Prepared b: M. S. KumarSwam, TGT(Maths) Page - 77 -
CHAPTER 4: DETERMINANTS QUICK REVISION (Important Concepts & Formulae) Determinant a b If A = c d, then determinant of A is written as A = a b = det (A) or Δ c d (i) For matri A, A is read as determinant of A and not modulus of A. (ii) Onl square matrices have determinants. MARKS WEIGHTAGE 0 marks Determinant of a matri of order one Let A = [a ] be the matri of order, then determinant of A is defined to be equal to a Determinant of a matri of order two a a Let A = a a be a matri of order, then the determinant of A is defined as: a a det (A) = A = Δ = = aa aa a a Determinant of a matri of order 3 3 Determinant of a matri of order three can be determined b epressing it in terms of second order determinants. This is known as epansion of a determinant along a row (or a column). There are si was of epanding a determinant of order 3 corresponding to each of three rows (R, R and R 3 ) and three columns (C, C and C 3 ) giving the same value as shown below. Consider the determinant of square matri A = [a ij ] 3 3 a a a i.e., A = 3 a a a 3 a a a 3 3 33 Epansion along first Row (R ) Step Multipl first element a of R b ( ) ( + ) [( ) sum of suffies in a ] and with the second order determinant obtained b deleting the elements of first row (R ) and first column (C ) of A as a lies in R and C, a a 3 i.e., ( ) a a a 3 33 Step Multipl nd element a of R b ( ) + [( ) sum of suffies in a ] and the second order determinant obtained b deleting elements of first row (R ) and nd column (C ) of A as a lies in R and C, a a 3 i.e., ( ) a a a 3 33 Step 3 Multipl third element a3 of R b ( ) + 3 [( ) sum of suffies in a3 ] and the second order determinant obtained b deleting elements of first row (R ) and third column (C 3 ) of A as a 3 lies in R and C 3, Prepared b: M. S. KumarSwam, TGT(Maths) Page - 78 -
i.e., ( ) a a 3 a3 a 3 a 3 Step 4 Now the epansion of determinant of A, that is, A written as sum of all three terms obtained in steps, and 3 above is given b a a3 a a3 a 3 a A ( ) a ( ) a ( ) a3 a a a a a a 3 33 3 33 3 3 or A = a (a a 33 a 3 a 3 ) a (a a 33 a 3 a 3 ) + a 3 (a a 3 a 3 a ) Epansion along second row (R ) A = a a a 3 a a a 3 a a a 3 3 33 Epanding along R, we get A ( ) a a a ( ) a a a ( ) a a a Epansion along first Column (C ) 3 3 3 3 a3 a33 a3 a33 a3 a3 A = a a a 3 a a a 3 a a a 3 3 33 B epanding along C, we get a a a a a a 3 A ( ) a ( ) a ( ) a3 a a a a a a 3 3 3 3 33 3 33 3 For easier calculations, we shall epand the determinant along that row or column which contains maimum number of zeros. While epanding, instead of multipling b ( ) i + j, we can multipl b + or according as (i + j) is even or odd. If A = kb where A and B are square matrices of order n, then A = k n B, where n =,, 3 Properties of Determinants Propert The value of the determinant remains unchanged if its rows and columns are interchanged. if A is a square matri, then det (A) = det (A ), where A = transpose of A. If R i = i th row and C i = i th column, then for interchange of row and columns, we will smbolicall write C i R i Propert If an two rows (or columns) of a determinant are interchanged, then sign of determinant changes. We can denote the interchange of rows b R i R j and interchange of columns b C i C j. Prepared b: M. S. KumarSwam, TGT(Maths) Page - 79 -
Propert 3 If an two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero. Propert 4 If each element of a row (or a column) of a determinant is multiplied b a constant k, then its value gets multiplied b k. o B this propert, we can take out an common factor from an one row or an one column of a given determinant. o If corresponding elements of an two rows (or columns) of a determinant are proportional (in the same ratio), then its value is zero. Propert 5 If some or all elements of a row or column of a determinant are epressed as sum of two (or more) terms, then the determinant can be epressed as sum of two (or more) determinants. Propert 6 If, to each element of an 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 appl the operation R i R i + kr j or C i C i + k C j. If Δ is the determinant obtained b appling R i kr i or C i kc i to the determinant Δ, then Δ = kδ. If more than one operation like R i R i + kr j 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. Area of triangle Area of a triangle whose vertices are (, ), (, ) and ( 3, 3 ), is given b the epression () 3 3 Since area is a positive quantit, we alwas take the absolute value of the determinant in (). If area is given, use both positive and negative values of the determinant for calculation. The area of the triangle formed b three collinear points is zero. Minors