a squa re matrix using determina nts & e l ementary transformations, Test of consistency & solution of simultaneous linear equations in two or three variables using determinants and matrices. Matrices and determinants are two very important tools in the hands of a mathematician. These have wide application in algebra, trigonometry, co-ordinate geometry & calculus. Problems can be handled in a much simplified way and lot- many calculations are cut short. DETERMINANTS OF 2ND & 3RD ORDER C H A P T E R CHAPTER INCLUDES : Determinants of 2 nd & 3rd order Minors and Cofactors a1 The symbol a2 b1 b2 is called a determinant of order 2. It has two rows and Properties for evaluation of determinants two columns. a b – a b a1 b1 is called expansion of determinant . These Determinant of 1 2 2 1 a2 b2 cofactors idea can be extended to higher order determinants. Determinant of 3rd order Multiplication of determinants Area of triangle System of linear a11 The symbol Δ = a21 a31 a12 a22 a32 a13 a23 a33 = | aij | is called the determinant of 3rd order equations Cramer's rule Matrices and its value is equal to the number: Types of matrices Algebra of matrices (−1)1+1a a22 a23 + (−1)1+2a a21 a23 + (−1)1+3 a a21 a22 Adjoint of matrix 11 a32 a33 12 a31 a33 13 a31 a32 Inverse of matrix = a (a a a a ) − a (a a a a ) + a (a a a a ) ...(A) Solution of 11 22 33 23 32 12 21 33 31 23 13 21 32 31 22 simultaneous linear equations is called expansion of determinant Δ along its first row. In fact, expansion can be carried along any row or column in a simmilar way. Solved examples MINORS AND CO-FACTORS OF DETERMINANT In the determinant Δ, the minor of the element aij (where i denotes number of row and j the number of columns, i.e., element common to ith row and jth column) is obtained by removing the ith row and jth column and it is given by Mij and For example, M12 = a21 a31 a23 a33 cofactor of a is defined as C = (–1)i + j M , for example, C = (–1)1+2 a21 a23 or C12 ij = − a21 a31 ij ij a23 . a33 12 a31 a33 We can also express Δ as follows : Δ = a11 C11 + a12 C12 + a13 C13 or Δ = a21 C21 + a22 C22 + a23 C23 or Δ = a31 C31 + a32 C32 + a33 C33 Notations for operations performed on rows/columns of determinants Ri (i = 1, 2, 3) and Cj (j = 1, 2, 3) are used to denote ith row and jth column respectively. The interchange of ith row and jth row : Ri ←→ Rj The addition of m times the element of ith row to the corresponding elements of jth row : Rj ⎯→ Rj + mRi The addition of m times the element of jth row and n times the element of kth row to the corresponding element of ith row : Ri ⎯→ Rj + mRj + nRk (Similar operation can be performed with respect to columns of Δ) PROPERTIES FOR EVALUATION OF DETERMINANTS The value of determinant is not changed when all rows and columns are interchanged i.e., a1 b1 c1 a1 a2 a3 a2 b2 c2 = b1 b2 b3 a3 b3 c3 c1 c2 c3 If any two rows (or any two columns) of a determinant are interchanged, the sign of the determinant is changed, but its value remains same (numerically), a1 a2 b1 b2 c1 c2 = − c1 c2 b1 b2 a1 a2 a3 b3 c3 c3 b3 a3 If all the elements of the same row (or the same column) are multiplied by a number, then the determinant (its