643 + Interview Questions in DETERMINANTS IN GENERAL AWARENESS Page 2 InterviewSolution

51.	if `(x_(1),x_(2))^(2)+(y_(1)-y_(2))^(2)=a^(2), (x_(2)-x_(3))^(2)+(y_(2)-y_(3))^(2)=b^(2) (x_(3)-x_(1))^(2)+(y_(3)-y_(1))^(2)=c^(2).` where a,b,c are positive then prove that `4 \|{:(x_(1),,y_(1),,1),(x_(2) ,,y_(2),,1),( x_(3),, y_(3),,1):}\| = (a+b+c) (b+c-a) (c+a-b)(a+b-c)`
Answer» we have `sqrt((x_(1)-x_(2))^(2)+(y_(1)-y_(2))^(2))=a` `sqrt((x^(2)-x_(3))^(2)+(y_(2)-y_(3))^(2))=b` `sqrt((x_(3)-x_(1))^(2)+(y_(3)-y_(1))^(2))=c` Consider triangle having vertices `A(x_(1),y_(1)),B(x_(2),y_(2))` and So , `AB=a,BC=b " and "AC= c.Also ,2s=a+b+c` where s is semi-perimetre. Area of triangle ABC `Delta=(1)/(2) \|\|{:(x_(1),,y_(1),,1),(x_(2),,y_(2),,1),(x_(3),,y_(3),,1):}\|\|` Also `Delta= sqrt(s(s-a)(s-b)(s-c))` `:. \|\|{:(x_(1),,y_(1),,1),(x_(2),,y_(2),,1),(x_(3),,y_(3),,1):}\|\|=sqrt(s(s-a)(s-b)(s-c))` Squaring and simplifying we get `4 \|{:(x_(1),,y_(1),,1),(x_(2),,y_(2),,1),(x_(3),,y_(3),,1):}\|^(2)` =`(a+b+c)(b+c-a)(c+a-b)(a+b-c)`

Discussion

52.	Find the value of `lambda " if " 2x^(2) +6xy +3y^(2) +8x +14y +lambda =0` represent a pair of straight lines.
Answer» `ax^(2) +2hxy+ by^(2) 2gx + 2fy+ c=0` represents a pair of lines if `\|{:(a,,h,,g),(h,,b,,f),(g,,f,,c):}\|=0` `rArr\|{:(2,,7//2,,4),(7//2,,3,,7),(4,,7,,lambda):}\|=0` `" or " 6lambda +2(7) (4) ((7)/(2)) -2(7)^(2) -3 (4)^(2) -lambda ((7)/(2))^(2) =0` `" or " 6 lambda +196 -98 -48 -(49lambda)/(4)=0` `" or " (49lambda)/(4)-6lambda =196 -146 =50` `" or " (25lambda)/(4)=50 " or " lambda =(200)/(25)=8`

Discussion

53.	Factorize the following`\|3a+b+c a^3+b^3+c^3a+b+c a^2+b^2+c^2a^4+b^4+c^4a^2+b^2+c^2a^3+b^3+c^3a^5+b^5+c^5\|`
Answer» `\|{:(3,,a+b+c,,a^(3)+b^(3)+c^(3)),(a+b+c,,a^(2)+b^(2)+c^(2),,a^(4)+b^(4)+c^(4)),(a^(2)+b^(2)+c^(2),,a^(3)+b^(3)+c^(3),,a^(5)+b^(5)+c^(5)):}\|` `= \|{:(3,,a+b+c,,a^(3)+b^(3)+c^(3)),(a+b+c,,a^(2)+b^(2)+c^(2),,a^(4)+b^(4)+c^(4)),(a^(2)+b^(2)+c^(2),,a^(3)+b^(3)+c^(3),,a^(5)+b^(5)+c^(5)):}\|` `=\|underset(a^(2)" "b^(2)" "c^(2))underset(a" "b" "c)(1" "1" "1)\|\|underset(a^(3)" "b^(3)" "c^(3))underset(a" "b" "c)(1" "1" "1)\|` `=[(a-b)(b-c)(c-a)][(a+b+c)(a-b)(b-c)(c-a)` `=(a-b)^(2)(b-c)^(2)(c-a)^(2)(a+b+c)`

Discussion

54.	` " if " Delta = \|{:(-x,,a,,b),(b,,-x,,a),(a,,b,,-x):}\|" then a factor of " Delta " is "`A. `a+b+x`B. `x^(2)-(a-b)x+a^(2)+b^(2)+ab`C. `x^(2)+(a+b)x+a^(2)+b^(2)-ab`D. `a+b-x`
Answer» Correct Answer - C::D Applying `C_(1) to C_(1)+C_(2)+C_(3)` we get `Delta =\|{:(a+b-x,,a,,b),(a+b-x,,-x,,a),(a+b-x,,b,,-x):}\|=(a+b-x) \|{:(1,,a,,b),(1,,-x,,a),(1,,b,,-x):}\|` `=(a+b-x) \|{:(1,,a,,b),(0,,-x-a,,a-b),(0,,b-a,,-x-b):}\|` `"[Applying "R_(2) to R_(2) -R_(1)" and " R_(3) to R_(3)-R_(1)"]"` `=(a+b-x)[(x+a)(x+b)+(a-b)^(2)]` [expanding along `C_(1)`] `=(a+b-x)[x^(2) +(a+b)x +a^(2) +b^(2) -ab]`

Discussion

55.	If `a != b != c`, are value of x which satisfies the equation `\|(0,x -a,x -b),(x +a,0,x -c),(x +b,x +c,0)\| = 0` is given byA. aB. bC. cD. 0
Answer» Correct Answer - D

Discussion

56.	If x, y , z are in A.P., then the value of the det (A) is , where `A = [(4,5,6,x),(5,6,7,y),(6,7,8,z),(x,y,z,0)]`
Answer» Correct Answer - A

Discussion

57.	`" if " f(0) = \|{:(sin 0 ,,cos 0,,sin 0),(cos0,,sin0,,cos0),(cos0,, sin 0,,sin 0):}\|` thenA. f(0)= 0 has exactly 2 real solutions in `[0 ,pi]`B. f(0) =0 has exactly 3 real solutions in `[0 ,pi]`C. range of function `(f(0))/(1-sin 2 0) " is " [-sqrt(2) , sqrt(2)]`D. range of fucntion .`(f(0))/(sin 2 0-1) " is " [-3,3]" is " [-3,3]`
Answer» Correct Answer - A::C `f(0) =sin^(3) 0 +cos^(3) 0 -cos 0 sin 0 (sin0 + cos 0)` `=(sin 0 + cos 0 )^(3) -4 sin 0 cos 0 (sin 0 + sin 0)` `=(sin 0 +cos 0) [1-sin 2 0]` `Now f(0)=0` `rArr tan 0 =-1 " or " sin 20 =1 ` `rArr f(0) =0` has 2 real solutions in `[0,pi]`

Discussion

58.	show that the determinant `\|{:(a^(2)+b^(2)+c^(2),,bc+ca+ab,,bc+ca+ab),(bc+ca+ab,,a^(2)+b^(2)+c^(2),,bc+ca+ab),(bc+ca+ab,,bc+ca+ab,,a^(2)+b^(2)+c^(2)):}\|` is always non- negative.
Answer» `\|{:(a^(2)+b^(2)+c^(2),,bc+ca+ab,,bc+ca+ab),(bc+ca+ab,,a^(2)+b^(2)+c^(2),,bc+ca+ab),(bc+ca+ab,,bc+ca+ab,,a^(2)+b^(2)+c^(2)):}\|` `=\|underset(c" "a" "b)underset(b" "c" "a)(a" "b" "c)\|\|underset(c" "a" "b)underset(b" "c" "a)(a" "b" "c)\|` Which is always non-negative.