and Cofactors Minor of an element a ij of a determinant is the determinant obtained b deleting its i th row and j th column in which element a ij lies. Minor of an element a= is denoted b M ij. Minor of an element of a determinant of order n(n ) is a determinant of order n. Cofactor of an element a ij, denoted b A ij is defined b Aij = ( ) i + j M ij, where M ij is minor of a ij. If elements of a row (or column) are multiplied with cofactors of an other row (or column), then their sum is zero. Adjoint and Inverse of a Matri The adjoint of a square matri A = [a ij ] n n is defined as the transpose of the matri [A ij ] n n, where A ij is the cofactor of the element a ij. Adjoint of the matri A is denoted b adj A. Prepared b: M. S. KumarSwam, TGT(Maths) Page - 80 -
a a For a square matri of order, given b a a The adj A can also be obtained b interchanging a and a and b changing signs of a and a, i.e., Theorem If A be an given square matri of order n, then A(adj A) = (adj A) A = A I, where I is the identit matri of order n A square matri A is said to be singular if A = 0. A square matri A is said to be non-singular if A 0 Theorem 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 If A is a square matri of order n, then adj(a) = A n. Theorem 4 A square matri A is invertible if and onl if A is nonsingular matri. Let A is a non-singular matri then we have A 0 A is invertible and A adja A Applications of Determinants and Matrices Application of determinants and matrices for solving the sstem of linear equations in two or three variables and for checking the consistenc of the sstem of linear equations: Consistent sstem A sstem of equations is said to be consistent if its solution (one or more) eists. Inconsistent sstem A sstem of equations is said to be inconsistent if its solution does not eist. Solution of sstem of linear equations using inverse of a matri Consider the sstem of equations a + b + c z = d a + b + c z = d a 3 + b 3 + c 3 z = d 3 a b c d Let A a b c, X and B d a3 b3 c 3 z d 3 Then, the sstem of equations can be written as, AX = B, i.e., Prepared b: M. S. KumarSwam, TGT(Maths) Page - 8 -
a b c d a b c d a3 b3 c 3 z d 3 Case I If A is a nonsingular matri, then its inverse eists. Now AX = B or A (AX) = A B (premultipling b A ) or (A A) X = A B (b associative propert) or I X = A B or X = A B This matri equation provides unique solution for the given sstem of equations as inverse of a matri is unique. This method of solving sstem of equations is known as Matri Method. Case II If A is a singular matri, then A = 0. In this case, we calculate (adj A) B. If (adj A) B O, (O being zero matri), then solution does not eist and the sstem of equations is called inconsistent. If (adj A) B = O, then sstem ma be either consistent or inconsistent according as the sstem have either infinitel man solutions or no solution. Prepared b: M. S. KumarSwam, TGT(Maths) Page - 8 -
CHAPTER 4: DETERMINANTS MARKS WEIGHTAGE 0 marks NCERT Important Questions & Answers 6. If, then find the value of. 8 8 6 6 Given that 8 8 6 On epanding both determinants, we get 8 = 6 6 8 36 = 36 36 36 = 0 = 36 = ± 6. Prove that a a b a b c a 3a b 4a 3b c a 3a 6a 3b 0a 6b 3c Prepared b: M. S. KumarSwam, TGT(Maths) Page - 83-3 Appling operations R R R and R 3 R 3 3R to the given determinant Δ, we have a a b a b c 0 a a b 0 3a 7a 3b Now appling R 3 R 3 3R, we get a a b a b c 0 a a b 0 0 a Epanding along C, we obtain a a b a 0 0 a( a 0) a( a ) a 0 a b c a a 3 3. Prove that b c a b 4abc c c a b b c a a Let b c a b c c a b Appling R R R R 3 to Δ, we get 0 c b b c a b c c a b Epanding along R, we obtain 0 c a b ( c) b b ( b) b c a c a b c a b c c
= c (a b + b bc) b (b c c ac) = abc + cb bc b c + bc + abc = 4 abc 4. If,, z are different and We have 3 3 z z z 3 3 3 z z z 0 then show that + z = 0 3 Now, we know that If some or all elements of a row or column of a determinant are epressed as sum of two (or more) terms, then the determinant can be epressed as sum of two (or more) determinants. 3 3 z z z z z 3 ( ) z (Using C 3 C and then C C ) z z z z ( z) z ( z) 0 0 z z z (Using R R R and R 3 R 3 R ) Taking out common factor ( ) from R and (z ) from R 3, we get ( z)( )( z ) 0 0 z = ( + z) ( ) (z ) (z ) (on epanding along C ) Since Δ = 0 and,, z are all different, i.e., 0, z 0, z 0, we get + z = 0 a 5. Show that b abc abc bc ca ab a b c c LHS a b c Taking out factors a,b,c common from R, R and R 3, we get Prepared b: M. S. KumarSwam, TGT(Maths) Page - 84 -
