Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

501.

Evalute the determinants in queations 1 and 2 : If `A=|{:(1,0,1),(0,1,2),(0,0,4):}|`

Answer» `A=|{:(1,0,1),(0,1,2),(0,0,4):}|`
`rArr" "A=|{:(1,0,1),(0,1,2),(0,0,4):}|=1|{:(1,2),(0,4):}|-0-0=4-0=4`
`rArr" "27|A|=27xx4=108`
Now, `3A=|{:(3,0,3),(0,3,6),(0,0,12):}|`
`|3A|=|{:(3,0,3),(0,3,6),(0,0,12):}|=3|{:(3,6),(0,12):}|-0+0`
=3(36-0)108`
`therefore" "|A|=27|A|`
Discussion
502.

Evalute the determinants in queations 1 and 2 : `[{:(2,4),(-5,-1):}]`

Answer» Let `A=[{:(2,4),(-5,-1):}]`
`rArr" "|A|=[{:(2,4),(-5,-1):}]=-2-(-20)=2=18`
Discussion
503.

Evalute the determinants in queations 1 and 2 : If `A=[{:(1,2),(4,2):}]`, then show that |2A|=4|A|.

Answer» `A=[{:(1,2),(4,2):}]=2-8=-6`
`therefore:" "R.H.S.=4|A|=4(-6)=-24`
`"and "2A=2[{:(1,2),(4,2):}]=[{:(2,4),(8,4):}]`
`rArr" "|2A|=[{:(2,4),(8,4):}]=8-32=-24`
`therefore" "L.H.S.=-24`
Now |2A|=4|A|
Discussion
504.

Evalute the determinants in queations 1 and 2 : If `A = |{:(1,1,-2),(2,1,-3),(5,4,-9):}|`, find |A|.

Answer» `|A|=|{:(1,1,-2),(2,1,-3),(5,4,-9):}|`
`rArr" "|A|=|{:(1,1,-2),(2,1,-3),(5,4,-9):}|`
`=1|{:(1,-3),(4,-9):}|-1|{:(2,-3),(5,-9):}|+(-2)|{:(2,1),(5,4):}|`
(Expanding along`R_(2)`)
=1(-9+12)-1(-18+15)-2(8-5)
=3+3-6=0
Discussion
505.

Choose the correct answer from the following : The value of `|{:(-costheta,sintheta),(-sontheta,-costheta):}|`is:A. `-1`B. 1C. 0D. None of these

Answer» Correct Answer - B
Discussion
506.

If `x , y in R ,`then the determinant `=|cosx-sinx1sinxcosx1cos(x+y)-sin(x+y)0|`lies in the interval`[-sqrt(2),sqrt(2)]`(b) `[-1,1]``[-sqrt(2),1]`(d) `[-1,-sqrt(2)]`

Answer» `|(cos x , -sin x,1),(sin x, cos x,1),(cos(x+y), -sin(x+y), 0)|`
`|(cos x, -sin x,1),(sin x, cos x,1),(cos x cos y - sin xsiny , -sinxcosy - cosx siny , 0)|`
`r_3 -> r_3 - cosy(r_1) + siny(r_2)`
`|(cosx, - sinx,1),(sin x, cos x,1),(0,0, siny - cosy)|`
`(sin y - cos y)[cos^2 x - (-sin^2x)]`
`= (siny - cos y)(cos^2x+sin^2 x)`
`= siny - cosy`
`= -[cosy - siny]`
`= - sqrt2[ 1/sqrt2 cos y - 1/sqrt2 sin y]`
`- sqrt2 sin(pi/4 - y)`
when it is `=1`
`/_= - sqrt2`
when it is `= -1`
`/_ = sqrt2`
a is correct option
Discussion
507.

Show that: `|a b-cc+b a+c b c-a a-bb+a c|=(a+b+c)(a^2+b^2+c^2)dot`

Answer» `= |(a,b-c,c+b),(a+c, b, c-a),(a-b,b+a, c)|`
`c_1-> ac_1`
`= 1/a |(a^2, b-c,c+b),(a^2+ac, b,c-a),(a^2 - ab, b+a, c)|`
`c_1-> c_1 + bc_2 + c c_3`
`=1/a |(a^2+b^2+c^2 , b-c, c+b),(a^2+b^2+c^2, b,c-a),(a^2+b^2+c^2, b+a,c)|`
`r_2-> r_2- r_1 & r_3-> r_3- r_1`
`= (a^2+b^2+c^2)/a |(1,b-c, c+b),(0,c,-(a+b)),(0,a+c,-b)|`
`= (a^2 + b^2+c^2)/a [ (a+b)(a+c)- bc]`
`= (a^2 + b^2 +c^2)/a [a^2 + ab+ac+bc-bc]`
`= (a^2+b^2+c^2)(a+b+c)`
hence proved
Discussion
508.

Evalute the determinants in queations 1 and 2 : (i) `[{:(costheta,-sintheta),(sintheta,costheta):}|` (ii) `|{:(x^(2),-x+1,x-1),(,x+1,x+1):}|`

Answer» (i) `[{:(costheta,-sintheta),(sintheta,costheta):}]=cos^(2)theta-(-0sin^(2)theta)`
`cos^(2)theta+sin^(2)theta=1`
(ii) `|{:(x^(2),-x+1,x-1),(,x+1,x+1):}|`
`(x^(2)-x+1)(x+1)-(x+1)(x-1)`
`=(x+1)(x^(2)-x+1-1)`
`=x^(3)+x^(2)-2x^(2)-2x+2x+2`
`=x^(3)-x^(2)2`
Discussion
509.

Write the value of the following determinant:`|102 18 36 1 3 4 17 3 6|`

Answer» Apply `R_(1) to R_(1) - 6R_(3)`
Discussion
510.

The maximum value of `Delta=|(1,1,1),(1,1+sintheta,1),(1+costheta,1,1)|` is (`theta` is real)(A) `1/2`(B) `(sqrt(3))/2`(C) `sqrt(2)`(D) `-(sqrt(3))/2`

Answer» `|(1,1,1),(1,1+sin theta, 1),(1+cos theta, 1,1)|`
`c_2-> c_2 - c_1`
`c_3 -> c_3 - c_1`
`|(1,0,0),(1,sin theta, 0),(1+cos theta, - cos theta, -cos theta)|`
`= 1(-sin theta cos theta)`
`= -(sin2 theta)/2`
`= 1/2`
Answer
Discussion
511.

`|[29, 26, 22], [25, 31, 27], [63, 54, 46]|`

Answer» Apply `C_(1) to C_(1) -C_(2) " and "C_(3) to C_(3) -C_(2)`
Take 3 common from `C_(1)` and -4 common from `C_(3)`
Apply `C_(2) to C_(2) - 26C_(3) " and " C_(3) to C_(3) -C_(1)`
Discussion
512.

