4.1 Introduction 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 matrices. This means, a system of linear equations like a 1 x + b 1 y = c 1 a2 x + b2 y = c2 can be represented as 1 1 1 2 2 2 a b x c a b y c           =         . Now, this system of equations has a unique solution or not, is determined by the number a1 b2 – a2 b1. (Recall that if 1 1 2 2 a b a b  or, a1 b2 – a2 b1  0, then the system of linear equations has a unique solution). The number a 1 b 2 – a 2 b 1 which determines uniqueness of solution is associated with the matrix 1 1 2 2 A a b a b   =    and is called the determinant of A or det A. Determinants have wide applications in 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. Chapter 4 DETERMINANTS P.S. Laplace (1749-1827) 2022-23 104 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 . If A = a b c d       , then determinant of A is written as |A| = a b c d = det (A) Remarks (i) For matrix A, |A| is read as determinant of A and not modulus of A. (ii) Only square matrices have determinants. 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 4.2.2 Determinant of a matrix of order two Let A = 11 12 21 22 a a a a       be a matrix of order 2 × 2, then the determinant of A is defined as: det (A) = |A| =  = = a11a22 – a21a12 Example 1 Evaluate 2 4 –1 2 . Solution We have 2 4 –1 2 = 2 (2) – 4(–1) = 4 + 4 = 8. Example 2 Evaluate 1 – 1 x x x x + Solution We have 1 – 1 x x x x + = x (x) – (x + 1) (x – 1) = x2 – (x2 – 1) = x2 – x2 + 1 = 1 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 2022-23 DETERMINANTS 1 05 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 i.e., | A | = 21 22 23 31 32 33 a a a a a a 11 12 13 a a a Expansion along first Row (R1) Step 1 Multiply first element a11 of R1 by (–1)(1 + 1) [(–1)sum of suffixes in a 11] and with the 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, i.e., (–1)1 + 1 a11 22 23 32 33 a a a a Step 2 Multiply 2nd element a12 of R1 by (–1)1 + 2 [(–1)sum of suffixes in a 12] and the second order determinant obtained by deleting elements of first row (R1) and 2nd column (C2) of | A | as a12 lies in R1 and C2, i.e., (–1)1 + 2 a 12 21 23 31 33 a a a a Step 3 Multiply third element a13 of R1 by (–1)1 + 3 [(–1)sum of suffixes in a 13] and the second order determinant obtained by deleting elements of first row (R1) and third column (C3) of | A | as a13 lies in R1 and C3, i.e., (–1)1 + 3 a13 21 22 31 32 a a a a Step 4 Now the expansion of determinant of A, that is, | A | written as sum of all three terms obtained in steps 1, 2 and 3 above is given by det A = |A| = (–1)1 + 1 a11 22 23 1 2 21 23 12 32 33 31 33 (–1) a a a a a a a a a

• + + 1 3 21 22 13 31 32 (–1) a a a a a + or |A| = a11 ( a22 a33 – a32 a23) – a12 ( a21 a33 – a31 a23)
• a 13 ( a 21 a 32 – a 31 a 22) 2022-23 106 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 (R 2) | A | = 11 12 13 31 32 33 a a a a a a 21 22 23 a a a Expanding along R2, we get | A | = 2 1 12 13 2 2 11 13 21 22 32 33 31 33 (–1) (–1) a a a a a a a a a a
• • + 2 3 11 12 23 31 32 (–1) a a a a a
• + = – a21 ( a12 a33 – a32 a13) + a22 ( a11 a33 – a31 a13) – a23 ( a11 a32 – a31 a12) | A | = – a21 a12 a33 + a21 a32 a13 + a22 a11 a33 – a22 a31 a13 – a23 a11 a32
• a23 a31 a12 = a11 a22 a33 – a11 a23 a32 – a12 a21 a33 + a12 a23 a31 + a13 a21 a32 – a13 a31 a22 … (2) Expansion along first Column (C1) | A | = 12 13 22 23 32 33 11 21 31 a a a a a a a a a By expanding along C1, we get | A | = 1 1 22 23 2 1 12 13 11 21 32 33 32 33 (–1) ( 1) a a a a a a a a a a
• • • − + 3 1 12 13 31 22 23 (–1) a a a a a + = a11 (a22 a33 – a23 a32) – a21 (a12 a33 – a13 a32) + a31 (a12 a23 – a13 a22) 2022-23 DETERMINANTS 1 07 | 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 – a 13 a 31 a 22 … (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 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 which contains maximum number of zeros. (ii) While expanding, instead of multiplying by (–1)i + j, we can multiply by +1 or –1 according as (i + j) is even or odd. (iii) Let A = 2 2 4 0       and B = 1 1 2 0       . Then, it is easy to verify that A = 2B. Also |A| = 0 – 8 = – 8 and |B| = 0 – 2 = – 2. Observe that, |A| = 4(– 2) = 22|B| or |A| = 2n|B|, where n = 2 is the order of square matrices A and B. In general, if A = kB where A and B are square matrices of order n, then | A| = kn | B |, where n = 1, 2, 3 Example 3 Evaluate the determinant  = 1 2 4 –1 3 0 4 1 0 . Solution Note that in the third column, two entries are zero. So expanding along third column (C3), we get  = –1 3 1 2 1 2 4 – 0 0 4 1 4 1 –1 3 + = 4 (–1 – 12) – 0 + 0 = – 52 Example 4 Evaluate  = 0 sin – cos – sin 0 sin cos – sin 0       . 2022-23 108 MATHEMATICS Solution Expanding along R1, we get  = 0 sin – sin sin –sin 0 0 – sin – cos – sin 0 cos 0 cos – sin           = 0 – sin  (0 – sin  cos ) – cos  (sin  sin  – 0) = sin  sin  cos  – cos  sin  sin  = 0 Example 5 Find values of x for which 3 3 2 1 4 1 x x = . Solution We have 3 3 2 1 4 1 x x = i.e. 3 – x2 = 3 – 8 i.e. x2 = 8 Hence x = ± 2 2 EXERCISE 4.1 Evaluate the determinants in Exercises 1 and 2. 1. 2 4 –5 –1