abc a a a b b b c c c Appling R R + R + R 3, we have a b c a b c a b c abc b b b c c c Now appling C C C, C 3 C 3 C, we get 0 0 abc a b c b 0 c 0 abc ( 0) a b c abc abc bc ca ab = RHS a b c 6. Using the propert of determinants and without epanding, prove that b c q r z a p c a r p z b q a b p q c r z b c q r z LHS c a r p z a b p q b c c a a b q r r p p q (interchange row and column) z z b c c a c q r r p r [usingc 3 C 3 (C + C )] z z z b c c a c ( ) q r r p r (taking common from C 3 ) z z z Prepared b: M. S. KumarSwam, TGT(Maths) Page - 85 -
b a c ( ) q p r (using C C C 3 and C C C 3 ) z a b c p q r (using z C C ) a p b q RHS (interchange row and column) c r z 7. Using the propert of determinants and without epanding, prove that a ab ac ba b bc 4a b c ca cb c a ab ac a b c LHS ba b bc abc a b c R 3 ] ca cb c a b c ( abc)( abc) a b c 0 0 0 Epanding corresponding to first row R, we get 0 a b c [taking out factors a from R, b from R and c from (taking out factors a from C, b from C and c from C 3 ) (using R R + R and R R R 3 ) a b c (0 ) 4a b c RHS 8. Using the propert of determinants and without epanding, prove that a a b b ( a b)( b c)( c a) c c LHS b b a c a c Appling R R R3 and R R R3, we get 0 a c a c 0 a c ( a c)( a c) 0 b c b c 0 b c ( b c)( b c) c c c c Taking common factors (a c) and (b c) from R and R respectivel, we get Prepared b: M. S. KumarSwam, TGT(Maths) Page - 86 -
0 ( a c) ( a c)( b c) 0 ( b c) c c Now, epanding corresponding to C, we get = (a c) (b c) (b + c a c) = (a b) (b c) (c a) = RHS 9. Using the propert of determinants and without epanding, prove that a b c ( a b)( b c)( c a)( a b c) a b c 3 3 3 LHS a b c a b c 3 3 3 Appling C C C and C C C3, we get 0 0 a b b c c a b b c c 3 3 3 3 3 0 0 a b b c c ( a b)( a ab b ) ( b c)( b bc c ) c 3 Taking common (a b) from C and (b c) from C, we get 0 0 ( a b)( b c) a b b c c ( a ab b ) ( b bc c ) c 3 Now, epanding along R, we get = (a b) (b c) [ (b + bc + c ) (a + ab + b )] = (a b) (b c) [b + bc + c a ab b ] = (a b) (b c) (bc ab + c a ) = (a b) (b c) [b(c a) + (c a) (c + a)] = (a b) (b c) (c a) (a + b + c)= RHS. 0. Using the propert of determinants and without epanding, prove that z z z z z z z z z LHS z z z ( )( )( )( ) Appling R R, R R and R3 zr3, we have 3 z 3 z z z z 3 z Prepared b: M. S. KumarSwam, TGT(Maths) Page - 87 -
z z z 3 3 (take out z common from C 3 ) z 3 3 3 3 z z 3 3 Epanding corresponding to C 3, we get 3 3 3 3 z z 0 0 (using R R R and R3 R3 R ) 3 3 3 3 ( )( z ) ( z )( ) = ( + ) ( ) (z ) (z + + z) (z + ) (z ) ( ) ( + + ) = ( ) (z ) [( + ) (z + + z) (z + ) ( + + )] = ( )(z )[z + + z + z + 3 + z z z z 3 ] = ( )(z )[z z + z ] = ( )(z )[z(z ) + (z )] = ( )(z )[z(z ) + (z )(z + )] = ( ) (z ) [(z ) ( + z + z)] = ( ) ( z) (z ) ( + z + z) = RHS.. Using the propert of determinants and without epanding, prove that 4 4 (5 4)(4 ) 4 4 LHS 4 4 5 4 5 4 4 5 4 4 (5 4) 4 4 (5 4) 0 4 0 0 0 4 Epanding along C, we get = (5 + 4) {(4 ) (4 )} (5 4)(4 ) = RHS. (using C C + C + C 3 ) [take out (5 + 4) common from C ]. (Using R R R and R 3 R 3 R ). Using the propert of determinants and without epanding, prove that k k k (3 k) k Prepared b: M. S. KumarSwam, TGT(Maths) Page - 88 -
k LHS k k 3 k 3 k k 3 k k (3 k) k k (using C C + C + C 3 ) [take out (5 + 4) common from C ]. (3 k) 0 k 0 (Using R R R and R 3 R 3 R ) 0 0 Epanding along C 3, we get (3 k) ( k 0) k (3 k) = RHS k 3. Using the propert of determinants and without epanding, prove that a b c a a b b c a b ( a b c) c c c a b a b c a a LHS b b c a b c c c a b a b c a b c a b c b b c a b (Using R R R R ) c c c a b Take out (a + b + c) common from R, we get ( a b c) b b c a b c c c a b 0 0 ( a b c) b b c a 0 c 0 c a b Epanding along R, we get = (a + b + c) {( b c a) ( c a b)} = (a + b + c) [ (b + c + a) ( ) (c + a + b)] 3 ( a b c)( a b c)( a b c) ( a b c) = RHS 3 3 (Using C C C and C 3 C 3 C ) Prepared b: M. S. KumarSwam, TGT(Maths) Page - 89 -
4. Using the propert of determinants and without epanding, prove that z z z ( z) z z z LHS z z z z ( z) ( z) z (using C C + C + C 3 ) ( z) z ( z) z [take out ( + + z) common from C ]. z ( z) 0 z 0 (Using R R R and R 3 R 3 R ) 0 0 z ( z)( z)( z) 0 0 0 0 Epanding along R 3, we get ( z)( z)( z) ( 0) 3 ( z)( z)( z) ( z) =RHS 3 5. Using the propert of determinants and without epanding, prove that LHS (using C C + C + C 3 ) ( ) [take out ( ) 0 0 3 ( ) common from C ]. (Using R R R and R 3 R 3 R ) Prepared b: M. S. KumarSwam, TGT(Maths) Page - 90 -
( ) 0 ( ) 0 ( ) Take out ( ) common from R and same from R 3, we get ( )( )( ) 0 Epanding along C, we get ( )( )( ) 0 ( )( )( )( ) 3 3 3 = RHS 6. Using the propert of determinants and without epanding, prove that a b ab b ab a b a ( a b ) 3 b a a b a b ab b LHS ab a b a b a a b 0 a b b 0 a b a b( a b ) a( a b ) a b 0 b ( a b ) 0 a b a a b 0 b ( a b ) 0 a 0 0 a b Epanding along R, we get ( a b ) ( a b ) 3 ( a b ) RHS ( R R br ar ) 3 3 (Using C C bc3 and C C ac3 ) 7. Using the propert of determinants and without epanding, prove that a ab ac ab b bc a b c ca cb c a ab ac LHS ab b bc ca cb c Prepared b: M. S. KumarSwam, TGT(Maths) Page - 9 -
