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601.

`|{:(1,1,1),(m_(C1),m+1_(C1),m+2_(C1)),(m_(C2),m+1_(C2),m+2_(C2)):}|=`

Answer» Correct Answer - B
Discussion
602.

If x+y+z=0=a+b+c, then `|{:(xa,yd,zc),(yc,za,xb),(zb,xc,ya):}|=`

Answer» Correct Answer - A
Discussion
603.

Prove the following identities :\(\begin{vmatrix}a & b& c \\[0.3em]a-b & b-c & c-a \\[0.3em]b+c & c+a & a+b\end{vmatrix}\) = a3 + b3 + c3 + 3abc

Answer»

\(\begin{vmatrix}a & b& c \\[0.3em]a-b & b-c & c-a \\[0.3em]b+c & c+a & a+b\end{vmatrix}\)

L.H.S = \(\begin{vmatrix}a & b& c \\[0.3em]a-b & b-c & c-a \\[0.3em]b+c & c+a & a+b\end{vmatrix}\)

Apply C1→C1 + C2 + C3

= (a + b + c) \(\begin{vmatrix}1 & b& c \\[0.3em]0 & b-c & c-a \\[0.3em]2 & c+a & a+b\end{vmatrix}\)

Applying, 

R3→R3 – 2R1

 = (a + b + c) \(\begin{vmatrix}1 & b& c \\[0.3em]0 & b-c & c-a \\[0.3em]0 & c+a-2b & a+b-2c\end{vmatrix}\)

= (a + b + c)[(b – c)(a + b – 2c) – (c – a)(c + a – 2b)] 

= a3 + b3 + c3 – 3abc 

As, 

L.H.S = R.H.S 

Hence, proved.
Discussion
604.

Prove the following identities :\(\begin{vmatrix} b+c & a-b& a \\[0.3em] c+a & b-c & b \\[0.3em] a+b & c-a & c \end{vmatrix}\) = 3abc - a3 - b3 - c3

Answer»

L.H.S = \(\begin{vmatrix} b+c & a-b& a \\[0.3em] c+a & b-c & b \\[0.3em] a+b & c-a & c \end{vmatrix}\)

As,

|A| = |A|T

So,

\(\begin{vmatrix} b+c & c + a& a+b \\[0.3em] a-b & b-c & c-a \\[0.3em] a &b & c \end{vmatrix}\)

If any two rows or columns of the determinant are interchanged, then determinant changes its sign

\(-\begin{vmatrix}a &b & c \\[0.3em] a-b & b-c & c-a \\[0.3em]b+c & c + a& a+b \end{vmatrix}\)

Apply C1→C1 + C2 + C3

\(-\begin{vmatrix}a+b+c & b& c \\[0.3em] 0 & b-c & c-a \\[0.3em] 2(a+b+c) &c+a & a+b \end{vmatrix}\)

Taking (a + b + c) common from C1 we get,

= - (a + b + c)\(\begin{vmatrix}1 & b& c \\[0.3em] 0 & b-c & c-a \\[0.3em] 2&c+a & a+b \end{vmatrix}\)

Applying,

R3→R3 – 2R1

 = - (a + b + c)\(\begin{vmatrix}1 & b& c \\[0.3em] 0 & b-c & c-a \\[0.3em] 0&c+a-2b & a+b-2c \end{vmatrix}\)

= – (a + b + c)[(b – c)(a + b – 2c) – (c – a)(c + a – 2b)] 

= 3abc – a3 – b3 – c3 

As, 

L.H.S = R.H.S, 

hence proved.
Discussion
605.

