Determinants PDF

Chapter 4 DETERMINANTS he ™ All Mathematical truths are relative and conditional. — C.P. STEINMETZ ™ 4.1 Introdu...

Chapter 4 DETERMINANTS he ™ All Mathematical truths are relative and conditional. — C.P. STEINMETZ ™ 4.1 Introduction is In the previous chapter, we have studied about matrices and algebra of matrices. We have also learnt that a system of algebraic equations can be expressed in the form of bl matrices. This means, a system of linear equations like a1 x + b1 y = c 1 pu a2 x + b2 y = c 2 ⎡a b ⎤ ⎡ x⎤ ⎡c ⎤ can be represented as ⎢ 1 1 ⎥ ⎢ ⎥ = ⎢ 1 ⎥. Now, this be T ⎣ a2 b2 ⎦ ⎣ y ⎦ ⎣ c2 ⎦ re system of equations has a unique solution or not, is o R determined by the number a1 b2 – a2 b1. (Recall that if a1 b1 P.S. Laplace ≠ or, a1 b2 – a2 b1 ≠ 0, then the system of linear tt E a2 b2 (1749-1827) equations has a unique solution). The number a1 b2 – a2 b1 ⎡a b ⎤ C which determines uniqueness of solution is associated with the matrix A = ⎢ 1 1 ⎥ ⎣ a2 b2 ⎦ and is called the determinant of A or det A. Determinants have wide applications in no N Engineering, Science, Economics, Social Science, etc. In this chapter, we shall study determinants up to order three only with real entries. Also, we will study various properties of determinants, minors, cofactors and applications of determinants in finding the area of a triangle, adjoint and inverse of a square matrix, © consistency and inconsistency of system of linear equations and solution of linear equations in two or three variables using inverse of a matrix. 4.2 Determinant 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.  104 MATHEMATICS This may be thought of as a function which associates each square matrix with a unique number (real or complex). If M is the set of square matrices, K is the set of numbers (real or complex) and f : M → K is defined by f (A) = k, where A ∈ M and k ∈ K, then f (A) is called the determinant of A. It is also denoted by | A | or det A or Δ. ⎡a b ⎤ a b If A = ⎢ ⎥ , then determinant of A is written as |A | = = det (A) ⎣c d ⎦ c d he Remarks (i) For matrix A, | A | is read as determinant of A and not modulus of A. (ii) Only square matrices have determinants. is 4.2.1 Determinant of a matrix of order one Let A = [a ] be the matrix of order 1, then determinant of A is defined to be equal to a bl 4.2.2 Determinant of a matrix of order two ⎡ a11 a12 ⎤ Let A= ⎢ ⎥ be a matrix of order 2 × 2, ⎣ a21 a22 ⎦ pu then the determinant of A is defined as: be T det (A) = |A| = Δ = = a11a22 – a21a12 re o R 2 4 Example 1 Evaluate. –1 2 tt E 2 4 Solution We have = 2 (2) – 4(–1) = 4 + 4 = 8. –1 2 C x x +1 Example 2 Evaluate no N x –1 x Solution We have x x +1 = x (x) – (x + 1) (x – 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1 © x –1 x 4.2.3 Determinant of a matrix of order 3 × 3 Determinant of a matrix of order three can be determined by expressing it in terms of second order determinants. This is known as expansion of a determinant along a row (or a column). There are six ways of expanding a determinant of order  DETERMINANTS 105 3 corresponding to each of three rows (R1, R2 and R3) and three columns (C1, C2 and C3) giving the same value as shown below. Consider the determinant of square matrix A = [aij]3 × 3 a11 a12 a13 i.e., | A | = a21 a22 a23 he a31 a32 a33 Expansion along first Row (R1) Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a11] and with the is second order determinant obtained by deleting the elements of first row (R1) and first column (C1) of | A | as a11 lies in R1 and C1, bl a22 a23 i.e., (–1)1 + 1 a11 a32 a33 Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a12] and the second pu order determinant obtained by deleting elements of first row (R1) and 2nd column (C2) of | A | as a12 lies in R1 and C2, be T a21 a23 i.e., (–1)1 + 2 a12 re a31 a33 o R Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a ] and the second 13 order determinant obtained by deleting elements of first row (R1) and third column (C3) tt E of | A | as a13 lies in R1 and C3, a21 a22 C i.e., (–1)1 + 3 a13 a a32 31 Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three no N terms obtained in steps 1, 2 and 3 above is given by a22 a23 a a det A = |A| = (–1)1 + 1 a11 a + (–1)1 + 2 a12 21 23 32 a33 a31 a33 © 1+ 3 a21 a22 + (–1) a13 a31 a32 or |A| = a11 (a22 a33 – a32 a23) – a12 (a21 a33 – a31 a23) + a13 (a21 a32 – a31 a22)  106 MATHEMATICS = a11 a22 a33 – a11 a32 a23 – a12 a21 a33 + a12 a31 a23 + a13 a21 a32 – a13 a31 a22... (1) $Note We shall apply all four steps together. Expansion along second row (R2) he a 11 a 12 a 13 | A | = a 21 a 22 a 23 a 31 a 32 a 33 is Expanding along R2, we get 2 +1 a12 a13 a a | A | = (–1) a21 + (–1)2 + 2 a22 11 13 bl a32 a33 a31 a33 a11 a12 + (–1) 2 + 3 a23 pu a31 a32 = – a21 (a12 a33 – a32 a13) + a22 (a11 a33 – a31 a13) be T – a23 (a11 a32 – a31 a12) | A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32 re o R + a23 a31 a12 = a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32 tt E – a13 a31 a22... (2) Expansion along first Column (C1) C a11 a12 a13 | A | = a21 a22 a23 a31 a32 a33 no N By expanding along C1, we get 1 + 1 a22 a23 a a13 | A | = a11 (–1) + a21 (−1) 2 + 1 12 © a32 a33 a32 a33 3 + 1 a 12 a13 + a31 (–1) a22 a23 = a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22)  DETERMINANTS 107 | A | = a11 a22 a33 – a11 a23 a32 – a21 a12 a33 + a21 a13 a32 + a31 a12 a23 – a31 a13 a22 = a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32 – a13 a31 a22... (3) Clearly, values of | A | in (1), (2) and (3) are equal. It is left as an exercise to the reader to verify that the values of |A| by expanding along R3, C2 and C3 are equal to the he value of | A | obtained in (1), (2) or (3). Hence, expanding a determinant along any row or column gives same value. Remarks (i) For easier calculations, we shall expand the determinant along that row or column is which contains maximum number of zeros. (ii) While expanding, instead of multiplying by (–1)i + j, we can multiply by +1 or –1 bl according as (i + j) is even or odd. ⎡2 2⎤ ⎡1 1⎤ (iii) Let A = ⎢ ⎥ and B = ⎢ ⎥. Then, it is easy to verify that A = 2B. Also pu ⎣4 0⎦ ⎣2 0⎦ | A | = 0 – 8 = – 8 and | B | = 0 – 2 = – 2. be T Observe that, | A | = 4 (– 2) = 22 | B | or | A | = 2n | B |, where n = 2 is the order of square matrices A and B. re o R In general, if A = kB where A and B are square matrices of order n, then | A| = kn | B |, where n = 1, 2, 3 1 2 4 tt E Example 3 Evaluate the determinant Δ = –1 3 0. 4 1 0 C Solution Note that in the third column, two entries are zero. So expanding along third column (C3), we get no N –1 3 1 2 1 2 Δ= 4 –0 +0 4 1 4 1 –1 3 © = 4 (–1 – 12) – 0 + 0 = – 52 0 sin α – cos α Example 4 Evaluate Δ = – sin α 0 sin β. cos α – sin β 0  108 MATHEMATICS Solution Expanding along R1, we get 0 sin β – sin α sin β – sin α 0 Δ= 0 – sin α – cos α – sin β 0 cos α 0 cos α – sin β = 0 – sin α (0 – sin β cos α) – cos α (sin α sin β – 0) = sin α sin β cos α – cos α sin α sin β = 0 he 3 x 3 2 Example 5 Find values of x for which =. x 1 4 1 3 x 3 2 = is Solution We have x 1 4 1 i.e. 3 – x2 = 3 – 8 bl i.e. x2 = 8 Hence x= ±2 2 pu EXERCISE 4.1 be T Evaluate the determinants in Exercises 1 and 2. re 2 4 o R 1. –5 –1 cos θ – sin θ x2 – x + 1 x – 1 tt E 2. (i) (ii) sin θ cos θ x +1 x +1 C ⎡1 2⎤ 3. If A = ⎢ ⎥ , then show that | 2A | = 4 | A | ⎣4 2⎦ no N ⎡1 0 1⎤ ⎢ ⎥ 4. If A = ⎢ 0 1 2 ⎥ , then show that | 3 A | = 27 | A | ⎢⎣ 0 0 4 ⎥⎦ © 5. Evaluate the determinants 3 –1 –2 3 –4 5 (i) 0 0 –1 (ii) 1 1 –2 3 –5 0 2 3 1  DETERMINANTS 109 0 1 2 2 –1 –2 (iii) –1 0 –3 (iv) 0 2 –1 –2 3 0 3 –5 0 ⎡ 1 1 –2 ⎤ ⎢ ⎥ 6. If A = ⎢ 2 1 –3 ⎥ , find | A | he ⎢⎣ 5 4 –9 ⎥⎦ 7. Find values of x, if 2 4 2x 4 2 3 x 3 = = is (i) (ii) 5 1 6 x 4 5 2x 5 x 2 6 2 = bl 8. If , then x is equal to 18 x 18 6 (A) 6 pu (B) ± 6 (C) – 6 (D) 0 4.3 Properties of Determinants In the previous section, we have learnt how to expand the determinants. In this section, we will study some properties of determinants which simplifies its evaluation by obtaining be T maximum number of zeros in a row or a column. These properties are true for determinants of any order. However, we shall restrict ourselves upto determinants of re o R order 3 only. Property 1 The value of the determinant remains unchanged if its rows and columns are interchanged. tt E a1 a2 a3 Verification Let Δ = b1 b2 b3 C c1 c2 c3 Expanding along first row, we get no N b2 b3 b b b b Δ = a1 − a2 1 3 + a3 1 2 c2 c3 c1 c3 c1 c2 = a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) © By interchanging the rows and columns of Δ, we get the determinant a1 b1 c1 Δ1 = a2 b2 c2 a3 b3 c3  110 MATHEMATICS Expanding Δ1 along first column, we get Δ1 = a1 (b2 c3 – c2 b3) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) Hence Δ = Δ 1 Remark It follows from above property that if A is a square matrix, then det (A) = det (A′), where A′ = transpose of A. he $ Note If R = ith row and C = ith column, then for interchange of row and i i columns, we will symbolically write C ↔ R i i Let us verify the above property by example. is 2 –3 5 Example 6 Verify Property 1 for Δ = 6 0 4 bl 1 5 –7 Solution Expanding the determinant along first row, we have 0 4 6 4 6 0 pu Δ= 2 5 –7 – (–3) 1 –7 +5 1 5 = 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0) be T = – 40 – 138 + 150 = – 28 re By interchanging rows and columns, we get o R 2 6 1 Δ1 = –3 0 5 (Expanding along first column) tt E 5 4 –7 0 5 6 1 6 1 C = 2 – (–3) +5 4 –7 4 –7 0 5 = 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0) no N = – 40 – 138 + 150 = – 28 Clearly Δ = Δ1 Hence, Property 1 is verified. © Property 2 If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes. a1 a2 a3 Verification Let Δ = b1 b2 b3 c1 c2 c3  DETERMINANTS 111 Expanding along first row, we get Δ = a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) Interchanging first and third rows, the new determinant obtained is given by c1 c2 c3 Δ1 = b1 b2 b3 he a1 a2 a3 Expanding along third row, we get Δ1 = a1 (c2 b3 – b2 c3) – a2 (c1 b3 – c3 b1) + a3 (b2 c1 – b1 c2) is = – [a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1)] Clearly Δ1 = – Δ bl Similarly, we can verify the result by interchanging any two columns. $ Note We can denote the interchange of rows by R ↔ R and interchange of columns by C ↔ C. pu i j i j 2 –3 5 be T Example 7 Verify Property 2 for Δ = 6 0 4. re 1 5 –7 o R 2 –3 5 tt E Solution Δ = 6 0 4 = – 28 (See Example 6) 1 5 –7 C Interchanging rows R2 and R3 i.e., R2 ↔ R3, we have 2 –3 5 no N Δ1 = 1 5 –7 6 0 4 Expanding the determinant Δ1 along first row, we have © 5 –7 1 –7 1 5 Δ1 = 2 – (–3) +5 0 4 6 4 6 0 = 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30) = 40 + 138 – 150 = 28  112 MATHEMATICS Clearly Δ1 = – Δ Hence, Property 2 is verified. Property 3 If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero. Proof If we interchange the identical rows (or columns) of the determinant Δ, then Δ does not change. However, by Property 2, it follows that Δ has changed its sign he Therefore Δ=–Δ or Δ=0 Let us verify the above property by an example. is 3 2 3 Example 8 Evaluate Δ = 2 2 3 bl 3 2 3 Solution Expanding along first row, we get pu Δ = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6) = 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0 be T Here R1 and R3 are identical. Property 4 If each element of a row (or a column) of a determinant is multiplied by a re o R constant k, then its value gets multiplied by k. a1 b1 c1 tt E Verification Let Δ = a2 b2 c2 a3 b3 c3 C and Δ1 be the determinant obtained by multiplying the elements of the first row by k. Then no N k a1 k b1 k c1 Δ1 = a2 b2 c2 a3 b3 c3 © Expanding along first row, we get Δ1 = k a1 (b2 c3 – b3 c2) – k b1 (a2 c3 – c2 a3) + k c1 (a2 b3 – b2 a3) = k [a1 (b2 c3 – b3 c2) – b1 (a2 c3 – c2 a3) + c1 (a2 b3 – b2 a3)] =k Δ  DETERMINANTS 113 k a1 k b1 k c1 a1 b1 c1 Hence a2 b2 c2 = k a2 b2 c2 a3 b3 c3 a3 b3 c3 Remarks he (i) By this property, we can take out any common factor from any one row or any one column of a given determinant. (ii) If corresponding elements of any two rows (or columns) of a determinant are proportional (in the same ratio), then its value is zero. For example is a1 a2 a3 Δ = b1 b2 b3 = 0 (rows R and R are proportional) bl 1 2 k a1 k a2 k a3 pu 102 18 36 Example 9 Evaluate 1 3 4 17 3 6 be T re 102 18 36 6(17) 6(3) 6(6) 17 3 6 o R Solution Note that 1 3 4 = 1 3 4 =6 1 3 4 =0 17 3 6 17 3 6 17 3 6 tt E (Using Properties 3 and 4) C Property 5 If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants. no N a1 + λ1 a2 + λ 2 a3 + λ 3 a1 a2 a3 λ1 λ 2 λ3 For example, b1 b2 b3 = b1 b2 b3 + b1 b2 b3 © c1 c2 c3 c1 c2 c3 c1 c2 c3 a1 + λ1 a2 + λ 2 a3 + λ 3 Verification L.H.S. = b1 b2 b3 c1 c2 c3  114 MATHEMATICS Expanding the determinants along the first row, we get Δ = (a1 + λ1) (b2 c3 – c2 b3) – (a2 + λ2) (b1 c3 – b3 c1) + (a3 + λ3) (b1 c2 – b2 c1) = a1 (b2 c3 – c2 b3) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) + λ1 (b2 c3 – c2 b3) – λ2 (b1 c3 – b3 c1) + λ3 (b1 c2 – b2 c1) he (by rearranging terms) a1 a2 a3 λ1 λ 2 λ3 = b1 b2 b3 + b1 b2 b3 = R.H.S. is c1 c2 c3 c1 c2 c3 Similarly, we may verify Property 5 for other rows or columns. bl a b c Example 10 Show that a + 2 x b + 2 y c + 2 z = 0 pu x y z a b c a b c a b c be T Solution We have a + 2 x b + 2 y c + 2 z = a b c + 2x 2 y 2z re x y z x y z x y z o R (by Property 5) =0+0=0 (Using Property 3 and Property 4) tt E Property 6 If, to each element of any row or column of a determinant, the equimultiples of corresponding elements of other row (or column) are added, then value of determinant C remains the same, i.e., the value of determinant remain same if we apply the operation Ri → Ri + kRj or Ci → Ci + k Cj. no N Verification a1 a2 a3 a1 + k c1 a2 + k c2 a3 + k c3 Let Δ = b1 b2 b3 and Δ = 1 b1 b2 b3 , © c1 c2 c3 c1 c2 c3 where Δ1 is obtained by the operation R1 → R1 + kR3. Here, we have multiplied the elements of the third row (R3) by a constant k and added them to the corresponding elements of the first row (R1). Symbolically, we write this operation as R1 → R1 + k R3.  DETERMINANTS 115 Now, again a1 a2 a3 k c1 k c2 k c3 Δ1 = b1 b2 b3 + b1 b2 b3 (Using Property 5) c1 c2 c3 c1 c2 c3 =Δ+0 (since R1 and R3 are proportional) he Hence Δ = Δ1 Remarks (i) If Δ1 is the determinant obtained by applying Ri → kRi or Ci → kCi to the is determinant Δ, then Δ1 = kΔ. (ii) If more than one operation like Ri → Ri + kRj is done in one step, care should be taken to see that a row that is affected in one operation should not be used in bl another operation. A similar remark applies to column operations. a a+b a+b+c pu Example 11 Prove that 2a 3a + 2b 4a + 3b + 2c = a 3. 3a 6a + 3b 10a + 6b + 3c be T Solution Applying operations R2 → R2 – 2R1 and R3 → R3 – 3R1 to the given re o R determinant Δ, we have a a+b a+b+c tt E Δ= 0 a 2a + b 0 3a 7 a + 3b C Now applying R3 → R3 – 3R2 , we get a a+b a+b+c no N Δ= 0 a 2a + b 0 0 a © Expanding along C1, we obtain a 2a + b Δ= a +0+0 0 a = a (a2 – 0) = a (a2) = a3  116 MATHEMATICS Example 12 Without expanding, prove that x+ y y+z z+x Δ= z x y =0 1 1 1 Solution Applying R1 → R1 + R2 to Δ, we get he x+ y+z x+ y+z x+ y+z Δ= z x y 1 1 1 is Since the elements of R1 and R3 are proportional, Δ = 0. Example 13 Evaluate bl 1 a bc Δ = 1 b ca 1 c ab pu Solution Applying R2 → R2 – R1 and R3 → R3 – R1, we get be T 1 a bc Δ = 0 b − a c ( a − b) re o R 0 c − a b (a − c) Taking factors (b – a) and (c – a) common from R2 and R3, respectively, we get tt E 1 a bc Δ = (b − a ) (c − a) 0 1 –c C 0 1 –b = (b – a) (c – a) [(– b + c)] (Expanding along first column) no N = (a – b) (b – c) (c – a) b+c a a Example 14 Prove that b c+a b = 4 abc © c c a+b b+c a a Solution Let Δ = b c+a b c c a+b  DETERMINANTS 117 Applying R1 → R1 – R2 – R3 to Δ, we get 0 –2c –2b Δ= b c+a b c c a+b Expanding along R1, we obtain he c+a b b b b c+a Δ= 0 – (–2 c ) + (–2b) c a+b c a+b c c = 2 c (a b + b2 – bc) – 2 b (b c – c2 – ac) is = 2 a b c + 2 cb2 – 2 bc2 – 2 b2c + 2 bc2 + 2 abc = 4 abc x2 1 + x3 bl x Example 15 If x, y, z are different and Δ = y y 2 1 + y 3 = 0 , then z z2 1 + z3 pu