DETERMINANT. = 0. The expression a 1. is called a determinant of the second order, and is denoted by : y + c 1

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1 NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65. INTRODUCTION : If the equtios x + b 0, x + b 0 re stisfied by the sme vlue of x, the b b 0. The expressio b b is clled determit of the secod order, d is deoted by : b b A determit of secod order cosists of two rows d two colums. Next cosider the system of equtios x + b y + c 0, x + b y + c 0, x + b y + c 0 If these equtios re stisfied by the sme vlues of x d y, the o elimitig x d y we get. (b c b c + b (c c + c ( b b 0 The expressio o the left is clled determit of the third order, d is deoted by A determit of third order cosists of three rows d three colums. Illustrtio : limite, m, from the equtios + cm + b 0, c + bm + 0, b + m + c 0 d express the result i the simplest form. Solutio : The give set of equtios c lso be writte s (if 0 : m c b 0 ; m c b 0 ; m b c 0 m The, let x ; y System of equtios : x + cy + b 0...(i cx + by (ii bx + y + c 0...(iii We hve to elimite x & y from these simulteous lier equtios. Sice these equtios re stisfied by the sme vlues of x d y, the elimitig x d y we get, c b c b 0 b c. VALU OF A DTRMINANT : b c c b b c b c b c c b D (b c b c b ( c c + c ( b b Note : Srrus digrm to get the vlue of determit of order three : D b c b c b c b c ve ve ve DTRMINANT b b b ( b c + b c + b c ( b c + b c + b c +ve +ve +ve Note tht the product of the terms i first brcket (i.e. b b b c c c is sme s the product of the terms i secod brcket.

2 Illustrtio : The vlue of is - (A (B (C (D 9 Solutio : (7 + 4 ( 6 + (8 6 Altertive : By srrus digrm ( ( (8 4 As. (C. MINORS & COFACTORS : The mior of give elemet of determit is the determit obtied by deletig the row & the colum i which the give elemet stds. For exmple, the mior of i is b b c c & the mior of b is c c. Hece determit of order three will hve 9 miors. If M ij represets the mior of the elemet belogig to i th row d j th colum the the cofctor of tht elemet is give by : C ij ( i + j. M ij Illustrtio : Solutio : Mior of Fid the miors d cofctors of elemets ' ', '5', ' ' & '7' i the determit Mior of 5 Mior of ; Cofctor of 9 ; Cofctor of ; Cofctor of Mior of 7 ; Cofctor of NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65

3 4. XPANSION OF A DTRMINANT IN TRMS OF TH LMNTS OF AN Y ROW OR COLUMN : Let D (i (ii The sum of the product of elemets of y row (colum with their correspodig cofctors is lwys equl to the vlue of the determit. D c be expressed i y of the six forms : A + b B + c C, A + A + A, A + b B + c C, b B + b B + b B, A + b B + c C, c C + c C + c C, where A i,b i & C i (i,, deote cofctors of i,b i & c i respectively. The sum of the product of elemets of y row (colum with the cofctors of other row (colum is lwys equl to zero. Hece, A + b B + c C 0, b A + b A + b A 0 d so o. where A i,b i & C i (i,, deote cofctors of i,b i & c i respectively. Do yourself - : (i Fid miors & cofctors of elemets '6', '5', '0' & '4' of the determit (ii Clculte the vlue of the determit NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65 b 0 (iii The vlue of the determit 0 b is equl to - b 0 (A b (B + b (C 0 (D oe of these 0 (iv Fid the vlue of 'k', if 4 k z y ( v Prove tht z x x y z y x 5. PROPRTIS OF DTRMINANTS : ( The vlue of determit remis ultered, if the rows & colums re iter-chged, e.g. if D b b b c c c

4 ( b If y two rows (or colums of determit be iterchged, the vlue of determit is chged i sig oly. e.g. b c The D D. Let D & D ( c If ll the elemets of row (or colum re zero, the the vlue of the determit is zero. ( d If ll the elemets of y row (or colum re multiplied by the sme umber, the the determit is multiplied by tht umber. e.g. If D d D K Kb Kc The D KD (e If ll the elemets of row (or colum re proportiol (or ideticl to the elemet of y other row, the the determit vishes, i.e. its vlue is zero. e.g. If D D 0 ; If D k kb kc D 0 Illustrtio 4 : Prove tht y b q x y z x p p q r z c r Solutio : D x y z p q r x p b y q c z r (By iterchgig rows & colums x p y b q (C C z c r y b q x p (R R z c r Illustrtio 5 : Fid the vlue of the determit Solutio : D c bc c b c 4 b b bc c bc c b c b b bc b b bc bc 0 c bc c Sice ll rows re sme, hece vlue of the determit is zero. NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65