Consider the system of equations `x-2y+3z=-1, x-3y+4z=1` and `-x+y-2z=k`Statement 1: The system of equation has no solution for `k!=3` andStatement 2: The determinant `|[1,3,-1], [-1,-2,k] , [1,4,1]| !=0` for `k!=0`A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1C. Statement 1 is true, Statement 2 is FalseD. Statement 1 is False, Statement 2 is true

Answer» Correct Answer - A
We have,
`D = |(1,-2,3),(-1,1,-2),(1,-3,4)|`
`rArr D = |(1,-2,3),(0,-1,1),(0,-1,1)| = 0 " " [("Applying " R_(2) rarr R_(2) - R_(1)),(R_(3) rarr R_(3) - R_(1))]`
and `D_(2) = |(1,-1,3),(-1,k,-2),(1,1,4)| = - |(1,3,-1),(-1,-2,k),(1,4,1)| " " ["Applying " C_(2) hArr C_(3)]`
`rArr D_(2) = - |(1,3,-1),(0,1,k -1),(0,1,2)| = k -3 " " [("Applying " R_(2) rarr R_(2) - R_(1)),(R_(3) rarr R_(3) - R_(1))]`
Clearly, `D_(2) != 0 " for " k != 3` and the determinant is not zero for `k != 3`.
Hence, both the statement are true and statement 2 is a correct explanation for statement 1
Discussion
606.

`|[x-3, x-4, x-alpha], [x-2, x-3, x-beta], [x-1, x-2, x-gamma]| =0,"where" alpha, beta, gamma "are in AP"`

Answer» Apply `R_(1) to (R_(1) + R_(2) + R_(3)) "and use"alpha + gamma = 2beta.`
Then, `R_(1) "and"R_(2) "are proportional and hence"Delta = 0.`
Discussion
607.

The value of the determinant `|[log_a(x/y), log_a(y/z), log_a(z/x)], [log_b (y/z), log_b (z/x), log_b (x/y)], [log_c (z/x), log_c (x/y), log_c (y/z)]|`A. 1B. -1C. 0D. `(1)/(6) log _(a) xyz`

Answer» Correct Answer - C
`Delta = |{:(log_(a)((x)/(y)),,log_(a)((y)/(z)),,log_(a)((z)/(x))),(log_(a^(2))((y)/(z)),,log_(a^(2))((z)/(x)),,log_(a^(2))((x)/(y))),(log_(a^(3))((z)/(y)),,log_(a^(3))((x)/(y)),,log_(a^(3))((y)/(z))):}|`
`=(1)/(2) xx(1)/(3) |{:(log_(a)((x)/(y)),,log_(a)((y)/(z)),,log_(a)((z)/(x))),(log_(a)((y)/(z)),,log_(a)((z)/(x)),,log_(a)((x)/(y))),(log_(a)((z)/(y)),,log_(a)((x)/(y)),,log_(a)((y)/(z))):}|`
Now ,`log_(a)((x)/(y)) +log_(a)((y)/(z))+log_(a)((z)/(y))`
`=log_(a)((x)/(y))((y)/(z))((z)/(x))=log_(a)1=0`
So applying operation `R_(1) to R_(1)+R_(2)+R_(3) " on " Delta ` we get
`Delta =(1)/(2)xx(1)/(3) |{:(0,,0,,0),(log_(a).((y)/(z)),,log_(a).((z)/(x)),,log_(a).((x)/(y))),(log_(a).((z)/(x)),,log_(a).((x)/(y)),,log_(a).((y)/(z))):}|=0`
Discussion
608.

Useproduct `[1-1 2 0 2-3 3-2 4] [-2 0 1 9 2-3 6 1-2]`to solve the system of equation:`x-y+2z=1``2y-3z=1``3x-2y+4z=2`