Discussion

59.	If `f(x)= \|{:(1,,x,,x+1),(2x,,x(x-1),,(x+1)x),(3x(x-1),,x(x-1)(x-2),,(x+1)x(x-1)):}\|` then the value of f(500) `"_____"`
Answer» Taking x common from `R_(2) " and " x(x-1)` common from`R_(3)` we get `f(x) =x^(2) (x-1) \|{:(1,,x,,x+1),(2,,x-1,,x+1),(3,,x-2,,x+1):}\|` Applying `C_(3) to C_(3)-C_(2)` we get `f(x) =x^(2) (x-1) \|{:(1,,x,,1),(2,,x-1,,2),(3,,x-2,,3):}\|=0` Thus `f(500) =0 `

Discussion

60.	If `\|(x, x+y, x+y+z),(2x+3x+2y, 4x+3y+2z),(3x+6x+3y, 10x+6y+3z)\|=64` then the real value of x is
Answer» Correct Answer - 4 `Delta= x \|{:(1,,x+y,,x+y+z),(2,,3x+2y,,4x+3y+2z),(3,,6x+3y,,10x+6y+3z):}\|` `=x^(2) \|{:(1,,1,,x+y),(2,,3,,4x+3y),(3,,6,,10x+6y):}\|[C_(3) to C_(3)-zC_(1),C_(2)to C_(2)-yC_(1)]` `=x^(3) \|{:(1,,1,,1),(2,,3,,4),(3,,6,,10):}\|[C_(3) to C_(3)-yC_(2)]` `=x^(3) (6-8+3) =64` `x^(3) 64" or " x=4`

Discussion

61.	Let `D_1=\|a b a+b c d c+d a b a-b\|a n dD_2=\|a c a+c b d b+d a c a+b+c\|`then the value of `\|(D_1)/(D_2)\|,w h e r eb!=0a n da d!=b c ,`is _____.
Answer» Correct Answer - 2 Using `C_(3) to C_(3) -(C_(1)+C_(2))" in " D_(1) " and" D_(2) ` we have `(D_(1))/(D_(2)) =(-2b(ad-bc))/(b(ad-bc)) =-2`

Discussion

62.	If `a^(2) + b^(2) + c^(3) + ab + bc + ca le 0` for all, `a, b, c in R`, then the value of the determinant `\|((a + b +2)^(2),a^(2) + b^(2),1),(1,(b +c + 2)^(2),b^(2) + c^(2)),(c^(2) + a^(2),1,(c +a +2)^(2))\|`, is equal toA. 65B. `a^(2) + b^(2) + c^(2) + 31`C. `4(a^(2) + b^(2) + c^(2))`D. 0
Answer» Correct Answer - A We have, `a^(2) + b^(2) + c^(3) + ab + bc + ca le 0` `rArr 2a^(2) + 2b^(2) + 2c^(2) + 2ab + 2bc + 2ca le 0` `rArr (a +b)^(2) + (b +c)^(2) + (c +a)^(2) le 0` `rArr a + b = 0, b +c = 0, c + a= 0` `rArr (a +b) + (b+c) + (c +a) = 0 rArr a + b + c = 0` Thus, we have `a + b = 0, b + c = 0, c + a = 0 and a + b + c = 0` `rArr a = b = c = 0` `:. [((a + b +2)^(2),a^(2) + b^(2),1),(1,(b + c =2)^(2),b^(2) + c^(2)),(c^(2) + a^(2),1,(c + a + 2)^(2))] = [(4,0,1),(1,4,0),(0,1,4)] = 65`

Discussion

63.	If a,b,c are all distinct and `\|[a,a^3,a^4-1],[b,b^3,b^4-1],[c,c^3,c^4-1]\|` =0, show that abc(ab+bc+ac) = a+b+c
Answer» we have `\|{:(a,,a^(3),,a^(4)),(b,,b^(3),,b^(4)),(c,,c^(3),,c^(4)):}\|-\|{:(a,,a^(3),,1),(b,,b^(3),,1),(c,,c^(3),,1):}\|=0 "(Splitting w.r.t " C_(3)")"` `:. Abc \|{:(1,,a^(2),,a^(3)),(1,,b^(2),,b^(3)),(1,,c^(2),,c^(3)):}\|-\|{:(1,,a,,a^(3)),(1,,b,,b^(3)),(1,,c,,c^(3)):}\|=0` `rArr abc(a -b) (b-c) (c-a) (ab+bc+ca)` `-(a-b)(b-c) (c-a)(a+b+c) =0` `rArr abc (ab+bc+ca) -(a+b+c)=0` `(As a,b,c " are distinct)"`

Discussion

64.	`[[a,b,ax+by],[b,c,bx+cy],[ax+by,bx+cy,0]]=(b^2-ac)(ax^2+2bxy+cy^2)`
Answer» Apply `R_(3) to (R_(3) -xR_(1) - yR_(2))`

Discussion

65.	Prove that `\|[a^2+1,ab,ac],[ab,b^2+1,bc],[ac,bc,c^2+1]\|=1+a^2+b^2+c^2`
Answer» `\|{:(a^(2)+1,,ab,,ac),(ab,,b^(2)+1,,bc),(ac,,bc,,c^(2)+1):}\|` `=(1)/(abc) \|{:(a(a^(2)+1),,ab^(2),,ac^(2)),(a^(2)b,,b(b^(2)+1),,bc^(2)),(a^(2),,b^(2),,c(c^(2)+1)):}\|` [Multiplying `C_(1)C_(2)C_(3)` by a,b,c respectively] `=(abc)/(abc) \|{:(a^(2)+1,,b^(2),,c^(2)),(a^(2),,b^(2)+1,,c^(2)),(a^(2),,b^(2),,c^(2)+1):}\|` [Taking common a,b , c from `R_(1)R_(2)R)_(3)` respectively] `\|{:(1+a^(2)+b^(2)+c^(2),,b^(2),,c^(2)),(1+a^(2)+b^(2)+c^(2),,b^(2)+1,,c^(2)),(1+a^(2)+b^(2)+c^(2),,b^(2),,c^(2)+1):}\|` `[C_(1) to C_(1) +C_(2) +C_(3)]` `= (1+a^(2) +b^(2) +c^(2)) \|{:(1,,b^(2),,c^(2)),(1,,b^(2)+1,,c^(2)),(1,,b^(2),,c^(2)+1):}\|` `=(1+a^(2)+b^(2)+c^(2)) \|{:(1,,b^(2),,c^(2)),(0,,1,,0),(0,,0,,1):}\|` [Applying `R_(2) to R_(2) -R_(1) " and " R_(3) to R_(3) to R_(3)-R_(1)]` `=(1+a^(2) +b^(2)+c^(2))`