5 Do yourself - : (i Without expdig the determit prove tht p r c b q m q m b 0 c r p (ii If D, the is equl to - (A D (B D (C 4D (D 6D (iii If p q r D x y z m, the KD is equl to - (A Kp q r x Ky z m K (B p q r x y z K Km K (C p Kx q Ky m r Kz (D Kp Kx K Kq Ky Km Kr Kz K ( f If ech elemet of y row (or colum is expressed s sum of two (or more terms, the the determit c be expressed s the sum of two (or more determits. e.g. x b y c z x y z Note tht : If r ƒ(r g(r h(r D where r N d,b,c,, b,c re costts, the NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65 r r ƒ(r g(r h(r r r r D ( g Row - colum opertio : The vlue of determit remis ultered uder colum (C i opertio of the form C i C i + C j + C k (j, k i or row (R i opertio of the form R i R i + R j + R k (j, k i. I other words, the vlue of determit is ot ltered by ddig the elemets of y row (or colum to the sme multiples of the correspodig elemets of y other row (or colum e.g. Let D D b b c c b b c c 5 (R R + R ; R R + R Note : (i By usig the opertio R i xr i + yr j + zr k (j, k i, the vlue of the determit becomes x times the origil oe. (ii While pplyig this property ATLAST ON ROW (OR COLUMN must remi uchged.

6 Illustrtio 6 : If r r D r ( ( (, fid D r. r 0 Solutio : r r r 0 r 0 r 0 D r r 0 ( ( ( ( ( ( ( ( ( 0 A s. Illustrtio 7 : Prove tht b c b b c b ( b c c c c b Solutio : b c D b b c b c c c b b c b c b c D b b c b c c c b (R R + R + R D ( b c b b c b c c c b 0 0 D ( b c b ( b c 0 c 0 ( b c (C C C ; C C C D ( + b + c b c ( ( b Illustrtio 8 : Determit ( c b c ( b is equl to - ( c ( c b (A ( + b + c (B ( + b + c (C ( ( + b + c (D oe of these Solutio : Applyig C C + (C + C ( ( b D ( + b + c D ( + b + c b c ( b ( c b ( ( b R R R 0 b c 0 R R R 0 0 b c ( + b + c As. (B 6 NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65

7 Illustrtio 9 : If k 4 k 4 k k 5 k k 0, the the vlue of k is- (A (B (C (D 0 Solutio : Applyig (C C C k 4 D 4 k k k (R R R ; R R R 9 k 0 k As. (B Do yourself - : (i (iii Fid the vlue of (ii Solve for x : 0 5 Usig row-colum opertios prove tht x 0 x x ( x x b x c y y b y c z z b z c 0 (b bc (b c c b(c 0 b c( b NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65 r (iv If D r, the fid the vlue of D r. r (h Fctor theorem : If the elemets of determit D re rtiol itegrl fuctios of x d two rows (or colums become ideticl whe x the (x is fctor of D. Note tht if r rows become ideticl whe is substituted for x, the (x r is fctor of D. x Illustrtio 0 : Prove tht m m m m(x (x b b x b Solutio : Usig fctor theorem, Put x D m m m 0 b b Sice R d R re proportiol which mkes D 0, therefore (x is fctor of D. Similrly, by puttig x b, D becomes zero, therefore (x b is fctor of D. 7

8 D x m m m (x (x b...(i b x b To get the vlue of put x 0 i equtio (i 0 m m m b b 0 b mb b m D m(x (x b Illustrtio : Prove tht (x (x b (x c (y (y b (y c (z (z b (z c (x y (y z (z x ( b (b c (c Solutio : D (x (x b (x c (y (y b (y c (z (z b (z c Usig fctor theorem, Put x y (y (y b (y c D (y (y b (y c (z (z b (z c R d R re ideticl which mkes D 0. Therefore, (x y is fctor of D. Similrly (y z & (z x re fctors of D Now put b (x b (x b (x c D (y b (y b (y c (z b (z b (z c C d C become ideticl which mkes D 0. Therefore, ( b is fctor of D. Similrly (b c d (c re fctors of D. Therefore, D (x y (y z (z x ( b (b c (c To get the vlue of put x, y 0 b d z c 0 4 D 0 ( ( (( ( ( D (x y (y z (z x ( b (b c (c 8 NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65