2. (i) cos – sin sin cos     (ii) 2 – 1 – 1 1 1 x x x x x +

• +

3. If A = 1 2 4 2       , then show that | 2A | = 4 | A |
4. If A = 1 0 1 0 1 2 0 0 4         , then show that | 3 A | = 27 | A |
5. Evaluate the determinants (i) 3 –1 –2 0 0 –1 3 –5 0 (ii) 3 –4 5 1 1 –2 2 3 1 2022-23 DETERMINANTS 1 09 (iii) 0 1 2 –1 0 –3 –2 3 0 (iv) 2 –1 –2 0 2 –1 3 –5 0
6. If A = 1 1 –2 2 1 –3 5 4 –9         , find | A |
7. Find values of x, if (i) 2 4 2 4 5 1 6 x x = (ii) 2 3 3 4 5 2 5 x x =
8. If 2 6 2 18 18 6 x x = , then x is equal to (A) 6 (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 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 order 3 only. Property 1 The value of the determinant remains unchanged if its rows and columns are interchanged. Verification Let  = 1 2 3 1 2 3 1 2 3 a a a b b b c c c Expanding along first row, we get  = 2 3 1 3 1 2 1 2 3 2 3 1 3 1 2 b b b b b b a a a c c c c c c − + = a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) By interchanging the rows and columns of , we get the determinant 1 = 1 1 1 2 2 2 3 3 3 a b c a b c a b c 2022-23 110 MATHEMATICS Expanding 1 along first column, we get 1 = a1 (b2 c3 – c2 b3) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1) Hence  = 1 Remark It follows from above property that if A is a square matrix, then det (A) = det (A), where A = transpose of A. Note If Ri = ith row and Ci = ith column, then for interchange of row and columns, we will symbolically write Ci  Ri Let us verify the above property by example. Example 6 Verify Property 1 for  = 2 –3 5 6 0 4 1 5 –7 Solution Expanding the determinant along first row, we have  = 0 4 6 4 6 0 2 – (–3) 5 5 –7 1 –7 1 5 + = 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0) = – 40 – 138 + 150 = – 28 By interchanging rows and columns, we get 1 = 2 6 1 –3 0 5 5 4 –7 (Expanding along first column) = 0 5 6 1 6 1 2 – (–3) 5 4 –7 4 –7 0 5 + = 2 (0 – 20) + 3 (– 42 – 4) + 5 (30 – 0) = – 40 – 138 + 150 = – 28 Clearly  = 1 Hence, Property 1 is verified. Property 2 If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes. Verification Let  = 1 2 3 1 2 3 1 2 3 a a a b b b c c c 2022-23 DETERMINANTS 1 11 Expanding along first row, we get  = 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 1 = 1 2 3 1 2 3 1 2 3 c c c b b b a a a Expanding along third row, we get 1 = a1 (c2 b3 – b2 c3) – a2 ( c1 b3 – c3 b1) + a3 (b2 c1 – b1 c2) = – [a1 (b2 c3 – b3 c2) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1)] Clearly 1 = –  Similarly, we can verify the result by interchanging any two columns. Note We can denote the interchange of rows by Ri  Rj and interchange of columns by Ci  Cj. Example 7 Verify Property 2 for  = 2 –3 5 6 0 4 1 5 –7 . Solution  = 2 –3 5 6 0 4 1 5 –7 = – 28 (See Example 6) Interchanging rows R2 and R3 i.e., R2  R3, we have 1 = 2 –3 5 1 5 –7 6 0 4 Expanding the determinant 1 along first row, we have 1 = 5 –7 1 –7 1 5 2 – (–3) 5 0 4 6 4 6 0 + = 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30) = 40 + 138 – 150 = 28 2022-23 112 MATHEMATICS Clearly 1 = –  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 , then  does not change. However, by Property 2, it follows that  has changed its sign Therefore  = –  or  = 0 Let us verify the above property by an example. Example 8 Evaluate  = 3 2 3 2 2 3 3 2 3 Solution Expanding along first row, we get  = 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6) = 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0 Here R1 and R3 are identical. Property 4 If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets multiplied by k. Verification Let  = 1 1 1 2 2 2 3 3 3 a b c a b c a b c and 1 be the determinant obtained by multiplying the elements of the first row by k. Then 1 = 1 1 1 2 2 2 3 3 3 k a k b k c a b c a b c Expanding along first row, we get 1 = 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  2022-23 DETERMINANTS 1 13 Hence 1 1 1 2 2 2 3 3 3 k a k b k c a b c a b c = k 1 1 1 2 2 2 3 3 3 a b c a b c a b c Remarks (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  = 1 2 3 1 2 3 1 2 3 a a a b b b k a k a k a = 0 (rows R1 and R2 are proportional) Example 9 Evaluate 102 18 36 1 3 4 17 3 6 Solution Note that 102 18 36 6(17) 6(3) 6(6) 17 3 6 1 3 4 1 3 4 61 3 4 0 17 3 6 17 3 6 17 3 6 = = = (Using Properties 3 and 4) 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. For example, 1 1 2 2 3 3 1 2 3 1 2 3 a a a b b b c c c

•  +  +  = 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 a a a b b b b b b c c c c c c    + Verification L.H.S. = 1 1 2 2 3 3 1 2 3 1 2 3 a a a b b b c c c
•  +  +  2022-23 114 MATHEMATICS Expanding the determinants along the first row, we get  = (a1 + 1) (b2 c3 – c2 b3) – (a2 + 2) (b1 c3 – b3 c1)
• (a 3 + 3) (b 1 c 2 – b 2 c 1) = a1 (b2 c3 – c2 b3) – a2 (b1 c3 – b3 c1) + a3 (b1 c2 – b2 c1)
• 1 (b2 c3 – c2 b3) – 2 (b1 c3 – b3 c1) + 3 (b1 c2 – b2 c1) (by rearranging terms) = 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 a a a b b b b b b c c c c c c   
• = R.H.S. Similarly, we may verify Property 5 for other rows or columns. Example 10 Show that 2 2 2 0 a b c a x b y c z x y z
• • • = Solution We have 2 2 2 a b c a x b y c z x y z
• • • = 2 2 2 a b c a b c a b c x y z x y z x y z + (by Property 5) = 0 + 0 = 0 (Using Property 3 and Property 4) 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 remains the same, i.e., the value of determinant remain same if we apply the operation Ri  Ri + kRj or Ci  Ci + k Cj . Verification Let  = 1 2 3 1 2 3 1 2 3 a a a b b b c c c and 1 = 1 1 2 2 3 3 1 2 3 1 2 3 a k c a k c a k c b b b c c c
• • + , where 1 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. 2022-23 DETERMINANTS 1 15 Now, again 1 = 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 a a a k c k c k c b b b b b b c c c c c c
• (Using Property 5) =  + 0 (since R1 and R3 are proportional) Hence  = 1 Remarks (i) If 1 is the determinant obtained by applying Ri  kRi or Ci  kCi to the determinant , then 1 = k. (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 another operation. A similar remark applies to column operations. Example 11 Prove that 3 2 3 2 4 3 2 3 6 3 10 6 3 a a b a b c a a b a b c a a a b a b c
• • +
• • • =
• • + . Solution Applying operations R2  R 2 – 2R1 and R3  R3 – 3R1 to the given determinant , we have  = 0 2 0 3 7 3 a a b a b c a a b a a b
• • + + + Now applying R3  R3 – 3R2 , we get  = 0 2 0 0 a a b a b c a a b a
• • + + Expanding along C1, we obtain  = 2 0 a a b a a +
• 0 + 0 = a (a2 – 0) = a (a2) = a3 2022-23 116 MATHEMATICS Example 12 Without expanding, prove that  = 0 1 1 1 x y y z z x z x y
• • + = Solution Applying R1  R1 + R2 to , we get  = 1 1 1 x y z x y z x y z z x y
• • • • • + Since the elements of R1 and R3 are proportional,  = 0. Example 13 Evaluate  = 1 1 1 a bc b ca c ab Solution Applying R2  R2 – R1 and R3  R3 – R1, we get  = 1 0 ( ) 0 ( ) a bc b a c a b c a b a c − − − − Taking factors (b – a) and (c – a) common from R2 and R3, respectively, we get  = 1 ( ) ( ) 0 1 – 0 1 – a bc b a c a c b − − = (b – a) (c – a) [(– b + c)] (Expanding along first column) = (a – b) (b – c) (c – a) Example 14 Prove that 4 b c a a b c a b abc c c a b +
• = + Solution Let  = b c a a b c a b c c a b + + + 2022-23 DETERMINANTS 1 17 Applying R1  R1 – R2 – R3 to , we get  = 0 –2c –2b b c a b c c a b + + Expanding along R1, we obtain  = 0 – (–2 ) c a b b b c c a b c a b +
• + (–2 ) b c a b c c + + = 2 c (a b + b2 – bc) – 2 b (b c – c2 – ac) = 2 a b c + 2 cb2 – 2 bc2 – 2 b2c + 2 bc2 + 2 abc = 4 abc Example 15 If x, y, z are different and 2 3 2 3 2 3 1 1 0 1 x x x y y y z z z + = + = + , then show that 1 + xyz = 0 Solution We have  = 2 3 2 3 2 3 1 1 1 x x x y y y z z z + + + = 2 2 3 2 2 3 2 2 3 1 1 1 x x x x x y y y y y z z z z z
• (Using Property 5) = 2 2 2 2 2 2 2 1 1 ( 1) 1 1 1 1 x x x x y y xyz y y z z z z − + (Using C3C2 and then C1  C2) = 2 2 2 1 1 (1 ) 1 x x y y xyz z z + 2022-23 118 MATHEMATICS =( ) 2 2 2 2 2 1 1 0 0 x x xyz y x y x z x z x
• − − − − (Using R2R2–R1 and R3  R3–R1) Taking out common factor (y – x) from R2 and (z – x) from R3, we get  = 2 1 (1+ ) ( – ) ( – ) 0 1 0 1 x x xyz y x z x y x z x + + = (1 + xyz) (y – x) (z – x) (z – y) (on expanding along C1) Since  = 0 and x, y, z are all different, i.e., x – y  0, y – z  0, z – x  0, we get 1 + xyz = 0 Example 16 Show that 1 1 1 1 1 1 1 1 1 1 1 1 1 a b abc abc bc ca ab a b c c +  
• =  + + +  = + + +   + Solution Taking out factors a,b,c common from R1, R2 and R3, we get L.H.S. = 1 1 1 1 1 1 1 1 1 1 1 1 a a a abc b b b c c c + + + Applying R1 R1 + R2 + R3, we have  = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 a b c a b c a b c abc b b b c c c
• • • • • • • • + + + 2022-23 DETERMINANTS 1 19 = 1 1 1 1 1 1 1 1 1 1+ 1 1 1 1 1 abc a b c b b b c c c    + +  +   + Now applying C2  C2 – C1, C3  C3 – C1, we get  = 1 0 0 1 1 1 1 1+ 1 0 1 0 1 abc a b c b c    + +    = ( ) 1 1 1 abc 1 1 1– 0 a b c    + + +      = 1 1 1 abc 1+ a b c    + +    = abc + bc + ca + ab = R.H.S. Note Alternately try by applying C1  C1 – C2 and C3  C3 – C2, then apply C1  C1 – a C3. EXERCISE 4.2 Using the property of determinants and without expanding in Exercises 1 to 7, prove that:

1. 0 x a x a y b y b z c z c +

• = +

2. 0 a b b c c a b c c a a b c a a b b c − − − − − − = − − − 3. 2 7 65 3 8 75 0 5 9 86 = 4. ( ) ( ) ( ) 1 1 0 1 bc a b c ca b c a ab c a b +

• = +

5. 2 b c q r y z a p x c a r p z x b q y a b p q x y c r z

• • +
• • • =
• • + 2022-23 120 MATHEMATICS 6. 0 0 0 0 a b a c b c − − − = 7. 2 2 2 2 2 2 4 a ab ac ba b bc a b c ca cb c − − = − By using properties of determinants, in Exercises 8 to 14, show that:

8. (i) ( )( )( ) 2 2 2 1 1 1 a a b b a b b c c a c c = − − − (ii) ( )( )( )( ) 3 3 3 1 1 1 a b c a b b c c a a b c a b c = − − − + + 9. 2 2 2 x x yz y y zx z z xy = (x – y) (y – z) (z – x) (xy + yz + zx)
9. (i) ( )( )2 4 2 2 2 4 5 4 4 2 2 4 x+ x x x x+ 2x x x x x x+ = + − (ii) ( ) 2 3 y+k y y y y+k y k y k y y y+k = +
10. (i) ( )3 2 2 2 2 2 2 a b c a a b b c a b a b c c c c a b − − − − = + + − − (ii) ( )3 2 2 2 2 x y z x y z y z x y x y z z x z x y

• +
• • = + +
• + 2022-23 DETERMINANTS 1 21

12. ( ) 2 2 2 3 2 1 1 1 1 x x x x x x x = −
13. ( ) 2 2 3 2 2 2 2 2 2 1 2 2 2 1 2 1 2 2 1 a b ab b ab a b a a b b a a b

• − − − + = + + − − − 14. 2 2 2 2 2 2 1 1 1 1 a ab ac ab b bc a b c ca cb c +
• = + + + + 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) k2 | A | (C) k3 | A | (D) 3k | A |
16. Which of the following is correct (A) Determinant is a square matrix. (B) Determinant is a number associated to a matrix. (C) Determinant is a number associated to a square matrix. (D) None of these 4.4 Area of a Triangle In earlier classes, we have studied that the area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y 3), is given by the expression 1 2 [x1(y2–y3) + x2 ( y3–y1) + x 3 (y 1–y 2)]. Now this expression can be written in the form of a determinant as  = 1 1 2 2 3 3 1 1 1 2 1 x y x y x y … (1) Remarks (i) Since area is a positive quantity, we always take the absolute value of the determinant in (1). 2022-23 122 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 1 4 2 1 2 5 1 1 – = ( ) ( ) ( ) 1 3 2 –1 –8 –4 – 5 1 –4 –10 2  +  = ( ) 1 61 3 72 14 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. Solution Let P (x, y) be any point on AB. Then, area of triangle ABP is zero (Why?). So 0 0 1 1 1 3 1 2 x y 1 = 0 This gives ( ) 1 3 2 y – x = 0 or y = 3x, which is the equation of required line AB. Also, since the area of the triangle ABD is 3 sq. units, we have 1 3 1 1 0 0 1 2 k 0 1 = ± 3 This gives, 3 3 2 − k = ± , i.e., k =  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) 2022-23 DETERMINANTS 1 23
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. (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 In this section, we will learn to write the expansion of a determinant in compact form using minors and cofactors. 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. Remark Minor of an element of a determinant of order n(n  2) is a determinant of order n – 1. Example 19 Find the minor of element 6 in the determinant 1 2 3 4 5 6 7 8 9  = Solution Since 6 lies in the second row and third column, its minor M23 is given by M23 = 1 2 7 8 = 8 – 14 = – 6 (obtained by deleting R2 and C3 in ). Definition 2 Cofactor of an element aij , denoted by Aij is defined by Aij = (–1)i + j Mij , where Mij is minor of a ij. Example 20 Find minors and cofactors of all the elements of the determinant 1 –2 4 3 Solution Minor of the element aij is Mij Here a 11 = 1. So M11 = Minor of a 11= 3 M12 = Minor of the element a12 = 4 M21 = Minor of the element a21 = –2 2022-23 124 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 A22 = (–1)2 + 2 M22 = (–1)4 (1) = 1 Example 21 Find minors and cofactors of the elements a11, a21 in the determinant  = 11 12 13 21 22 23 31 32 33 a a a a a a a a a Solution By definition of minors and cofactors, we have Minor of a11 = M11 = 22 23 32 33 a a a a = a22 a33– a23 a32 Cofactor of a11 = A11 = (–1)1+1 M 11 = a22 a33 – a 23 a32 Minor of a 21 = M21 = 12 13 32 33 a a a a = a 12 a 33 – a 13 a 32 Cofactor of a21 = A21 = (–1)2+1 M 21 = (–1) (a12 a33 – a13 a32) = – a12 a33 + a13 a32 Remark Expanding the determinant , in Example 21, along R1, we have  = (–1)1+1 a 11 22 23 32 33 a a a a

• (–1)1+2 a 12 21 23 31 33 a a a a
• (–1)1+3 a 13 21 22 31 32 a a a a = a 11 A11 + a 12 A12 + a 13 A13, where Aij is cofactor of a ij = sum of product of elements of R1 with their corresponding cofactors Similarly,  can be calculated by other five ways of expansion that is along R2, R3, C1, C2 and C3. Hence  = 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, 2022-23 DETERMINANTS 1 25  = a 11 A21 + a 12 A22 + a 13 A23 = a 11 (–1)1+1 12 13 32 33 a a a a
• a 12 (–1)1+2 11 13 31 33 a a a a
• a 13 (–1)1+3 11 12 31 32 a a a a = 11 12 13 11 12 13 31 32 33 a a a a a a a a a = 0 (since R1 and R2 are identical) Similarly, we can try for other rows and columns. Example 22 Find minors and cofactors of the elements of the determinant 2 3 5 6 0 4 1 5 7 – – and verify that a 11 A31 + a 12 A32 + a 13 A33= 0 Solution We have M11 = 0 4 5 –7 = 0 –20 = –20; A11 = (–1)1+1 (–20) = –20 M12 = 6 4 1 –7 = – 42 – 4 = – 46; A12 = (–1)1+2 (– 46) = 46 M13 = 6 0 1 5 = 30 – 0 = 30; A13 = (–1)1+3 (30) = 30 M21 = 3 5 5 7 – – = 21 – 25 = – 4; A21 = (–1)2+1 (– 4) = 4 M22 = 2 5 1 –7 = –14 – 5 = –19; A22 = (–1)2+2 (–19) = –19 M23 = 2 3 1 5 – = 10 + 3 = 13; A23 = (–1)2+3 (13) = –13 M31 = 3 5 0 4 – = –12 – 0 = –12; A31 = (–1)3+1 (–12) = –12 2022-23 126 MATHEMATICS M32 = 2 5 6 4 = 8 – 30 = –22; A32 = (–1)3+2 (–22) = 22 and M33 = 2 3 6 0 – = 0 + 18 = 18; A33 = (–1)3+3 (18) = 18 Now a11 = 2, a12 = –3, a13 = 5; A31 = –12, A32 = 22, A33 = 18 So a 11 A31 + a 12 A32 + a 13 A33 = 2 (–12) + (–3) (22) + 5 (18) = –24 – 66 + 90 = 0 EXERCISE 4.4 Write Minors and Cofactors of the elements of following determinants:

1. (i) 2 4 0 3 – (ii) a c b d
2. (i) 1 0 0 0 1 0 0 0 1 (ii) 1 0 4 3 5 1 0 1 2 –
3. Using Cofactors of elements of second row, evaluate  = 5 3 8 2 0 1 1 2 3 .
4. Using Cofactors of elements of third column, evaluate  = 1 1 1 x yz y zx z xy .
5. If  = 11 12 13 21 22 23 31 32 33 a a a a a a a a a and Aij is Cofactors of aij, then value of  is given by (A) a11 A31+ a12 A32 + a13 A33 (B) a11 A11+ a 12 A21 + a13 A31 (C) a21 A11+ a22 A12 + a23 A13 (D) a11 A11+ a 21 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. 2022-23 DETERMINANTS 1 27 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. Let 11 12 13 21 22 23 31 32 33 A = a a a a a a a a a         Then 11 12 13 21 22 23 31 32 33 A A A A =Transposeof A A A A A A adj         11 21 31 12 22 32 13 23 33 A A A = A A A A A A         Example 23 2 3 Find A for A = 1 4 adj       Solution We have A11 = 4, A12 = –1, A21 = –3, A22 = 2 Hence adj A = 11 21 12 22 A A 4 –3 = A A –1 2             Remark For a square matrix of order 2, given by A = 11 12 21 22 a a a a       The adj A can also be obtained by interchanging a11 and a22 and by changing signs of a12 and a21, i.e., 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 2022-23 128 MATHEMATICS Verification Let A = 11 12 13 21 22 23 31 32 33 a a a a a a a a a         , then adj A = 11 21 31 12 22 32 13 23 33 A A A A A A A A A         Since sum of product of elements of a row (or a column) with corresponding cofactors is equal to |A| and otherwise zero, we have A (adj A) = A 0 0 0 A 0 0 0 A         = A 1 0 0 0 1 0 0 0 1         = A I Similarly, we can show (adj A) A = A I Hence A (adj A) = (adj A) A = A I Definition 4 A square matrix A is said to be singular if A = 0. For example, the determinant of matrix A = 1 2 4 8       is zero Hence A is a singular matrix. Definition 5 A square matrix A is said to be non-singular if A  0 Let A = 1 2 3 4       . Then A = 1 2 3 4 = 4 – 6 = – 2  0. Hence A is a nonsingular matrix We state the following theorems without proof. Theorem 2 If A and B are nonsingular matrices of the same order, then AB and BA 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 Remark We know that (adj A) A = A I = A A A A 0 0 0 0 0 0 0         ,  2022-23 DETERMINANTS 1 29 Writing determinants of matrices on both sides, we have (adjA)A = A 0 0 0 A 0 0 0 A i.e. |(adj A)| |A| = 3 1 0 0 A 0 1 0 0 0 1 (Why?) 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. 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. 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) This gives A  0. Hence A is nonsingular. Conversely, let A be nonsingular. Then A  0 Now A (adj A) = (adj A) A = A I (Theorem 1) or A 1 1 A A A I | A | | A | adj adj      =   =     or AB = BA = I, where B = 1 A | A | adj Thus A is invertible and A–1 = 1 A | A | adj Example 24 If A = 1 3 3 1 4 3 1 3 4         , then verify that A adj A = |A| I. Also find A–1. Solution We have A = 1 (16 – 9) –3 (4 – 3) + 3 (3 – 4) = 1  0 2022-23 130 MATHEMATICS Now A11 = 7, A12 = –1, A13 = –1, A21 = –3, A22 = 1,A23 = 0, A31 = –3, A32 = 0, A33 = 1 Therefore adj A = 7 3 3 1 1 0 1 0 1  − −    −  −  Now A (adj A) = 1 3 3 7 3 3 1 4 3 1 1 0 1 3 4 1 0 1    − −        −    −  = 7 3 3 3 3 0 3 0 3 7 4 3 3 4 0 3 0 3 7 3 4 3 3 0 3 0 4  − − − + + − + +     − − − + + − + +   − − − + + − + +  = 1 0 0 0 1 0 0 0 1         = (1) 1 0 0 0 1 0 0 0 1         = A . I Also A–1 1 A A = a d j = 7 3 3 1 1 1 0 1 1 0 1  − −    −  −  = 7 3 3 1 1 0 1 0 1  − −    −  −  Example 25 If A = 2 3 1 2 and B 1 4 1 3    −    =    −  −  , then verify that (AB)–1 = B–1A–1. Solution We have AB = 2 3 1 2 1 5 1 4 1 3 5 14    −  −      =   −  −   −  Since, AB = –11  0, (AB)–1 exists and is given by (AB)–1 = 1 1 14 5 (AB) AB 11 5 1 adj − −  =−    − −  1 14 5 11 5 1   =     Further, A = –11  0 and B = 1  0. Therefore, A–1 and B–1 both exist and are given by A–1 = − − − −       =       − 1 11 4 3 1 2 3 2 1 1 1 ,B 2022-23 DETERMINANTS 1 31 Therefore B A − − =−       − − −       1 1 1 11 3 2 1 1 4 3 1 2 =− − − − −       1 11 14 5 5 1 1 14 5 11 5 1   =     Hence (AB)–1 = B–1 A–1 Example 26 Show that the matrix A = 2 3 1 2       satisfies the equation A2 – 4A + I = O, where I is 2 × 2 identity matrix and O is 2 × 2 zero matrix. Using this equation, find A–1. Solution We have 2 2 3 2 3 7 12 A A.A 1 2 1 2 4 7      = =     =       Hence 2 7 12 8 12 1 0 A 4A I 4 7 4 8 0 1       − + =   −   +         0 0 O 0 0   =   =   Now A2 – 4A + I = O Therefore A A – 4A = – I or 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 or AI – 4I = – A–1 or A–1 = 4I – A = 4 0 2 3 2 3 0 4 1 2 1 2      −    −   =       −  Hence 1 2 3 A 1 2 −  −  =  −  EXERCISE 4.5 Find adjoint of each of the matrices in Exercises 1 and 2. 1. 1 2 3 4       2. 1 1 2 2 3 5 2 0 1 − −         Verify A (adj A) = (adj A) A = |A| I in Exercises 3 and 4 3. 2 3 −4 −6       4. 1 1 2 3 0 2 1 0 3 − −           2022-23 132 MATHEMATICS Find the inverse of each of the matrices (if it exists) given in Exercises 5 to 11. 5. 2 2 4 3  −      6. − −       1 5 3 2 7. 1 2 3 0 2 4 0 0 5         8. 1 0 0 3 3 0 5 2 1 −         9. 2 1 3 4 1 0 7 2 1 − −         10. 1 1 2 0 2 3 3 2 4 − − −         11. 1 0 0 0 cos sin 0 sin cos           − 