Discussion

66.	prove that `\|{:((b+c)^(2),,bc,,ac),(ba,,(c+a)^(2),,cb),(ca,,cb,,(a+b)^(2)):}\|` `\|{:((b+c)^(2),,a^(2),,a^(2)),(b^(2),,(c+a)^(2),,b^(2)),(c^(2),,c^(2),,(a+b)^(2)):}\| =2abc (a+b+c)^(3)`
Answer» Multiplying `R_(1)R_(2),R_(3)` by a,b,c respectively and dividing by abc we get `Delta =(1)/(abc)xx \|{:(a(b+c)^(2),,ba^(2),,ca^(2)),(ab^(2),,b(c+a)^(2),,cb^(2)),(ac^(2),,bc^(2),,c(a+b)^(2)):}\|` Taking a,b,c common from` C_(1),C_(2) " and " C_(3) ` respectively we get `Delta = \|{:((b+c)^(2),,a^(2),,a^(2)),(b^(2),,(c+a)^(2),,b^(2)),(c^(2),,c^(2),,(a+b)^(2)):}\|` ` " Now applying " C_(2) to C_(2) -C_(1),C_(3) to C_(3) -C_(1)` gives `Delta =\|{:((b+c)^(2),,(a+b+c)(a-b-c),,(a-b-c)(a+b+c)),(b^(2),,(c+a+c)(c+a-b),,0),(c^(2),,0,,(a+b+c)(a+b-c)):}\|` ` (a+b+c)^(2) xx \|{:((b+c)^(2),,a-b-c,,a-b-c),(b^(2),,c+a-b,,0),(c^(2),,0,,a+b-c):}\|` Applying `R_(1) to R_(1) -(R_(2)+R_(3))` and then taking 2 common from `R_(1)` we get `Delta =2(a+b+c)^(2) xx \|{:(bc,,-c,,-b),(b^(2),,c+a-b,,0),(c^(2),,0,,a+b-c):}\|` `" Now applying "C_(2) to bC_(2) +C_(1), C_(3) to cC_(3) +C_(1)` gives `Delta =(2(a+b+c)^(2))/(bc) xx \|{:(bc,,0,,0),(b^(2),,b(c+a),,b^(2)),(c^(2),,c^(2),,c(a+b)):}\|` `=(2(a+b+c)^(2))/(bc) bc[(bc +ba)(ca+cb)-b^(2)c^(2)]` `=2(a+b+c)^(2) [bc(ac+bc+ab+a^(2)-bc)]` `2abc (a+b+c)^(3)`

Discussion

67.	If `A , Ba n dC`are theangels of a triangle, show that`\|-1+cos B cos C+cos B cos B cos C+cos A-1+cos A cos A-1+cos B-1+cos A-1\|=0`
Answer» `Delta= \|{:(-1+cos B,,cos C+cos B,,cos B),(cos C+cos A,,-1+cos A,,cos A),(-1+cos B,,-1+cos A,,-1):}\|` Applying `C_(1) to C_(1) -C_(3):C_(2)to C_(2)-C_(3)` `\|{:(-1,,cos C,,cos B),(cos C,,-1,,cos A),(cos B,,cos A,,-1):}\|" "underset("A,B and C respectvely")underset("opposite to angles")("where a, b and c are sides")` Applying `C_(1) to C_(1)+bC_(2)+cC_(3)` `=(1)/(a) \|{:(-a+b cos C+c cos B,,cos C,,cos B),(a cos C-b + cos A,,-1,,cos A),(a cos B +b cos A-c,,cos A,,-1):}\|` `=(1)/(a) \|{:(0,,cos C,,cos B),(0,,-1,,cos A),(0,,cos A,,-1):}\|=0` (using projection rule in trianble)

Discussion

68.	A triangle has vertices `A_(i) (x_(i),y_(i))` for i= 1,2,3,. If the orthocenter of triangle is (0,0) then prove that `\|{:(x_(2)-x_(3),,y_(2)-y_(3),,y_(1)(y_(2)-y_(3))+x_(1)(x_(2)-x_(3))),(x_(3)-x_(1) ,,y_(3)-y_(1),,y_(2)(y_(3)-y_(1))+x_(2)(x_(3)-x_(1))),( x_(1)-x_(2),,y_(1)-y_(2),,y_(3)(y_(1)-y_(2))+x_(3)(x_(1)-x_(2))):}\|=0`
Answer» Altitudes of triangle are concurrent at `H(0,0)` Then `A_(1) H bot A_(2)A_(3)` `rArr (y_(1)-0)/(x_(1)-0).(y_(2)-y_(3))/(x^(2)-x_(3))=-1` `" or " y_(1)(y_(2)-y_(3))+x_(1)(x_(2)-x_(3))=0` Similarly `A_(2)H bot A_(3)A_(1) " and " A_(3) H bot A_(1)A_(2)` `rArr y_(2)(y_(3)-y_(1)) +x_(2)(x_(3)-x_(1))=0` `" and " y_(3)(y_(1)-y_(2))+x_(3)(x_(1)-x_(2))=0` `Dela =\|{:(x_(2)-x_(3),,y_(2)-y_(3),,y_(1)(y_(2)-y_(3))+x_(1)(x_(2)-x_(3))),(x_(3)-x_(1) ,,y_(3)-y_(1),,y_(2)(y_(3)-y_(1))+x_(2)(x_(3)-x_(1))),( x_(1)-x_(2),,y_(1)-y_(2),,y_(3)(y_(1)-y_(2))+x_(3)(x_(1)-x_(2))):}\|` `= \|{:(x_(2)-x_(3),,y_(2)-y_(3),,0),(x_(3)-x_(1),,y_(3)-y_(1),,0),(x_(1)-x_(2),,y_(1)-y_(2),,0):}\|=0`

Discussion

69.	If `A (x_(1), y_(1)), B (x_(2), y_(2)) and C (x_(3), y_(3))` are vertices of an equilateral triangle whose each side is equal to a, then prove that `\|(x_(1),y_(1),2),(x_(2),y_(2),2),(x_(3),y_(3),2)\|` is equal toA. `2a^(2)`B. `2a^(4)`C. `3a^(2)`D. `3a^(4)`
Answer» Correct Answer - D Let `Delta` be the area of triangle ABC. Then, `Delta = (1)/(2) \|(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)\| rArr 2 Delta = \|(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)\|` `rArr 4 Delta = \|(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)\| = \|(x_(1),y_(1),2),(x_(2),y_(2),2),(x_(3),y_(3),2)\|` `rArr 16Delta^(2) = \|(x_(1),y_(1),2),(x_(2),y_(2),2),(x_(3),y_(3),2)\|`..(i) But, the area of an equilateral triangle with each side equal to a is `(sqrt3)/(4) a^(2)` `:. Delta = (sqrt3)/(4) a^(2) rArr 16 Delta^(2) = 3a^(4)`..(ii) From (i) and (ii), we obtain `\|(x_(1),y_(1),2),(x_(2),y_(2),2),(x_(3),y_(3),2)\|^(2) = 3a^(4)`

Discussion

70.	If `f(alpha,beta)=\|(cos alpha,-sin alpha,1),(sin alpha,cos alpha,1),(cos(alpha+beta),-sin(alpha+beta),1)\|,`thenA. f(300,200=f(400,200)B. f(200,400)=f(200,600)C. f(100,200)=f(200,200)D. none of these
Answer» Correct Answer - A::C `\|{:(cos alpha,,-sin alpha,,1),(sin alpha,,cos alpha,,1),(cos (alpha+beta),,-sin(alpha+beta),,1):}\|` `=\|{:(cos alpha,,-sinalpha,,1),(sin alpha,,coa alpha,,1),(0,,0,,1+ sin beta -cos beta):}\|` `"[Applying " R_(3) toR_(3)-R_(1) (cos beta) +R_(2) (sin beta)"]"` `=(1+sin beta -cos beta) (cos^(2) alpha+ sin^(2)alpha)` `=1+ sin beta- cos beta `which is independent of `alpha`