value) becomes multiplied by the number ma1 a2 a3 mb1 b2 b3 mc1 c2 c3 a1 = m a2 a3 b1 c1 b2 c2 b3 c3 If any two rows (or columns) are identical or their respective elements are proportional the the value of a determinant is zero. a1 ma1 a2 ma2 a3 ma3 c1 c2 = 0 c3 The value of a determinant does not change when any row (or column) is multiplied by a number or an expression and is added to or subtracted from any other row (or column) a1 b1 c1 a1 b1 + m1c1 c1 a2 b2 c2 = a2 b2 + m1c2 c2 a3 b3 c3 a3 b3 + m1c3 c3 If every element of some row (or column) is the sum of two numbers, then determinant can be expressed as the sum of two determinants of the same order; one containing only the first number in place of sum, the other only the second number. The remaining element of both determinant are the same as in the given determinant. a1 + α1 + β1 b1 c1 a1 b1 c1 α1 b1 c1 β1 b1 c1 a2 + α2 + β2 b2 c2 = a2 b2 c2 + α2 b2 c2 + β2 b2 c2 a3 + α3 + β3 b3 c3 a3 b3 c3 α3 b3 c3 β3 b3 c3 (In this example, we have shown it for sum of three numbers) Remarks : The value of determinant remains unaltered under an operation of the form Ri ⎯→ Ri + mRj + nRk (j, k ≠ i). These operation enable us to convert maximum elements into zeros for making evaluation of determinant simpler. If a determinant Δ(x) becomes zero on putting x = α, then x – α is a factor of Δ(x). Determinant which have all elements equal to zero except the diagonal elements is equal to the product of the diagonal elements i.e. a 0 0 0 b 0 = abc 0 0 c Also a α β a 0 b γ = α 0 0 c β 0 0 b 0 = abc γ c 0 a Determinant − a 0 b − c b c = 0. (note that upper triangular is reflected along diagonal with opposite sign) 0 Without expanding the determinant, prove that a + b b + c c + a b + c c + a a + b c + a a + b b + c a b c = 2 b c a c a b Solution : LHS = 2(a + b + c) 2 (a + b + c) 2(a + b + c) b + c c + a a + b c + a a + b b + c (C1 → C1 + C2 + C3 ) a + b + c = 2 a + b + c a + b + c b + c c + a a + b c + a a + b b + c (taking 2 common from 1st - column) = 2 = 2 a = 2 b c b c a b + c c + a b c + a a + b (C1 → C1 – C2 ) c a + b b + c a b + c c c + a a (C3 → C3 – C1) a + b b c a (C2 → C2 – C3 ) a b = RHS DETERMINANT OF CO-FACTORS If Δ ≠ 0 be the determinant and Δc denotes determinant of cofactors then Δc = Δn-1 (where n > 0) and n is order of determinant Δ. In particular if Δ is determinant of 3rd order and Δ= a11 a21 a31 a12 a22 a32 a13 a23 a33 ≠ 0 then Δc = C11 C21 C31 C12 C22 C32 C13 C23 C33 = Δ2 2. Δ = 0 then Δc = 0 Product of two Determinants a1 a2 Let D1 = 1 2 c1 c2 a3 α1 α2 α3 3 and D2 = 1 2 3 c3 γ1 γ 2 γ 3 then product of D1 and D2 is defined as follows : D1 D2 = ∑ai αi ∑bi αi ∑ci αi ∑ai βi ∑bi βi ∑ci βi ∑aiγi ∑biγi ∑ciγi (Where Σ runs from 1 to 3 i.e. ∑ai αi = a1α1 + a2α2 + a3α3 ) Remark : 1. Since value of determinant remains same when rows and column are interchanged therefore, in finding product of two determinants, we can also multiply rows by columns or column by columns. 