6. Let A = 3 7 2 5       and B = 6 8 7 9       . Verify that (AB)–1 = B–1 A–1.
7. If A = 3 1 −1 2       , show that A2 – 5A + 7I = O. Hence find A–1.
8. For the matrix A = 3 2 1 1       , find the numbers a and b such that A2 + aA + bI = O.
9. For the matrix A = 1 1 1 1 2 3 2 1 3 − −           Show that A3– 6A2 + 5A + 11 I = O. Hence, find A–1.
10. If A = 2 1 1 1 2 1 1 1 2 − − − −           Verify that A3 – 6A2 + 9A – 4I = O and hence find A–1
11. 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|
12. If A is an invertible matrix of order 2, then det (A–1) is equal to (A) det (A) (B) 1 det (A) (C) 1 (D) 0 2022-23 DETERMINANTS 1 33 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. 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 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 inverse of the coefficient matrix. Consider the system of equations a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a 3 x + b 3 y + c 3 z = d 3 Let A = 1 1 1 1 2 2 2 2 3 3 3 3 , X and B a b c x d a b c y d a b c z d               =   =        Then, the system of equations can be written as, AX = B, i.e., 1 1 1 2 2 2 3 3 3 a b c x a b c y a b c z                 = 1 2 3 d d d         Case I If A is a nonsingular matrix, then its inverse exists. Now AX = B or A–1 (AX) = A–1 B (premultiplying by A–1) or (A–1A) X = A–1 B (by associative property) or I X = A–1 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. 2022-23 134 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. Example 27 Solve the system of equations 2x + 5y = 1 3x + 2y = 7 Solution The system of equations can be written in the form AX = B, where A = 2 5 1 ,X and B 3 2 7 x y         =   =         Now, A = –11  0, Hence, A is nonsingular matrix and so has a unique solution. Note that A–1 = − − −       1 11 2 5 3 2 Therefore X = A–1B = – 1 11 2 5 3 2 1 7 − −           i.e. x y       = − −      = −      1 11 33 11 3 1 Hence x = 3, y = – 1 Example 28 Solve the following system of equations by matrix method. 3x – 2y + 3z = 8 2x + y – z = 1 4x – 3y + 2z = 4 Solution The system of equations can be written in the form AX = B, where 3 2 3 8 A 2 1 1 , X and B 1 4 3 2 4 x y z  −            =  −  =   =    −      We see that A = 3 (2 – 3) + 2(4 + 4) + 3 (– 6 – 4) = – 17  0 2022-23 DETERMINANTS 1 35 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 Therefore A–1 = 1 5 1 1 8 6 9 17 10 1 7  − − −    −  − −  −  So X = –1 1 5 1 8 1 A B = 8 6 9 1 17 10 1 7 4  − − −        −  − −    −    i.e. x y z         = 17 1 1 34 2 17 51 3 −        − −  =   −    Hence x = 1, y = 2 and z = 3. Example 29 The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the 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. Then, according to given conditions, we have x + y + z = 6 y + 3z = 11 x + z = 2y or x – 2y + z = 0 This system can be written as A X = B, where A = 1 1 1 0 1 3 1  2 1           , X = x y z           and B = 6 11 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 2022-23 136 MATHEMATICS Hence adj A = 7 –3 2 3 0 –3 –1 3 1         Thus A –1 = 1 A adj (A) = 7 3 2 1 3 0 3 9 1 3 1 – – –         Since X = A–1 B X = 7 3 2 6 1 3 0 3 11 9 1 3 1 0 – – –                 or x y z         = 1 9 42 33 0 18 0 0 6 33 0  − +     + +  − + +  = 1 9 9 18 27         = 1 2 3         Thus x = 1, y = 2, z = 3 EXERCISE 4.6 Examine the consistency of the system of equations in Exercises 1 to 6.
13. x + 2y = 2 2. 2x – y = 5 3. x + 3y = 5 2x + 3y = 3 x + y = 4 2x + 6y = 8
14. 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 ax + ay + 2az = 4 3x – 5y = 3 5x – 2y + 6z = –1 Solve system of linear equations, using matrix method, in Exercises 7 to 14.
15. 5x + 2y = 4 8. 2x – y = –2 9. 4x – 3y = 3 7x + 3y = 5 3x + 4y = 3 3x – 5y = 7
16. 5x + 2y = 3 11. 2x + y + z = 1 12. x – y + z = 4 3x + 2y = 5 x – 2y – z = 3 2 2x + y – 3z = 0 3y – 5z = 9 x + y + z = 2
17. 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 2022-23 DETERMINANTS 1 37
18. If A = 2 –3 5 3 2 –4 1 1 –2         , find A–1. Using A–1 solve the system of equations 2x – 3y + 5z = 11 3x + 2y – 4z = – 5 x + y – 2z = – 3
19. The cost of 4 kg onion, 3 kg wheat and 2 kg rice is  60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is  90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is  70. Find cost of each item per kg by matrix method. Miscellaneous Examples Example 30 If a, b, c are positive and unequal, show that value of the determinant  = a b c b c a c a b is negative. Solution Applying C1  C1 + C2 + C3 to the given determinant, we get  = a b c b c a b c c a a b c a b