Discussion

71.	Which of the following has`//`have value equal to zero ?A. `\|{:(8,,2,,7),(12,,3,,5),(16,,4,,3):}\|`B. `\|{:(1//a,,a^(2),,bc),(1//b,,b^(2),,ac),(1//c,,c^(2),,ab):}\|`C. `\|{:(a+b,,2a+b,,3a+b),(2a+b,,3a+b,,4a+b),(4a+b,,5a+b,,6a+b):}\|`D. `\|{:(2,,43,,6),(7,,35,,4),(3,,17,,2):}\|`
Answer» Correct Answer - A::B::C `\|{:(8,,2,,7),(12,,3,,5),(16,,4,,3):}\|=\|{:(8,,2,,7),(4,,1,,-2),(4,,1,,-2):}\|" " underset(R_(2) to R_(2)-R_(1))(R_(3) to R^(3) -R_(2)" and ")` `\|{:(1//a,,a^(2),,bc),(1//b,,b^(2),,ac),(1//c,,c^(2),,ab):}\|=(1)/(abc) \|{:(1,,a^(3),,abc),(1,,b^(3),,abc),(1,,c^(3),,abc):}\|` `[R_(1) to aR_(1),R_(2) to bR_(2) ,R_(3) to cR_(3)]` `=(abc)/(abc) \|{:(1,,a^(3),,1),(1,,b^(3),,1),(1,,c^(3),,1):}\|` [taking abc common from `C_(3)`] `\|{:(a+b,,2a+b,,3a+b),(2a+b,,3a+b,,4a+b),(4a+b,,5a+b,,6a+b):}\|=\|{:(a+b,,2a+b,,3a+b),(a,,a,,a),(2a,,2a,,2a):}\|` `[R_(3) to R_(3) -R_(2),R_(2) to R_(2)-R_(1)]` `=0` `\|{:(2,,43,,6),(7,,35,,4),(3,,17,,2):}\|=\|{:(2,,1,,6),(7,,7,,4),(3,,3,,2):}\|" "[C_(2) to C_(2) -7 C_(3)]` `=\|{:(1,,1,,6),(0,,7,,4),(0,,3,,2):}\|" "[C_(2) to C_(1) -C_(2)]` `=2`

Discussion

72.	if the system of equations `(a-t)x+by +cz=0` `bx+(c-t) y+az=0` `cx+ay+(b-t)z=0` has non-trivial solutions then product of all possible values of t isA. `\|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}\|`B. `a+b+c`C. `a^(2)+b^(2)+c^(2)`D. `1`
Answer» Correct Answer - A The given system of equations will have a non-trivial solutions if the determinant of coefficient is 0. `Delta= \|{:(a-t,,b,,c),(b,,c-t,,a),(c,,a,,b-t):}\|=0` `Delta=0` is a cubic equation in t, so it has 3 solutions say `t_(1), t_(2)" and " t_(3)` Let `Delta =p_(0)t_(3)+p_(1)t^(2) +p_(2)t+p_(3)` Clearly ,Po= coeff . of `t^(3)` which is equal to -1 , so `t_(1) t_(2)t_(3) =-(P_(3))/((-1))=P_(3)` = constant term in the expansion of `Delta i.e, Delta _((t=0))` hence `t_(1)t_(2)t_(3)= \|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}\|`

Discussion

73.	If `l_(1), m_(1), n_(1), l_(2), m_(2), n_(2) and l_(3), m_(3), n_(3)` are direction cosines of three mutuallyy perpendicular lines then, the value of `\|(l_(1),m_(1),n_(1)),(l_(2),m_(2),n_(2)),(l_(3),m_(3),n_(3))\|` isA. `l_(3) m_(3) n_(3)`B. `+- 1`C. `l_(1) m_(1) n_(1)`D. `l_(2) m_(2) n_(2)`
Answer» Correct Answer - B Since, `l_(1), m_(1), n_(1), l_(2), m_(2), n_(2) and l_(3), m_(3), n_(3)` are direction cosines of three mutually perpendicular lines `:. l_(1)^(2) + m_(1)^(2) + n_(1)^(2) -1, l_(2)^(2) + m_(2)^(2) + n_(2)^(2) = 1` `l_(3)^(2) + m_(3)^(2) + n_(3)^(2) = 1, l_(1) l_(2) + m_(1) m_(2) + n_(1) n_(2) = 0` `l_(1) l_(3) + m_(1) m_(3) + n_(1) n_(3) = 0, l_(2) l_(3) + m_(2) m_(3) + n_(2) n_(3) = 0` Let `Delta = \|(l_(1),m_(1),n_(1)),(l_(2),m_(2),n_(2)),(l_(3),m_(3),n_(3))\|`. Then, `Delta^(2) = \|(l_(1) ,m_(1),n_(1)),(l_(2),m_(2),n_(2)),(l_(3),m_(3),n_(3))\|\|(l_(1),m_(1),n_(1)),(l_(2),m_(2),n_(3)),(l_(3),m_(3),n_(3))\|` `= \|(l_(1)^(2) + m_(1)^(2) + n_(1)^(2),l_(1)l_(2) + m_(1) m_(2) + n_(1) n_(2),l_(1) l_(3) + m_(1) m_(3) + n_(1) n_(3)),(l_(1) l_(2) + m_(1) m_(2) + n_(1) n_(2),l_(1)^(2) + m_(2)^(2) + n_(2)^(2),l_(2) l_(3) + m_(2) m_(3) + n_(2) n_(3)),(l_(1) l_(3) + m_(1) m_(3) + n_(1) n_(3),l_(1) l_(3) + m_(2) m_(3) + n_(2) n_(3),l_(3)^(2) + m_(3)^(2) + n_(3)^(2))\|` `= \|(1,0,0),(0,1,0),(0,0,1)\| =1` `:. Delta = +- 1`

Discussion

74.	Let `lambda` and `alpha` be real. Then the numbers of intergral values `lambda` for which the system of linear equations `lambdax +(sin alpha) y+ (cos alpha) z=0` `x + (cos alpha) y+ (sin alpha) z=0` `-x+(sin alpha) y -(cos alpha) z=0` has non-trivial solutions is
Answer» Correct Answer - D The given system has non=trivial solution if `\|{:(lambda,,sin alpha,,cos alpha),(1,,cos alpha,,sin alpha),(-1,,sin alpha,,-cos alpha):}\|=0` By expanding the determinant along first column we get `lambda =sin 2alpha +cos 2alpha` We know that `-sqrt(2) le sin 2alpha + cos 2 alpha le sqrt(2)` `:. -sqrt(2) le lambda le sqrt(2)` hence integral values of `lambda` are -1 ,0 and 1

Discussion

75.	`\|(1,cos(beta-alpha),cos(gamma-alpha)), (cos(alpha-beta),1,cos(gamma-beta)), (cos(beta-alpha), cos(beta - gamma),1)\| = `A. `cos alpha cos beta cos gamma`B. `cos alpha + cos beta + cos gamma`C. 1D. 0
Answer» Correct Answer - D We have, `= Delta= \|(1,cos (beta- alpha),cos (gamma - alpha)),(cos (alpha - beta),1,cos (gamma - beta)),(cos (alpha - gamma),cos (beta - gamma),1)\|` `= \|(cos^(2) alpha + sin^(2) alpha,cos beta cos alpha + sin beta alpha,cos alpha cos gamma + sin alpha sin gamma),(cos alpha cos beta + sin alpah sin beta,"cos"^(2)beta + sin^(2) beta,cos beta cos gamma + sin beta sin gamma),(cos alpha cos gamma + sin alpha sin gamma,cos beta cos gamma + sin beta sin gamma,cos^(2) beta + sin^(2) beta)\|` `= \|(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)\|.\|(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)\| = 0`