Taking out common factors a, b and c from R, R and R 3 respectivel, we get a a b c a b b c a b c c a a b c a b 0 (Using R R R and R 3 R 3 R ) 0 a c Multipl and divide C b a, C b b and C 3 b c and then take common out from C, C and C 3 respectivel, we get a b c a b c abc 0 0 abc 0 Epanding along R 3, we get ( c ) ( a ) ( b ) a b c RHS 0 8. Find values of k if area of triangle is 4 sq. units and vertices are (i) (k, 0), (4, 0), (0, ) (ii) (, 0), (0, 4), (0, k) k 0 (i) We have Area of triangle = 4 0 4 0 k(0 ) + (8 0) = 8 k(0 ) + (8 0) = ± 8 On taking positive sign k + 8 = 8 k = 0 k = 0 On taking negative sign k + 8 = 8 k = 6 k = 8 k =0, 8 0 (ii) We have Area of triangle = 0 4 4 0 k (4 k) + (0 0) = 8 (4 k) + (0 0) = ± 8 [ 8 + k] = ± 8 On taking positive sign, k 8 = 8 k = 6 k = 8 On taking negative sign, k 8 = 8 k = 0 k = 0 k =0, 8 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 9 -
9. If area of triangle is 35 sq units with vertices (, 6), (5, 4) and (k, 4). Then find the value of k. 6 We have Area of triangle = 5 4 35 k 4 (4 4) + 6(5 k) + (0 4k) = 70 (4 4) + 6 (5 k) + (0 4k) = ± 70 30 6k + 0 4k = ± 70 On taking positive sign, 0k + 50 = 70 0k = 0 k = On taking negative sign, 0k + 50 = 70 0k = 0 k = k =, 0. Using Cofactors of elements of second row, evaluate Given that 5 3 8 0 3 Cofactors of the elements of second row 3 8 A ( ) (9 6) 7 3 A 5 8 3 ( ) (5 8) 7 5 3 8 0 3 3 5 3 and A3 ( ) (0 3) 7 Now, epansion of Δ using cofactors of elements of second row is given b aa a A a3 A3 = 7 + 0 7 + ( 7) = 4 7 = 7 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 93-3. If A =, show that A 5A + 7I = O. Hence find A. 3 Given that A = Now, A 5A + 7I = O 3 3 9 3 8 5 A A. A 3 4 5 3 8 5 3 0 5 7 5 3 0 8 5 5 5 7 0 5 3 5 0 0 7 8 5 7 5 5 0 0 0 O 5 5 0 30 3 0 0
A 5A + 7I = O 3 A 6 7 0 A eists. Now, A.A 5A = 7I Multipling b A on both sides, we get A.A (A ) 5A(A ) = 7I(A ) AI 5I = 7A (using AA = I and IA = A ) A ( A 5 I) 5I A 5 0 3 7 7 7 0 5 7 3 A 7 3 3. For the matri A =, find the numbers a and b such that A + aa + bi = O. 3 Given that A = 3 3 9 6 8 A A. A 3 4 3 Now, A aa bi O 8 3 0 a b O 4 3 0 8 3a a b 0 O 4 3 a a 0 b 3a b 8 a 0 0 4 a 3 a b 0 0 If two matrices are equal, then their corresponding elements are equal. + 3a + b = 0 (i) 8 + a = 0 (ii) 4 + a = 0 (iii) and 3 + a + b = 0 (iv) Solving Eqs. (iii) and (iv), we get 4 + a = 0 a = 4 and 3 + a + b = 0 3 4 + b = 0 b = Thus, a = 4 and b = 3. For the matri A = 3, Show that A 3 6A + 5A + I = O. Hence, find A. 3 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 94 -
Given that A = 3 3 3 3 4 A A. A 3 3 6 4 3 6 9 3 8 4 3 3 6 3 3 9 7 3 4 4 4 4 4 4 6 3 3 and A A. A 3 8 4 3 3 8 8 3 6 4 3 4 4 7 3 4 3 7 3 8 7 6 4 7 9 4 8 7 3 7 69 3 3 58 3 A 6A 5A I 8 7 4 0 0 3 7 69 6 3 8 4 5 3 0 0 3 3 58 7 3 4 3 0 0 8 7 4 6 5 5 5 0 0 3 7 69 8 48 84 5 0 5 0 0 3 3 58 4 8 84 0 5 5 0 0 8 4 5 7 5 0 6 5 0 3 8 5 0 7 48 0 69 84 5 0 3 4 0 0 38 5 0 58 84 5 0 0 0 0 0 0 O 0 0 0 A 3 (6 3) (3 6) ( 4) 3 9 5 0 3 A eist 3 Now, A 6A 5A I O AA( AA ) 6 A( AA ) 5( AA ) ( IA ) O AAI 6AI 5I A O A 6A 5I A ( A A 6 A 5 I ) ( A A 6 A 5 I ) 4 0 0 A 3 8 4 6 3 5 0 0 7 3 4 3 0 0 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 95 -
4 6 6 6 5 0 0 A 3 8 4 6 8 0 5 0 7 3 4 6 8 0 0 5 4 6 5 6 0 6 0 A 3 6 0 8 5 4 8 0 7 0 3 6 0 4 8 5 3 4 5 A 9 4 5 3 4. Solve sstem of linear equations, using matri method, + + z = z = 3 3 5z = 9 The given sstem can be written as AX = B, where A 4, X and B 3 0 3 5 z 9 A 4 (0 6) ( 0 0) (6 0) 0 3 5 = 5 + 0 + 6 = 68 0 Thus, A is non-singular, Therefore, its inverse eists. Therefore, the given sstem is consistent and has a unique solution given b X = A B. Cofactors of A are A = 0 + 6 = 6, A = ( 0 + 0) = 0, A 3 = 6 + 0 = 6 A = ( 5 3) = 8, A = 0 0 = 0, A 3 = (6 0) = 6 A 3 = ( + 4) =, A 3 = ( 4 ) = 6, A 33 = 8 = 0 6 0 6 6 8 adj( A) 8 0 6 0 0 6 6 0 6 6 0 6 8 A ( adja) 0 0 6 A 68 6 6 0 6 8 Now, X A B 0 0 6 3 68 z 6 6 0 9 T Prepared b: M. S. KumarSwam, TGT(Maths) Page - 96 -
6 4 8 68 0 30 54 34 68 68 z 6 8 90 0 3 3 Hence,, and z 5. Solve sstem of linear equations, using matri method, + z = 4 + 3z = 0 + + z = The given sstem can be written as AX = B, where 4 A 3, X and B 0 z Here, A 3 = ( + 3) ( ) ( + 3) + ( ) = 4 + 5 + = 0 0 Thus, A is non-singular, Therefore, its inverse eists. Therefore, the given sstem is consistent and has a unique solution given b X = A B. Cofactors of A are A = + 3 = 4, A = ( + 3) = 5, A 3 = =, A = ( ) =, A = = 0, A 3 = ( + ) =, A 3 = 3 =, A 3 = ( 3 ) = 5, A 33 = + = 3 T 4 5 4 adj( A) 0 5 0 5 5 3 3 4 A ( adja) 5 0 5 A 0 3 4 4 Now, X A B 5 0 5 0 0 z 3 6 0 4 0 0 0 0 0 0 0 z 4 0 6 0 Hence, =, = and z =. Prepared b: M. S. KumarSwam, TGT(Maths) Page - 97 -