2. D1 D2 = D2 D1 Important Observation a1 a2 If D1 = 1 2 c1 c2 a3 A1 A2 A3 3 and D2 = 1 2 3 c3 C1 C2 C3 where Ai = cofactor of ai Bi = cofactor of bi and C = cofactor of c then D =(D )2 i i 2 1 If t = ar + br + cr, show that t0 t1 t1 t2 t2 t3 t2 t3 = (a – b)2 (b – c)2 (c – a)2 t4 Solution : t0 t1 We have t1 t2 t2 t3 t2 t3 = t4 1 1 1 1 1 1 = a b c a b c a2 b2 c 2 a2 b2 c 2 1 1 1 2 = a b c a2 b2 c 2 1 0 = a b − a a2 b2 − a2 0 2 c − a c 2 − a2 Applying (C2 → C2 – C1, C3 → C3 – C1 ) 1 = (b – a)2 (c – a)2 a a2 0 1 b + a 0 2 1 c + a = (a – b)2 (c – a)2 [c + a – b – a]2 = (a – b)2 (b – c)2 (c – a)2 AREA OF TRIANGLE x1 If ABC is a triangle with A(x , y ), B(x ,y ) and C(x , y ) then area of Δ = x2 x3 y1 1 y 2 1 y3 1 The area may come out to be either positive or negative, hence modulus of above would give the magnitude of area. x1 If x2 x3 y1 1 y 2 1 turns out to be zero, it means points A, B & C are collinear and vice-versa. y3 1 SYSTEM OF LINEAR EQUATIONS System of Linear equatins : a11x1 + a12 x2 + ....... + a1n xn = b1 a21x1 + a22 x2 + ....... + a2n xn = b2 an1x1 + an2 x2 + .......+ ann xn = bn is said to be consistent if it has at least one solution. Otherwise, it is said to be inconsistent. I. System of homogeneous equations (linear) in three variables. a1x + b1y + c1 z = 0, a2 x + b2 y + c2 z = 0, a3 x + b3 y + c3 z = 0 Note that x = y = z = 0 (trivial solution) is always a solution. Therefore, this system is always consistent. a1 Case I : If Δ = a2 a3 b1 c1 b2 c2 b3 c3 ≠ 0, there is unique solution x = y = z = 0 Case II : If Δ = 0, the system has infinite number of solutions (existence of non-trivial solutions) CRAMER'S RULE Consider the system of equations : a1 x + b1 y + c1 z = d1, a2 x + b2 y + c2 z = d2, a3 x + b3 y + c3 z = d3 We define a1 The denominator determinant = Δ = a2 a3 b1 c1 b2 c2 b3 c3 ≠ 0 ...(1) d1 b1 c1 The numerator determinant of x = Δ = d2 b2 c2 (Replace C of (1) by d , d , d ) x d3 b3 c3 1 1 2 3 a1 d1 c1 The numerator determinant of y = Δ = a2 d2 c2 a3 d3 c3 a1 and the numerator determinant of z = Δ = a2 a3 b1 d1 b2 d2 b3 d3 Case I : If Δ ≠ 0 then system is called consistent and have unique solution given by x = Δ x ,y = Δ y and z = Δ z .(Δ ≠ 0) Δ Δ Δ Case II : If Δ = 0 and but if Δ , Δ , Δ atleast one is non-zero, then system in called inconsistent and have no solution. Case III : If Δ = Δ = Δ = Δ = 0, then system have infinite no. of solutions. ⎧ 1 + 2 + ⎪ ⎪ ⎪ Solve the system ⎨ ⎪ y − 4 − 2 = 1 y z using determinants. ⎪ 2 + 5 − 2 = 3 ⎪⎩ x y z Solution : Let 1 = u, 1 = v, 1 = w , then given system can be expressed as u + 2v + w = 2, 3u – 4v – 2w = 1, x y z 2u + 5v – 2w = 3 1 2 Now the denominator Δ = 3 − 4 2 5 1 − 2 = 45 − 2 2 2 Δ , the numerator of u = 1 − 4 3 5 1 − 2 = 45 − 2 1 2 Δ , the numerator of v = 3 1 2 3 1 − 2 = 15 − 2 1 Δ , the numerator of w = 3 2 2 2 − 4 1 5 3 = 15 Then, u = Δ1 = 45 = 1, v = Δ2 = 1 , w = Δ3 = 1 Δ 45 Δ 3 Δ 3 ∴ Solution of system is given by x = 1, y = 3, z = 3. MATRICES A set of mn numbers (real or complex) arranged in the form of a rectangular array having ‘m’ rows and ‘n’ columns is called an m × n matrix. It is usually written as ⎡ a11 ⎢ a21 A a12 a22 ... ... a1n ⎤ a2n ⎥ ⎢ Μ Μ ⎢a a Μ Μ ⎥ ... a ⎥ ⎣⎢ m1 m 2 mn ⎥⎦ It is represented as A = [aij]m × n. The numbers a11, a12, ... etc. are called elements of the matrix. The element aij belong to ith row and jth column. ⎡3 2 e.g., A = ⎢5 − 4 7 ⎤ 6 ⎥ is a 3 × 