Answer» `"Let "A=[{:(1,-1,2),(0,2,-3),(3,-2,4):}]" and "B=[{:(-2,0,1),(9,2,-3),(6,1,-2):}]`
`therefore" AB"=[{:(1,-1,2),(0,2,-3),(3,-2,4):}][{:(-2,0,1),(9,2,-3),(6,1,-2):}]`
`=[{:(-2-9+12,0-2+2,1+3-4),(0+18-18,0+4-3,0-6+6),(-6-18+24,0-4+4,3+6-8):}]`
`=[{:(1,0,0),(0,1,0),(0,0,1):}]=I`
write the given equations in matrix form
`=[{:(1,-1,2),(0,2,-3),(3,-2,4):}][{:(x),(y),(z):}]=[{:(1),(1),(2):}]`
AX=C
`rArr" "X=A^(-1)C=BC`
`=[{:(x),(y),(z):}]=[{:(-2,0,1),(9,2,-3),(6,1,-2):}][{:(1),(1),(2):}]=[{:(-2,+0,+2),(9,+2,-6),(6,+1,-4):}]=[{:(0),(5),(3):}]`
`therefore" "x=0, y=5, z=3`.
Discussion
609.

Show that `| (-a(b^2 + c^2 - a^2), 2b^3, 2c^3), (2a^3, -b(c^2 + a^2 - b^2), 2c^3), (2a^3, 2b^3, -c(a^2 + b^2 - C^2))|` = abc`(a^2 + b^2 + c^2)^3`

Answer» Take a, b, c common from `C_(1), C_(2) "and " C_(3)` respectively
Now, apply `C_(1) to (C_(1) + C_(2) + C_(3))`
Discussion
610.

Using properties of determinants, show the following:`|(a+C)^2a b c a a b(a+c)^2b c a c b c(a+b)^2|=2a b c(a+b+c)^3`

Answer» Apply `R_(1) to aR_(1), R_(2) to bR_(2), R_(3) to cR_(3) "and divide "Delta "by abc"`
Take a, b, c common from `C_(1), C_(2) "and"C_(3)` respectively
Discussion
611.

The determinant `Delta = |(a^(2) + x^(2),ab,ac),(ab,b^(2) + x^(2),bc),(ac,bc,c^(2) + x^(2))|` is divisibleA. `x^(5)`B. `x^(4)`C. `x^(4) +1`D. `x^(4) -1`

Answer» Correct Answer - B
Discussion
612.

` " if " |{:(b+c,,c+a,,a+b),(a+b,,b+c,,c+a),(c+a,,a+b,,b+c):}|=k |{:(a,,b,,c),(c,,a,,b),(b,,c,,a):}|` then the value of k isA. 1B. 2C. 3D. 4

Answer» Correct Answer - B
we have
`|{:(b+c,,c+a,,a+b),(a+b,,b+c,,c+a),(c+a,,a+b,,b+c):}|=k |{:(a,,b,,c),(c,,a,,b),(b,,c,,a):}|`
`" or " |{:(2(a+b+c),,c+a,,a+b),(2(a+b+c),,b+c,,c+a),(2(a+b+c),,a+b,,b+c):}|=k |{:(a,,b,,c),(c,,a,,b),(b,,c,,a):}|`
`"[Applying " C_(1) to C_(1)+(C_(2)+C_(3)) " on " L.H.S"]"`
`" or " 2|{:(a+b+c,,-b,,-c),(a+b+c,,-a,,-b),(a+b+c,,-c,,-a):}|=k |{:(a,,b,,c),(c,,a,,b),(b,,c,,a):}|`
`"[Applying" C_(2) to C_(2) -C_(1),C_(3) to C_(3)-C_(1)" on " L.H.S "]"`
`" or " |{:(a,,-b,,-c),(c,,-a,,-b),(b,,-c,,-a):}|=k |{:(a,,b,,c),(c,,a,,b),(b,,c,,a):}|`
`"[Applying" C_(1) to C_(1) +C_(2) +C_(3) " on " L.H.S "]"`
`" or " 2|{:(a,,b,,c),(c,,a,,b),(b,,c,,a):}|=k |{:(a,,b,,c),(c,,a,,b),(b,,c,,a):}|`
`:. k=2`
Discussion
613.