Discussion

76.	The value of `\|(2 y_(1)z_(1),y_(1) z_(2) + y_(2) z_(1),y_(1) z_(3) + y_(3) z_(1)),(y_(1) z_(2) y_(2) z_(1),2y_(2) z_(2),y_(2) z_(3) + y_(3) z_(2)),(y_(1) z_(3) + y_(3) z_(1),y_(2) z_(3) + y_(3) z_(2),2y_(3) z_(3))\|`, isA. `y_(1) y_(2) y_(3) z_(1) z_(2) z_(3)`B. `y_(1) + y_(2) + y_(3)`C. `z_(1) + z_(2) + z_(3)`D. 0
Answer» Correct Answer - D We have, `\|(2 y_(1) z_(1),y_(1) z_(2) + y_(2) z_(1),y_(1) z_(3) + y_(3) z_(1)),(y_(1) z_(2) + y_(2) z_(1),2y_(2) z_(2),y_(2) z_(3) + y_(3) z_(2)),(y_(1) z_(3) + y_(3) z_(1),y_(2) z_(3) + y_(3) z_(2),2y_(3) z_(3))\|` `= \|(y_(1) z_(1) + y_(1) z_(1),y_(1) z_(2) + y_(2) z_(1),y_(1) z_(3) + y_(3) z_(1)),(y_(1) z_(2) + y_(2) z_(1),y_(2) z_(2) + y_(2) z_(2),y_(2) z_(3) + y_(3) z_(2)),(y_(1) z_(3) + y_(3) z_(1),y_(2) z_(3) + y_(3) z_(2),y_(3) z_(3) + y_(3) z_(3))\|` `= \|(y_(1),z_(1),0),(y_(2),z_(2),0),(y_(3),z_(3),0)\|\|(z_(1),y_(1),0),(z_(2),y_(2),0),(z_(3),y_(3),0)\| = 0 xx 0 = 0`

Discussion

77.	One root of the equation `\|(3x-8, 3, 3),(3,3x-8, 3),(3,3,3x-8)\|=0 isA. `{(2)/(3), (8)/(3)}`B. `{(2)/(3), (11)/(3)}`C. `{(3)/(2), (8)/(3)}`D. none of these
Answer» Correct Answer - B Apply `C_(1) to (C_(1) + C_(2) +C_(3))` and take (3x-2) common from `C_(1)`

Discussion

78.	it `x_(1)^(2) +2y_(1)^(2)+3z_(1)^(2)=x_(2)^(2)+2y_(2)^(2)+3z_(2)^(2)=x_(3)^(2)+2y_(3)^(2)+3z_(3)^(2)=2 " and " x_(2)x_(3) +2y_(2)y_(3)+3z_(2)z_(3)=x_(3)x_(1)+2y_(3)y_(1)+3z_(3)z_(1)=x_(1)x_(2)+2y_(1)y_(2)+3z_(1)z_(2)=1` Then find the value of `\|{:(x_(1),,y_(1),,z_(1)),(x_(2),,y_(2),,z_(2)),(x_(3),,y_(3),,z_(3)):}\|`
Answer» Let `Delta= \|{:(x_(1),,y_(1),,z_(1)),(x_(2),,y_(2),,z_(2)),(x_(3),,y_(3),,z_(3)):}\|` `:. Delta^(2)= Delta xx Delta` `=\|{:(x_(1),,y_(1),,z_(1)),(x_(2),,y_(2),,z_(2)),(x_(3),,y_(3),,z_(3)):}\|xx \|{:(x_(1),,y_(1),,z_(1)),(x_(2),,y_(2),,z_(2)),(x_(3),,y_(3),,z_(3)):}\|` ` =(1)/(1xx2xx3) \|{:(x_(1),,2y_(1),,z_(1)),(x_(2),,2y_(2),,z_(2)),(x_(3),,2y_(3),,z_(3)):}\|xx \|{:(x_(1),,y_(1),,z_(1)),(x_(2),,y_(2),,z_(2)),(x_(3),,y_(3),,z_(3)):}\|` `= (1)/(6) \|{:(x_(1)^(2)+2y_(1)^(2)+3z_(1)^(2),,x_(1)x_(2)+2y_(1)y_(2)+3z_(1)z_(2),,x_(3)x_(1)+2y_(2)y_(3)+3z_(2)z_(3)),(x_(1)x_(2)+2y_(1)y_(2)+3z_(1)z_(2),,x_(2)^(2)+2y_(2)^(2)+3z_(2)^(2),,x_(2)x_(3)+2y_(2)y_(3)+3z_(3)z_(3)),(x_(3)x_(1)+2y_(3)y_(1)+3z_(3)z_(1),,x_(3)^(2)+2y_(3)^(2)+3z_(3)^(2),,):}\|` `=(1)/(6) \|{:(2,,1,,1),(1,,2,,1),(1,,1,,2):}\|` `=(2)/(3)`

Discussion

79.	If the points A(3, -2), B(k, 2) and C(8, 8) are collinear then the value of k isA. 2B. `-3`C. 5D. `-4`
Answer» Correct Answer - C If `Delta = \|[3, -2, 1], [k, 2, 1], [8, 8, 1]\|` then we must have `Delta = 0`

Discussion

80.	Solve the following equations:\(\begin{vmatrix}x-1&x&x-2\\0&x-2&x-3\\0&0&x-3\end{vmatrix}=0\)[(x-1, x, x-2) (0, x-2, x-3) (0, 0, x-3)] = 0
Answer» \(\begin{vmatrix}x-1&x&x-2\\0&x-2&x-3\\0&0&x-3\end{vmatrix}=0\) Applying R₂ → R₂ – R₃ , we get \(\begin{vmatrix}x-1&x&x-2\\0&x-2&0\\0&0&x-3\end{vmatrix}=0\) ∴ (x – 1)(x – 2)(x – 3) – 0] – x(0 – 0) + (x – 2)(0 – 0) = ∴ (x – 1)(x – 2)(x – 3) = 0 ∴ x — 1 = 0 or x-2 = 0 or x-3 = 0 ∴ x = 1 or x = 2 or x = 3

80.

Solve the following equations:\(\begin{vmatrix}x-1&x&x-2\\0&x-2&x-3\\0&0&x-3\end{vmatrix}=0\)[(x-1, x, x-2) (0, x-2, x-3) (0, 0, x-3)] = 0

Answer»

\(\begin{vmatrix}x-1&x&x-2\\0&x-2&x-3\\0&0&x-3\end{vmatrix}=0\)

Applying R₂ → R₂ – R₃ , we get

\(\begin{vmatrix}x-1&x&x-2\\0&x-2&0\\0&0&x-3\end{vmatrix}=0\)

∴ (x – 1)(x – 2)(x – 3) – 0] – x(0 – 0) + (x – 2)(0 – 0) =

∴ (x – 1)(x – 2)(x – 3) = 0

∴ x — 1 = 0 or x-2 = 0 or x-3 = 0

∴ x = 1 or x = 2 or x = 3

Discussion

81.	Solve the following equations.\(\begin{vmatrix}x+2&x+6&x-1\\x+6&x-1&x+2\\x-1&x+2&x+6\end{vmatrix}\)=0[(x+2, x+6, x-1) (x+6, x-1 x+2) (x-1, x+2, x+6)] =0
Answer» \(\begin{vmatrix}x+2&x+6&x-1\\x+6&x-1&x+2\\x-1&x+2&x+6\end{vmatrix}\) Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₁ , we get \(\begin{vmatrix}x+2&x+6&x-1\\4&-7&3\\-3&-4&7\end{vmatrix}=0\) ∴ (x + 2)(- 49 + 12) – (x + 6)(28 + 9) + (x- 1)(- 16 – 21) = 0 ∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0 ∴ -37(x + 2+ x + 6 + x – 1) = 0 ∴ 3x + 7 = 0 ∴ x = -7/3

81.

