## 12 CBSE Maths-Matrices

Published in: Mathematics
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This PPT was covered basics of matrices, types of matrices, examples, definition of Determinant , properties of determinant and matrices, etc.

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Mae ices Columns, Rows, Entries Columns 594 Rows s 218 2 by 3 matrix Entry
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Session Objectives Determinant of a Square Matrix Minors and Cofactors Properties of Determinants Applications of Determinants Area of a Triangle Solution of System of Linear Equations (Cramer's Rule) Class Exercise
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Determinants If A then IAI all If A = a21 is a square matrix of order 1, all is a square matrix of order 2, then a22 alla22 a21a12
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Example 4 Evaluate the determinant : 2 5 4 Solution : 2 — 20+6 = 26 5
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Solution If A = all a31 all a31 a22 a22 an a13 a23 a33 al 3 a23 a33 is a square matrix of order 3, then a22 = all a32 a23 a21 a23 a21 a22 a33 a31 a33 a31 a32 [Expanding along first row = all (a 22 a33 )-a12 a a a3F23 21 33 31 23 21 32 a31a22 11 22 33 12 31 23 13 21 32 11 23 32 12 21 33 13 31 2
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Example Evaluate the determinant : Solution : 7 -3 4 [Expanding along first row] = + 8) -3(7 - 6) - +3) = 18-3-155
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Minors -1 If A = 2 4 then 3 = Minor of all Mil — Minor of a21 M21 - = Minor of an M12 — Minor of an = -1 M22 -
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Minors If A = Mll - , then — Minor of all = determinant of the order 2 x 2 square sub-matrix is obtained by leaving first row and first column of A Similarly, M23 = Minor of a23 4 2 7 3 = 12-14=-2 4 — Minor of an -9 8 = +72 = 72 etc.
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Cofactors Cij = Cofactor of ajj in A = (-1) + J MiJ , where Mij is minor of ajj in A
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Cofactors (Con.) 11 ( — — Cofactor of a — Cofactor of a = Cofactor of a 23 = (-1)2+3M 11 23 4 2 4 -9 3 7 3 4 32 3 + 2 (VI 32 81=-72 etc.
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Terms of Minors and Cof ctors all a12 a13 If a21 a22 a23 , then a31 a32 a33 3 3 ij = ailCi1 + ai2Ci2 + ai3Ci3, for I - •—1 or i = 2 or i = 3
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Properties of IDqtevmÅnqnts e va ueo a e erminant remains unchanged, if its rows and columns are interchanged. 2. If any two rows (or columns) of a determinant are interchanged, then the value of the determinant is changed by minus sign. al al Cl [Applying R 2 RI] C3
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Properties (Con.) 3. If all the elements of a row (or column) is multiplied by a non-zero number k, then the value of the new determinant is k times the value of the original determinant. kal kbl b2 b3 I
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Properties (Con.) 4. 5. If each element of any row (or column) consists of two or more terms, then the determinant can be expressed as the sum of two or more determinants. a2+Y b2 = a2 b2 + Y b2 a3+z b3 c3 The value of a determinant is unchanged, if any row (or column) is multiplied by a number and then added to any other row (or column). al bl 3 b al +mbl - ncl 1 +mb2 - nc2 b 2 a3 + mb3 - nC3 b 3 [Applying Cl + Cl + mC2 -nC3]
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Properties (Con.) 6. 7. If any two rows (or columns) of a determinant are identical, then its value is zero. If each element of a row (or column) of a determinant is zero, then its value is zero. O O O c2 = O
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Properties (Con.) Let A = IA a O a O O O b O O O b O be a diagonal matrix, then O c O O = abc c