The determinat `Delta=|(b^2-ab,b-c,bc-ac),(ab-a^2,a-b,b^2-ab),(bc-ac,c-a,ab-a^2)|` equals

Answer» `Delta = |[b(b-a), b-c, c(b-a)], [a(b-a), a-b, b(b-a)], [c(b-a), c-a, a(b-a)]|`
`= (b-a)^(2) * |[b, b-c, c], [a, a-b, b], [c, c-a, a]| " " ["taking (b-a) common from each of "C_(1) " and "C_(3)]`
Now, apply `C_(2) to (C_(2) -C_(1) + C_(3))`
Discussion
614.

if `a_(1)b_(1)c_(1), a_(2)b_(2)c_(2)" and " a_(3)b_(3)c_(3)` are three-digit even natural numbers and `Delta = |{:(c_(1),,a_(1),,b_(1)),(c_(2),,a_(2),,b_(2)),(c_(3),,a_(3),,b_(3)):}|" then " Delta ` isA. divisible by 2 but not necessarily by 4B. divisible by 4 but not necessarily by 8C. divisible by 8D. none of these

Answer» Correct Answer - A
As ` a_(1) ,b_(1),C_(1) , a,b,c " and " a_(3) ,b_(3) ,c_(3) ` are even natural numbers , each of `C_(1),C_(2),C_(3)` is divisible by 2, Let `C_(i) = 2k_(i) " for " i=1,2,3` thus
`Delta = 2 |{:(k_(1),,a_(1),,b_(1)),(k_(2),,a_(2),,b_(2)),(k_(3),,a_(3),,b_(3)):}|=2m`
Where m is some natural number. thus `Delta ` is divisible by 2. That `Delta ` may not be divisible by 4 can be seen by taking the three numbers as 112, 122 and 134 Note that
`Delta =|{:(2,,1,,1),(2,,1,,2),(4,,1,,3):}|=2`
Which is divisible by 2 but not 4
Discussion
615.

if `Delta (x)= |{:(a_(1)+x,,b_(1)+x,,c_(1)+x),(a_(2)+x,,b_(2)+x,,c_(2)+x),(a_(3)+x,,b_(3)+x,,c_(3)+x):}|` then show that `Delta (x)=0` and that `Delta (x)=Delta(0)+sx.` where s denotes the sum of all the cofactors of all the elements in `Delta (0)`

Answer» `Delta (x)= |{:(a_(1)+x,,b_(1)+x,,c_(1)+x),(a_(2)+x,,b_(2)+x,,c_(2)+x),(a_(3)+x,,b_(3)+x,,c_(3)+x):}|`
` :. Delta (x) = |{:(a_(1)+x,,b_(1)+x,,c_(1)+x),(a_(2)+x,,b_(2)+x,,c_(2)+x),(a_(3)+x,,b_(3)+x,,c_(3)+x):}|+|{:(a_(1)+x,,1,,c_(1)+x),(a_(2)+x,,1,,c_(2)+x),(a_(3)+x,,1,,c_(3)+x):}|`
`+|{:(a_(1)+x,,b_(1)+x,,1),(a_(2)+x,,b_(2)+x,,1),(a_(3)+x,,b_(3)+x,,1):}|`
Applying `C_(2) to C_(2)-xC_(1),C_(3) to C_(3) xC_(1)` in the first det.
`C_(1) to C_(1) -xC_(2),C_(3) to C_(3) -xC_(3)-xC_(2)` in the second det.
`" and " C_(1) to C_(1)-xC_(3),C_(2)to C_(2)-xC_(3)` in the third det. we get
`Delta (x)= |{:(1,,b_(1),,C_(1)),(1,,b_(2),,c_(2)),(1,,b_(3),,c_(3)):}|+|{:(a_(1),,1,,c_(1)),(a_(2),,1,,c_(2)),(a_(3),,1,,c_(3)):}|+|{:(a_(1),,b_(1),,1),(a_(2),,b_(2),,1),(a_(3),,b_(3),,1):}|`
`Delta (0)= |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}|`
which are `b_(2) C_(3) =b_(3)C_(2),C_(2)a_(3)-C_(3)a_(2),a_(2)b_(3)-b_(2)a_(3)` etc
Clearly
` |{:(1,,b_(1),,c_(1)),(1,,b_(2),,c_(2)),(1,,b_(3),,c_(3)):}| = (b_(2)b_(3) -b_(3)c_(2))+(c_(1)b_(3)-c_(3)b_(1))+(b_(1)c_(2)-b_(2)c_(1))`
Which is the sum of cofactors of the first row elements of `Delta (0)`
`" similarly " |{:(a_(1),,1,,c_(1)),(a_(2),,1,,c_(2)),(a_(3),,1,,c_(3)):}|" and " |{:(a_(1),,b_(1),,1),(a_(2),,b_(2),,1),(a_(3),,b_(3),,1):}|` are the sum of
cofactors of 2nd row and 3rd elements respectively of `Delta (0)` .
Hence `Delta(x)=s` where S denotes the sum of all cofactors of elements of `Delta(0)`
`:. Delta '(x) =0`
since `Delta (x) = s Delta (x) sx+k`
So `Delta (0) =k`
Hence `Delta(x)= x S+Delta (0)`
Discussion
616.