show that 1 + xyz = 0 Solution We have be T x x2 1 + x3 re o R Δ= y y 2 1 + y3 z z2 1 + z3 tt E x x2 1 x x2 x3 y2 1 + y y2 y 3 (Using Property 5) C = y z z2 1 z z2 z3 no N 1 x x2 1 x x2 = (−1) 1 y y 2 + xyz 1 y 2 y2 (Using C3 ↔ C2 and then C1 ↔ C2) 1 z z2 1 z z2 © 1 x x2 = 1 y y 2 (1+ xyz ) 1 z z2  118 MATHEMATICS 1 x x2 = (1 + xyz ) 0 y−x y 2 − x2 (Using R2 → R2–R1 and R3 → R3– R1) 0 z−x z 2 − x2 Taking out common factor (y – x) from R2 and (z – x) from R3, we get he 1 x x2 Δ = (1+xyz ) (y –x) (z –x) 0 1 y+x 0 1 z+x is = (1 + xyz) (y – x) (z – x) (z – y) (on expanding along C1) Since Δ = 0 and x, y, z are all different, i.e., x – y ≠ 0, y – z ≠ 0, z – x ≠ 0, we get bl 1 + xyz = 0 Example 16 Show that pu 1+ a 1 1 ⎛ 1 1 1⎞ 1 1+ b 1 = abc ⎜1 + + + ⎟ = abc + bc + ca + ab ⎝ a b c⎠ be T 1 1 1+ c re Solution Taking out factors a,b,c common from R1, R2 and R3, we get o R 1 1 1 +1 tt E a a a 1 1 1 L.H.S. = abc +1 b b b C 1 1 1 +1 c c c no N Applying R1→ R1 + R2 + R3, we have 1 1 1 1 1 1 1 1 1 1+ + + 1+ + + 1+ + + © a b c a b c a b c 1 1 1 Δ = abc +1 b b b 1 1 1 +1 c c c  DETERMINANTS 119 1 1 1 ⎛ 1 1 1⎞ 1 1 1 = abc ⎜ 1+ + + ⎟ +1 ⎝ a b c⎠ b b b 1 1 1 +1 c c c he Now applying C2 → C2 – C1, C3 → C3 – C1, we get 1 0 0 ⎛ 1 1 1⎞ 1 Δ = abc ⎜ 1+ + + ⎟ 1 0 ⎝ a b c⎠ is b 1 0 1 c bl ⎛ 1 1 1⎞ = abc ⎜1 + + + ⎟ ⎡⎣1(1 – 0 ) ⎤⎦ ⎝ a b c⎠ pu ⎛ 1 1 1⎞ = abc ⎜1+ + + ⎟ = abc + bc + ca + ab = R.H.S. ⎝ a b c⎠ be T $ Note Alternately try by applying C → C – C and C → C – C , then apply re 1 1 2 3 3 2 C →C –aC. o R 1 1 3 EXERCISE 4.2 tt E Using the property of determinants and without expanding in Exercises 1 to 7, prove that: C x a x+a a−b b−c c−a 1. y b y +b = 0 2. b−c c−a a−b = 0 z+c c−a a−b b−c no N z c 2 7 65 1 bc a ( b + c ) 3. 3 8 75 = 0 4. 1 ca b ( c + a ) = 0 © 5 9 86 1 ab c ( a + b ) b+c q+r y+z a p x 5. c+a r+ p z+x = 2 b q y a+b p+q x+ y c r z  120 MATHEMATICS 0 a −b −a 2 ab ac 6. − a 0 −c = 0 7. ba −b 2 bc = 4 a 2 b 2 c 2 b c 0 ca cb −c 2 By using properties of determinants, in Exercises 8 to 14, show that: he 1 a a2 8. (i) 1 b b = ( a − b )( b − c )( c − a ) 2 1 c c2 is 1 1 1 a b c = ( a − b )( b − c )( c − a )( a + b + c ) bl (ii) pu a3 b3 c3 x x2 yz 9. y y2 zx = (x – y) (y – z) (z – x) (xy + yz + zx) be T 2 z z xy re o R x + 4 2x 2x 10. (i) 2 x x + 4 2x = ( 5 x + 4 )( 4 − x ) 2 tt E 2x 2x x + 4 y+k y y C (ii) y y+k y = k 2 (3 y + k ) y y y+k no N a−b−c 2a 2a 2b = ( a + b + c ) 3 11. (i) 2b b−c−a c−a−b © 2c 2c x + y + 2z x y = 2( x + y + z) 3 (ii) z y + z + 2x y z x z + x + 2y  DETERMINANTS 121 1 x x2 ( ) 2 12. x2 1 x = 1 − x3 x x2 1 1 + a 2 − b2 −2b he 2ab ( ) 3 13. 2ab 1− a + b 2 2 2a = 1 + a2 + b2 2b −2a 1 − a2 − b2 is a2 + 1 ab ac ab b2 + 1 bc =1 + a 2 + b 2 + c 2 bl 14. ca pu cb c2 + 1 Choose the correct answer in Exercises 15 and 16. 15. Let A be a square matrix of order 3 × 3, then | kA | is equal to (A) k| A | (B) k 2 | A | (C) k 3 | A | (D) 3k | A | be T 16. Which of the following is correct re (A) Determinant is a square matrix. o R (B) Determinant is a number associated to a matrix. (C) Determinant is a number associated to a square matrix. tt E (D) None of these 4.4 Area of a Triangle C In earlier classes, we have studied that the area of a triangle whose vertices are 1 (x1, y1), (x2, y2) and (x3, y3), is given by the expression [x (y –y ) + x2 (y3–y1) + no N 2 1 2 3 x3 (y1–y2)]. Now this expression can be written in the form of a determinant as x1 y1 1 1 © Δ= x2 y2 1... (1) 2 x3 y3 1 Remarks (i) Since area is a positive quantity, we always take the absolute value of the determinant in (1).  122 MATHEMATICS (ii) If area is given, use both positive and negative values of the determinant for calculation. (iii) The area of the triangle formed by three collinear points is zero. Example 17 Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1). Solution The area of triangle is given by 3 8 1 he 1 Δ= –4 2 1 2 5 1 1 1 ⎡3 ( 2 – 1) – 8 ( – 4 – 5 ) + 1( – 4 – 10 ) ⎤⎦ is 2⎣ = 1 61 ( 3 + 72 – 14 ) = bl = 2 2 Example 18 Find the equation of the line joining A(1, 3) and B (0, 0) using determinants and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units. pu Solution Let P (x, y) be any point on AB. Then, area of triangle ABP is zero (Why?). So 0 0 1 be T 1 1 3 1 =0 re 2 o R x y 1 1 This gives ( y – 3 x ) = 0 or y = 3x, tt E 2 which is the equation of required line AB. Also, since the area of the triangle ABD is 3 sq. units, we have C 1 3 1 1 0 0 1 =±3 no N 2 k 0 1 − 3k This gives, = ± 3 , i.e., k = m 2. 2 © EXERCISE 4.3 1. Find area of the triangle with vertices at the point given in each of the following : (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (–2, –3), (3, 2), (–1, –8)  DETERMINANTS 123 2. Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear. 3. Find values of k if area of triangle is 4 sq. units and vertices are (i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k) 4. (i) Find equation of line joining (1, 2) and (3, 6) using determinants. he (ii) Find equation of line joining (3, 1) and (9, 3) using determinants. 