For what value of x, the given matrix `A = [[3-2x, x+1],[2, 4]]` is a singular matrix?

Answer» Correct Answer - x =1
A is singular `hArr |A| = 0`
`hArr |[3-2x, x+1],[2, 4]| = 0`
`hArr 4(3-2x)-2(x+1) = 0 hArr 10x = 10 hArr x= 1`
Discussion
513.

Let `A = [(1,sin theta,1),(- sin theta,1,sin theta),(-1,-sin theta,1)], " where " 0 le theta lt 2 pi`. then, which of the following is not correct ?A. `D = 0`B. `D in (0, oo)`C. `D in [2, 4]`D. `D in [2, oo)`

Answer» Correct Answer - C
We have,
`D = |(1,sin theta,1),(- sin theta,1,sin theta),(-1,- sin theta,1)|`
`rArr D = |(2,sin theta,1),(0,1,sin theta),(0,- sin theta,1)| " " ["Applying " C_(1) rarr C_(1) + C_(3)]`
`rArr D = 2 (1 + sin^(2) theta)`
Now, `0 le sin^(2) theta le 1`
`rArr 2 le 2 (1 + sin^(2) theta) le 4 rArr D in [2, 4]`
Discussion
514.

Prove that the determinant `[(x,sintheta,costheta),(-sintheta,-x,1),(costheta,1,x)]` is independent of `theta`.

Answer» Expanding it across R3:-`costheta(sintheta+xcostheta)-1(x+sinthetacostheta)+x(-x^2+(sintheta)^2)``=>sinthetacostheta+x(costheta)^2+x(sintheta)^2-x-sinthetacostheta-x^3`=`-x^3`
Discussion
515.

Find the Determinant :-`|[67, 19, 21], [39, 13, 14], [81, 24, 26]|`

Answer» Apply `C_(1) to C_(1) -3C_(2), C_(3) to C_(3) -C_(2)`
Now, apply `C_(2) to C_(2) - 13C_(3). "Expand by"R_(2)`
Discussion
516.

Evaluate the following determinant :\(\begin{vmatrix} 67& 19& 21 \\[0.3em] 39& 13 &14 \\[0.3em] 81 &24 & 26 \end{vmatrix}\)

Answer»

Let Δ =  \(\begin{vmatrix} 67& 19& 21 \\[0.3em] 39& 13 &14 \\[0.3em] 81 &24 & 26 \end{vmatrix}\)

Applying, C1 → C1 – 4 C3, we get,

⇒ Δ =  \(\begin{vmatrix} 4& 19& 21 \\[0.3em] -3& 13 &14 \\[0.3em] -3&24 & 26 \end{vmatrix}\) 

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

⇒ Δ =  \(\begin{vmatrix} 1& 32& 35 \\[0.3em] -3& 13 &14 \\[0.3em] 0&11 & 12 \end{vmatrix}\) 

Now, applying R2 → R2 + 3R1, we get,

⇒ Δ =  \(\begin{vmatrix} 1& 32& 35 \\[0.3em] 0& 109 &119 \\[0.3em] 0&11 & 12 \end{vmatrix}\) 

= 1[(109)(12) – (119)(11)] 

= 1308 – 1309

= – 1 

So, 

Δ = – 1
Discussion
517.

if determinant `|{:( cos (theta + phi),,-sin (theta+phi),,cos 2phi),(sin theta,,cos theta,,sin phi),(-cos theta,,sintheta,,cos phi):}|` isA. non-negativeB. independent of 0C. independent of `phi`D. none of these

Answer» Correct Answer - A::B
Applying `R_(1) to R_(1) +sin phi (R_(2)) +cos phi (R_(3))`
`f(x)= Delta = |{:(0,,0,,cos 2phi+1),(sin0,,cos0,,sinphi),(-cos phi,,sin 0 ,,cos phi):}|`
`=(cos 2phi + 1) (sin^(2) 0+ cos^(2) 0)`
`=(1+cos 2phi)`
Hence `Delta` is independent of 0
Discussion
518.

Prove that the determinant `Delta =|{:(x,,sintheta,,cos theta),(-sin theta,,-x,,1),(cos theta,,1,,x):}|` is independent of `theta`.

Answer» Expanding along the first row we get
`Delta =|{:(x,,sin0,,cos 0),(-sin 0,,-x,,1),(cos 0,,1,,x):}|`
`=x(-x^(2)-1) -sin 0( -x sin 0- cos 0)`
` + cos 0 (-sin 0 +x cos 0)`
`=-x^(3) +x + x sin^(2)0 + sin 0 cos 0 -sin 0 cos + x cos^(2) 0`
`= -x^(3) + x+ x(sin^(2) 0+ cos^(2)0 )`
`= -x^(3) + 2x`
Hence `Delta ` is independent of 0.
Discussion
519.

Prove that the determinant `[xsinthetacostheta-sintheta-x1costheta1x]`is independent of 0.

Answer» `[{:(x,"sin"theta,"cos"theta),("-sin"theta,-x,1),("cos"theta,1,x):}]`
`[{:(-x,1),(1,x):}]"-sin"theta[{:("-sin"theta,1),("cos"theta,x):}]"+cos"theta|{:("-sin"theta,-x),("cos"theta,1):}|`
`=x(-x^(2)-1)-"sin"theta(-x"sin"theta-"cos"theta)+"cos"theta(-"sin"theta"+x"cos"theta)`
`=-x^(3)-x+x"sin"^(2)theta+"sin"theta"cos"theta-"sin"theta"cos"theta+x"cos"^(2)theta`
`=-x^(3)-x+x(sin^(2)theta+cos^(2)theta)`
`=-x^(3)-x+x=-x^(3)`
Discussion
520.

Let `|(x,2,x),(x^(2),x,6),(x,x,6)| = ax^(4) + bx^(3) + cx^(2) + dx + e` Then, the value of `5a + 4b + 3c + 2d + e` is equal to

Answer» Correct Answer - D
We have,
`|(x,2,x),(x^(2),x,6),(x,x,6)|`
`= |(x,2,x),(x^(2),x,6),(x -x^(2),0,0)| " " ["Applying " R_(3) rarr R_(3) - R_(2)]`
`= (x -x^(2)) (12 -x^(2))`
`= 12 x - x^(3) - 12x^(2) + x^(4)`
`:. a =1, b = -1, c = -12, d = 12 and e = 0`
`:. 5a + 4b + 3c + 2d + e = 5 - 4 - 36 + 24 + 0 = -11`
Discussion
521.