If `Delta = |(3,4,5,x),(4,5,6,y),(5,6,7,z),(x,y,z,0)|, " then " Delta` equalsA. `(y - 2z + 3x)^(2)`B. `(x - 2y + z)^(2)`C. `(x + y + z)^(2)`D. `x^(2) + y^(2) + z^(2) - zy - yz - zx`

Answer» Correct Answer - B
Discussion
617.

if `x gt m,y gt n,z gt r (x,y,zgt 0)` such that `|{:(x,,n,,r),(m,,y,,r),(m,,n,,z):}|=0` The value of `(x)/(x-m)+(y)/(y-n)+(z)/(z-r)`isA. 1B. -1C. 2D. -2

Answer» Correct Answer - C
`|{:(x,,n,,r),(m,,y,,r),(m,,n,,z):}|=0`
Applying `R_(1) to R_(1)-R_(2) " and " R_(2) to R_(2)-R_(3)` we get
`|{:(x-m,,n-y,,0),(0,,y-n,,r-z),(m,,n,,z):}| =0`
`rArr (x-m)(y-n)(n-y)(r-z)m-n(r-z)(x-m) =0`
Dividing by `(x-m)(y-n)(z-r)` we have
`(z)/(z-r)+(m)/(x-m) +(n)/(y-n) =0`
`" or " (z)/(z-r) +(m)/(x-m) +1 +(m)/(y-n) +1=2`
` " or " (z)/(z-r) +(x)/(x-m) +(y)/(y-n) =2`
`" or " (m)/(x-m)+(n)/(y-n)+(r)/(z-r)=-1`
Now `,A.M ge G.M.`
`rArr ((z)/(z-r)+(x)/(x-m)+(y)/(y-n))/(3)ge ((z)/((z-r))(x)/((x-m))(y)/((y-n)))^(1//3)`
`" or " (z)/(z-r)(x)/(x-m) (y)/(y-n) le (8)/(27)`
Discussion
618.

if `x gt m,y gt n,z gt r (x,y,zgt 0)` such that `|{:(x,,n,,r),(m,,y,,r),(m,,n,,z):}|=0` the value of `(m)/(x-m)+(n)/(y-n) +(r )/(z-r)`isA. -2B. -4C. 0D. -1

Answer» Correct Answer - D
`|{:(x,,n,,r),(m,,y,,r),(m,,n,,z):}|=0`
Applying `R_(1) to R_(1)-R_(2) " and " R_(2) to R_(2)-R_(3)` we get
`|{:(x-m,,n-y,,0),(0,,y-n,,r-z),(m,,n,,z):}| =0`
`rArr (x-m)(y-n)(n-y)(r-z)m-n(r-z)(x-m) =0`
Dividing by `(x-m)(y-n)(z-r)` we have
`(z)/(z-r)+(m)/(x-m) +(n)/(y-n) =0`
`" or " (z)/(z-r) +(m)/(x-m) +1 +(m)/(y-n) +1=2`
` " or " (z)/(z-r) +(x)/(x-m) +(y)/(y-n) =2`
`" or " (m)/(x-m)+(n)/(y-n)+(r)/(z-r)=-1`
Now `,A.M ge G.M.`
`rArr ((z)/(z-r)+(x)/(x-m)+(y)/(y-n))/(3)ge ((z)/((z-r))(x)/((x-m))(y)/((y-n)))^(1//3)`
`" or " (z)/(z-r)(x)/(x-m) (y)/(y-n) le (8)/(27)`
Discussion
619.