5. If area of triangle is 35 sq units with vertices (2, – 6), (5, 4) and (k, 4). Then k is (A) 12 (B) –2 (C) –12, –2 (D) 12, –2 4.5 Minors and Cofactors is In this section, we will learn to write the expansion of a determinant in compact form using minors and cofactors. bl Definition 1 Minor of an element aij of a determinant is the determinant obtained by deleting its ith row and jth column in which element aij lies. Minor of an element aij is denoted by Mij. pu Remark Minor of an element of a determinant of order n(n ≥ 2) is a determinant of order n – 1. be T 1 2 3 re Example 19 Find the minor of element 6 in the determinant Δ = 4 5 6 o R 7 8 9 tt E Solution Since 6 lies in the second row and third column, its minor M23 is given by 1 2 = 8 – 14 = – 6 (obtained by deleting R2 and C3 in Δ). C M23 = 7 8 Definition 2 Cofactor of an element aij , denoted by Aij is defined by no N Aij = (–1)i + j Mij , where Mij is minor of aij. 1 –2 Example 20 Find minors and cofactors of all the elements of the determinant 4 3 © Solution Minor of the element aij is Mij Here a11 = 1. So M11 = Minor of a11= 3 M12 = Minor of the element a12 = 4 M21 = Minor of the element a21 = –2  124 MATHEMATICS M22 = Minor of the element a22 = 1 Now, cofactor of aij is Aij. So A11 = (–1)1 + 1 M11 = (–1)2 (3) = 3 A12 = (–1)1 + 2 M12 = (–1)3 (4) = – 4 A21 = (–1)2 + 1 M21 = (–1)3 (–2) = 2 he A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1 Example 21 Find minors and cofactors of the elements a11, a21 in the determinant a11 a12 a13 is Δ = a21 a22 a23 a31 a32 a33 bl Solution By definition of minors and cofactors, we have pu a22 a23 Minor of a11 = M11 = = a22 a33– a23 a32 a32 a33 be T Cofactor of a11 = A11 = (–1)1+1 M11 = a22 a33 – a23 a32 re a12 a13 o R Minor of a21 = M21 = = a12 a33 – a13 a32 a32 a33 Cofactor of a21 = A21 = (–1)2+1 M21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32 tt E Remark Expanding the determinant Δ, in Example 21, along R1, we have a22 a23 a21 a23 a21 a22 C Δ = (–1) a11 a 1+1 1+2 a + (–1) a12 a 1+3 a + (–1) a13 a31 a32 32 33 31 33 = a11 A11 + a12 A12 + a13 A13, where Aij is cofactor of aij no N = sum of product of elements of R1 with their corresponding cofactors Similarly, Δ can be calculated by other five ways of expansion that is along R2, R3, C1, C2 and C3. © Hence Δ = sum of the product of elements of any row (or column) with their corresponding cofactors. $ Note If elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero. For example,  DETERMINANTS 125 Δ = a11 A21 + a12 A22 + a13 A23 a12 a13 a11 a13 a a = a11 (–1)1+1 + a12 (–1)1+2 + a13 (–1)1+3 11 12 a32 a33 a31 a33 a31 a32 a11 a12 a13 he = a11 a12 a13 = 0 (since R1 and R2 are identical) a31 a32 a33 Similarly, we can try for other rows and columns. is Example 22 Find minors and cofactors of the elements of the determinant 2 –3 5 bl 6 0 4 and verify that a11 A31 + a12 A32 + a13 A33= 0 1 5 –7 pu 0 4 Solution We have M11 = = 0 –20 = –20; A11 = (–1)1+1 (–20) = –20 5 –7 be T re 6 4 o R M12 = = – 42 – 4 = – 46; A12 = (–1)1+2 (– 46) = 46 1 –7 6 0 tt E M13 = = 30 – 0 = 30; A13 = (–1)1+3 (30) = 30 1 5 C –3 5 M21 = = 21 – 25 = – 4; A21 = (–1)2+1 (– 4) = 4 5 –7 no N 2 5 M22 = = –14 – 5 = –19; A22 = (–1)2+2 (–19) = –19 1 –7 © 2 –3 M23 = = 10 + 3 = 13; A23 = (–1)2+3 (13) = –13 1 5 –3 5 M31 = = –12 – 0 = –12; A31 = (–1)3+1 (–12) = –12 0 4  126 MATHEMATICS 2 5 M32 = = 8 – 30 = –22; A32 = (–1)3+2 (–22) = 22 6 4 2 –3 and M33 = = 0 + 18 = 18; A33 = (–1)3+3 (18) = 18 6 0 Now a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18 he So a11 A31 + a12 A32 + a13 A33 = 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0 is EXERCISE 4.4 Write Minors and Cofactors of the elements of following determinants: bl 2 –4 a c 1. (i) (ii) pu 0 3 b d 1 0 0 1 0 4 2. (i) 0 1 0 (ii) 3 5 –1 0 0 1 0 1 2 be T re 5 3 8 o R 3. Using Cofactors of elements of second row, evaluate Δ = 2 0 1. 1 2 3 tt E 1 x yz 4. Using Cofactors of elements of third column, evaluate Δ = 1 y zx. C 1 z xy a11 a12 a13 no N 5. If Δ = a21 a22 a23 and Aij is Cofactors of aij , then value of Δ is given by a31 a32 a33 (A) a11 A31+ a12 A32 + a13 A33 (B) a11 A11+ a12 A21 + a13 A31 © (C) a21 A11+ a22 A12 + a23 A13 (D) a11 A11+ a21 A21 + a31 A31 4.6 Adjoint and Inverse of a Matrix In the previous chapter, we have studied inverse of a matrix. In this section, we shall discuss the condition for existence of inverse of a matrix. To find inverse of a matrix A, i.e., A–1 we shall first define adjoint of a matrix.  DETERMINANTS 127 4.6.1 Adjoint of a matrix Definition 3 The adjoint of a square matrix A = [aij]n × n is defined as the transpose of the matrix [Aij]n × n, where Aij is the cofactor of the element aij. Adjoint of the matrix A is denoted by adj A. ⎡ a11 a12 a13 ⎤ A = ⎢⎢ a21 a23 ⎥⎥ he Let a22 ⎢⎣ a31 a32 a33 ⎥⎦ ⎡ A11 A12 A13 ⎤ ⎡ A11 A 21 A 31 ⎤ adj A = Transpose of ⎢⎢ A 21 A 23 ⎥⎥ = ⎢⎢ A12 A32 ⎥⎥ is Then A 22 A 22 ⎢⎣ A 31 A 32 A 33 ⎥⎦ ⎢⎣ A13 A 23 A 33 ⎥⎦ bl ⎡ 2 3⎤ Example 23 Find adj A for A = ⎢ ⎥ ⎣1 4⎦ pu Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2 ⎡ A11 A 21 ⎤ ⎡ 4 –3⎤ be T Hence adj A = ⎢ ⎥ =⎢ ⎥ ⎣ A12 A 22 ⎦ ⎣ –1 2 ⎦ re Remark For a square matrix of order 2, given by o R ⎡ a11 a12 ⎤ A= ⎢ ⎥ ⎣ a21 a22 ⎦ tt E The adj A can also be obtained by interchanging a11 and a22 and by changing signs of a12 and a21, i.e., C no N © We state the following theorem without proof. Theorem 1 If A be any given square matrix of order n, then A(adj A) = (adj A) A = A I , where I is the identity matrix of order n  128 MATHEMATICS Verification ⎡ a11 a12 a13 ⎤ ⎡ A11 A 21 A 31 ⎤ ⎢a ⎥ ⎢ Let A = ⎢ 21 a22 a23 ⎥ , then adj A = ⎢ A12 A 22 A 32 ⎥⎥ ⎢⎣ a31 a32 a33 ⎥⎦ ⎢⎣ A13 A 23 A33 ⎥⎦ Since sum of product of elements of a row (or a column) with corresponding he cofactors is equal to | A | and otherwise zero, we have ⎡A 0 0⎤ ⎡1 0 0 ⎤ ⎢ ⎥ A (adj A) = ⎢ 0 A 0 ⎥ = A ⎢0 1 0 ⎥ = A I ⎢ ⎥ is ⎢⎣ 0 0 A ⎥⎦ ⎢⎣0 0 1 ⎥⎦ Similarly, we can show (adj A) A = A I bl Hence A (adj A) = (adj A) A = A I pu Definition 4 A square matrix A is said to be singular if A = 0. 