3 matrix ⎢⎣4 8 − 12⎥⎦ The number of matrices having 12 elements is (1) 3 (2) 1 (3) 6 (4) None of these Solution : Ans. (3) The matrix will be any one of the following type 1 × 12, 12 × 1, 2 × 6, 6 × 2, 3 × 4, 4 × 3 TYPES OF MATRICES Square Matrix : A matrix having equal number of rows and columns is called a square matrix. If the matrix A has n rows and n columns, it is said to be square matrix of order n. Example: ⎡1 2 ⎢ 3 ⎢⎣3 4 3⎤ ⎥ ⎥ 5⎥⎦ is a square matrix of order 3. The diagonal of this matrix containing the elements 1, 3, 5 is called the leading or principal diagonal. Trace of matrix : The sum of the diagonal elements of a square matrix A is called trace of A. Trace (A) = ∑aii i =1 = a11 + a22 + a33 ... + ann Example : ⎡2 A = ⎢0 ⎢⎣8 − 7 9⎤ 3 ⎥ 9 4⎥⎦ Trace (A) = 2 + 3 + 4 = 9 Diagonal Matrix : A square matrix in which all the elements, except those in the leading diagonal are zero is called a diagonal matrix. Thus square matrix is called diagonal matrix if aij = 0 for i ≠ j. e.g., ⎡2 0 A = ⎢0 1 ⎢⎣0 0 0⎤ ⎥ ⎥ 4⎥⎦ ⎡d1 0 or ⎢ 0 d2 ⎢⎣ 0 0 0 ⎤ ⎥ ⎥ d3 ⎥⎦ Above are diagonal matrices of the type 3 × 3. These are in short written as diag [2, 1, 4] or diag [d1, d2 d3] If the diagonal elements are d1, d2, dn, then it is written as diag [d1, d2, , dn]. Scalar Matrix : A diagonal matrix in which all the elements in the leading diagonal are equal is called scalar matrix. It is equal to KIn for some scalar K. ⎡3 0 e.g., A = ⎢0 3 ⎢⎣0 0 0⎤ ⎥ ⎥ 3⎥⎦ ⎡K 0 or ⎢ 0 K ⎢⎣ 0 0 0 ⎤ ⎥ ⎥ K ⎥⎦ In general for a scalar matrix, a = 0 for i ≠ j and a = K for i = j Unit or Identity Matrix : A square matrix in which all the elements in the leading diagonal are 1 and remaining elements are zero is called a unit or Identity Matrix. A unit matrix of order n is usually denoted by In. Thus for an Identity Matrix a = 1 and a = 0 for i ≠ j ⎡1 e.g., I3 = ⎢0 0 0⎤ 1 0⎥ & I = ⎡1 0⎤ ⎢ ⎢⎣0 0 ⎥ 1⎥⎦ 2 ⎣0 1⎦ Row matrix : A matrix A which has only one row and n columns is called a row matrix. It is evidently 1 × n matrix. It is also called a row vector. e.g., A = [1 2 3] is a row matrix of type 1 × 3 Column matrix : A matrix A which has m rows and only one column is called a column matrix. It is evidently m × 1 matrix. It is also called a column vector. ⎡1⎤ ⎢2⎥ e.g., A = ⎢ ⎥ is a column matrix of type 4 × 1 ⎢0⎥ ⎢4⎥ ⎣ ⎦ Null Matrix or Zero Matrix : A (m × n) matrix in which all the elements are zero is called a zero matrix or null matrix of the type m × n and is denoted by Om × n or O. ⎡0 Thus ⎢0 0⎤ ⎡0 0⎥ , ⎢0 0 0⎤ 0 0⎥ , ⎡0 0 0 0⎤ ⎢ ⎢⎣0 ⎥ 0⎥⎦ ⎢ ⎢⎣0 0 0⎥⎦ ⎣ 0 0 ⎦ All the above are zero or null matrices of type 3 × 2, 3 × 3 and 2 × 4 respectively. Horizontal matrix : A m × n matrix is called a horizontal matrix if m < n ⎡1 0 2⎤ e.g., A = ⎣ 4 ⎦ Vertical matrix : A m × n matrix is called a vertical matrix of m > n. ⎡1 e.g., A = ⎢2 ⎢⎣3 0⎤ ⎥ ⎥ 5⎥⎦ Triangular matrix : Let A be a square matrix ⎡a11 ⎢a21 A = ⎢a31 ⎢ ... ⎢⎣an1 a1n ⎤ a2n ⎥ a3n ⎥ ... ⎥ ann ⎥⎦ Any element of A is aij and for i = j we get the diagonal elements. All those elements which are above the principal diagonal are of the type a12, a13, a23, a34, a3n or in general aij where i < j and all those elements which are below