6. Solve sstem of linear equations, using matri method, + 3 +3 z = 5 + z = 4 3 z = 3 The given sstem can be written as AX = B, where 3 3 5 A, X and B 4 3 z 3 3 3 Here, A 3 = (4 + ) 3 ( 3) + 3 ( + 6) = 0 + 5 + 5 = 40 0 Thus, A is non-singular. Therefore, its inverse eists. Therefore, the given sstem is consistent and has a unique solution given b X = A B Cofactors of A are A = 4 + = 5, A = ( 3) = 5, A 3 = ( + 6) = 5, A = ( 6 + 3) = 3, A = ( 4 9) = 3, A 3 = ( 9) =, A 3 = 3 + 6 = 9, A 3 = ( 3) =, A 33 = 4 3 = 7 T 5 5 5 5 3 9 adj( A) 3 3 5 3 9 7 5 7 5 3 9 A ( adja) 5 3 A 40 5 7 5 3 9 5 Now, X A B 5 3 4 40 z 5 7 3 5 7 40 5 5 3 80 40 40 z 5 44 40 Hence, =, = and z =. 7. Solve sstem of linear equations, using matri method, + z = 7 3 + 4 5z = 5 + 3z = The given sstem can be written as AX = B, where Prepared b: M. S. KumarSwam, TGT(Maths) Page - 98 -
7 A 3 4 5, X and B 5 3 z Here, A 3 4 5 = ( 5) ( ) (9 + 0) + ( 3 8) 3 = 7 + 9 = 4 0 Thus, A is non-singular. Therefore, its inverse eists. Therefore, the given sstem is consistent and has a unique solution given b X = A B Cofactors of A are A = 5 = 7, A = (9 + 0) = 9, A 3 = 3 8 =, A = ( 3 + ) =, A = 3 4 =, A 3 = ( + ) =, A 3 = 5 8 = 3, A 3 = ( 5 6) =, A 33 = 4 + 3 = 7 T 7 9 7 3 adj( A) 9 3 7 7 7 3 A ( adja) 9 A 4 7 7 3 5 Now, X A B 9 4 4 z 7 3 49 5 36 8 33 5 3 4 4 4 z 77 5 84 3 Hence, =, = and z = 3. 3 5 8. If A = 3 4 find A. Using A,Solve sstem of linear equations: 3 + 5z = 3 + 4z = 5 + z = 3 The given sstem can be written as AX = B, where 3 5 A 3 4, X and B 5 z 3 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 99 -
3 5 Here, A 3 4 = ( 4 + 4) ( 3) ( 6 + 4) + 5 (3 ) = 0 6 + 5 = 0 Thus, A is non-singular. Therefore, its inverse eists. Therefore, the given sstem is consistent and has a unique solution given b X = A B Cofactors of A are A = 4 + 4 = 0, A = ( 6 + 4) =, A 3 = 3 =, A = (6 5) =, A = 4 5 = 9, A 3 = ( + 3) = 5, A 3 = ( 0) =, A 3 = ( 8 5) = 3, A 33 = 4 + 9 = 3 T 0 0 adj( A) 9 5 9 3 3 3 5 3 0 0 A ( adja) 9 3 9 3 A 5 3 5 3 0 Now, X A B 9 3 5 z 5 3 3 0 5 6 45 69 z 5 39 3 Hence, =, = and z = 3. 9. The cost of 4 kg onion, 3 kg wheat and kg rice is Rs 60. The cost of kg onion, 4 kg wheat and 6 kg rice is Rs 90. The cost of 6 kg onion kg wheat and 3 kg rice is Rs 70. Find cost of each item per kg b matri method. Let the prices (per kg) of onion, wheat and rice be Rs., Rs. and Rs. z, respectivel then 4 + 3 + z = 60, + 4 + 6z = 90, 6 + + 3z = 70 This sstem of equations can be written as AX = B, where 4 3 60 A 4 6, X and B 90 6 3 z 70 4 3 Here, A 4 6 = 4( ) 3(6 36) + (4 4) 6 3 = 0 + 90 40 = 50 0 Thus, A is non-singular. Therefore, its inverse eists. Therefore, the given sstem is consistent and has a unique solution given b X = A B Cofactors of A are, Prepared b: M. S. KumarSwam, TGT(Maths) Page - 00 -
A = = 0, A = (6 36) = 30, A 3 = 4 4 = 0, A = (9 4) = 5, A = = 0, A 3 = (8 8) = 0, A 3 = (8 8) = 0, A 3 = (4 4) = 0, A 33 = 6 6 = 0 T 0 30 0 0 5 0 adj( A) 5 0 0 30 0 0 0 0 0 0 0 0 0 5 0 A ( adja) 30 0 0 A 50 0 0 0 0 5 0 60 Now, X A B 30 0 0 90 50 z 0 0 0 70 0 450 700 50 5 800 0 400 400 8 50 50 z 00 900 700 400 8 = 5, = 8 and z = 8. Hence, price of onion per kg is Rs. 5, price of wheat per kg is Rs. 8 and that of rice per kg is Rs. 8. 30. Without epanding the determinant, prove that a a bc LHS b b ca c c ab Appling R ar, R br and R3 cr3, we get 3 a a abc 3 b b abc abc c c 3 abc a abc b abc c 3 3 b [Taking out factor abc from C 3 ] a c 3 3 a a bc a a 3 b b ca b b 3 c c ab c c 3 ( ) b b (using C C 3 and C C 3 ) a c a c 3 3 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 0 -
3 a a 3 b b RHS c c 3 b c c a a b 3. If a, b and c are real numbers, and c a a b b c 0. Show that either a + b + c = 0 or a = b = c. b c c a a b c a a b b c a b b c c a ( a b c) c a a b ( a b c) a b b c ( a b c) b c c a ( a b c) a b b c c a a b b c c a ( a b c) 0 b c c a 0 c a a b b a c b Epanding along C, we get ( ) b a b c c c a b a c b ( a b c) ( b c)( c b) ( c a)( b a) a b b c c a ( a b c) bc b c bc ( bc ac ab a ) ( a b c) bc b c bc bc ac ab a ( a b c) ab bc ac a b c It is given that Δ= 0, ( a b c) ab bc ac a b c 0 Either a b c 0 or ab bc ac a b c 0 ab bc ac a b c 0 ab bc ac a b c 0 a b c ab bc ac 0 a b ab b c bc c a ac 0 ( a b) ( b c) ( c a) 0 (using C C + C + C 3 ) [take out ( a b c) common from C ]. (Using R R R and R 3 R 3 R ) ( a b) ( b c) ( c a) 0 [since square of an real number is never negative] ( a b) ( b c) ( c a) 0 a b, b c, c a a b c Prepared b: M. S. KumarSwam, TGT(Maths) Page - 0 -