Solve the following equations.\(\begin{vmatrix}x+2&x+6&x-1\\x+6&x-1&x+2\\x-1&x+2&x+6\end{vmatrix}\)=0[(x+2, x+6, x-1) (x+6, x-1 x+2) (x-1, x+2, x+6)] =0

Answer»

\(\begin{vmatrix}x+2&x+6&x-1\\x+6&x-1&x+2\\x-1&x+2&x+6\end{vmatrix}\)

Applying R₂ → R₂ – R₁ and R₃ → R₃ – R₁ , we get

\(\begin{vmatrix}x+2&x+6&x-1\\4&-7&3\\-3&-4&7\end{vmatrix}=0\)

∴ (x + 2)(- 49 + 12) – (x + 6)(28 + 9) + (x- 1)(- 16 – 21) = 0

∴ (x + 2) (-37) – (x + 6) (37) + (x – 1) (-37) = 0

∴ -37(x + 2+ x + 6 + x – 1) = 0

∴ 3x + 7 = 0

∴ x = -7/3

Discussion

82.	`\|[2a_1b_1, a_1b_2+a_2b_1, a_1b_3+a_3b_1] , [a_1b_2+a_2b_1, 2a_2b_2, a_2b_3+a_3b_2] , [a_1b_3+a_3b_1, a_3b_2+a_2b_3, 2a_3b_3]\|=`A. 1B. `-1`C. 0D. `a_(1) a_(2) a_(3) b_(1) b_(2) b_(3)`
Answer» Correct Answer - C

Discussion

83.	Show that `\|{:("sin"10^(@), -"cos"10^(@)), ("sin"80^(@), "cos"80^(@)):}\| =1.`
Answer» We have `\|{:("sin"10^(@), -"cos"10^(@)), ("sin"80^(@), "cos"80^(@)):}\|` `=("sin"10^(@))("cos" 80^(@)) - ("sin" 80^(@))(-"cos" 10^(@))` `= ("sin" 10^(@) "cos" 80^(@) + "cos" 10^(@) "sin"80^(@))` `= "sin" (10^(@) +80^(@)) " "[because "sin" A"cos" B + "cos" A"sin" B = "sin (A+B)]` `="sin" 90^(@) =1.`

Discussion

84.	If `\|{:(x+1, x-1), (x-3, x+2):}\| = \|{:(4, -1), (1, 3):}\|`, find the value of x.
Answer» We have `\|{:(x+1, x-1), (x-3, x+2):}\| = \|{:(4, -1), (1, 3):}\|` `hArr (x+1) (x+2)-(x-3)(x-1) = (4 xx 3)-1 xx (-1)` `hArr (x^(2) +3x +2) -(x^(2) -4x +3) = 12+1` `hArr 7x -1 = 13 hArr 7x = 14 hArr x =2` Hence, x =2.

Discussion

85.	If `\|{:(3x, 7), (-2, 4):}\| = \|{:(8, 7), (6, 4):}\|`, find the value of x.
Answer» We have `\|{:(3x, 7), (-2, 4):}\| = \|{:(8, 7), (6, 4):}\|` `hArr (3x xx 4) -(-2) xx 7 = (8 xx 4) - (6 xx 7)` `hArr 12x + 14 = 32-42` `hArr 12x + 14 =-10` `hArr 12x =-24 hArr x =-2` Hence, x = 2.

Discussion

86.	Examine the consistency of the system of equations`2x - y = 5``x + y = 4`
Answer» Given system of equations, 2x-y=5 x+y=5 `rArr" "[{:(2,-1),(1,1):}][{:(x),(y):}]=[{:(5),(4):}]rArrAX=B` `therefore" "A[{:(2,-1),(1,1):}]` `rArr" "\|A\|[{:(2,-1),(1,1):}]=2-(-1)=3ne0` `therefore`A is invertible. `rArr` Given system of equation is consistent.

Discussion

87.	Let A = [aij] be a square matrix of order 3 × 3 and Cij denote cofactor of aij in A. If \|A\| = 5, write the value of a31C31 + a32C32 + a33C33.
Answer» The value of determinant \|A\|, of matrix A = [a_ij] of order 3 × 3, is given to be 5. The value of determinant written in the form of cofactors is equal to the sum of products of elements of that row (or column) multiplied by their corresponding cofactors. The value of \|A\| expanded along row 3 will be \|A\| = a₃₁× C₃₁+ a₃₂× C₃₂+ a₃₃× C₃₃, which is the required expression Hence, the value of required expression is equal to \|A\| = 5.

87.

Let A = [aij] be a square matrix of order 3 × 3 and Cij denote cofactor of aij in A. If |A| = 5, write the value of a31C31 + a32C32 + a33C33.

Answer»

The value of determinant |A|, of matrix A = [a_ij] of order 3 × 3, is given to be 5.

The value of determinant written in the form of cofactors is equal to the sum of products of elements of that row (or column) multiplied by their corresponding cofactors.

The value of |A| expanded along row 3 will be

|A| = a₃₁× C₃₁+ a₃₂× C₃₂+ a₃₃× C₃₃, which is the required expression

Hence, the value of required expression is equal to |A| = 5.

Discussion

88.	Show that each of the following systems of linear equations is inconsistent : 2x – y = 5 4x – 2y = 7
Answer» Given : - Two equation 2x – y = 5 and 4x – 2y = 7 Tip : - We know that For a system of 2 simultaneous linear equation with 2 unknowns (i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by x = \(\frac{D_1}{D}\), y = \(\frac{D_2}{D}\) (ii) If D = 0 and D₁ = D₂ = 0, then the system is consistent and has infinitely many solution. (iii) If D = 0 and one of D₁ and D₂ is non – zero, then the system is inconsistent. Now, We have, 2x – y = 5 4x – 2y = 7 Lets find D ⇒ D = \(\begin{vmatrix}2& -1 \\[0.3em]4 &-2 \\[0.3em]\end{vmatrix}\) ⇒ D = – 4 + 4 ⇒ D = 0 Again, D₁ by replacing 1^st column by B Here, B = \(\begin{vmatrix}5 \\[0.3em]7\\[0.3em]\end{vmatrix}\) ⇒ D₁ = \(\begin{vmatrix}5& -1 \\[0.3em]7 &-2 \\[0.3em]\end{vmatrix}\) ⇒ D₁ = – 10 + 7 ⇒ D₁ = – 3 And, D₂ by replacing 2^nd column by B Here, B = \(\begin{vmatrix}5 \\[0.3em]7\\[0.3em]\end{vmatrix}\) ⇒ D₂ = \(\begin{vmatrix}2& 5 \\[0.3em]4 &7 \\[0.3em]\end{vmatrix}\) ⇒ D₂ = 14 – 20 ⇒ D₂ = – 6 So, here we can see that D = 0 and D₁ and D₂ are non – zero Hence the given system of equation is inconsistent.