the principal diagonal are of the type a21, a31, an3, so on or in general aij where i > j. Upper Triangular Matrix : A square matrix (aij) is called an upper triangular matrix if aij = 0 when i > j. In other words, all elements below the principal diagonal be zero. Lower Triangular Matrix : A square matrix (aij) is called lower triangular matrix if aij = 0 when i < j. In other words, all the elements above the principal diagonal be zero. ⎡3 for e.g., ⎢0 ⎢⎣0 − 2 4 ⎤ 2 − 3⎥ & 0 7 ⎥⎦ ⎡2 0 ⎢ 4 ⎢⎣2 8 0⎤ ⎥ ⎥ 6⎥⎦ are upper and lower triangular matrices respectively. Transpose of a Matrix : A matrix obtained by interchanging rows and columns of a matrix A is called transpose of A and is denoted by A′ or AT. Note : If A is a m × n matrix, then A′ is an n × m matrix. ⎡1 2 3⎤ ⎡1 4 7⎤ e.g., A = ⎢4 5 6⎥ , then A′ = ⎢2 5 8⎥ ⎢ ⎢⎣7 ⎥ 8 9⎥⎦ ⎢ ⎢⎣3 6 ⎥ 9⎥⎦ Properties of Transpose (i) (A′)′ = A (ii) (A + B)′ = A′ + B′ a12 a13 ... a22 a23 ... a32 a33 ... ... ... ... an2 an3 ... (iii) (KA)′ = KA′ (iv) (AB)′ = B′ A′ Symmetric Matrix : A square matrix A = [a ] is said to be symmetric when a = a for all i and j. In other words, A′ = A ⎡1 4 e.g., ⎢4 2 ⎢⎣5 − 6 5 ⎤ − 6⎥ 3 ⎥⎦ is 3 × 3 matrix Here a12 = a21, a13 = a31, a23 = a32 Skew-symmetric matrix : A square matrix is said to be skew-symmetric if aij = –aji for all i and j so that all the leading diagonal elements are zero. In other words A′ = –A ⎡ 0 2 e.g., ⎢− 2 0 ⎢⎣ 3 − 5 − 3⎤ ⎥ ⎥ 0 ⎥⎦ is 3 × 3 matrix here a12 = –a21, a13 = –a31, a23 = –a32 a11 = a22 = a33 = 0 Every square matrix can be uniquely expressed as a sum of a symmetric and skew-symmetric matrix. Let A be the given square matrix, then A = 1 (A + A′) + 2 1 (A – A′) = P + Q 2 where P is a symmetric matrix and Q is a skew-symmetric matrix. Singular matrix : A square matrix A is said to be singular if | A | = 0. Non-singular matrix : A square matrix A is said to be non-singular if | A | ≠ 0. Orthogonal Matrix : A matrix A is said to be orthogonal if AA′ = I. (Identity matrix) Equality of Matrices Two matrices A = [aij]m × n , B = [bij]m × n are said to be equal and written as A = B if and only if they are of the same order and each element of A is equal to the corresponding element of B. ⎡1 2 0⎤ ⎡1 2 0⎤ Let A = ⎢3 2 4⎥ , B = ⎢3 2 4⎥ ⎣ ⎦ ⎣ ⎦ Here A and B are equal matrices and we write A = B Find value of x, y, z and w which satisfy the matrix equation ⎡x + 3 2y + x ⎤ ⎡− x − 1 0 ⎤ ⎢ z − 1 4w − 8⎥ = ⎢ 3 2w ⎥ Solution : As the two matrices are equal and are of order 2 × 2, so x + 3 = –x –1 ...(1) 2y + x = 0 ...(2) z –1 = 3 ...(3) 4w –8 = 2w ...(4) on solving above equations, we get x = –2, y = 1, z = 4, w = 4 ALGEBRA OF MATRICES Let A and B be two matrices of same order m × n. Then their sum is defined to be the matrix of order m × n obtained by adding the corresponding elements of A and B ⎡a11 a12 ⎤ ⎡b11 b12 ⎤ e.g., If A = ⎢a21 a22 ⎥ and B = ⎢ ⎣ 21 ⎥ , b22 ⎦ then A + B = ⎡a11 + b11 ⎢a21 + b21 a12 + b12 ⎤ a22 + b22 ⎥ Properties : Matrix addition is commutative If A and B be two m × n matrices, then A + B = B + A Matrix addition is associative If A, B and C be three matrices each of order m × n, then (A + B) + C = A + (B + C) Existence of additive Identity : If O be the m × n matrix each of whose elements is zero, then A + O = A = O + A for every m × n matrix A.O is called additive Identity. Existence of additive Inverse: Let A = [aij]m × n. Then the negative of the matrix A is defined as the matrix [–aij]m × n and is denoted by – A. A + (– A) = 0 e.g., If A = ⎡1 − 2 ⎣ 2 0 ⎤ − 5⎥ then –A = ⎡ − 1 ⎣ 2 0⎤ − 2 ⎦ Cancellation laws hold good for addition : If A, B, C be any three m × n matrices, then A + B = A + C ⇒ B = C k(A + B) = kA + kB, where k is scalar. Subtraction of two matrices Let A and B be two m × n matrices. Then the difference of A and B is denoted by A – B and is defined by A – B = A + (–B) A – B will also be a m × n matrix. In order to find A – B, the elements of B must be subtracted from the corresponding elements of A. ⎡1 2 3⎤ ⎡2 − 1 4⎤ e.g., A = ⎢3 − 1 0⎥ , B = ⎢5 0 6⎥ ⎣ ⎦ ⎣ ⎦ ⎡1− 2 then A – B = ⎢3 − 5 2 + 1 − 1− 0 3 − 4⎤ 0 − 6⎥ ⎡ − 1 = ⎢− 2 3 − 1⎤ − 1 − 6⎥ ⎡1 4⎤ ⎡− 1 − 2⎤ If A = ⎢3 2⎥ , B = ⎢ 0 5 ⎥ , find the matrix D such that A + B –D = 0 i.e., zero matrix. ⎢ ⎢⎣2 ⎥ 5⎥⎦ ⎢ ⎢⎣ 3 ⎥ 1 ⎥⎦ Solution : ⎡a b⎤ Let the matrix D be ⎢c ⎢⎣e ⎡ 1− 1− a ⎥ ⎥ f ⎥⎦ 4 − 2 − b⎤ ∴ A + B –D = ⎢3 + 0 − c ⎢⎣2 + 3 − e 2 + 5 − d ⎥ 5 + 1− f ⎥⎦ ⎡ − a 2 − b⎤ ⎡0 0⎤ = ⎢3 − c 7 − d ⎥ = ⎢0 0⎥ given ⎢ ⎢⎣5 − e ⎥ 6 − f ⎥⎦ ⎢ ⎢⎣0 ⎥ 0⎥⎦ From the definition of equality of matrices, – a = 0 ⇒ a = 0 2 – b = 0 ⇒ b = 2 3 – c = 0 ⇒ c = 3 7 – d = 0 ⇒ d = 7 5 – e = 0 ⇒ e = 5 6 – f = 0 ⇒ f = 6 ⎡0 ∴ D = ⎢3 ⎢⎣5 2⎤ ⎥ ⎥ 6⎥⎦ which is clearly equal to A + B Multiplication of Matrix by a scalar The product of a matrix A by a scalar k is a matrix whose each element is k times the corresponding elements of A. ⎡a1 Thus k ⎢a2 b1 c1 ⎤ b2 c2 ⎥ = ⎡ ka1 ⎢ka2 kb1 kb2 kc1 ⎤ kc2 ⎥ Properties of Scalar Multiplication If A and B be any two m × n matrices and k is any scalar, then k(A + B) = kA + kB If A is any m × n matrix and a and b are any two scalars, then (a + b)A = aA + bA If A be any m × n matrix and k be any scalar, then (– k)A = – (kA) = k(– A) Multiplication of Matrices Let A = [aij] be a m × n matrix and B = [bjk] be a n × p matrix such that number of columns in A is equal to number of rows in B. Then product of matrices A and B denoted by AB is a matrix of m × p matrix. ⎡a1 b1 c1 ⎤ ⎡ l l ⎤ ⎢a b c ⎥ ⎢ 1 2 ⎥ For instance, let A = ⎢ 2 2 2 ⎥ and B = ⎢m1 m2 ⎥ ⎢a3 b3 c3 ⎥ ⎢ n n ⎥ ⎢ ⎣a4 b4 ⎥ 4 ⎦ 4 × 3 ⎣ 1 2 ⎦3×2 then matrix AB is a matrix of order 4 × 2 AB = ⎡ a1l1 + b1m1 + c1n1 ⎢a2l1 + b2m1 + c2n1 ⎢a3l1 + b3m1 + c3n1 a1l2 + b1m2 + c1n2 ⎤ a2l2 + b2m2 + c2n2 ⎥ a3 l2 + b3m2 + c3 n2 ⎥ ⎢a l + b m + c n a l + b m + c n ⎥ ⎣ 4 1 4 1 4 1 4 2 4 2 4 2 ⎦ In other words, product AB is a matrix C = [Cik]m × p = ∑aij bjk j =1 Properties : If AB = O (null matrix), it does not necessarily imply that A or B is a null matrix. ⎡1 1⎤ ⎡ 1 − 1⎤ For instance, if A = ⎢1 1⎥ and B = ⎢− 1 1 ⎥ , then the product AB is a null matrix, although neither A ⎣ ⎦ ⎣ ⎦ nor B is a null matrix. Multiplication of matrices is not always commutative AB may or may not be equal to BA Multiplication of matrices is associative (AB)C = A(BC) If A, B, C are m × n, n × p and p × q matrices respectively. Multiplication of matrices is distributive w.r.t. addition of matrices: If A