Let `|(x^2+3x,x-1,x+3),(x+1,-2x,x-4),(x-3,x+4,3x)| = ax^4 + bx^3 + cx^2 + dx + e `be an identity in x, where a, b, c, d, e are independent of x. Then, the value of e isA. 4B. 0C. 1D. none of these

Answer» Correct Answer - B
Clearly, e is the value of L.H.S. of the given identity at x = 0
For x = 0, We obtain
`LHS = |(0,-1,3),(1,0,-4),(-3,4,0)|`
Note 1 Only square matrices have determinants. The matrices which are not square do not have determinants
Note2 The determinant of a square matrix of order 3 can be expanded along any row or column.
Note 3 If a row or a column of a determinant consists of all zeros, then the value of the determinant is zero.
To evaluate the determinant of a square matrix of order 4 or more we follow the same procedure as discussed in evaluating the determinant of a square matrix of order 3.
Discussion
522.

If `A = int_(1)^(sintheta) (t)/(1 + r^(2)) dt and B = int_(1)^("cosec"theta) (1)/(t(1 +t^(2))) dt`, then the value of the determinant `|(A,A^(2),B),(e^(A +B),B^(2),-1),(1,A^(2) + B^(2) ,-1)|` isA. `sin theta`B. `cosec theta`C. 0D. 1

Answer» Correct Answer - C
We have,
`A + B = underset(1)overset(sin theta) int (t)/(1 +t^(2)) dt + underset(1)overset("cosec" theta)int (1)/(t(1 + t^(2))) dt`
`rArr A + B = underset(1)overset(sin theta)int (t)/(1 +t^(2)) dt + underset(1)overset(sin theta)int - (u)/(1 + u^(2)) du`, where `u = (1)/(t)`
`rArr A + B = 0`
`rArr B = - A`
`:. |(A,A^(2),B),(e^(A +B),B^(2),-1),(1,A^(2) + B^(2),-1)|`
`= |(A,A^(2),-A),(1,A^(2),-1),(1,2A^(2),-1)| = -|(A,A^(2),A),(1,A^(2),1),(1,2A^(2),1)| = 0`
Discussion
523.

The value of the determinant `|(10!,11!,12!),(11!,12!,13!),(12!,13!,14!)|`, isA. `2(10! 11!)`B. `2(10! 13!)`C. `2(10! 11! 12!)`D. `2(11 ! 12! 13!)`

Answer» Correct Answer - C
We have,
`|(10!,11!,12!),(11!,12!,13!),(12!,13!,14!)|`
`= 10! xx 11! xx12! |(1,11,132),(1,12,156),(1,13,182)|`
`= 10! xx 11! xx 12! |(1,11,132),(0,1,24),(0,1,26)|[("Applying " R_(2) rarr R_(2)- R_(1)),(" "R_(3) rarr R_(3) - R_(2))]`
`= 2 (10! xx 11! xx 12!)`
Discussion
524.

If `Delta_(1) = |(1,1,1),(a,b,c),(a^(2),b^(2),c^(2))|, Delta_(2) = |(1,bc,a),(1,ca,b),(1,ab,c)|`, thenA. `Delta_(1) + Delta_(2) =0`B. `Delta_(1) + 2Delta_(2) = 0`C. `Delta_(1) = Delta_(2)`D. none of these

Answer» Correct Answer - A
We have,
`Delta_(2) = |(1,bc,a),(1,ca,b),(1,ab,c)|` ltrbgt `rArr Delta_(2) = (1)/(abc)|(a,abc,a^(2)),(b,abc,b^(2)),(c,abc,c^(2))| "Applying " R_(1) rarr R_(1) (a) " " R_(2) rarr R_(2) (b) " " R_(3) rarr R_(3) (c)`
`rArr Delta_(2) = |(a,1,a^(2)),(b,1,b^(2)),(c,1,c^(2))|` [Taking abc common from `C_(2)`]
`rArr Delta_(2) = -|(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2))|` [Interchanging `C_(1) and C_(2)`]
`rArr Delta_(2) = - |(1,1,1),(a,b,c),(a^(2),b^(2),c^(2))| = - Delta_(1) rArr Delta_(1) + Delta_(2) = 0`
Discussion
525.

If `|(-12,0,lamda),(0,2,-1),(2,1,15)| = -360`, then the value of `lamda` isA. `-1`B. `-2`C. `-3`D. 4

Answer» Correct Answer - C
Discussion
526.

Given, `2x - y + 2z = 2, x - 2y + z = -4, x + y+ lamda z = 4`,then the value of `lambda` such that the given system of equations has no solution, isA. 3B. 1C. 0D. `-3`

Answer» Correct Answer - B
The given system of equations will have no solution, if `D = 0` and at least one of `D_(1), D_(2), D_(3)` is non-zero, where
`D = |(2,-1,2),(1,-2,1),(1,1,lamda)|, D_(1) = |(2,-1,2),(-4,-2,1),(4,1,lamda)|, D_(2) = |(2,2,2),(1,-4,1),(4,1,lamda)| and, D_(3) = |(2,-1,2),(1,-2,-4),(1,1,4)|`
Now,
`D = 0`
`rArr |(2,-1,2),(1,-2,1),(1,1,lamda)| = 0`
`rArr 2(-2 lamda -1) + (lamda - 1) + 2 (1 +2 ) = 0`
`rArr -3 lamda + 3 = 0`
`rArr lamda = 1`
For this value of `lamda`, weh ave
`D_(1) = |(2,-1,2),(-4,-2,1),(4,1,1)| = 6 - 8 + 8 != 0`
Hence, the system has no solution for `lamda = 1`
Discussion
527.

Let `|(1 +x,x,x^(2)),(x,1 +x,x^(2)),(x^(2),x,1 +x)| = ax^(5) + bx^(4) + cx^(3) + dx^(2) + lamdax +mu` be an identity in x, where a, b, c, `lamda, mu` are independent of x. Then, the value of `lamda` isA. 3B. 2C. 4D. none of these

Answer» Correct Answer - A
Let `Delta = |(1 +x,x,x^(2)),(x,1 +x,x^(2)),(x^(2),x,1 +x)|`
Applying `C_(1) rarr C_(1) + C_(2) + C_(3)`, we get
`Delta = (1 +x)^(2) |(1,x,x^(2)),(1,1 +x,x^(2)),(1,x,1 +x)|`
Clearly, `lamda = ((d Delta)/(dx))_(x = 0)`
We have,
`(d Delta)/(dx) = 2 (1 + x) |(1,x,x^(2)),(1,1 +x,x^(2)),(1,x,1 +x)| + (1 +x)^(2) |(1,1,x^(2)),(1,1,x^(2)),(1,1,1 +x)| + (1 +x)^(2) |(1,x,2x),(1,1 +x,2x),(1,x,1)|`
`:. ((d Delta)/(dx))_(x =0) = 2 |(1,0,0),(1,1,0),(1,0,1)| + 0 + |(1,0,0),(1,1,0),(1,0,1)| = 2 + 1 = 3`.
Hence, `lamda = 3`
Discussion
528.