3. Prove that a bc ac c a ab b ac 4a b c ab b bc c a bc ac c LHS a ab b ac ab b bc c Taking out a from C, b from C and c from C 3, we get a b a c abc a b b a 0 b b c c c a c abc b b a b b c c 0 c a c abc 0 c a c b b c c Epanding along C,we get = (abc) [ ( b) { c(a c) + c (a + c) } ] = (ab c) (ac) = 4 a b c = RHS. [Using C C + C C 3] [Using R R R 3] 33. Using properties of determinants, prove that LHS (using C 3 C 3 + C ) ( ) (Taking out (α + β + γ) common from C ) ( ) 0 0 (Using R R R and R 3 R 3 R ) ( )( )( )( ) Epanding along C 3, we get = (α + β + γ) [(β α)(γ α ) (γ α)(β α )] = (α + β + γ) [(β α)(γ α)(γ + α) (γ α)(β α)(β + α)] = (α + β + γ) (β α)(γ α)[γ + α β α] = (α + β + γ) (β α)(γ α)(γ β) = (α + β + γ) (α β)(β γ)(γ α) = RHS Prepared b: M. S. KumarSwam, TGT(Maths) Page - 03 -
34. Using properties of determinants, prove that 3a a b a c LHS b a 3b b c c a c b 3c a b c a b a c a b c 3b b c a b c c b 3c a b a c ( a b c) 3b b c c b 3c Now appling R R R, R 3 R 3 R, we get a b a c ( a b c) 0 b a a b 0 a c c a 3a a b a c b a 3b b c 3( a b c)( ab bc ca) c a c b 3c (using C C + C + C 3 ) (Taking out (a + b + c) common from C ) Epanding along C, we get = (a + b + c)[(b + a) (c + a) (a b) (a c)] = (a + b + c)[4bc + ab + ac + a a + ac + ba bc] = (a + b + c) (3ab + 3bc + 3ac) = 3(a + b + c)(ab + bc + ca) = RHS 35. Solve the sstem of equations: 3 0 4 z 4 6 5 z 6 9 0 z Let p, q and r, then the given equations become z p + 3q + 0r = 4, 4p 6q + 5r =, 6p + 9q 0r = This sstem can be written as AX = B, where 3 0 p 4 A 4 6 5, X q and B 6 9 0 r 3 0 Here, A 4 6 5 (0 45) 3( 80 30) 0(36 36) 6 9 0 = 50 + 330 + 70 = 00 0 Thus, A is non-singular. Therefore, its inverse eists. Therefore, the above sstem is consistent and has a unique solution given b X = A B Cofactors of A are A = 0 45 = 75, Prepared b: M. S. KumarSwam, TGT(Maths) Page - 04 -
A = ( 80 30) = 0, A 3 = (36 + 36) = 7, A = ( 60 90) = 50, A = ( 40 60) = 00, A 3 = (8 8) = 0, A 3 = 5 + 60 = 75, A 3 = (0 40) = 30, A 33 = = 4 75 0 7 75 50 75 adj( A) 50 00 0 0 00 30 75 30 4 7 0 4 75 50 75 A ( adja) 0 00 30 A 00 7 0 4 75 50 75 4 X A B 0 00 30 00 z 7 0 4 300 50 50 600 440 00 60 400 00 00 3 z 88 0 48 40 5 p, q, r 3 5,, 3 z 5 =, = 3 and z = 5. 36. If a, b, c, are in A.P, then find the determinant of 3 a Let A 3 4 b 4 5 c 3 a 0 0 ( b a c) 4 5 c T 3 a 3 4 b 4 5 c (using R R R R 3 ) But a,b, c are in AP. Using b = a + c, we get 3 a A 0 0 0 0 [Since, all elements of R are zero] 4 5 c Prepared b: M. S. KumarSwam, TGT(Maths) Page - 05 -
3 37. Show that the matri A satisfies the equation A 4A + I = O, where I is identit matri and O is zero matri. Using this equation, find A. 3 Given that A A 3 3 7 AA 4 7 7 3 0 Hence, A 4A I 4 4 7 0 7 8 0 7 8 0 0 0 4 7 4 8 0 4 4 0 7 8 O 0 0 Now, A 4A I O AA 4A I AA( A ) 4AA IA (Post multipling b A because A 0) A( AA ) 4I A AI 4I A 0 3 4 0 3 3 A 4I A 4 0 0 4 3 A 38. Solve the following sstem of equations b matri method. 3 + 3z = 8 + z = 4 3 + z = 4 The sstem of equation can be written as AX = B, where 3 3 8 A, X and B 4 3 z 4 3 3 Here, A 4 3 = 3 ( 3) + (4 + 4) + 3 ( 6 4) = 7 0 Hence, A is nonsingular and so its inverse eists. Now A =, A = 8, A 3 = 0 A = 5, A = 6, A 3 = A 3 =, A 3 = 9, A 33 = 7 8 0 5 adj( A) 5 6 8 6 9 9 7 0 7 5 A ( adja) 8 6 9 A 7 0 7 T Prepared b: M. S. KumarSwam, TGT(Maths) Page - 06 -
5 8 X A B 8 6 9 7 z 0 7 4 7 34 7 z 5 3 Hence =, = and z = 3. Given that ( ) z z 39. Show that ( z) z 3 z( z) z z ( ) ( z) z ( z) z z z ( ) Appling R R, R R,R 3 zr 3 to Δ and dividing b z, we get ( z) z ( z) z z z z z( ) Taking common factors,, z from C, C and C 3 respectivel, we get ( z) z ( z) z z z ( ) Appling C C C, C 3 C 3 C, we have ( z) ( z) ( z) ( z) 0 z 0 ( ) z Taking common factor ( + + z) from C and C 3, we have ( z) ( z) ( z) ( z) ( z) 0 z 0 ( ) z Appling R R (R + R 3 ), we have z z ( z) z 0 z 0 z Appling C (C + C ) and C 3 C 3 + z C, we get Prepared b: M. S. KumarSwam, TGT(Maths) Page - 07 -
z 0 0 ( ) z z z z z Finall epanding along R, we have Δ = ( + + z) (z) [( + z) ( + ) z] = ( + + z) (z) ( + + z) = ( + + z) 3 (z) 0 40. Use product 0 3 9 3 to solve the sstem of equations 3 4 6 + z = 3z = 3 + 4z = 0 Consider the product 0 3 9 3 3 4 6 9 0 3 4 0 0 0 8 8 0 4 3 0 6 6 0 0 6 8 4 0 4 4 3 6 8 0 0 0 Hence, 0 3 9 3 3 4 6 Now, given sstem of equations can be written, in matri form, as follows 0 3 3 4 z 0 0 3 9 3 z 3 4 6 0 0 9 6 5 6 4 3 Hence = 0, = 5 and z = 3 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 08 -