• +
• +
• + = (a + b + c) 1 1 1 b c c a a b = (a + b + c) 1 0 – – 0 – – b c c b a c a b b c (ApplyingR2R2–R1,andR3R3–R1) = (a + b + c) [(c – b) (b – c) – (a – c) (a – b)] (Expanding along C1) = (a + b + c)(– a2 – b2 – c2 + ab + bc + ca) = 1 2 – (a + b + c) (2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca) = 1 2 – (a + b + c) [(a – b)2 + (b – c)2
• (c – a)2] which is negative (since a + b + c > 0 and (a – b)2 + (b – c)2
• (c – a)2

0) 2022-23 138 MATHEMATICS Example 31 If a, b, c, are in A.P, find value of 2 4 5 7 8 3 5 6 8 9 4 6 7 9 10 y y y a y y y b y y y c

• • +
• • +
• • + Solution Applying R1  R1 + R3 – 2R2 to the given determinant, we obtain 0 0 0 3 5 6 8 9 4 6 7 9 10 y y y b y y y c
• • +
• • + = 0 (Since 2b = a + c) Example 32 Show that  = ( ) ( ) ( ) 2 2 2 y z xy zx xy x z yz xz yz x y + + + = 2xyz (x + y + z)3 Solution Applying R1  xR1, R2  yR2 , R3  zR3 to  and dividing by xyz, we get  = ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 1 + + + x y z x y x z xy y x z y z xyz xz yz z x y Taking common factors x, y, z from C1 C2 and C3, respectively, we get  = ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 y z x x xyz y x z y xyz z z x y + + + Applying C2  C2– C1, C3  C3– C1, we have  = ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 – 0 0 – y z x y z x y z y x z y z x y z
• • − +
• − + 2022-23 DETERMINANTS 1 39 Taking common factor (x + y + z) from C2 and C3, we have  = (x + y + z)2 ( ) ( ) ( ) ( ) ( ) 2 2 2 y 0 0 z x – y z x – y z y x z – y z x y – z
• • + + + Applying R1  R1 – (R2 + R3), we have  = (x + y + z)2 2 2 2 2 2
• 0 0 –z yz – z – y y x y z z x y − + Applying C2  (C2 + 1 y C1) and 3 3 1 1 C C C  + z , we get  = (x + y + z)2 2 2 2 2 2 0 0 + + yz y y x z z z z x y y Finally expanding along R1, we have  = (x + y + z)2 (2yz) [(x + z) (x + y) – yz] = (x + y + z)2 (2yz) (x2 + xy + xz) = (x + y + z)3 (2xyz) Example 33 Use product 1 1 2 0 2 3 3 2 4 2 0 1 9 2 3 6 1 2                       to solve the system of equations x – y + 2z = 1 2y – 3z = 1 3x – 2y + 4z = 2 Solution Consider the product 1 1 2 2 0 1 0 2 3 9 2 3 3 2 4 6 1 2 – – – – – –                 2022-23 140 MATHEMATICS = 2 9 12 0 2 2 1 3 4 0 18 18 0 4 3 0 6 6 6 18 24 0 4 4 3 6 8 − − + − + + −   + − + − − +    − − + − + + −  = 1 0 0 0 1 0 0 0 1           Hence 1 1 2 0 2 3 3 2 4 2 0 1 9 2 3 6 1 2 1 – – – – – – –           =           Now, given system of equations can be written, in matrix form, as follows 1 –1 2 0 2 –3 3 –2 4 x y z                 = 1 1 2         or x y z = 1 1 1 2 1 0 2 3 1 3 2 4 2 −  −     −         −    =    2 0 1 9 2 3 6 1 2 1 1 2                     = 2 0 2 0 9 2 6 5 6 1 4 3 − + +       
• − =      + −    Hence x = 0, y = 5 and z = 3 Example 34 Prove that  = 2 (1 ) a bx c dx p qx a c p ax b cx d px q x b d q u v w u v w
• • +
• • • = − Solution Applying R1  R1 – x R2 to , we get  = 2 2 2 a(1 x ) c(1 x ) p(1 x ) ax b cx d px q u v w − − −
• • + = 2 (1 ) a c p x ax b cx d px q u v w − + + + 2022-23 DETERMINANTS 1 41 Applying R2  R2 – x R1, we get  = 2 (1 ) a c p x b d q u v w − Miscellaneous Exercises on Chapter 4

1. Prove that the determinant sin cos –sin – 1 cos 1 x x x     is independent of .
2. Without expanding the determinant, prove that 2 2 3 2 2 3 2 2 3 1 1 1 = a a bc a a b b ca b b c c ab c c .
3. Evaluate cos cos cos sin – sin – sin cos 0 sin cos sin sin cos             .
4. If a, b and c are real numbers, and  = b c c a a b c a a b b c a b b c c a

• • +
• • +
• • + = 0, Show that either a + b + c = 0 or a = b = c.

5. Solve the equation 0 x a x x x x a x x x x a +

• = + , a  0

6. Prove that 2 2 2 2 2 2 + + + a bc ac c a ab b ac ab b bc c = 4a2b2c2
7. If A–1 = ( )1 3 1 1 1 2 2 15 6 5 and B 1 3 0 , find AB 5 2 2 0 2 1 – – – – – – – –           =       2022-23 142 MATHEMATICS
8. Let A = 1 2 1 2 3 1 1 1 5             . Verify that (i) [adj A]–1 = adj (A–1) (ii) (A–1)–1 = A
9. Evaluate x y x y y x y x x y x y + + +
10. Evaluate 1 1 1 x y x y y x x+ y + Using properties of determinants in Exercises 11 to 15, prove that: 11. 2 2 2   +     +    + = ( – ) ( – ) ( – ) ( +  + ) 12. 2 3 2 3 2 3 1 1 1 + + + x x px y y py z z pz = (1 + pxyz) (x – y) (y – z) (z – x), where p is any scalar. 13. 3 3 a 3c a –a+b –a+c –b a b –b c –c –c+b

• + + = 3(a + b + c) (ab + bc + ca) 14. 1 1 1 2 3 2 4 3 2 3 6 3 10 6 3 p p q p p q p p q
• • +
• • +
• • + = 1 15. ( ) ( ) ( ) sin cos cos sin cos cos 0 sin cos cos   +    +  =    +

16. Solve the system of equations 2 3 10

• • = 4 x y z 2022-23 DETERMINANTS 1 43 4 6 5 – + =1 x y z 6 9 20
• – = 2 x y z Choose the correct answer in Exercise 17 to 19.