9 Do yourself - 4 : (i Without expdig the determit prove tht bc b c ( b(b c(c c b (ii Usig fctor theorem, fid the solutio set of the equtio x 5x 6. MULTIPLICATION OF TWO DTRMINANTS : b l m l b l b l m l b l m b m m b m Similrly two determits of order three re multiplied. ( Here we hve multiplied row by colum. We c lso multiply row by row, colum by row d colum by colum. ( b If D is the determit formed by replcig the elemets of determit D of order by their correspodig cofctors the D D Illustrtio : If, b, c x, y, z R, the prove tht ( x (b x (c x ( x ( bx ( cx ( y (b y (c y ( y ( by ( cy ( z (b z (c z ( z ( bz ( cz NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65 Solutio : L.H.S. ( x (b x (c x ( y (b y (c y ( z (b z (c z x x y y b b (Row by Row z z c c x x y y ( b b z z c c x x y y ( ( b b (C C z z c c 9 x x b bx x c cx x y y b by y c cy y z z b bz z c z z

10 x x y y b b z z c c Multiplyig row by row x x bx b x cx c x y y by b y cy c y z z bz b z cz c z ( x ( bx ( cx ( y ( by ( cy ( z ( bz ( cz R.H.S. Illustrtio : Let be the roots of equtio x + bx + c 0 d S + for. vlute the vlue of the determit S S S S S S S S 4. Solutio : D S S S S S S S S [( ( ( ] D ( ( re roots of the equtio x + bx + c 0 b & c 0 b (b 4c b c D (b 4c( b c y z (z y x z (x z x y (y x Illustrtio 4 : If D y z (y z xz (z x xy (x y y z (z y xz (x z xy (y x 4 (A D (B D (C D 4c d x y z D x y z A s The D D is equl to x y z (D oe of these Solutio : The give determit D is obtied by correspodig cofctors of determit D. Hece D D D D D D D As. (A NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65

11 Do yourself - 5 : (i If the determit D D such tht D D. 0 0 d D 0 0, the fid the determit D (ii If b c bc b c b c D c b b bc bc c c b c b & D, the D D is equl to - bc c b (A 0 (B D (C D (D D 7. SPCIAL DTRMINANTS : ( Cyclic Determit : The elemets of the rows (or colums re i cyclic rrgemet. b c ( bc c b ( + b + c ( + b + c b bc c ( {( b (b c (c } ( + b + c ( + b + c ( + b + c, where, re cube roots of uity ( b Other Importt Determits : NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65 (i (ii (iii (iv (v 0 b c b 0 0 c 0 ( b(b c(c bc c b ( b(b c(c ( b c ( b (b c(c (b bc c ( b (b c (c ( b c b bc c 4 4 4

12 Illustrtio 5 : Prove tht (. Solutio : This is cyclic determit. ( + ( ( + + ( ( + + ( ( + + ( ( + + ( Do yourself - 6 : k k (i The vlue of the determit kb k b kc k c is (A k( + b(b + c(c + (B kbc( + b + c (C k( b(b c(c (D k( + b c(b + c (c + b b c c (ii Fid the vlue of the determit 0 c b c b. (iii Prove tht bc c b ( b c( b(b c(c b c c b 8. CR A MR'S RUL (SYSTM OF LIN AR QUATIONS : Trivil solutio All vrible zero is the oly solutio xctly oe solutio or Uique solutio Cosistet (t lest oe solutio No trivil solutio At lest oe o zero vrible stisfies the system Simulteous lier equtios Ifiite solutios Icosistet (o solutio NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65

13 ( qutios ivolvig two vribles : (i Cosistet qutios : Defiite & uique solutio (Itersectig lies (ii Icosistet qutios : No solutio (Prllel lies (iii Depedet qutios : Ifiite solutios (Ideticl lies Let, x + b y + c 0 x + b y + c 0 the : ( ( ( b b Give equtios re cosistet with uique solutio b c Give equtios re icosistet b c Give equtios re cosistet with ifiite solutios ( b qutios Ivolvig Three vribles : Let x + b y + c z d... (i x + b y + c z d... (ii x + b y + c z d... (iii D The, x D, y D D, z D D. Where D d b c ; D d b c d b c d c ; D d c d c b d & D b d b d NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65 Note : (i (ii (iii (iv If D 0 d tlest oe of D, D, D 0, the the give system of equtios is cosistet d hs uique o trivil solutio. If D 0 & D D D 0, the the give system of equtios is cosistet d hs trivil solutio oly. If D D D D 0, the the give system of equtios is cosistet d hs ifiite solutios. Note tht I cse x by cz d x by cz d x by cz d (Atlest two of d, d & d re ot equl D D D D 0. But these three equtios represet three prllel ples. Hece the system is icosistet. If D 0 but tlest oe of D, D, D is ot zero the the equtios re icosistet d hve o solutio. ( c Homogeeous system of lier equtios : Let x + b y + c z 0... (i x + b y + c z 0... (ii x + b y + c z 0... (iii D D D 0