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Row(Column) Operations Following are the notations to evaluate a determinant: (i) Ri to denote ith row (ii) to denote the interchange of ith and jth rows. (iii) Ri
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Evaluation of Determinants If a determinant becomes zero on putting x = u, then (x - a)is the factor of the determinant. For example, if A = X X x 5 9 16 2 4, then at x = 2 8 A = 0 , because Cl and C2 are identical at x = Hence, (x - 2) is a factor of determinantA . 2
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bign Expansion of PetepxVnqn! ign ys em or or er and order 3 are given by
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Example-I Find the value of the following determinants 42 1 (i) 28 7 14 3 Solution : 6 4 2 6 2 1 7 3 6 2 -10 6 4 2 42 (i) 28 14 = 74 7 = 7 xo 1 7 3 6 4 2 [Taking out 7 common from q] Cl and C3 are identical]
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Example 6 -10 5 2 3x(-2) -3 1 —3 5 5 -3 -1 5 -1 2 2 2 2 2 2 [Taking out-2 common from Cl ] Cl and C2 are identical]
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Example - 2 Evaluate the determinant 1 1 C Solution : 1 c a-I-b 1 = (a-I-b+c) I 1 1 1 1 a b c 1 1 1 [Applying —>C2 -I-C3] [Taking (a+b+ c) common from [o: Cl and C3 are identical]
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Example - 3 Evaluate the determinant: a a bc c 2 c ab b ca Solution: a We have a2 bc (a-b) -c(a-b) b ca 1 -c c 2 c ab b-c -a(b-c) c 2 c ab [Applying — C2 and — 1 -a c 2 c ab Taking (a-b) and (b-c) common from Cl and C2 respectively
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Solution Cont. O 1 c -Cc-a) o 1 O -a ab 1 c b+c c2 -a ab 1 [Applying c —c c =-(a-b)(b-c)(c-a)0 a+b+c c2 -ab [Applying R 2 2 -R3] 1 -a ab Now expanding along Cl , we get (a-b) (b-c) (c-a) [- (c2 - ab - ac - bc - c2)] = (a-b) (b-c) (c-a) (ab + bc + ac)
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Example-4 Without expanding the determinant, prove that 4х+3у 3х 3х 5х+6у 4х 6х Solution : 3х + у 2х L.H.S= 4х+Зу 3х х 1 6 3х 4х 1 З 6 2х 4х х у 6х бу 2х 3х 4х х З 6 2 З 4 з 5 з 4 5 2 4 1 6 С1 and С2 аге identical in II determinant]
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Solution Cont. 321 1 1 1 o 2 3 4 2 1 1 1 3 [Applying Cl Cl — C2 6 1 2 [ApplyingR2 -»R2-Rl and R3 -R2] 3 [Expanding along Cl] = R.H.S.
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Example -5 1 Prove that : o Solution : O 1 O 1 o 1 O O 1 O 1 1 L.H.S=o O O o = O , where o is cube root of unity. 03.0 03.0 03.0 03.0 Cl and C2 are identical]
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Example-6 b c c Prove that : a Solution : L.H.S= b b c c b c c [Applying C 1 .+C2 ] b 1 c c 1 [Taking (x+a+b+c) common from
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Solution cont. [Applying R2 -RI and R 3 ->R3-R ] Expanding along Cl , we get - R.H.S
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Example -7 Using properties of determinants, prove that Solution : L.H.S= c + a -c2). [Applying Ri .+R2 + R 3] 1 1 1
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Solution Cont. O O 1 =2(a+b+c) (c-b) (a-c) b+c [Applying Cl •Cl -C2 and C2 —>C2 -C ] (a-c) (b-a) c+a Now expanding along RI , we get 2 (a-I-b+ -a) - (a - c) 2 =R.H.S
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Example - 8 Using properties of determinants prove that х+4 2х 2х х+4 2х Solution : 2х 2х Х 4-4 2х 2х 2х 2х 2х 2х 5х +4 = 5х+4 5х +4 2х L.H.S= Х 4-4 2х 1 1 2х 2х [Applying С1 э с 1 -.FC2 +
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Solution Cont. 2x 1 2x O [Applying R 2 -RI and R 3 Now expanding along Cl , we get =R.H.S
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Example -9 Using properties of determinants, prove that x+9 x x x x x+9 x x x+9 = 243 (x+3) Solution : L.H.S= X x x X x X x x x x [Applying C 1 .+C2 + C 3]