Prove that all values of theta: `|(sintheta, costheta, sin2theta),(sin(theta+(2pi)/3), cos(theta(2pi)/3), sin (2theta+(4pi)/3)),(sin (theta- (2pi)/3), cos (theta- (2pi)/3), sin (2theta- (4pi)/3))|=0`A. `sin theta`B. `cos theta`C. `sin theta cos theta`D. none of these

Answer» Correct Answer - D
Applying `R_(2) rarr R_(2) + R_(3)`, we have,
`Delta = |(sin theta,cos theta,sin 2 theta),(- sin theta,- cos theta,- sin 2 theta),(sin (theta - (2pi)/(3)),cos (theta - (2pi)/(3)),sin (2 theta - (4pi)/(3)))|= 0`
Discussion
529.

If `f(x) = |(x + lamda,x,x),(x,x + lamda,x),(x,x,x + lamda)|, " then " f(3x) - f(x) =`A. `3x lamda^(2)`B. `6x lamda^(2)`C. `x lamda^(2)`D. none of these

Answer» Correct Answer - B
We have,
`f(x) = |(x + lamda,x,x),(x,x + lamda,x),(x,x,x + lamda)|`
`rArr f(x) |(3x + lamda,x,x),(3x + lamda,x + lamda,x),(3x + lamda,x,x + lamda)| " " [(" Applying "),(C_(1) rarr C_(1) + C_(2) + C_(3))]`
`rArr f(x) = (3x + lamda) |(1,x,x),(1,x + lamda,x),(1,x,x + lamda)|`
`rArr f(x) = (3x + lamda) |(1,x,x),(0,lamda,0),(0,0,lamda)| " " [("Applying " R_(2) rarr R_(2) - R_(1)),(" "R_(3) rarr R_(3) - R_(1))]`
`rArr f(x) = (3x + lamda) lamda^(2)`
`:. f(3x) - f(x) = (9x + lamda) lamda^(2) - (3x + lamda) lamda^(2) = 6x lamda^(2)`
Discussion
530.

The value of the determinant `|(bc,ca,ab),(p,q,r),(1,1,1)|`, where a, b and c respectively the pth,qth and rth terms of a H.P., is

Answer» Correct Answer - A
Let D be the common difference of the corresponding A.P. and A be its first term. Then,
`(1)/(a) = A + (p -1) D, (1)/(b) = A + (q -1) D and (1)/(c) = A + (r - 1) D`
Now,
`|(bc,ca,ab),(p,q,r),(1,1,1)|`
`= abc |((1)/(a),(1)/(b),(1)/(c)),(p,q,r),(1,1,1)| " " ["Applying" R_(1) rarr R_(1) ((1)/(ac))]`
`= abc |((1)/(a),(1)/(b),(1)/(c)),(p -1,q -1,r -1),(1,1,1)| " " ["Applying" R_(2) rarr R_(2) - R_(3)]`
`= abc |((1)/(a) - (p -1) D,(1)/(b) - (q -1) D,(1)/(c) - (r -1) D),(p -1,q -1,r -1),(1,1,1)| " " ["Applying " R_(1) rarr R_(1) - DR_(2)]`
`= abc |(A,A,A),(p -1,q -1,r -1),(1,1,1)| = abc xx 0 = 0`
Discussion
531.

`" if " a_(a) ,a_(2), a_(3)….." are in A.P, then find the value of the following determinant:"` `|{:(a_(p)+a_(p+m) +a_(p+2m),,2a_(p)+3a_(p+m)+4a_(p+2m),,4a_(p)+9a_(p+m)+16a_(p+2m)),(a_(p)+a_(q+m)+a_(q+2m),,2a_(q)+3a_(q+m) +4a_(q+2m),,4a_(q)+9a_(q+m)+16a_(q+2m)),(a_(r)+a_(r+m)+a_(r+2m),,2a_(r)+3a_(r+m)+4a_(r+2m),,4a_(r) +9a_(r+m)+16a_(r+2m)):}|`

Answer» `|{:(a_(p)+a_(p+m) +a_(p+2m),,2a_(p)+3a_(p+m)+4a_(p+2m),,4a_(p)+9a_(p+m)+16a_(p+2m)),(a_(p)+a_(q+m)+a_(q+2m),,2a_(q)+3a_(q+m) +4a_(q+2m),,4a_(q)+9a_(q+m)+16a_(q+2m)),(a_(r)+a_(r+m)+a_(r+2m),,2a_(r)+3a_(r+m)+4a_(r+2m),,4a_(r) +9a_(r+m)+16a_(r+2m)):}|`
Applying `C_(2) to C_(2) -2C_(1) " and " C_(3) to C_(3) -4C_(1), ` we get
`|{:(a_(p)-a_(p+2m),,a_(p+m)+2a_(p+2m),,2a_(p+2m)),(a_(q)-a_(q+2m),,a_(q+m)+2a_(q+2m),,2a_(q+2m)),(a_(r) -a_(r+2m),,a_(r+m)+2a_(r+2m),,2a_(r+2m)):}|`
Applying `C_(2) to C_(2) -C_(3) to C_(1) +(1)/(2) C_(3)` and then taking 2 common from `C_(3)` we get
`=2 |{:(a_(p),,a_(p+m),,a_(p+2m)),(a_(q),,a_(q+m),,a_(p+2m)),(a_(r),,a_(r+m),,a_(r+2m)):}|`
Applying `C_(1) to C_(1)+C_(3)+2C_(2)`
`=2 |{:(a_(p)+a_(p+2m),,a_(p+m),,a_(p+2m)),(a_(q)+a_(q+2m),,a_(q+m),,a_(p+2m)),(a_(r)+a_(r+2m),,a_(r+m),,a_(r+2m)):}|`
`=2 |{:(0,,a_(p+m),,a_(p+2m)),(0,,a_(q+m),,a_(p+2m)),(0,,a_(r+m),,a_(r+2m)):}|`
`(Ax a_(p)a_(p+m),a_(p+2m)` are in A.P.)
`=0`
Discussion
532.

Using properties ofdeterminants, prove the following:`|(1,x,x^2),(x^2, 1,x),(x,x^2, 1)|=(1-x^3)^2`

Answer» Let the value of the given determinant be `Delta`. Then,
`Delta = |{:(1, x, x^(2)), (x^(2), 1, x),(x, x^(2), 1):}|`
`= |{:(1+x+x^(2), x, x^(2)), (1+x+x^(2), 1, x),(1+x+x^(2), x^(2), 1):}| ["applying"C_(1) to (C_(1) + C_(2) +C_(3))]`
`=(1+x+x^(2)) |{:(1, x, x^(2)), (1, 1, x),(1, x^(2), 1):}| ["taking"(1+x+x^(2))"common from"C_(1)]`
`=(1+x+x^(2)) |{:(1, " "x, " "x^(2)), (0, 1-x, x(1-x)),(0, x(x-1), (1-x)(1+x)):}| ["applying"R_(2) to (R_(2)-R_(1)) "and"R_(3) to (R_(3)-R_(1))]`
`=(1+x+x^(2))(1-x)^(2)* |{:(1, x, x^(2)), (0, 1, x),(0, -x, (1+x)):}| ["taking(1-x) common from each of "R_(2)" and "R_(3)]`
` =(1+x+x^(2))(1-x)^(2)*1* |{:(1, x), (-x, (1+x)):}| ["expanded by"C_(1)]`
`=(1+x+x^(2))(1-x)^(2) * (1+x+x^(2))`
`={(1-x)(1+x+x^(2))}^(2) = (1-x^(3))^(2)`
Hence, `Delta = (1-x^(3))^(2)`
Discussion
533.