88.

Show that each of the following systems of linear equations is inconsistent : 2x – y = 5 4x – 2y = 7

Answer»

Given : -

Two equation 2x – y = 5 and 4x – 2y = 7

Tip : - We know that

For a system of 2 simultaneous linear equation with 2 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

x = \(\frac{D_1}{D}\), y = \(\frac{D_2}{D}\)

(ii) If D = 0 and D₁ = D₂ = 0, then the system is consistent and has infinitely many solution.

(iii) If D = 0 and one of D₁ and D₂ is non – zero, then the system is inconsistent.

Now,

We have,

2x – y = 5

4x – 2y = 7

Lets find D

⇒ D = \(\begin{vmatrix}2& -1 \\[0.3em]4 &-2 \\[0.3em]\end{vmatrix}\)

⇒ D = – 4 + 4

⇒ D = 0

Again,

D₁ by replacing 1^st column by B

Here,

B = \(\begin{vmatrix}5 \\[0.3em]7\\[0.3em]\end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix}5& -1 \\[0.3em]7 &-2 \\[0.3em]\end{vmatrix}\)

⇒ D₁ = – 10 + 7

⇒ D₁ = – 3

And,

D₂ by replacing 2^nd column by B

Here,

B = \(\begin{vmatrix}5 \\[0.3em]7\\[0.3em]\end{vmatrix}\)

⇒ D₂ = \(\begin{vmatrix}2& 5 \\[0.3em]4 &7 \\[0.3em]\end{vmatrix}\)

⇒ D₂ = 14 – 20

⇒ D₂ = – 6

So, here we can see that

D = 0 and D₁ and D₂ are non – zero

Hence the given system of equation is inconsistent.

Discussion

89.	Mark the correct alternative in the following :If \(\begin{vmatrix}2x & 5 \\[0.3em]8 & x \\[0.3em]\end{vmatrix}\) = \(\begin{vmatrix}6 & -2 \\[0.3em]7 & 3 \\[0.3em]\end{vmatrix}\), then x =A. 3 B. ± 3 C. ± 6 D. 6
Answer» Correct answer : (C) \(\begin{vmatrix}2x & 5 \\[0.3em]8 & x \\[0.3em]\end{vmatrix}\) = \(\begin{vmatrix}6 & -2 \\[0.3em]7 & 3 \\[0.3em]\end{vmatrix}\) 2x² - 40 = 18+14 x = ± 6

89.

Mark the correct alternative in the following :If \(\begin{vmatrix}2x & 5 \\[0.3em]8 & x \\[0.3em]\end{vmatrix}\) = \(\begin{vmatrix}6 & -2 \\[0.3em]7 & 3 \\[0.3em]\end{vmatrix}\), then x =A. 3 B. ± 3 C. ± 6 D. 6

Answer»

Correct answer : (C)

\(\begin{vmatrix}2x & 5 \\[0.3em]8 & x \\[0.3em]\end{vmatrix}\) = \(\begin{vmatrix}6 & -2 \\[0.3em]7 & 3 \\[0.3em]\end{vmatrix}\)

2x² - 40 = 18+14

x = ± 6

Discussion

90.	Mark the correct alternative in the following : The value of the determinant \(\begin{vmatrix}a-b &b+c & a \\[0.3em]b-a & c+a & b \\[0.3em]c-a & a+b & c\end{vmatrix}\) isA. a3 + b3 + c3 B. 3bc C. a3 + b3 + c3 – 3abc D. none of these
Answer» Correct answer : (D) Assume a = 1,b = 2, c = 3 (put in determinant) Δ = \(\begin{vmatrix}-1 & 5& 1 \\[0.3em]1 & 4 &2 \\[0.3em]2 &3&3\end{vmatrix}\) ∆ =[-1(12 - 6) - 5(3 - 4)+1(3 - 6)] ∆ =- 4 Put a = 1,b = 2, c = 3 in option A,B,C,D ANSWER = D (none of these )

90.

Mark the correct alternative in the following : The value of the determinant \(\begin{vmatrix}a-b &b+c & a \\[0.3em]b-a & c+a & b \\[0.3em]c-a & a+b & c\end{vmatrix}\) isA. a3 + b3 + c3 B. 3bc C. a3 + b3 + c3 – 3abc D. none of these

Answer»

Correct answer : (D)

Assume a = 1,b = 2, c = 3

(put in determinant)

Δ = \(\begin{vmatrix}-1 & 5& 1 \\[0.3em]1 & 4 &2 \\[0.3em]2 &3&3\end{vmatrix}\)

∆ =[-1(12 - 6) - 5(3 - 4)+1(3 - 6)]

∆ =- 4

Put a = 1,b = 2, c = 3 in option A,B,C,D

ANSWER = D (none of these )

Discussion

91.	If `A`is `a3x3`matrix, `\|A\|``!=``0`and `\|3A\|=k\|A\|`, then write the value of `k`.
Answer» Correct Answer - k = 27 Since A is a `3 xx 3` matrix, so `\|3A\| = (3 xx 3 xx 3) * \|A\|`

Discussion

92.	Manjit wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50 m and breadth is increased by 50m, then its area will remain same, but if length is decreased by 10m and breadth is decreased by 20m, then its area will decrease by 5300 m2Based on the information given above, answer the following questions:1. The equations in terms of X and Y area. x - y = 50, 2x - y = 550b. x - y = 50, 2x + y = 550c. x + y = 50, 2x + y = 550d. x + y = 50, 2x + y = 5502. Which of the following matrix equation is represented by the given information3. The value of x (length of rectangular field) isa. 150mb. 400mc. 200m y xd. 320m4. The value of y (breadth of rectangular field) isa. 150mb. 200mc. 430m.d. 350m5. How much is the area of rectangular field?a. 60000Sq.mb. 30000Sq.m.c. 30000md. 3000m
Answer» 1. b) x - y = 50, 2x + y = 550 2. a) \(\begin{bmatrix}1&-1\\2&1\end{bmatrix}\)\(\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}50\\550\end{bmatrix}\) 3. c) 200m 4. a) 150m 5. b) 30000Sq.m

92.

Manjit wants to donate a rectangular plot of land for a school in his village. When he was asked to give dimensions of the plot, he told that if its length is decreased by 50 m and breadth is increased by 50m, then its area will remain same, but if length is decreased by 10m and breadth is decreased by 20m, then its area will decrease by 5300 m2Based on the information given above, answer the following questions:1. The equations in terms of X and Y area. x - y = 50, 2x - y = 550b. x - y = 50, 2x + y = 550c. x + y = 50, 2x + y = 550d. x + y = 50, 2x + y = 5502. Which of the following matrix equation is represented by the given information3. The value of x (length of rectangular field) isa. 150mb. 400mc. 200m y xd. 320m4. The value of y (breadth of rectangular field) isa. 150mb. 200mc. 430m.d. 350m5. How much is the area of rectangular field?a. 60000Sq.mb. 30000Sq.m.c. 30000md. 3000m

Answer»

1. b) x - y = 50, 2x + y = 550

2. a) \(\begin{bmatrix}1&-1\\2&1\end{bmatrix}\)\(\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}50\\550\end{bmatrix}\)

3. c) 200m

4. a) 150m 5. b) 30000Sq.m

Discussion

93.	If A is a `2 xx 2` matrix such that `\|A\| ne 0 " and " \|A\| = 5, ` write the value of \|4A\|.
Answer» Correct Answer - 80 Since A is a `2 xx 2` matrix, so `\|4A\| = (4 xx 4) * \|A\|.`

Discussion

94.	If A is a 3 × 3 matrix, \|A\| ≠ 0 and \|3A\| = k\|A\| then write value of k.
Answer» We are given that, Order of matrix = 3 \|A\| ≠ 0 \|3A\| = k\|A\| We need to find the value of k. In order to find k, we need to solve \|3A\|. Using property of determinants, \|kA\| = kⁿ\|A\| Where, order of A is n × n. Similarly, \|3A\| = 3³\|A\| [∵, order of A is 3] ⇒ \|3A\| = 27\|A\| …(i) As, according to the question \|3A\| = k\|A\| Using (i), ⇒ 27\|A\| = k\|A\| Comparing the left hand side and right hand side, we get k = 27 Thus, the value of k is 27.