CHAPTER 3: DETERMINANTS Previous Years Board Eam (Important Questions & Answers) MARKS WEIGHTAGE 0 marks 9. Let A be a square matri of order 3 3. Write the value of A, where A = 4. Since A = n A where n is order of matri A. Here A = 4 and n = 3 A = 3 4 = 3 0. Write the value of the following determinant: 0 8 36 Given that 3 4 7 3 6 0 8 36 3 4 7 3 6 Appling R R 6R 3 0 0 0 3 4 0 (Since R is zero) 7 3 6. If A is a square matri and A =, then write the value of AA', where A' is the transpose of matri A. AA ' = A. A' = A. A = A = = 4. [since, AB = A. B and A = A', where A and B are square matrices.]. If A is a 3 3 matri, A 0 and 3A = k A, then write the value of k. Here, 3A = k A 3 3 A = k A [ ka = kn A where n is order of A] 7 A = k A k = 7 a ib c id 3. Evaluate: c id a ib a ib c id ( a ib)( a ib) ( c id)( c id) c id a ib ( a ib)( a ib) ( c id)( c id) a i b c i d a b c d 3 4. If 3, find the value of. 5 4 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 09 -
3 Given that 3 5 4 4 + 8 3 5 = 3 7 = 3 = 0 5. If = 5 3 8 0, write the minor of the element a 3. 3 Minor of a 3 = 5 3 = 0 3 = 7. 6. Evaluate: cos5 sin 75 sin5 0 0 cos75 0 0 Epanding the determinant, we get cos 5. cos 75 - sin 5. sin 75 = cos (5 + 75 ) = cos 90 = 0 [since cos (A + B) = cos A. cos B sina. sin B] 7. Using properties of determinants, prove the following: a a b a b Let a b a a b a b a b a Appling R R + R + R 3, we have 3( a b) 3( a b) 3( a b) a b a a b a b a b a Taking out 3(a + b) from st row, we have 3( a b) a b a a b a b a b a a a b a b a b a a b b a b 9 ( ) a b a b a Appling C C C and C C C 3 0 0 3( a b) b b a b b b a Epanding along first row, we have D = 3(a + b) [. (4b b )] = 3 (a + b) 3b = 9b (a + b) Prepared b: M. S. KumarSwam, TGT(Maths) Page - 0 -
8. Write the value of the determinant Given determinant A = 3 4 3 4 5 6 8 3 5 6 8 = 0 ( R = R 3 ) 3 4 6 9 3 4 5 6 8 6 9 9. Two schools P and Q want to award their selected students on the values of Tolerance, Kindness and Leadership. The school P wants to award Rs. each, Rs. each and Rs. z each for the three respective values to 3, and students respectivel with a total award mone of Rs.,00. School Q wants to spend Rs. 3,00 to award its 4, and 3 students on the respective values (b giving the same award mone to the three values as school P). If the total amount of award for one prize on each value is Rs.,00, using matrices, find the award mone for each value. Apart from these three values, suggest one more value that should be considered for award. According to question, 3 + + z = 00 4 + + 3z = 300 + + z = 00 The above sstem of equation ma be written in matri form as AX = B 3 00 A 4 3, X and B 300 z 00 3 Here, A 4 3 3( 3) (4 3) (4 ) 6 3 5 0 A eists. Now, A = ( 3) =, A = (4 3) =, A 3 = (4 ) = 3, A = ( ) =, A = (3 ) =, A 3 = (3 ) = A 3 = (6 ) = 5, A 3 = (9 4) = 5, A 33 = (3 8) = 5 T 3 5 adj( A) 5 5 5 5 3 5 5 5 A ( adja) 5 5 A 5 5 3 5 3 5 Prepared b: M. S. KumarSwam, TGT(Maths) Page - -
5 00 Now, X A B 5 300 5 z 3 5 00 4400 300 6000 500 300 00 600 6000 000 400 5 5 z 6600 300 6000 500 500 = 300, = 400, z = 500 i.e., Rs. 300 for tolerance, Rs. 400 for kindness and Rs. 500 for leadership are awarded. One more value like punctualit, honest etc ma be awarded. 30. Using properties of determinants, prove that a z LHS a z a z Appling C C + C + C 3, we get a z z a z a z a z a z Appl R R R, we get z ( a z) a z a z 0 a 0 ( a z) a z a z Epanding along R, we get = (a + + + z) {0 + a (a + z z)} = a (a + + + z) = RHS 3. 0 students were selected from a school on the basis of values for giving awards and were divided into three groups. The first group comprises hard workers, the second group has honest and law abiding students and the third group contains vigilant and obedient students. Double the number of students of the first group added to the number in the second group gives 3, while the combined strength of first and second group is four times that of the third group. Using matri method, find the number of students in each group. Apart from the values, hard work, honest and respect for law, vigilance and obedience, suggest one more value, which in our opinion, the school should consider for awards. Let no. of students in Ist, nd and 3rd group to,, z respectivel. From the statement we have + + z = 0 + =3 + 4z = 0 The above sstem of linear equations ma be written in matri form as AX = B where 0 A 0, X and B 3 4 z 0 Prepared b: M. S. KumarSwam, TGT(Maths) Page - -