Consider the system of equations `a_(1) x + b_(1) y + c_(1) z = 0` `a_(2) x + b_(2) y + c_(2) z = 0` `a_(3) x + b_(3) y + c_(3) z = 0` If `|(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))| =0`, then the system hasA. more than two solutionsB. one trivial and one non-trivial solutionsC. no solutionD. only trivial solution `(0, 0,0)`

Answer» Correct Answer - A
Discussion
620.

if x,y and z are not all zero and connected by the equations `a_(1)x+b_(1)y+c_(1)z=0,a_(z)x+b_(2)y+c_(2)z=0` and `(p_(1)+lambdaq_(1))x+(p_(2)+lambdaq_(2))y+(p_(3)+lambdaq_(3))z=0` show that `lambda =-|{:(a_(1),,b_(1),,c_(1)),(a_(2) ,,b_(2),,c_(2)),(p_(1) ,, p_(2),,p_(3)):}|-:|{:(a_(1),,b_(1),,c_(1)),(a_(2) ,,b_(2),,c_(2)),(q_(1) ,, q_(2),,q_(3)):}|`

Answer» Since x,y and z are not all zero the determinant of the coefficient of the given set of equation must satisfy
` |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(p_(1)+lambdaq_(1),,p_(2)+lambdaq_(2),,p_(3)+lambdaq_(3)):}|=0`
` rArr lambda= |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(p_(1),,p_(2),,p_(3)):}|-: |{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(q_(1),,q_(2),,q_(3)):}|`
Discussion
621.

If `|(x^(n),x^(n+2),x^(n+3)),(y^(n),y^(n+2),y^(n+3)),(z^(n),z^(n+2),z^(n+3))|` `= (x -y) (y -z) (z -x) ((1)/(x) + (1)/(y) + (1)/(z))`, then n equalsA. 1B. `-1`C. 2D. `-2`

Answer» Correct Answer - B
The degree of the determinant is `n + (n+2) + (n+3) = 3n +5` and the degree of the expression on RHS is 2.
`:. 3n + 5 = 2 rArr n = -1`
Discussion
622.

If `a, b gt 0 and Delta (x)= |(x,a,a),(b,x,a),(b,b,x)|`, thenA. `Delta(x)` is increasing on `(-sqrt(ab), sqrt(ab))`B. `Delta (x)` is decreasing on `(sqrt(ab), oo)`C. `Delta(x)` has a local maximum at `x = sqrt(ab)`D. none of these

Answer» Correct Answer - C
Discussion
623.

If `|(1 +ax,1 +bx,1 + bx),(1 +a_(1) x,1 +b_(1) x,1 + c_(1) x),(1 + a_(2) x,1 + b_(2) x,1 + c_(2) x)| = A_(0) + A_(1) x + A_(2) x^(2) + A_(3) x^(3)`, then `A_(1)` is equal toA. abcB. 0C. 1D. none of these

Answer» Correct Answer - B
Discussion
624.

If `alpha, beta, gamma` are the roots of `x^(3) + ax^(2) + b = 0`, then the value of `|(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta)|`, isA. `-a^(3)`B. `a^(3) -3b`C. `a^(3)`D. `a^(2) - 3b`

Answer» Correct Answer - C
Since, `alpha, beta , gamma` are the roots of the given equations
`:. alpha + beta + gamma = -a, alpha beta + beta gamma + gamma alpha = 0 and alpha beta gamma = -b`
`Delta = |(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta)|`
`Delta = - (alpha + beta + gamma) (alpha^(2) + beta^(2) + gamma^(2) - alpha beta - beta gamma - gamma alpha)`
`Delta = - (alpha + beta + gamma) {(alpha + beta + gamma)^(2) - 3(alpha beta + beta gamma + gamma beta)}`
`Delta = - (-a) {(a^(2) - 0)} = a^(3)`
Discussion
625.