1 2 be T For example, the determinant of matrix A = is zero 4 8 re Hence A is a singular matrix. o R Definition 5 A square matrix A is said to be non-singular if A ≠ 0 ⎡1 2 ⎤ tt E 1 2 Let A= ⎢ ⎥. Then A = = 4 – 6 = – 2 ≠ 0. ⎣3 4 ⎦ 3 4 Hence A is a nonsingular matrix C We state the following theorems without proof. Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA no N are also nonsingular matrices of the same order. Theorem 3 The determinant of the product of matrices is equal to product of their respective determinants, that is, AB = A B , where A and B are square matrices of © the same order A 0 0 Remark We know that (adj A) A = A I = 0 A 0 0 0 A  DETERMINANTS 129 Writing determinants of matrices on both sides, we have A 0 0 ( adj A) A = 0 A 0 0 0 A he 1 0 0 3 i.e. |(adj A)| |A| = A 0 1 0 (Why?) 0 0 1 is i.e. |(adj A)| |A| = | A |3 (1) i.e. |(adj A)| = | A | 2 In general, if A is a square matrix of order n, then | adj (A) | = | A |n – 1. bl Theorem 4 A square matrix A is invertible if and only if A is nonsingular matrix. Proof Let A be invertible matrix of order n and I be the identity matrix of order n. pu Then, there exists a square matrix B of order n such that AB = BA = I Now AB = I. So AB = I or A B =1 (since I =1, AB = A B ) be T This gives A ≠ 0. Hence A is nonsingular. re o R Conversely, let A be nonsingular. Then A ≠ 0 Now A (adj A) = (adj A) A = A I (Theorem 1) tt E ⎛ 1 ⎞ ⎛ 1 ⎞ or A⎜ adj A ⎟ = ⎜ adj A ⎟ A = I ⎝ | A | ⎠ ⎝ | A | ⎠ C 1 or AB = BA = I, where B = adj A |A| no N 1 Thus A is invertible and A–1 = adj A |A| © 1 3 3 Example 24 If A = 1 4 3 , then verify that A adj A = | A | I. Also find A–1. 1 3 4 Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1 ≠ 0  130 MATHEMATICS Now A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0, A33 = 1 ⎡ 7 −3 −3⎤ ⎢ ⎥ Therefore adj A = ⎢ −1 1 0 ⎥ ⎢⎣ −1 0 1 ⎥⎦ he ⎡1 3 3 ⎤ ⎡ 7 −3 −3⎤ ⎢ ⎥⎢ ⎥ Now A (adj A) = ⎢1 4 3 ⎥ ⎢ −1 1 0 ⎥ ⎢⎣1 3 4 ⎥⎦ ⎢⎣ −1 0 1 ⎥⎦ is ⎡ 7 − 3 − 3 −3 + 3 + 0 −3 + 0 + 3 ⎤ ⎢ ⎥ = ⎢ 7 − 4 − 3 −3 + 4 + 0 − 3 + 0 + 3 ⎥ bl ⎢⎣ 7 − 3 − 4 −3 + 3 + 0 −3 + 0 + 4 ⎥⎦ ⎡1 0 0 ⎤ ⎡1 0 0 ⎤ pu ⎢ ⎥ = ⎢0 1 0 ⎥ = (1) ⎢0 1 0 ⎥ ⎢ ⎥ = A.I ⎢⎣0 0 1 ⎥⎦ ⎢⎣0 0 1 ⎥⎦ be T ⎡ 7 −3 −3⎤ ⎡ 7 −3 −3⎤ re 1 1⎢ ⎥ ⎢ ⎥ o R Also A–1 a d j A = ⎢ −1 1 0 ⎥ = ⎢ −1 1 0 ⎥ A 1 ⎢⎣ −1 0 1 ⎥⎦ ⎢⎣ −1 0 1 ⎥⎦ tt E ⎡2 3 ⎤ ⎡ 1 −2 ⎤ Example 25 If A = ⎢ ⎥ and B = ⎢ ⎥ , then verify that (AB) = B A. –1 –1 –1 ⎣1 − 4 ⎦ ⎣ −1 3 ⎦ C ⎡2 3⎤ ⎡ 1 −2 ⎤ ⎡ − 1 5 ⎤ Solution We have AB = ⎢ ⎥ ⎢ −1 3 ⎥ = ⎢ 5 −14 ⎥ ⎣1 − 4 ⎦ ⎣ ⎦ ⎣ ⎦ no N Since, AB = –11 ≠ 0, (AB)–1 exists and is given by 1 1 ⎡ −14 −5⎤ = 1 ⎡14 5⎤ adj (AB) = − ⎢ ⎢ ⎥ 11 ⎣ −5 −1⎥⎦ 11 ⎣ 5 1 ⎦ (AB)–1 = © AB Further, A = –11 ≠ 0 and B = 1 ≠ 0. Therefore, A–1 and B–1 both exist and are given by 1 4 3 3 2 A–1 = ,B 1 11 1 2 1 1  DETERMINANTS 131 1 3 2 4 3 1 14 5 1 ⎡14 5⎤ Therefore B 1A 1 = 11 1 1 1 2 11 5 1 11 ⎢⎣ 5 1 ⎥⎦ Hence (AB)–1 = B–1 A–1 2 3 Example 26 Show that the matrix A = satisfies the equation A2 – 4A + I = O, he 1 2 where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. Using this equation, find A–1. ⎡ 2 3 ⎤ ⎡ 2 3 ⎤ ⎡ 7 12 ⎤ Solution We have A 2 = A.A = ⎢ ⎥⎢ ⎥ =⎢ ⎥ ⎣1 2⎦ ⎣1 2 ⎦ ⎣ 4 7 ⎦ is ⎡ 7 12 ⎤ ⎡8 12 ⎤ ⎡1 0 ⎤ ⎡0 0 ⎤ Hence A 2 − 4A + I = ⎢ ⎥− ⎢ ⎥+⎢ ⎥ =⎢ ⎥=O ⎣ 4 7 ⎦ ⎣ 4 8 ⎦ ⎣0 1 ⎦ ⎣0 0 ⎦ bl Now A2 – 4A + I = O Therefore A A – 4A = – I or pu A A (A–1) – 4 A A–1 = – I A–1 (Post multiplying by A–1 because |A| ≠ 0) or A (A A–1) – 4I = – A–1 be T or AI – 4I = – A–1 re ⎡ 4 0 ⎤ ⎡2 3⎤ ⎡ 2 −3 ⎤ o R or A–1 = 4I – A = ⎢ ⎥ −⎢ ⎥ = ⎢ ⎥ ⎣ 0 4⎦ ⎣1 2⎦ ⎣ −1 2 ⎦ ⎡ 2 −3 ⎤ tt E Hence A −1 = ⎢ ⎥ ⎣ −1 2 ⎦ C EXERCISE 4.5 Find adjoint of each of the matrices in Exercises 1 and 2. no N 1 1 2 1 2 1. 2. 2 3 5 3 4 2 0 1 © Verify A (adj A) = (adj A) A = | A | I in Exercises 3 and 4 1 1 2 2 3 3 0 2 3. 4. 4 6 1 0 3  132 MATHEMATICS Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. 1 2 3 2 2 1 5 5. 6. 7. 0 2 4 4 3 3 2 0 0 5 1 0 0 2 1 3 1 1 2 he 8. 3 3 0 9. 4 1 0 10. 0 2 3 5 2 1 7 2 1 3 2 4 ⎡1 0 0 ⎤ is ⎢ 0 cos α sin α ⎥⎥ 11. ⎢ ⎢⎣ 0 sin α − cos α ⎥⎦ bl 3 7 6 8 12. Let A = and B =. Verify that (AB)–1 = B–1 A–1. 2 5 7 9 pu 3 1 13. If A = , show that A2 – 5A + 7I = O. Hence find A–1. 1 2 be T 3 2 re 14. For the matrix A = , find the numbers a and b such that A2 + aA + bI = O. o R 1 1 1 1 1 tt E 15. For the matrix A = 1 2 3 2 1 3 C Show that A3– 6A2 + 5A + 11 I = O. Hence, find A–1. 