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Solution Cont. 1 1 x x x x x O [Applying R2 -»R2 -RI and R3 -R2 ] 9 [Expanding along Cl]
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Example - 10 Show that (c+a)2 a c bc ca ab bc ca ab a c bc ca ab = (a2 +b2 +c2)(a-b)(b-c)(c-a)(a+b+c) Solution : L.H.S.= (c+a)2 a c a c bc ca [Applying Cl ab -2C31 a 1 1 [Applying Cl bc ca ab
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Solution Cont. 1 = (a2 + b2 + c2)O 1 bc c(a-b) a(b-c) [Applying R 2 bc c -RI and R 3 +b2 +c2)(a-b)(b-c)(-ab-a2 +bc+c2) [Expanding along q]
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Determinants (Area qf aeTrjanglß) s e vertices are YD, (x2, Y2) and (x3, '13) is given by the expression 1 2
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Example Find the area of a triangle whose vertices are (-1, 8), (-2, -3) and (3, 2). Solution : Area of triangle = 1 2 1 1 1 1 -1 2 3 — [5+40+5] = 25 sq. units 2
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condition or Collinearity of Three ( 'L) and C ( ')/3) are three points, then A, B, C are collinear Area of triangle ABC — Xl Yl 1 1 1 1
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Example If the points (x, -2) , (5, 2), (8, 8) are collinear, find x , using determinants. Solution : Since the given points are collinear. x 5 8 -2 2 8 1 1 1 - 6 + 24 = 0 = 18 X = 3
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2 Linear Equations be al x + blY = Cl + b2Y = c2 Then x where D = provided D O, and D2 =
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Cramer's Rule (Con.) Note : (1) If O, then the system is consistent and has unique solution. (2) If D = O and DI -O, then the system is consistent and has infinitely many solutions. (3) If and one of DI, D2 then the system is inconsistent and has no solution.
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Example Using Cramer's rule , solve the following system of equations 2x-3y=7, 3x+y=5 Solution : DI = 2 3 7 5 2 3 -3 = 2+9 = 11 O 1 -3 = 7+15=22 1 7 = 10-21=-11 5 . By Cramer's Rule and D 11 D 11
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solution or system or 3 Linear Equations (LCramep4snRrude)ns be al x + biy + q z = dl a2X + b2Y + c2z = d2 + b3Y + c3z = d3 DI = Then x = where D = a2 and D3 provided D O, C2
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Cramer's Rule (Gpn.) (1) If D O, then the system is consistent and has a unique solution. (2) If and DI = D2 = D3 = O, then the system has infinite solutions or no solution. (3) If D = O and one of DI, D2, D3 O, then the system is inconsistent and has no solution. (4) If dl = d2— d = O, then the system is called the system of 3 homogeneous linear equations. If D O, then the system has only trivial solution x = y = z (ii) If D = O, then the system has infinite solutions.
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Example Using Cramer's rule , solve the following system of equations 5x - 2y + 6z Solution : 5 2 5 5 2 -1 -1 3 -2 -1 3 -2 4 5 6 4 5 6 — 5(18+10) + - 140 -13 -76 =140 - 89 = 51 - 140 +17 -4 - 153 -15)
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Solution Cont. 3, y - = -1.92=2 5 -2 5 5 5 5 2 -1 -1 3 -2 4 5 6 5 2 -1 - 5(12 - 25)+ 4(-2 - 10) = 85 + 65 -48 = 150 -48 - 102 - 5(-3 - 5 - 107 - 102 . By Cramer's Rule D 51 and z = D -102=-2 51
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Example Solve the following system of homogeneous linear equations: x + y -z = O, x- 2y + z = O, 3x + 6y + -5z = O Solution: We have D = 1 1 3 1 -2 6 -1 1 = 4+8-12=0 The system has infinitely many solutions. z = k, in first two equations, we get Putting = k, x - 2y = -k
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Solution (Con.) k By Cramer's rule x = -2-1 -2-1 2k These values of x, y and z = k satisfy (iii) equation. z = k, where k e R

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