If `omega`is an imaginary cube root of unity, then the value of the determinant `|(1+omega,omega^2,-omega),(1+omega^2,omega,-omega^2),(omega+omega^2,omega,-omega^2)|`

Answer» Correct Answer - D
Using `1 + omega + omega^(2) = 0`, we get
`|(1 + omega,omega^(2),-omega),(1 + omega^(2),omega,- omega^(2)),(omega^(2) + omega,omega,- omega^(2))|`
`= |(1 + omega + omega^(2),omega^(2),- omega),(1 + omega^(2) + omega,omega,- omega^(2)),(omega^(2) + 2 omega,omega,- omega^(2))| " " ["Applying" C_(1) rarr C_(1) + C_(2)]`
`= |(0,omega^(2),-omega),(0,omega,-omega^(2)),(omega -1,omega,-omega^(2))|`
`= (omega -1)|(omega^(2),-omega),(omega,- omega^(2))|`
`= (omega -1) (-omega^(4) + omega^(2)) = (omega -1)(- omega + omega^(2))`
`= - omega^(2) + omega^(3) + omega - omega^(2) = -omega^(2) + (1 + omega) - omega^(2) = -3 omega^(2)`
Discussion
626.

Let `f(x) = ax^(2) + bx + c, a, b, c, in R` and equation `f(x) - x = 0` has imaginary roots `alpha, beta`. If r, s be the roots of `f(f(x)) - x = 0`, then `|(2,alpha,delta),(beta,0,alpha),(gamma,beta,1)|` is

Answer» Correct Answer - B
Discussion
627.

The non-zero roots of the equation `Delta =|(a,b,ax+b),(b,c,bx+c),(ax+b,bx+c,c)|=0` areA. a,b ,c are in A.PB. a, b, c are in G.PC. a, b, c are in H.PD. `alpha` is a root of `ax^(2) + bx + c = 0`

Answer» Correct Answer - B
Discussion
628.

The non-zero roots of the equation `Delta =|(a,b,ax+b),(b,c,bx+c),(ax+b,bx+c,c)|=0` are

Answer» `C_1->xC_1`
`1/x[[ax,b,ax+b],[bx,c,bx+c],[ax^2+bx,bx+c,c]]=0`
`C_1->C_1+c_2`
`1/x[[ax+b,b,ax+b],[bx+c,c,bx+c],[ax^2+2bx+c,bx+c,c]]=0`
`C_1->C_1-C_3`
`1/x[[0,b,ax+b],[0,c,bx+c],[ax^2+2bx,bx+c,c]]=0`
`1/x[ax^2+2bx(b^2x+bc-acx-bc)]=0`
`1/x(ax^2+2bx)(b^2x-acx)=0`
`1/x*x(ax+2b)x(b^2-ac)=0`
`(ax+2b)=0`
`x=(-2b)/a`.
Discussion
629.

In triangle ABC, if `|[1,1,1], [cot (A/2), cot(B/2), cot(C/2)], [tan(B/2)+tan(C/2), tan(C/2)+ tan(A/2), tan(A/2)+ tan(B/2)]|` then the triangle must be (A) Equilateral (B) Isoceless (C) Right Angle (D) none of theseA. equilateralB. isoscelesC. obtuse angledD. none of these