Using properties of determinants, prove that`|(a+x, y, z),( x, a+y, z),( x, y, a+z)|=a^2(a+x+y+z)`

Answer» Let the value of the given determinant be `Delta`. Then.
`Delta =|{:(a+x, " "y, " "z), (" "x, a+y, " "z), (" "x, " "y, a+z):}|`
`=|{:(a+x+y+z, " "y, " "z), (a+x+y+z, a+y, " "z), (a+x+y+z, " "y, a+z):}| " "[C_(1) to C_(1) + C_(2) + C_(3)]`
`=(a +x +y+z)*|{:(1, y, z), (1, a+y, z), (1, y, a+z):}| " "["taking (a+x+y+z) common from"C_(1)]`
`=(a +x +y+z)*|{:(1, y, z), (0, a, 0), (0, 0, a):}| " "[R_(2) to R_(2) - R_(1) " and"R_(3) to R_(3)-R_(1)]`
`=(a +x +y+z)*1*|{:(a, 0), (0,a):}| " "["expanded by"C_(1)]`
` =a^(2) (a+x+y+z).`
Hence, `Delta =a^(2) (a+x+y+z)`
Discussion
534.

Prove that: `|(alpha,beta,gamma),(alpha^2,beta^2,gamma^2),(beta+gamma,gamma+alpha,alpha+beta)|``=(alpha-beta)(beta-gamma)(gamma-alpha)(alpha+beta+gamma)`.

Answer» Let the value of the given determinant be `Delta`. Then.
`Delta = |{:(alpha, beta, gamma), (alpha^(2), beta^(2), gamma^(2)), (beta + gamma, gamma + alpha, alpha + beta):}|`
`=|{:(alpha, beta, gamma), (alpha^(2), beta^(2), gamma^(2)), (alpha +beta + gamma, alpha + beta +gamma, alpha + beta +gamma):}|["applying "R_(3) to (R_(3) + R_(1))]`
`=(alpha + beta +gamma) * |{:(alpha, beta, gamma), (alpha^(2), beta^(2), gamma^(2)), (1, 1, 1):}|["taking" (alpha + beta + gamma)"common from "R_(3)]`
`=(alpha + beta +gamma) * |{:(alpha-gamma, beta-gamma, gamma), (alpha^(2)-gamma^(2), beta^(2)-gamma^(2), gamma^(2)), (0, 0, 1):}|["applying"C_(1)to (C_(1)-C_(3))"and "C_(2) to (C_(2)-C_(3))]`
`=(alpha + beta +gamma)(alpha-gamma)(beta-gamma) * |{:(1, 1, gamma), (alpha+gamma, beta+gamma, gamma^(2)), (0, 0, 1):}|`
`["taking" (alpha-gamma) "common from "C_(1) " and "(beta-gamma) "common from"C_(2)]`
`= (alpha + beta + gamma)(alpha - gamma)(beta -gamma) *1* |{:(1, 1), (alpha + gamma, beta + gamma):}| " "["expanded by"R_(3)]`
`=(alpha + beta + gamma) (alpha- gamma) (beta -gamma) [(beta + gamma) - (alpha + gamma)]`
`=(alpha + beta + gamma)(alpha -gamma)(beta-gamma)(beta -alpha)`
`=(alpha -beta)(beta-gamma)(gamma-alpha) (alpha + beta +gamma).`
Hence, `Delta = (alpha-beta)(beta-gamma)(gamma-alpha)(alpha + beta + gamma).`
Discussion
535.

Prove that `|{:(1, 1, 1),(a, b, c),(a^(3), b^(3), c^(3)):}|=(a-b)(b-c)(c-a)(a+b+c)`

Answer» Let the value of the given determinant be `Delta`. Then,
`Delta =|{:(1, 1, 1),(a, b, c),(a^(3), b^(3), c^(3)):}|`
`=|{:(0, 0, 1),(a-c, b-c, c),(a^(3)-c^(3), b^(3)-c^(3), c^(3)):}|" "["applying"C_(1) to (C_(1) -C_(3))" and " C_(2) to (C_(2) -C_(3))]`
`=(a-c)(b-c) *|{:(0, 0, 1),(1, 1, c),(a^(2)+ac+c^(2), b^(2)+bc+c^(2), c^(3)):}|" "["taking out (a-c) and (b-c) common from"C_(1) " and " C_(2)]` ltbr `=(a-c)(b-c) *1*|{:(1, 1),(a^(2)+ac+c^(2), b^(2) + ac +c^(2)):}| " "["expanded by"R_(1)]`
`=(a-c)(b-c) * [(b^(2) +bc+c^(2)) -(a^(2) +ac+c^(2))]`
`=(a-c)(b-c)[(b^(2)-a^(2)) + (b-a)c]`
`= (a-c)(b-c)(b-a)(b+a+c)`
`=(a-b)(b-c)(c-a)(a+b+c)`
Hence, `Delta = (a-b)(b-c)(c-a) (a+b+c)`.
Discussion
536.

Prove that `|(1,a,a^2),(1,b,b^2),(1,c,c^2)|=(a-b)(b-c)(c-a)`

Answer» Let the given determinant be `Delta`. Then,
`Delta =|{:(1, a, a^(2)),(1, b, b^(2)),(1, c, c^(2)):}|`
`=|{:(1, a, a^(2)),(0, b-a, b^(2)-a^(2)),(0, c-a, c^(2)-a^(2)):}| ["applying" R_(2) to (R_(2)-R_(1))"and" R_(3) to (R_(3) -R_(1))]`
` = (b-a)(c-a) * |{:(1, a, a^(2)),(0, 1, b+a),(0, 1, c+a):}|`
`["taking (b-a) common from "R_(2) "and (c-a) common from "R_(3)]`
`=(b-a)(c-a) xx 1 * |{:(1, b+a), (1, c+a):}| " "["expanded by"C_(1)]`
`=(b-a)(c-a){(c+a)-(b+a)}`
`=(b-a)(c-a)(c-b) = (a-b)(b-c)(c-a)`
Hence, `Delta = (a-b)(b-c)(c-a)`
Discussion
537.