14 The system lwys possesses tlest oe solutio x 0, y 0, z 0, which is clled Trivil solutio, i.e. this system is lwys cosistet. Check vlue of D D 0 D 0 Uique Trivil solutio Trivil & No-Trivil solutios (ifiite solutios Note tht if give system of lier equtios hs Oly Zero solutios for ll its vribles the the give equtios re sid to hve TRIVIAL SOLUTION. Also, ote tht if the system of equtios x + b y + c 0; x + b y + c 0; x + b y + c 0 is lwys cosistet the 0 but coverse is NOT true. 9. APPLICATION OF DTRMINANTS IN GOMTRY : ( The lies : x + b y + c 0... (i re cocurret if x + b y + c 0... (ii x + b y + c 0... (iii 0. This is the coditio for cosistecy of three simulteous lier equtios i vribles. ( b qutio x² + hxy + by² + gx + fy + c 0 represets pir of stright lies if : bc + fgh f² bg² ch² 0 h g h b f g f c ( c Are of trigle whose vertices re (x r, y r ; r,, is D If D 0 the the three poits re collier. 4 x y x y x y ( d qutio of stright lie pssig through poits (x, y & (x, y is x y x y Illustrtio 6 : Fid the ture of solutio for the give system of equtios : Solutio : D x + y + z ; x + y + 4z ; x + 4y + 5z Now, D x y D 0 but D 0 Hece o solutio. A s. 0 NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65

15 Illustrtio 7 : Fid the vlue of, if the followig equtios re cosistet : x + y 0; ( + x + ( + y 8 0; x ( + y + ( + 0 Solutio : The give equtios i two ukows re cosistet, the i.e. ( Applyig C C C d C C + C (5 ( 5( 0 5 0, 5 / Illustrtio 8 : If the system of equtios x + y + 0, x + y + 0 & x + y + 0. is cosistet the fid Solutio : the vlue of. For cosistecy of the give system of equtios D or + 0 ( ( + 0 or As. Illustrtio 9 : If x, y, z re ot ll simulteously equl to zero, stisfyig the system of equtios si( x y + z 0; cos(x + 4y + z 0; x + 7y + 7z 0, the fid the vlues of (0. NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65 Solutio : Give system of equtios is system of homogeeous lier equtios which posses o-zero solutio set, therefore D 0. si D cos 4 D 7 7 si 0 D cos (R R 7 4 D 4 si cos D 0 si + cos 0 5 si 0 cos R si 4si 4si (si(4si + 4si 0 (C C + C

16 (si(si (si + 0 si 0 ; si ; si si 0 0,, ; 5 si, ; 6 6 si o solutio. 5 0,,,, A s. 6 6 Do yourself -7 : (i (ii (iii Fid ture of solutio for give system of equtios x + y + z ; x + y + z 4 ; x + z If the system of equtios x + y + z, x + y z & x + y + kz 4 hs uique solutio the (A k 0 (B < k < (C < k < (D k 0 The system of equtios x + y + z 0, x + y + z 0 0 & x y + z 0 hs o-trivil solutio, the possible vlues of re - (A 0 (B (C (D ANSWRS FOR DO YOURSLF. (i miors : 4,, 4, 4 ; cofctors : 4,, 4, 4 (ii 98 (iii B (iv 0. (ii C (iii B,C. (i 0 (ii (iv 0 4. (ii x, 5. (i (ii D 6. (i C (ii 0 7. (i ifiite solutios (ii A (iii A 6 NOD6 (\Dt\04\Kot\J-Advced\SMP\Mths\Uit#0\NG\Prt-\0.Determits\0.Theory.p65

[ 20 ] 1. Inequality exists only between two real numbers (not complex numbers). 2. If a be any real number then one and only one of there hold.

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