2 1 1 no N 16. If A = 1 2 1 1 1 2 Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1 © 17. Let A be a nonsingular square matrix of order 3 × 3. Then | adj A | is equal to (A) | A | (B) | A | 2 (C) | A | 3 (D) 3 | A | –1 18. If A is an invertible matrix of order 2, then det (A ) is equal to 1 (A) det (A) (B) det (A) (C) 1 (D) 0  DETERMINANTS 133 4.7 Applications of Determinants and Matrices In this section, we shall discuss 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. he Inconsistent system A system of equations is said to be inconsistent if its solution does not exist. $ Note In this chapter, we restrict ourselves to the system of linear equations is having unique solutions only. 4.7.1 Solution of system of linear equations using inverse of a matrix Let us express the system of linear equations as matrix equations and solve them using bl inverse of the coefficient matrix. Consider the system of equations pu a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d 2 a3 x + b3 y + c3 z = d 3 be T ⎡ a1 b1 c1 ⎤ ⎡x⎤ ⎡ d1 ⎤ re ⎢ a b c ⎥ , X = ⎢ y ⎥ and B = ⎢ d ⎥ o R Let A = ⎢ 2 2 2⎥ ⎢ ⎥ ⎢ 2⎥ ⎢⎣ a3 b3 c3 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ d3 ⎥⎦ tt E Then, the system of equations can be written as, AX = B, i.e., ⎡ a1 b1 c1 ⎤ ⎡ x⎤ ⎡ d1 ⎤ ⎢a c2 ⎥⎥ ⎢ y⎥ ⎢ ⎥ C ⎢ 2 b2 ⎢ ⎥ = ⎢d2 ⎥ ⎢⎣ a3 b3 c3 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ d3 ⎥⎦ no N Case I If A is a nonsingular matrix, then its inverse exists. Now AX = B or A (AX) = A–1 B –1 (premultiplying by A–1) © or (A–1A) X = A–1 B (by associative property) –1 or IX=A B or X = A–1 B This matrix equation provides unique solution for the given system of equations as inverse of a matrix is unique. This method of solving system of equations is known as Matrix Method.  134 MATHEMATICS Case II If A is a singular matrix, then | A | = 0. In this case, we calculate (adj A) B. If (adj A) B ≠ O, (O being zero matrix), then solution does not exist and the system of equations is called inconsistent. If (adj A) B = O, then system may be either consistent or inconsistent according as the system have either infinitely many solutions or no solution. he Example 27 Solve the system of equations 2x + 5y = 1 3x + 2y = 7 is Solution The system of equations can be written in the form AX = B, where ⎡2 5⎤ ⎡ x⎤ ⎡1 ⎤ A= ⎢ ⎥ , X = ⎢ ⎥ and B = ⎢ ⎥ bl ⎣ 3 2⎦ ⎣ y⎦ ⎣7 ⎦ Now, A = –11 ≠ 0, Hence, A is nonsingular matrix and so has a unique solution. pu 1 2 5 Note that A–1 = 11 3 2 be T 1 2 5 1 Therefore X = A–1B = – re 11 3 2 7 o R ⎡x⎤ 1 33 3 i.e. ⎢ y⎥ = ⎣ ⎦ 11 11 1 tt E Hence x = 3, y = – 1 Example 28 Solve the following system of equations by matrix method. C 3x – 2y + 3z = 8 2x + y – z = 1 no N 4x – 3y + 2z = 4 Solution The system of equations can be written in the form AX = B, where ⎡ 3 −2 3 ⎤ ⎡ x⎤ ⎡8 ⎤ © A = ⎢ 2 1 −1⎥ , X = ⎢⎢ y ⎥⎥ and B = ⎢ ⎥ ⎢1 ⎥ ⎢ ⎥ ⎢⎣ 4 −3 2 ⎥⎦ ⎢⎣ z ⎥⎦ ⎢⎣ 4⎥⎦ We see that A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17 ≠ 0  DETERMINANTS 135 Hence, A is nonsingular and so its inverse exists. Now A11 = –1, A12 = – 8, A13 = –10 A21 = –5, A22 = – 6, A23 = 1 A31 = –1, A32 = 9, A33 = 7 ⎡ −1 − 5 −1⎤ 1 ⎢ ⎥ he Therefore A = − ⎢ −8 − 6 9 ⎥ –1 17 ⎢⎣ −10 1 7 ⎥⎦ ⎡ −1 − 5 −1⎤ ⎡8 ⎤ 1 ⎢ is ⎥ ⎢ ⎥ X = A B = − ⎢ −8 − 6 9 ⎥ ⎢1 ⎥ –1 So 17 ⎢⎣ −10 1 7 ⎥⎦ ⎢⎣ 4 ⎥⎦ bl ⎡x⎤ ⎡ −17 ⎤ ⎡ 1 ⎤ ⎢ y⎥ 1 ⎢ ⎥ ⎢ ⎥ i.e. ⎢ ⎥ = − 17 ⎢ −34 ⎥ = ⎢ 2 ⎥ pu ⎢⎣ z ⎥⎦ ⎢⎣ −51⎥⎦ ⎢⎣ 3 ⎥⎦ Hence x = 1, y = 2 and z = 3. be T Example 29 The sum of three numbers is 6. If we multiply third number by 3 and add re second number to it, we get 11. By adding first and third numbers, we get double of the o R second number. Represent it algebraically and find the numbers using matrix method. Solution Let first, second and third numbers be denoted by x, y and z, respectively. tt E Then, according to given conditions, we have x+y+z=6 y + 3z = 11 C x + z = 2y or x – 2y + z = 0 This system can be written as A X = B, where no N 1 1 1 x 6 A= 0 1 3 , X = y and B = 11 1 –2 1 z 0 © Here A 1 1 6 – (0 – 3) 0 –1 9 0. Now we find adj A A11 = 1 (1 + 6) = 7, A12 = – (0 – 3) = 3, A13 = – 1 A21 = – (1 + 2) = – 3, A22 = 0, A23 = – (– 2 – 1) = 3 A31 = (3 – 1) = 2, A32 = – (3 – 0) = – 3, A33 = (1 – 0) = 1  136 MATHEMATICS ⎡ 7 –3 2 ⎤ ⎢ ⎥ Hence adj A = ⎢ 3 0 –3⎥ ⎢⎣ –1 3 1 ⎥⎦ ⎡ 7 –3 2 ⎤ 1 1 ⎢ ⎥ he Thus A –1 = adj (A) = ⎢ 3 0 –3⎥ A 9 ⎢⎣ –1 3 1 ⎥⎦ Since X = A–1 B ⎡ 7 –3 2 ⎤ ⎡ 6 ⎤ is 1⎢ ⎥⎢ ⎥ X = ⎢ 3 0 –3⎥ ⎢11⎥ 9 ⎢⎣ –1 3 1 ⎥⎦ ⎢⎣ 0 ⎥⎦ bl x ⎡ 42 − 33 + 0 ⎤ 9 1 y = 1 ⎢ 18 + 0 + 0 ⎥ 1 18 = 2 ⎢ ⎥ = or pu z 9 ⎢ −6 + 33 + 0 ⎥ 9 ⎣ ⎦ 27 3 Thus x = 1, y = 2, z = 3 be T EXERCISE 4.6 re o R Examine the consistency of the system of equations in Exercises 1 to 6. 1. x + 2y = 2 2. 2x – y = 5 3. x + 3y = 5 tt E 2x + 3y = 3 x+y=4 2x + 6y = 8 4. x + y + z = 1 5. 3x–y – 2z = 2 6. 5x – y + 4z = 5 2x + 3y + 2z = 2 2y – z = –1 2x + 3y + 5z = 2 C ax + ay + 2az = 4 3x – 5y = 3 5x – 2y + 6z = –1 Solve system of linear equations, using matrix method, in Exercises 7 to 14. no N 7. 5x + 2y = 4 8. 2x – y = –2 9. 4x – 3y = 3 7x + 3y = 5 3x + 4y = 3 3x – 5y = 7 10. 5x + 2y = 3 11. 2x + y + z = 1 12. x – y + z = 4 3 © 3x + 2y = 5 x – 2y – z = 2x + y – 3z = 0 2 3y – 5z = 9 x+y+z=2 13. 2x + 3y +3 z = 5 14. x – y + 2z = 7 x – 2y + z = – 4 3x + 4y – 5z = – 5 3x – y – 2z = 3 2x – y + 3z = 12  DETERMINANTS 137 ⎡ 2 –3 5⎤ ⎢ 3 2 – 4⎥ –1 –1 15. If A = ⎢ ⎥ , find A. Using A solve the system of equations ⎢⎣1 1 –2 ⎥⎦ 2x – 3y + 5z = 11 he 3x + 2y – 4z = – 5 x + y – 2z = – 3 16. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg is rice is Rs 70. Find cost of each item per kg by matrix method. Miscellaneous Examples bl Example 30 If a, b, c are positive and unequal, show that value of the determinant a b c pu Δ = b c a is negative. c a b be T Solution Applying C1 → C1 + C2 + C3 to the given determinant, we get re a+b+c b c 1 b c o R Δ = a + b + c c a = (a + b + c) 1 c a a+b+c a b 1 a b tt E 1 b c C = (a + b + c) 0 c – b a – c (Applying R2→ R2–R1,and R3 →R3 –R1) 0 a–b b– c no N = (a + b + c) [(c – b) (b – c) – (a – c) (a – b)] (Expanding along C1) = (a + b + c)(– a2 – b2 – c2 + ab + bc + ca) –1 © = (a + b + c) (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca) 2 –1 = (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2] 2 which is negative (since a + b + c > 0 and (a – b)2 + (b – c)2 + (c – a)2 > 0)  138 MATHEMATICS Example 31 If a, b, c, are in A.P, find value of 2y + 4 5y + 7 8y + a 3y + 5 6 y + 8 9 y + b 4 y + 6 7 y + 9 10 y + c he Solution Applying R1 → R1 + R3 – 2R2 to the given determinant, we obtain 0 0 0 3y + 5 6 y + 8 9 y + b = 0 (Since 2b = a + c) is 4 y + 6 7 y + 9 10 y + c Example 32 Show that bl ( y+ z ) 2 xy zx ( x+ z ) 2 Δ= xy yz = 2xyz (x + y + z)3 pu xz yz ( x+ y ) 2 Solution Applying R1 → xR1, R2 → yR2 , R3 → z R3 to Δ and dividing by xyz, we get be T re 2 x y z x2 y x2 z o R 1 2 Δ= xy 2 y x z y2 z xyz 2 xz 2 yz 2 z x y tt E Taking common factors x, y, z from C1 C2 and C3, respectively, we get C ( y + z) 2 x2 x2 xyz ( x + z) 2 Δ= y2 y2 no N xyz ( x + y) 2 z2 z2 Applying C2 → C2– C1, C3 → C3– C1, we have © ( y + z )2 x2 – ( y + z ) 2 x2 − ( y + z ) 2 Δ= y2 ( x + z )2 − y 2 0 z2 0 ( x + y )2 – z 2  DETERMINANTS 139 Taking common factor (x + y + z) from C2 and C3, we have (y + z) x – ( y + z) x – ( y + z) 2 Δ = (x + y + z) 2 y 2 ( x + z) – y 0 z2 0 ( x + y) – z Applying R1 → R1 – (R2 + R3), we have he 2 yz –2z –2y Δ = (x + y + z)2 y2 x− y+z 0 is z2 0 x+ y –z bl 1 1 Applying C2 → (C2 + C ) and C3 C3 C1 , we get pu y 1 z 2 yz 0 0 be T y2 Δ = (x + y + z)2 y2 x z z re o R z2 z2 x y y Finally expanding along R1, we have tt E Δ = (x + y + z)2 (2yz) [(x + z) (x + y) – yz] = (x + y + z)2 (2yz) (x2 + xy + xz) = (x + y + z)3 (2xyz) C 1 –1 2 –2 0 1 Example 33 Use product 0 2 –3 9 2 –3 to solve the system of equations no N 3 –2 4 6 1 –2 x – y + 2z = 1 2y – 3z = 1 © 3x – 2y + 4z = 2 ⎡1 –1 2⎤ ⎡ –2 0 1 ⎤ ⎢ – 3 ⎥⎥ ⎢9 – 3 ⎥⎥ Solution Consider the product ⎢ 0 2 ⎢ 2 ⎢⎣ 3 –2 4 ⎥⎦ ⎢⎣ 6 1 – 2 ⎥⎦  140 MATHEMATICS ⎡ − 2 − 9 + 12 0 − 2 + 2 1 + 3 − 4⎤ 1 0 0 ⎢ 0 + 18 − 18 0 + 4 − 3 0 − 6 + 6⎥ = ⎢ ⎥ = 0 1 0 ⎢⎣ − 6 − 18 + 24 0 − 4 + 4 3 + 6 − 8⎥⎦ 0 0 1 –1 1 –1 2 –2 0 1 Hence he 0 2 –3 9 2 –3 3 –2 4 6 1 –2 Now, given system of equations can be written, in matrix form, as follows ⎡1 –1 2 ⎤ ⎡ x ⎤ ⎡1 ⎤ is ⎢ 0 2 –3⎥ ⎢ y ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢1 ⎥ ⎢⎣ 3 –2 4 ⎦⎥ ⎢⎣ z ⎥⎦ ⎢⎣ 2⎥⎦ bl −1 x ⎡1 −1 2 ⎤ ⎡ 1 ⎤ –2 0 1 1 ⎢ y = 0 ⎥ ⎢ ⎥ 2 −3⎥ ⎢ 1 ⎥ = 9 2 –3 1 or pu z ⎢ ⎢⎣ 3 −2 4 ⎥⎦ ⎢⎣ 2 ⎥⎦ 6 1 –2 2 ⎡ −2 + 0 + 2 ⎤ ⎡0 ⎤ be T ⎢ ⎥ ⎢ ⎥ = ⎢ 9 + 2 − 6 ⎥ = ⎢5 ⎥ re o R ⎢⎣ 6 + 1 − 4 ⎦⎥ ⎢⎣ 3⎥⎦ Hence x = 0, y = 5 and z = 3 Example 34 Prove that tt E a + bx c + dx p + qx a c p Δ = ax + b cx + d px + q = (1 − x ) b d q 2 C u v w u v w no N Solution Applying R1 → R1 – x R2 to Δ, we get a (1 − x 2 ) c (1 − x 2 ) p (1 − x 2 ) Δ= ax + b cx + d px + q © u v w a c p = (1 − x ) ax + b cx + d px + q 2 u v w  DETERMINANTS 141 Applying R2 → R2 – x R1, we get a c p Δ = (1 − x ) b d q 2 u v w he Miscellaneous Exercises on Chapter 4 x sin θ cos θ 1. Prove that the determinant – sin θ – x 1 is independent of θ. is cos θ 1 x a a2 bc 1 a2 a3 bl 2. Without expanding the determinant, prove that b b 2 ca 1 b2 b3. pu c c2 ab 1 c2 c3 cos α cos β cos α sin β – sin α 3. Evaluate – sin β cos β 0. be T sin α cos β sin α sin β cos α re 4. If a, b and c are real numbers, and o R b+ c c+ a a+ b Δ = c + a a + b b + c = 0, tt E a+ b b+ c c+ a Show that either a + b + c = 0 or a = b = c. C x+a x x 5. Solve the equation x x+a x = 0, a ≠ 0 no N x x x+a a2 bc ac c 2 2 6. Prove that a ab b2 ac = 4a2b2c2 © 2 ab b bc c2 ⎡ 3 –1 1 ⎤ ⎡ 1 2 –2 ⎤ ⎢ ⎥ ⎢ ⎥ 7. If A–1 = ⎢ –15 6 –5⎥ and B = ⎢ –1 3 0 ⎥ , find ( AB ) –1 ⎢⎣ 5 –2 2 ⎥⎦ ⎢⎣ 0 –2 1 ⎥⎦  142 MATHEMATICS 1 –2 1 8. Let A = –2 3 1. Verify that 1 1 5 (i) [adj A]–1 = adj (A–1) (ii) (A–1)–1 = A x y x+ y he 9. Evaluate y x+ y x x+ y x y is 1 x y 10. Evaluate 1 x + y y bl 1 x x+ y Using properties of determinants in Exercises 11 to 15, prove that: pu α α2 β+γ 11. β β2 γ + α = (β – γ) (γ – α) (α – β) (α + β + γ) γ γ2 α +β be T re o R x x 2 1 px 3 12. y y 2 1 py 3 = (1 + pxyz) (x – y) (y – z) (z – x), where p is any scalar. tt E z z 2 1 pz 3 3a – a+ b – a+ c C 13. –b+ a 3b – b + c = 3(a + b + c) (ab + bc + ca) – c + a – c+ b 3c no N 1 1+ p 1+ p + q sin α cos α cos ( α + δ ) 14. 2 3 + 2 p 4 + 3 p+ 2q = 1 15. sin β cos β cos ( β + δ ) = 0 © 3 6 + 3 p 10 + 6 p + 3 q sin γ cos γ cos ( γ + δ ) 16. Solve the system of equations 2 3 10 4 x y z  DETERMINANTS 143 4 6 5 – 1 x y z 6 9 20 – 2 x y z he Choose the correct answer in Exercise 17 to 19. 17. If a, b, c, are in A.P, then the determinant x + 2 x + 3 x + 2a is x + 3 x + 4 x + 2b is x + 4 x + 5 x + 2c bl (A) 0 (B) 1 (C) x (D) 2x ⎡ x 0 0⎤ pu 18. If x, y, z are nonzero real numbers, then the inverse of matrix A = ⎢⎢ 0 y 0 ⎥⎥ is ⎢⎣ 0 0 z ⎥⎦ be T re ⎡ x −1 0 ⎤ ⎡ x −1 0 ⎤ o R 0 0 ⎢ ⎥ ⎢ ⎥ (A) ⎢ 0 y −1 0 ⎥ (B) xyz ⎢ 0 y −1 0 ⎥ ⎢ ⎥ ⎢ ⎥ z −1 ⎦ z −1 ⎦ tt E ⎣ 0 0 ⎣ 0 0 C ⎡ x 0 0⎤ ⎡1 0 0 ⎤ 1 ⎢ 0 y 0 ⎥⎥ 1 ⎢ (C) (D) 0 1 0 ⎥⎥ xyz ⎢ xyz ⎢ ⎢⎣ 0 0 z ⎥⎦ ⎢⎣0 0 1 ⎥⎦ no N ⎡ 1 sin θ 1 ⎤ ⎢ − sin θ 1 sin θ ⎥⎥ , where 0 ≤ θ ≤ 2π. Then © 19. Let A = ⎢ ⎢⎣ −1 − sin θ 1 ⎥⎦ (A) Det (A) = 0 (B) Det (A) ∈ (2, ∞) (C) Det (A) ∈ (2, 4) (D) Det (A) ∈ [2, 4]  144 MATHEMATICS Summary  Determinant of a matrix A = [a11]1 × 1 is given by | a11| = a11 a11 a12  Determinant of a matrix A a21 a22 is given by he a11 a12 A = = a11 a22 – a12 a21 a21 a22 a1 b1 c1 is  Determinant of a matrix A a2 b2 c2 is given by (expanding along R1) a3 b3 c3 bl a1 b1 c1 b c2 a2 c2 a2 b2 pu A = a2 b2 c2 = a1 2 b3 c3 − b1 a3 c3 + c1 a3 b3 a3 b3 c3 be T For any square matrix A, the |A| satisfy following prop