Q. `|(x+y,x,x),(15x+4y,4x,2x),(10x +8y,8x,3x)|=x^3`

Answer» Let the given determinant be `Delta`. Then,
`Delta = |{:(x+y, x, x),(5x+4y,4x, 2x),(10x+8y, 8x, 3x):}|`
`= |{:(x+y, x, x),(3x+2y, 2x, 0),(7x+5y, 5x, 0):}| [R_(2) to (R_(2)-2R_(1)) " and"R_(3)to(R_(3)to 3R_(1))]`
`=x* |{:(3x+2y, 2x),(7x+5y, 5):}| ["expanded by"C_(3)]`
`=x^(2)* |{:(3x+2y, 2x),(7x+5y, 5):}| ["taking x common from "C_(2)]`
`=x^(2) *[5(3x +2y)-2(7x+5y)] =(x^(2)*x) =x^(3).`
Hence, `Delta = x^(3)`
Discussion
538.

Using properties of determinants, prove that `|{:(0, ab^(2), ac^(2)),(a^(2)b, 0, bc^(2)),(a^(2)c, cb^(2), 0):}|=2a^(3)b^(3)c^(3)`

Answer» Let the given determinant be `Delta`. Then,
`Delta = |{:(0, ab^(2), ac^(2)),(a^(2)b, 0, bc^(2)),(a^(2)c, cb^(2), 0):}|`
`=(a^(2)b^(2)c^(2))* |{:(0, a, a),(b, 0, b),(c, c, 0):}| " "["taking"a^(2), b^(2), c^(2) "common from"C_(1), C_(2) " and "C_(3)"respectively"]`
`=(a^(3)b^(3)c^(3))* |{:(0, 1, 1),(1, 0, 1),(1, 1, 0):}| " "["taking a, b, c common from"R_(1), R_(2) " and "R_(3)"respectively"]`
`=(a^(3)b^(3)c^(3))* |{:(0, 1, 1),(1, 0, 1),(0, 1, -1):}| " "[R_(3)to R_(3) - R_(2)]`
`=(a^(3)b^(3)c^(3))*(-1)* |{:(1, 1),(1, -1):}| " "["expanded by"C_(1)]`
`= -(a^(3)b^(3)c^(3))(-1-1) = 2a^(3)b^(3)c^(3).`
Hence, `Delta = 2a^(3)b^(3)c^(3).`
Discussion
539.

Prove that : (i) `|{:(a,c,a+c),(a+b,b,a),(b,b+c,c):}|=2 abc` (ii) Prove that : `|{:(a^(2),bc,ac+c^(2)),(a^(2)+ab,b^(2),ac),(ab,b^(2)+bc,c^(2)):}|=4a^(2)b^(2)c^(2)`

Answer» `|{:(a^(2), bc, c^(2)+ac), (a^(2)+ab, b^(2), ac), (ab, b^(2)+bc, c^(2)):}|`
`=(abc)*|{:(" "a, " "c, ca), (a+b, " "b, a), (" "b, b+c, c):}|`
`["taking a, b, c common from"C_(1), C_(2) " and "C_(3) "respectively"]`
`=(abc)*|{:(a, " "c, c+a), (0, -2c, -2c), (b, b+c, " "c):}| " "[R_(2)to R_(2) -(R_(1) + R_(3))]`
`=(abc)*|{:(a, -a, c+a), (0, 0, -2c), (b, b, c):}| " "[C_(2)to C_(2) -C_(3)]`
`=(abc)(2c)*|{:(a, -a), (b, b):}| " "["expanded by"R_(2)]`
` = 2abc^(2) * (ab+ab) = 2abc^(2)(2ab)`
`=4a^(2)b^(2)c^(2)`.
Discussion
540.

Let a, b, c be positive and not all equal. Show that the value of the determinant ` |[a,b,c],[b,c,a],[c,a,b]| ` is negative

Answer» ` Delta =|{:(a,,b,,c),(b ,,c,,a),(c,,a,,b):}|`
Expanding using Sarrus rule we get
`Delta =abc + abc +abc -a^(3) -b^(3)-c^(3)`
`=- (a^(3) =b^(3) +c^(3) - 3abc)`
`=-(a+b+c ) [a^(2) +b^(2) +c^(2) -ab -bc -ca]`
` =-(1)/(2) (a+b+c) [2a^(2) + 2b^(2) + 2c^(2) - 2ab- 2bc - 2ca]`
`=-(1)/(2) (a + b+ c)[(a-b)^(2) + (b -c)^(2) + (c-a)^(2)]`
Since a,b,c ` gt 0 , a + b+ c gt0`
Also ,` a ne b ne c`
`:. Delta lt 0`
Discussion
541.

Let `Delta = |{:(-bc,,b^(2)+bc,,c^(2)+bc),(a^(2)+ac,,-ac,,c^(2)+ac),(a^(2)+ab,,b^(2)+ab,,-ab):}|` and the equation `px^(3) +qx^(2) +rx+s=0` has roots a,b,c where `a,b,c in R^(+)` The value of `Delta` isA. `le 9r^(2)//p^(2)`B. `ge 27s^(2)//p^(2)`C. `le 27s^(3)//p^(3)`D. none of these

Answer» Correct Answer - B
Multiplying `R_(1) ,R_(2),R_(3) " by " a,b,c ` respectively ,and then taking a,b,c common from` C_(1),C_(2)" and " C_(3)` we get
`Delta = |{:(-bc,,ab+ac,,ac+ab),(ab+bc,,-ac,,bc+ab),(ac+bc,,bc+ac,,-ab):}|`
Now using `C_(2) to C_(2)-C_(1)" and " C_(3) to C_(3) -C_(1) ` and then taking (ab+bc+ca) common from `C_(2) " and " C_(3)` we get
`Delta =|{:(-bc,,1,,1),(ab+bc,,-1,,0),(ac+bc,,0,,-1):}|xx (ab +bc +ca)^(2)`
Now applying `R_(2) to R_(2) +R_(1)` we get
`Delta = |{:(-bc,,1,,1),(ab,,0,,1),(ac+bc,,0,,-1):}| (ab+bc+ca)^(2)`
Expanding along `c_(2)` we get
`Delta =(ab+bc+ca)^(2)[ac+bc+ca)^(2)`
`=(ab+bc+ca)^(2)`
`=(r//p)^(3) =r^(3)//p^(3)`
Now given a,b,c are all positive then
`A.M ge G.M.`
`rArr (ab+bc+ac)/(3) ge (abxx bcxx ac)^(1//3)`
`" or " (ab+bc+ac)^(3) ge 27a^(2)b^(2)c^(2)`
`" or " (ab+bc+ca)^(3) ge 27(s^(2)//p^(2))`
if `Delta =27` then ab+bc+ca =3 and given that `a^(2) +b^(2)+c^(2)=3`
From `(a+b+c)^(2) =a^(2) +b^(2)+c^(2) +2 (ab+bc+ca)` we have
`a+b+c = ne 3`
`rArr a+b+c =3`
`rArr 3p+q=0`
Discussion
542.