94.

If A is a 3 × 3 matrix, |A| ≠ 0 and |3A| = k|A| then write value of k.

Answer»

We are given that,

Order of matrix = 3

|A| ≠ 0

|3A| = k|A|

We need to find the value of k.

In order to find k, we need to solve |3A|.

Using property of determinants,

|kA| = kⁿ|A|

Where, order of A is n × n.

Similarly,

|3A| = 3³|A|

[∵, order of A is 3]

⇒ |3A| = 27|A| …(i)

As, according to the question

|3A| = k|A|

Using (i),

⇒ 27|A| = k|A|

Comparing the left hand side and right hand side, we get

k = 27

Thus, the value of k is 27.

Discussion

95.	If A is a 2 × 2 matrix such that \|A\| ≠ 0 and \|A\| = 5, write the value of \|4A\|.
Answer» It is given that A is a 2 × 2 matrix We know that \|4A\| = 4². \|A\| By substituting the values \|4A\| = 16 (5) = 80

95.

If A is a 2 × 2 matrix such that |A| ≠ 0 and |A| = 5, write the value of |4A|.

Answer»

It is given that

A is a 2 × 2 matrix

We know that

|4A| = 4². |A|

By substituting the values

|4A| = 16 (5) = 80

Discussion

96.	If A is a 3 × 3 matrix such that \|A\| ≠ 0 and \|3A\| = k\|A\| then write the value of k.
Answer» Theorem: If Let A be k × k matrix then \|pA\|=p^k \|A\|. Given: k=3 and p=3. \|3A\|=3³ × \|A\| =27\|A\|. Comparing above with k\|A\| gives k=27.

96.

If A is a 3 × 3 matrix such that |A| ≠ 0 and |3A| = k|A| then write the value of k.

Answer»

Theorem: If Let A be k × k matrix then |pA|=p^k |A|.

Given: k=3 and p=3.

|3A|=3³ × |A|

=27|A|.

Comparing above with k|A| gives k=27.

Discussion

97.	If A is a 3 × 3 matrix such that \|A\| ≠ 0 and \|3A\| = k \|A\| then write the value of k.
Answer» It is given that A is a 3 × 3 matrix We know that \|3A\| = k \|A\| It can be written as 3² \|A\| = k \|A\| So we get k = 27

97.

If A is a 3 × 3 matrix such that |A| ≠ 0 and |3A| = k |A| then write the value of k.

Answer»

It is given that A is a 3 × 3 matrix

We know that

|3A| = k |A|

It can be written as

3² |A| = k |A|

So we get

k = 27

Discussion

98.	Let A be a square matrix of order 3, write the value of \|2A\|, where \|A\| = 4.
Answer» It is given that A is a square matrix of order 3 So we get \|2A\| = 2³ . \|A\| By substituting the values \|2A\| = 8 (4) = 32

98.

Let A be a square matrix of order 3, write the value of |2A|, where |A| = 4.

Answer»

It is given that A is a square matrix of order 3

So we get

|2A| = 2³ . |A|

By substituting the values

|2A| = 8 (4) = 32

Discussion

99.	Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}0& x & y \\[0.3em]-x & 0 & z \\[0.3em]-y & -z &0\end{vmatrix}\)
Answer» Let Δ =\(\begin{vmatrix}0& x & y \\[0.3em]-x & 0 & z \\[0.3em]-y & -z &0\end{vmatrix}\) Multiplying C₁, C₂ and C₃ with z, y and x respectively we get, ⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& xy & yx \\[0.3em]-xz & 0 & zx \\[0.3em]-yz & -zy &0\end{vmatrix}\) Now, Taking y, x and z common from R₁,R₂ and R₃ gives, ⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& x & x \\[0.3em]-z & 0 & z \\[0.3em]-y & -y &0\end{vmatrix}\) Applying C₂ → C₂ – C₃ gives, ⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& x & x \\[0.3em]-z & -z & z \\[0.3em]-y & -y &0\end{vmatrix}\) As, C₁ = C₂, Therefore determinant is zero.

99.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}0& x & y \\[0.3em]-x & 0 & z \\[0.3em]-y & -z &0\end{vmatrix}\)

Answer»

Let Δ =\(\begin{vmatrix}0& x & y \\[0.3em]-x & 0 & z \\[0.3em]-y & -z &0\end{vmatrix}\)

Multiplying C₁, C₂ and C₃ with z, y and x respectively we get,

⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& xy & yx \\[0.3em]-xz & 0 & zx \\[0.3em]-yz & -zy &0\end{vmatrix}\)

Now,

Taking y, x and z common from R₁,R₂ and R₃ gives,

⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& x & x \\[0.3em]-z & 0 & z \\[0.3em]-y & -y &0\end{vmatrix}\)

Applying C₂ → C₂ – C₃ gives,

⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& x & x \\[0.3em]-z & -z & z \\[0.3em]-y & -y &0\end{vmatrix}\)

As,

C₁ = C₂,

Therefore determinant is zero.

Discussion

100.	Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}2& 3 & 7 \\[0.3em]13& 17 &5 \\[0.3em]15 &20 & 12\end{vmatrix}\)
Answer» Let Δ = \(\begin{vmatrix}2& 3 & 7 \\[0.3em]13& 17 &5 \\[0.3em]15 &20 & 12\end{vmatrix}\) Applying C₃ → C₃ – C₂, gives ⇒ Δ = \(\begin{vmatrix}2& 3 & 7 \\[0.3em]13& 17 &5 \\[0.3em]2 &3 & 7\end{vmatrix}\) As, R₁ = R₃, So value so determinant is zero.

100.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}2& 3 & 7 \\[0.3em]13& 17 &5 \\[0.3em]15 &20 & 12\end{vmatrix}\)

Answer»

Let Δ = \(\begin{vmatrix}2& 3 & 7 \\[0.3em]13& 17 &5 \\[0.3em]15 &20 & 12\end{vmatrix}\)

Applying C₃ → C₃ – C₂, gives

⇒ Δ = \(\begin{vmatrix}2& 3 & 7 \\[0.3em]13& 17 &5 \\[0.3em]2 &3 & 7\end{vmatrix}\)

As,

R₁ = R₃,

So value so determinant is zero.

Discussion

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If A is a 3 × 3 matrix such that |A| ≠ 0 and |3A| = k|A| then write the value of k.

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Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}2& 3 & 7 \\[0.3em]13& 17 &5 \\[0.3em]15 &20 & 12\end{vmatrix}\)