Here, A 0 ( 4 0) ( 8 0) ( ) 4 8 5 0 4 A eists. Now, A = 4 0 = 4 A = ( 8 0) = 8 A 3 = = A = ( 4 ) = 5 A = 4 = 5 A 3 = ( ) = 0 A 3 = 0 = A 3 = (0 ) = A 33 = = 4 8 4 5 adj( A) 5 5 0 8 5 0 4 5 A ( adja) 8 5 A 5 0 4 5 0 Now, X A B 8 5 3 5 z 0 0 40 65 5 5 80 65 5 3 5 5 z 0 0 = 5, = 3, z = T 3. The management committee of a residential colon decided to award some of its members (sa ) for honest, some (sa ) for helping others and some others (sa z) for supervising the workers to keep the colon neat and clean. The sum of all the awardees is. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honest is 33. If the sum of the number of awardees for honest and supervision is twice the number of awardees for helping others, using matri method, find the number of awardees of each categor. Apart from these values, namel, honest, cooperation and supervision, suggest one more value which the management of the colon must include for awards. According to question + + z = + 3 + 3z = 33 + z = 0 The above sstem of linear equation can be written in matri form as AX = B where A 3 3, X and B 33 z 0 Here, A 3 3 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 3 -
= (3 + 6) ( 3) + ( 4 3) = 9 + 7 = 3 A eists. A = 9, A =, A 3 = 7 A = 3, A = 0, A 3 = 3 A 3 = 0, A 3 =, A 33 = T 9 7 9 3 0 adj( A) 3 0 3 0 0 7 3 9 3 0 A ( adja) 0 A 3 7 3 9 3 0 Now, X A B 0 33 3 z 7 3 0 08 99 9 3 0 0 4 3 3 z 84 99 5 5 = 3, = 4, z = 5 No. of awards for honest = 3 No. of awards for helping others = 4 No. of awards for supervising = 5. The persons, who work in the field of health and hgiene should also be awarded. 33. Using properties of determinants, prove the following: 3 z 3 z 3( z)( z z) z z 3z 3 z LHS 3 z z z 3z Appling C C + C + C 3 z z z 3 z z z 3z Taking out ( + + z) along C, we get z ( z) 3 z z 3z Appling R R R ; R 3 R 3 R z ( z) 0 0 z z Appling C C C 3 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 4 -
z z ( z) 0 3 0 3z z Epanding along I column, we get D = ( + + z)[(3 ( + z) + 3z ( )] = 3( + + z)[ + z + z + z z] = 3( + + z)( + z + z) = R.H.S. 34. A school wants to award its students for the values of Honest, Regularit and Hard work with a total cash award of Rs. 6,000. Three times the award mone for Hardwork added to that given for honest amounts to `,000. The award mone given for Honest and Hardwork together is double the one given for Regularit. Represent the above situation algebraicall and find the award mone for each value, using matri method. Apart from these values, namel, Honest, Regularit and Hardwork, suggest one more value which the school must include for awards. Let, and z be the awarded mone for honest, Regularit and hardwork. From the statement + +z = 6000 (i) + 3z =000 (ii) +z = +z = 0 (iii) The above sstem of three equations ma be written in matri form as AX = B, where 6000 A 0 3, X and B 000 0 z 0 Here, A 0 3 (0 6) ( 3) ( 0) 6 6 0 0 Hence A eist If Aij is co-factor of aij then A = 0 + 6 = 6 A = ( 3) = ; A 3 = ( 0) = A = ( + ) = 3 A = 0 A 3 = ( ) = 3 A 3 = 3 0 = 3 A 3 = (3 ) = ; A 33 = 0 = T 6 6 3 3 adj( A) 3 0 3 0 3 3 6 3 3 A ( adja) 0 A 6 3 6 3 3 6000 Now, X A B 0 000 6 z 3 0 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 5 -
36000 33000 0 3000 500 000 0 0 000 000 6 6 z 000 33000 0 000 3500 =500, = 000, z = 3500 Ecept above three values, school must include discipline for award as discipline has great importance in student s life. 4 35. If, then write the value of. 3 3 4 Given that 3 3 ( +) ( + ) ( )( 3) = + + + + + 3 + 3 =3 7 =3 7 =4 = 36. Using properties of determinants, prove that a a b a b c LHS a 3a b 4a 3b c 3a 6a 3b 0a 6b 3c a a a b c a b a b c a 3a 4a 3b c a b 4a 3b c 3a 6a 0a 6b 3c 3a 3b 0a 6b 3c a b c a b c a 3 4a 3b c ab 4a 3b c 3 6 0a 6b 3c 3 3 0a 6b 3c a b c a 3 4a 3b c ab.0 3 6 0a 6b 3c a b c a 3 4a 3b c 3 6 0a 6b 3c a b c a 3 4a 3 3b 3 c 3 6 0a 3 6 6b 3 6 3c a. a 3 4 b 3 3 c 3 3 6 0 3 6 6 3 6 3 a a b a b c a 3a b 4a 3b c a 3a 6a 3b 0a 6b 3c [since C = C in second determinant] Prepared b: M. S. KumarSwam, TGT(Maths) Page - 6-3
a. a 3 4 b.0 c.0 [since C = C 3 in second determinant and C = C 3 in third 3 6 0 determinant] a 3 3 4 3 6 0 Appling C C C and C 3 C 3 C we get 0 0 a 3 3 3 7 Epanding along R we get = a 3.(7 6) 0 + 0 = a 3. 37. Using matrices, solve the following sstem of equations: + z = 4; + 3z = 0; + + z = Given equations + z = 4 + 3z = 0 + + z = We can write this sstem of equations as AX = B where 4 A 3, X and B 0 z Here, A 3 = ( + 3) - (- ) ( + 3) + ( - ) = 4 + 5 + = 0 A eists. A = 4, A = 5, A 3 = A =, A = 0, A 3 = A 3 =, A 3 = 5, A 33 = 3 T 4 5 4 adj( A) 0 5 0 5 5 3 3 4 A ( adja) 5 0 5 A 0 3 4 4 Now, X A B 5 0 5 0 0 z 3 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 7 -
6 0 4 0 0 0 0 0 0 0 z 4 0 6 0 The required solution is =, = -, z = 3 38. If A = 5 6 5 and B = 3 0, find (AB). 5 0 For B B 3 0 0 i.e., B is invertible matri B eist. A = 3, A =, A 3 = A =, A =, A 3 = A 3 = 6, A 3 =, A 33 = 5 = (3 0) ( 0) ( 0 ) = 3 + 4 = 0 T 3 3 6 adj( B) 6 5 5 3 6 3 6 B ( adjb) B 5 5 Now (AB) = B. A 3 6 3 5 6 5 5 5 9 30 30 3 30 3 5 0 6 4 5 4 6 30 5 0 0 0 9 3 5 0 0 Prepared b: M. S. KumarSwam, TGT(Maths) Page - 8 -