If a, b, c are positive and unequal, show that value of the determinant `Delta=|(a,b,c),(b,c,a),(c,a,b)|` is negative

Answer» The given determinant
`=|{:(a,b,c),(b,c,a),(c,a,b):}|`
`=|{:(a+b+c,b,c),(a+b+c,c,a),(a+b+c,a,b):}|" "[C_(1) to (C_(1) + C_(2) + C_(3))]`
`=(a+b+c)*|{:(1,b,c),(1,c,a),(1,a,b):}|`
`=(a+b+c)*|{:(1,b,c),(0,c-b,a-c),(0,a-b,b-c):}| [R_(2) to (R_(2)-R_(1)) " and "R_(3) to (R_(3) -R_(1))]`
`=(a+b+c0 * [(c-b)(b-c)-(a-b)(a-c)] " " ["expanded by"C_(1)]`
`(a+b+c)*(-a^(2)-b^(2)-c^(2)+ab+bc+ca)`
`=-(1)/(2) (a+b+c)(2a^(2) +2b^(2)+2c^(2)-2ab-2bc-2ca)`
`=-(1)/(2) (a+b+c)[(a-b)^(2) + (b-c)^(2) + (c-a)^(2)],` which is negative
`[because (a+b+c) gt 0, (a-b)^(2) gt 0, (b-c)^(2) gt 0 " and " (c-a)^(2) gt 0]`
Discussion
543.

Prove that: `|((b+c)^2,a^2,a^2),(b^2,(c+a)^2,b^2),(c^2,c^2,(a+b)^2)|=2a b c(a+b+c)^3`

Answer» Let the given determinant be `Delta`. Then.
`Delta =|{:((b+c)^(2), " "a^(2), a^(2)),(b^(2),(c+a)^(2), b^(2)),(c^(2)," "c^(2) ,(a+b^(2))):}|`
`=|{:((b+c)^(2)-a^(2), " "0, " "a^(2)),(" "0,(c+a)^(2)-b^(2), " "b^(2)),(c^(2)-(a+b)^(2),c^(2)-(a+b)^(2),(a+b^(2))):}| [C_(1) to C_(1)-C_(3) " and "C_(2) to C_(2) -C_(3)]`
`=|{:((a+b+c)(b+c-a), " "0, " "a^(2)),(" "0,(a+b+c)(c+a-b), " "b^(2)),((a+b+c)(c-a-b),(a+b+c)(c-a-b),(a+b^(2))):}|`
`=(a+b+c)^(2) *|{:((b+c-a), " "0, " "a^(2)),(" "0,c+a-b, " "b^(2)),(c-a-b,c-a-b,(a+b^(2))):}| ["taking (a + b+c) common from"C_(1) "and"C_(2) "both"]`
`=(a+b+c)^(2) *|{:((b+c-a), " "0, " "a^(2)),(" "0,c+a-b, " "b^(2)),(-2b,-2a, 2ab):}| [R_(3) to R_(3) -(R_(1) + R_(2))]`
`=(a+b+c)^(2)[(b+c-a){(c+a-b) *2b+2ab^(2)} + a^(2){0 +2b(c+a-b)}]`
`=(a+b+c)^(2)[(b+c-a)*2ab{(c+a-b+b)} + 2a^(2)b(c+a-b)]`
`=2ab(a+b+c)^(2){(b+c-a)(c+a)+a(c+a-b)}`
`=2ab(a+b+c)^(2) * {bc+ab+c^(2)+ac-ac-a^(2)+ac+a^(2)-ab}`
`=2ab(a+b+c)^(2){bc+c^(2)+ac} =2abc(a+b+c)^(3)`
Hence, `Delta = 2abc(a+b+c)^(3)`
Discussion
544.

Let `Delta = |{:(-bc,,b^(2)+bc,,c^(2)+bc),(a^(2)+ac,,-ac,,c^(2)+ac),(a^(2)+ab,,b^(2)+ab,,-ab):}|` and the equation `px^(3) +qx^(2) +rx+s=0` has roots a,b,c where `a,b,c in R^(+)` the value of `Delta ` isA. `r^(2)//p^(2)`B. `r^(3)//p^(3)`C. `-s//p`D. none of these

Answer» Correct Answer - B
Multiplying `R_(1) ,R_(2),R_(3) " by " a,b,c ` respectively ,and then taking a,b,c common from` C_(1),C_(2)" and " C_(3)` we get
`Delta = |{:(-bc,,ab+ac,,ac+ab),(ab+bc,,-ac,,bc+ab),(ac+bc,,bc+ac,,-ab):}|`
Now using `C_(2) to C_(2)-C_(1)" and " C_(3) to C_(3) -C_(1) ` and then taking (ab+bc+ca) common from `C_(2) " and " C_(3)` we get
`Delta =|{:(-bc,,1,,1),(ab+bc,,-1,,0),(ac+bc,,0,,-1):}|xx (ab +bc +ca)^(2)`
Now applying `R_(2) to R_(2) +R_(1)` we get
`Delta = |{:(-bc,,1,,1),(ab,,0,,1),(ac+bc,,0,,-1):}| (ab+bc+ca)^(2)`
Expanding along `c_(2)` we get
`Delta =(ab+bc+ca)^(2)[ac+bc+ca)^(2)`
`=(ab+bc+ca)^(2)`
`=(r//p)^(3) =r^(3)//p^(3)`
Now given a,b,c are all positive then
`A.M ge G.M.`
`rArr (ab+bc+ac)/(3) ge (abxx bcxx ac)^(1//3)`
`" or " (ab+bc+ac)^(3) ge 27a^(2)b^(2)c^(2)`
`" or " (ab+bc+ca)^(3) ge 27(s^(2)//p^(2))`
if `Delta =27` then ab+bc+ca =3 and given that `a^(2) +b^(2)+c^(2)=3`
From `(a+b+c)^(2) =a^(2) +b^(2)+c^(2) +2 (ab+bc+ca)` we have
`a+b+c = ne 3`
`rArr a+b+c =3`
`rArr 3p+q=0`
Discussion
545.

Let `f (n) = |{:(n,,n+1,,n+1),(.^(n)P_(n),,.^(n+1)P_(n+1),,.^(n+2)P_(n+2)),(.^(n)C_(n),,.^(n+1)C_(n+1),,.^(n+2)C_(n+2)):}|` where the sysmbols have their usual neanings .then f(n) is divisible byA. `n^(2)+n+1`B. `(n+1)!`C. `n!`D. none of these

Answer» Correct Answer - A::C
`f(n) = |{:(n,,n+1,,n+2),(n!,,(n+1)!,,(n+2)!),(1,,1,,1):}|=|{:(n,,1,,1),(n!,,nn!,,(n+1)(n+1)!),(1,,0,,0):}|`
[Applying `C_(3) to C_(3) -C_(2) " and " C_(2) to C_(2)-C_(1)]`
`=(n+1)(n+1)!-nn! =n![(n+1)^(2)-n]`
`=n!(n^(2) +n+1)` ltbr. Thus f(n) is divisible by n! and `n^(2) +n+1`
Discussion
546.