is a m × n matrix and B and C are both n × p matrices, then A(B + C) = AB + AC Multiplication of Matrix by Unit matrix, I AI = IA = A if conformability for multiplication is satisfied. Multiplication of Matrix by Null Matrix : OA = AO = O where O = null matrix Whenever AB and BA both exist and are matrices of same order, it is not necessary that AB = BA ⎡1 0 ⎤ ⎡0 1⎤ e.g., A = ⎢0 − 1⎥ , B = ⎢1 0⎥ ⎣ ⎦ ⎣ ⎦ ⎡1 0 ⎤ ⎡0 1⎤ ⎡ 0 1⎤ then AB = ⎢0 − 1⎥ ⎢1 0⎥ = ⎢− 1 0⎥ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎡0 BA = ⎢ ⎣ 1⎤ ⎡1 ⎥ ⎢ ⎦ ⎣ 0 ⎤ ⎡0 − 1⎥ = ⎢ − 1⎤ ⎥ ⎦ Thus AB ≠ BA ⎡0 1 2⎤ ⎡ 1 − 2⎤ If A = ⎢1 2 3⎥ and B = ⎢− 1 0 ⎥ , then ⎢ ⎢⎣2 3 ⎥ 4⎥⎦ ⎢ ⎢⎣ 2 ⎥ − 1⎥⎦ find AB. Is BA defined ? Solution : Matrix A is of the order 3 × 3 and matrix B is of the order 3 × 2. Since the number of columns of A = number of rows of B (each being = 3). ∴ Product AB is defined. AB = ⎡0 1 ⎢ 2 ⎢⎣2 3 2⎤ ⎥ ⎥ 4⎥⎦ ⎡ 1 ⎢− 1 ⎢⎣ 2 − 2⎤ ⎥ ⎥ − 1⎥⎦ ⎡0.1+ 1. − 1+ 2.2 = ⎢1.1+ 2. − 1+ 3.2 ⎢⎣2.1+ 3. − 1+ 4.2 0. − 2 + 1.0 + 2. − 1⎤ 1. − 2 + 2.0 + 3. − 1⎥ 2. − 2 + 3.0 + 4. − 1⎥⎦ ⎡3 ⎢ = ⎢ ⎢⎣7 − 2⎤ − 5⎥ − 8⎥⎦ Again, since number of columns of B ≠ number of rows of A. ∴ Product BA is not possible. ADJOINT OF MATRIX Let A = [aij] be a square matrix. Let B = [Aij] where Aij is the cofactor of the element aij in the det. A. Then adjoint of matrix A is transpose B′ of matrix B and is denoted by adj. A. Find the adjoint of the matrix ⎡1 A = ⎢1 ⎢⎣2 1 1 ⎤ 2 − 3⎥ − 1 3 ⎥⎦ Solution : First we have to find cofactors A , A , ... etc. A = (–1)1+1 1 − 3 2 − 3 − 1 3 = 3 A12 A13 = (–1)1+2 2 = (–1)1+3 1 2 3 = –9 2 = –5 − 1 Similarly, we can find all cofactors A21, A22, A23, A31, A32, A33 ⎡ A11 Adj A = ⎢A12 ⎢⎣A13 A21 A22 A23 A31 ⎤ A32 ⎥ A33 ⎥⎦ ⎡ 3 − 4 = ⎢− 9 1 ⎢⎣− 5 3 − 5⎤ ⎥ ⎥ 1 ⎥⎦ INVERSE OF A SQUARE MATRIX A square matrix A is said to be invertible if there exists a square matrix B such that AB = BA = In (identity matrix) Here B = A–1 or A = B–1. The necessary and sufficient condition for a square matrix A to possess the inverse is that |A| ≠ 0. Adj (A) Inverse of matrix A is denoted by A–1 and A–1 = | A | ⎡1 If A = ⎢2 ⎢⎣1 − 1 1⎤ − 1 0⎥ , find A2 and show that A2 = A–1 0 0⎥⎦ Solution : A2 = A.A ⎡1 − 1 = ⎢2 − 1 ⎢⎣1 0 1⎤ ⎡1 ⎥ ⎢ ⎥ ⎢ 0⎥⎦ ⎢⎣1 − 1 1⎤ − 1 ⎥ 0 0⎥⎦ ⎡0 A2 = ⎢0 ⎢⎣1 0 1⎤ − 1 ⎥ − 1 1⎥⎦ ... (i) | A | = 1 ... (ii) A11 − 1 = (– 1)1+1 0 0 = 0 0 A12 − 2 = (– 1)1+2 1 0 = 0 0 Similarly we can find A13, A21, A22, A23, A31, A32, A33 ⎡0 Adj. A = ⎢0 ⎢⎣1 0 1⎤ − 1 ⎥ − 1 1⎥⎦ A−1 = Adj. A | A | ⎡0 = ⎢ ⎢⎣1 0 1⎤ − 1 ⎥ − 1 1⎥⎦ ... (iii) So from (i) and (iii), A2 = A–1 Important Results : A.(adj. A) = (adj A). A = | A | In, where A is any square matrix of order n. AA–1 = A–1A = I (A–1)–1 = A Inverse of a square matrix, if it exists, is unique The reciprocal of the product of two matrices is the product of their reciprocals taken in the reverse order. (AB)–1 = B–1A–1. | adj A | = | A |n–1 | A. (adj A) | = | A |n (adj AB) = (adj B).