17. If a, b, c, are in A.P, then the determinant 2 3 2 3 4 2 4 5 2 x x x a x x x b x x x c

• • +
• • +
• • + is (A) 0 (B) 1 (C) x (D) 2x

18. If x, y, z are nonzero real numbers, then the inverse of matrix 0 0 A 0 0 0 0 x y z     =    is (A) 1 1 1 0 0 0 0 0 0 x y z − − −           (B) 1 1 1 0 0 0 0 0 0 x xyz y z − − −           (C) 0 0 1 0 0 0 0 x y xyz z         (D) 1 0 0 1 0 1 0 0 0 1 xyz        
19. Let A = 1 sin 1 sin 1 sin 1 sin 1      −    − −   , where 0  2. Then (A) Det (A) = 0 (B) Det (A)  (2, ) (C) Det (A)  (2, 4) (D) Det (A)  [2, 4] 2022-23 144 MATHEMATICS Summary  Determinant of a matrix A = [a11]1×1 is given by |a11| = a11  Determinant of a matrix A =       a a a a 11 12 21 22 is given by 11 12 21 22 A a a a a = = a 11 a 22 – a 12 a 21  Determinant of a matrix A =         a b c a b c a b c 1 1 1 2 2 2 3 3 3 is given by (expanding along R1) 1 1 1 2 2 2 2 2 2 2 2 2 1 1 1 3 3 3 3 3 3 3 3 3 A a b c b c a c a b a b c a b c b c a c a b a b c = = − + For any square matrix A, the |A| satisfy following properties.  |A| = |A|, where A = transpose of A.  If we interchange any two rows (or columns), then sign of determinant changes.  If any two rows or any two columns are identical or proportional, then value of determinant is zero.  If we multiply each element of a row or a column of a determinant by constant k, then value of determinant is multiplied by k.  Multiplying a determinant by k means multiply elements of only one row (or one column) by k.  If 3 3 3 A , then .A A ij a k k × = =  If elements of a row or a column in a determinant can be expressed as sum of two or more elements, then the given determinant can be expressed as sum of two or more determinants.  If to each element of a row or a column of a determinant the equimultiples of corresponding elements of other rows or columns are added, then value of determinant remains same. 2022-23 DETERMINANTS 1 45  Area of a triangle with vertices (x 1, y 1), (x 2, y

2. and (x 3, y
3. is given by 1 1 2 2 3 3 1 1 1 2 1 x y x y x y =  Minor of an element aij of the determinant of matrix A is the determinant obtained by deleting ith row and jth column and denoted by Mij.  Cofactor of aij of given by Aij = (– 1)i+ j Mij  Value of determinant of a matrix A is obtained by sum of product of elements of a row (or a column) with corresponding cofactors. For example, A = a11 A11 + a12 A12 + a13 A13.  If elements of one row (or column) are multiplied with cofactors of elements of any other row (or column), then their sum is zero. For example, a11 A21 + a12 A22 + a 13 A23 = 0  If 11 12 13 21 22 23 31 32 33 A , a a a a a a a a a     =     then 11 21 31 12 22 32 13 23 33 A A A A A A A A A A adj     =     , where Aij is cofactor of a ij  A (adj A) = (adj A) A = |A| I, where A is square matrix of order n.  A square matrix A is said to be singular or non-singular according as |A| = 0 or |A|  0.  If AB = BA = I, where B is square matrix, then B is called inverse of A. Also A–1 = B or B–1 = A and hence (A–1)–1 = A.  A square matrix A has inverse if and only if A is non-singular.  –1 1 A ( A) A = adj  If a1 x + b1 y + c1 z = d1 a2 x + b2 y + c2 z = d2 a3 x + b3 y + c3 z = d3, then these equations can be written as A X = B, where 1 1 1 1 2 2 2 2 3 3 3 3 A ,X= and B= a b c x d a b c y d a b c z d             =            2022-23 146 MATHEMATICS  Unique solution of equation AX = B is given by X = A–1 B, where A  0 .  A system of equation is consistent or inconsistent according as its solution exists or not.  For a square matrix A in matrix equation AX = B (i) |A|  0, there exists unique solution (ii) |A| = 0 and (adj A) B  0, then there exists no solution (iii) |A| = 0 and (adj A) B = 0, then system may or may not be consistent. Historical Note The Chinese method of representing the coefficients of the unknowns of several linear equations by using rods on a calculating board naturally led to the discovery of simple method of elimination. The arrangement of rods was precisely that of the numbers in a determinant. The Chinese, therefore, early developed the idea of subtracting columns and rows as in simplification of a determinant Mikami, China, pp 30, 93. Seki Kowa, the greatest of the Japanese Mathematicians of seventeenth century in his work ‘Kai Fukudai no Ho’ in 1683 showed that he had the idea of determinants and of their expansion. But he used this device only in eliminating a quantity from two equations and not directly in the solution of a set of simultaneous linear equations. T. Hayashi, “The Fakudoi and Determinants in Japanese Mathematics,” in the proc. of the Tokyo Math. Soc., V. Vendermonde was the first to recognise determinants as independent functions. He may be called the formal founder. Laplace (1772), gave general method of expanding a determinant in terms of its complementary minors. In 1773 Lagrange treated determinants of the second and third orders and used them for purpose other than the solution of equations. In 1801, Gauss used determinants in his theory of numbers. The next great contributor was Jacques - Philippe - Marie Binet, (1812) who stated the theorem relating to the product of two matrices of m-columns and nrows, which for the special case of m = n reduces to the multiplication theorem. Also on the same day, Cauchy (1812) presented one on the same subject. He used the word ‘determinant’ in its present sense. He gave the proof of multiplication theorem more satisfactory than Binet’s. The greatest contributor to the theory was Carl Gustav Jacob Jacobi, after this the word determinant received its final acceptance.