Evaluate `|{:(""^(m)C_(1),""^(m)C_(2),""^(m)C_(3)),(""^(n)C_(1),""^(n)C_(2),""^(n)C_(3)),(""^(p)C_(1),""^(p)C_(2), ""^(p)C_(3)):}|`

Answer» Let the given determinant be `Delta`. Then,
`Delta = |{:(m, (1)/(2)m(m-1),(1)/(6)*m(m-1)(m-2)),(n, (1)/(2)n(n-1),(1)/(6)*n(n-1)(n-2)),(p,(1)/(2)p(p-1),(1)/(6)*p(p-1)(p-2)):}|`
`=((1)/(2) xx (1)/(6) xx mnp)* |{:(1,(m-1),(m-1)(m-2)),(1,(n-1),(n-1)(n-2)),(1,(p-1),(p-1)(p-2)):}|`
`=(1)/(12)*mnp* |{:(1,m-1,(m-1)(m-2)),(0,n-m,(n-m)(n+m-3)),(0,p-m,(p-m)(p+m-3)):}| [R_(2) to (R_(2)-R_(1)) "and "R_(3) to (R_(3) -R_(1))]`
`=(1)/(12)*(mnp)(n-m)(p-m)* |{:(1,m-1,(m-1)(m-2)),(0," "1,(n+m-3)),(0," "1,(p+m-3)):}|`
`["taking (n-m) common from "R_(2) "and (p-m) common from"R_(3)]`
` =(1)/(12) * (mnp)(n-m)(p-m) *1*|{:(1, (n+m-3)), (1, (p+m-3)):}|`
`=(1)/(12) * (mnp)(n-m)(p-m)[(p + m-3)-(n+m-3)]`
`=(1)/(12) *(mnp)(n-m)(p-m)(p-n)`
`=(1)/(12) * (mnp)(m-n)(n-p)(p-m)`
Hence, `Delta = (1)/(12) * (mnp)(m-n)(n-p)(p-m)`
Discussion
547.

Prove that `Delta = |{:(1,,1,,1),(a,,b,,c),(bc+1^(2),,ac+b^(2),,ab+c^(2)):}|` `= 2(a-b)(b-c)(c-1)`

Answer» `Delta = |{:(1,,1,,1),(a,,b,,c),(bc+a^(2),,ac+b^(2),,ab+c^(2)):}|`
` |{:(1,,1,,1),(a,,b,,c),(bc,,ac,,ab):}|+|{:(1,,1,,1),(a,,b,,c),(a^(2),,b^(2),,b^(2)):}|`
`" Now " |{:(1,,1,,1),(a,,b,,c),(bc,,ac,,ab):}|`
`=(1)/(abc) |{:(a,,b,,c),(a^(2),,b^(2),,c^(2)),(abc,,abc,,abc):}|" "underset("a,b,c respectively]")("[Multiplying "C_(1),C_(2),C_(3)" by")`
`=|{:(a,,b,,c),(a^(2),,b^(2),,c^(2)),(1,,1,,1):}|`
`=|{:(1,,1,,1),(a,,b,,c),(a^(2),,b^(2),,c^(2)):}|`
`rArr Delta =2 |{:(1,,1,,1),(a,,b,,c),(a^(2),,b^(2),,c^(2)):}| = 2(a-b)(b-c)(c-a)`
Discussion
548.

the determinant `|{:(a,,b,,aalpha+b),(b,,c,,balpha+c),(aalpha+b,,balpha+c,,0):}|=0` is equal to zero ifA. a,b,c are in A.PB. a,b,c are in G.P.C. `alpha` is a root of the equation `ax^(2) +bx+c=0`D. `(x-alpha)` is a factor fo `ax^(2) +2bx+c`

Answer» Correct Answer - B::D
Given that `|{:(a,,b,,aalpha+b),(b,,c,,balpha+c),(aalpha+b,,balpha+c,,0):}|=0`
Operating `C_(3) to C_(3) -C_(1) alpha-C_(2)` we get
`|{:(a,,b,,0),(b,,c,,0),(aalpha+b,,balpha+c,,-(aalpha^(2)+balpha+balpha+c)):}|=0`
`rArr (aalpha^(2) +2balpha +c) |{:(a,,b,,0),(b,,c,,0),(aalpha +b,,balpha+c,,1):}|=0`
`rArr (ac-b^(2)) (aalpha^(2) +2balpha+c)=0`
so , either `ac-b^(2)=0 " or " aalpha^(2) +2balpha +c=0`
This means that either a,b,c are in G.P. or `(x-alpha)` is a factor of `ax^(2) +2bx +C`
Discussion
549.

The value of the determinant `|(1,1,1),(.^(m)C_(1),.^(m +1)C_(1),.^(m+2)C_(1)),(.^(m)C_(2),.^(m +1)C_(2),.^(m+2)C_(2))|` is equal toA. 1B. `-1`C. 0D. none of these

Answer» Correct Answer - A
Discussion
550.

Evaluate `|{:(.^(x)C_(1),,.^(x)C_(2),,.^(x)C_(3)),(.^(y)C_(1),,.^(y)C_(2),,.^(y)C_(3)),(.^(x)C_(1),,.^(z)C_(2),,.^(z)C_(3)):}|`

Answer» Correct Answer - `(1)/(12) xyz (x-y) (y-z) (z-y)`
`Delta=|{:(x,,(1)/(2)x(x-1),,(1)/(6)x(x-1)(x-2)),(y,,(1)/(2)y(y-1),,(1)/(6)y(y-1)(y-2)),(z,,(1)/(2)z(z-1),,(1)/(6)z(z-1)(z-2)):}|`
`=(1)/(12)xyz = |{:(1,,x-1,,x^(2)-3x+2),(1,,y-1,,y^(2)-3y+2),(1,,z-1,,z^(2)-3z+2):}|`
`=(1)/(12)xyz = |{:(1,,x,,x^(2)-3x+2),(1,,y,,y^(2)-3y+2),(1,,z,,z^(2)-3z+2):}|" "(C_(2) to C_(2) +C_(1))`
`=(1)/(12)xyz =|{:(1,,x,,x^(2)),(1,,y,,y^(2)),(1,,z,,z^(2)):}|" "(C_(3) to C_(3)+3C_(2)-2C_(1))`
`=(1)/(12)xyz (x-y) (y-z)(z-x)`
Discussion