(adj A) adj (A′ ) = (adj A)′ If A is non-singular square matrix, then adj. (adj A) = |A|n–2A 1 If A is a non-singular matrix, then |A–1| = |A|–1 = | A | det (kA) = k n (det A) if A is of order n × n. det (An) = (det A)n if n is a +ve integer. The operation of transposing and inverting are commutative i.e., (A′ )–1 = (A–1)′ SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS USING MATRICES Consider a system of n linear equations in n unknowns x1 x2, ..., xn. i.e., ai1 x1 + ai2 x2 + ... + ain xn = bi , i = 1, 2, 3 ..., n If b1 = b2 = ... = bn = 0, then the system of equations is called a system of homogenous linear equations and if atleast one of b1, b2, ..., bn is non-zero, then it is called a system of non-homogeneous linear equations. These equations can be written in the form of a single matrix equation of AX = B Where ⎡a11 a12 … a1n ⎤ ⎡ x1 ⎤ ⎡ b1 ⎤ ⎢a21 a22 … a2n ⎥ ⎢x ⎥ ⎢b ⎥ ⎢ ⎥ ⎢ 2 ⎥ ⎢ 2 ⎥ A = ⎢ … ⎢ … … … … ⎥ … … … ⎥ , X = ⎢…⎥ ⎢…⎥ and B = ⎢…⎥ ⎢…⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣an1 an 2 … ann ⎥⎦n × n ⎢⎣xn ⎥⎦ n ×1 ⎢⎣bn ⎥⎦n ×1 When system of equations is non-homogeneous If | A | ≠ 0, then system of equations is consistent and has a unique solution given by X = A–1 B. If | A | = 0, and (adj A).B ≠ 0, then system of equations is inconsistent and has no solution. | A | = 0, and (adj A).B = 0, then system of equations is consistent and has infinite number of solutions. When system of equations is homogeneous If | A | ≠ 0, system of equations have only one solution i.e. x = y = z = 0 If | A | = 0, system of equations have infinite number of solutions. SOLVED EXAMPLES Example : 1 Write down in matrix form the system of equations 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2 and solve it Solution : The given system of equations can be written in matrix form as AX = B ⎡2 Where A = ⎢1 − 1 3⎤ 1 1⎥ , ⎡x ⎤ X = ⎢y ⎥ , ⎡9⎤ B = ⎢6⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣1 − 1 1⎦⎥ ⎢⎣z ⎥⎦ ⎢⎣2⎥⎦ We have | A | = – 2 ≠ 0 i.e. A is non-singular and therefore A–1 exists ⎡− 1 1 A−1 = adj A = ⎢ 0 1 2 ⎤ −1 ⎥ | A | ⎢ ⎢⎣ 1 2 2 ⎥ −12 −3 2⎥⎦ ∴ X = A–1 B = ⎡− 1 1 ⎢ 0 2 ⎤ ⎡9⎤ −1 ⎥ ⎢6⎥ ⎢ 2 ⎥ ⎢ ⎥ ⎢⎣ 1 −12 −3 2⎥⎦ ⎢⎣2⎥⎦ ⎡x ⎤ ⎡1⎤ X = ⎢y ⎥ = ⎢2⎥ ⇒ ⎢ ⎥ ⎢ ⎥ ⎢⎣z ⎥⎦ ⎢⎣3⎥⎦ i.e., (x, y, z) ≡ (1, 2, 3). Example 2 : Solve the system of equations x + 3y – 2z = 0 2x – y + 4z = 0 x – 11y + 14z = 0 Solution : The given system of equations in the matrix form are written as below ⎡1 3 − 2⎤ ⎡x⎤ ⎡0⎤ ⎢2 − 1 4 ⎥ ⎢y ⎥ = ⎢0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢⎣1 − 11 14 ⎥⎦ ⎢⎣z⎥⎦ ⎢⎣0⎥⎦ AX = 0 | A | = 1(–14 + 44) – 3(28 – 4) –2(– 22 + 1) = 30 – 72 + 42 = 0 (i) and therefore the system has infinitely many solutions. Taking first two equations, x + 3y = 2z (ii) 2x – y = – 4z (iii) and solving in terms of z, we get x = −10z , y = 8 z ... (iv) 7 7 Put these values of x and y in third equation x – 11y + 14z = 0, its value comes to be equal to zero. Now if z = 7k, then x = – 10k, y = 8k where k is any arbitrary constant. So then required solutions are x = –10k, y = 8k & z = 7k ⎡ 4 2 Express the matrix A as the sum of a symmetric and a skew-symmetric matrix, where A = ⎢ 1 3 ⎢⎣− 5 0 − 3⎤ − 6⎥ − 7⎥⎦ Solution : ⎡ 4 A′ = ⎢ 2 ⎢⎣− 3 1 − 5⎤ 3 ⎥ − 6 − 7⎥⎦ ⎡ 8 3 Then A + A′ = ⎢ 3 6 ⎢⎣− 8 − 6 − 8 ⎤ 6 ⎥ and − 14⎥⎦ ⎡ 0 1 A – A′ = ⎢ − 1 0 ⎢⎣− 2 6 2 ⎤ − 6⎥ 0 ⎥⎦ ∴ A = 1 (A + A′ ) + 2 1 (A – A′ ) 2 ⎡ 4 1.5 − 4⎤ ⎡ 0 0.5 1 ⎤ ⎡ 4 2 − 3⎤ = ⎢1.5 3 − 3⎥ + ⎢− 0.5 0 − 3⎥ = ⎢ 3 − 6⎥ ⎢⎣− 4 − 3 − 7⎥⎦ ⎣⎢ − 1 3 0 ⎥⎦ ⎢⎣− 5 0 − 7⎥⎦ ❑ ❑ ❑