Answer» Correct Answer - B
Applying `C_(1) to C_(1)-C_(2),C_(2) to C_(2)-C_(3)` we get
`|{:(0,,0,,1),(cot .(A)/(2)-cot.(B)/(2),,cot.(B)/(2)-cot.(C)/(2),,cot.(C)/(2)),(tan.(B)/(2)-tan.(A)/(2),,tan.(C)/(2)-tan.(B)/(2),,tan.(A)/(2)+tan.(B)/(2)):}|`
`|{:(0,,0,,1),(cot.(A)/(2)-cot.(B)/(2),,cot.(B)/(2)-cot.(C )/(2),,cot.(C)/(2)),((cot.(A)/(2)-cot.(B)/(2))/(cot.(A)/(2)-cot.(A)/(2)),,(cot.(B)/(2)-cot.(C)/(2))/(cot.(B)/(2)cot.(C)/(2)),,tan.(A)/(2)+tan.(B)/(2)):}|`
`=( cot .(A)/(2)-cot.(B)/(2))(cot.(B)/(2)-cot.(C)/(2))`
`xx |{:(0,,0,,0),(1,,1,,cot.(C)/(2)),(tan.(A)/(2)-cot.(B)/(2),,tan.(B)/(2)tan.(C)/(2),,tan.(A)/(2).tan.(B)/(2)):}|`
`=(cot.(A)/(2)-cot.(B)/(2))(cot.(B)/(2)-cot.(C)/(2))(tan.(C)/(2)-tan.(A)/(2))tan.(B)/(2)`
Since `Delta =0 ` we have
:. `cot .(A)/(2)=cot.(B)/(2) " or " cor .(B)/(2) -cot .(C)/(2)" or " tan.(A)/(2) =tan .(C)/(2)`
Hence the triangle is definitely isosceles.
Discussion
630.

If `x ,y ,z`are different from zero and `"Delta"=a b-y c-z a-x b c-z a-x b-y c=0,`then the value of the expression `a/x+b/y+c/z`is`0`b. `-1`c. `1`d. `2`

Answer» Correct Answer - D
Applying `R_(2) to R_(2)-R_(1),R_(3) to R_(3) -R_(1)` we get
`Delta = |{:(a,,b-y,,c-z),(-x,,y,,0),(-x,,0,,z:}|=0`
Expanding along `C_(3) ` we get
`" or " (c-z)(xy)+z(ay+bx-xy)=0`
`" or " cxy -xyz +ayz +bxz -xyz =0`
`" or " ayz +bzx + cxz =2xyz`
`" or " (a)/(x)+(b)/(y)+(c )/(z)=2`
Discussion
631.

Let `|{:(y^(5)z^(6)(z^(3)-y^(3)),,x^(4)z^(6)(x^(3)-z^(3)),,x^(4)y^(5)(y^(3)-x^(3))),(y^(2)z^(3)(y^(6)-z^(6)),,xz^(3)(z^(6)-x^(6)) ,,xy^(2)(x^(6)-y^(6))),(y^(2)^(3)(z^(3)-y^(3)),,xz^(3)(x^(3)-z^(3)),,xy^(2)(y^(3)-x^(3))):}| " and " Delta_(2)= |{:(x,,y^(2),,z^(3)),(x^(4),,y^(5) ,,z^(6)),(x^(7),,y^(8),,z^(9)):}| `.Then `Delta_(1)Delta_(2)` is equal toA. `Delta_(2)^(6)`B. `Delta_(2)^(4)`C. `Delta_(2)^(3)`D. `Delta_(2)^(2)`

Answer» Correct Answer - C
the given determinant `Delta_(1)` is obtained by corresponding cofactors or determinant `Delta_(2)` and hence `Delta_(1)= Delta_(2)^(2)`
Now `Delta_(1)Delta_(2) =Delta_(2)^(2) Delta_(2) =Delta_(2)^(2)`
Discussion
632.

The number of positive integral solutions of the equation `|(x^3+1,x^2y,x^2z),(xy^2,y^3+1,y^2z),(xz^2,z^2y,z^3+1)|=11` is

Answer» Correct Answer - B
`|{:(x^(3)+1,,x^(2)y,,x^(2)z),(xy^(2),,y^(3)+1,,y^(2)z),(xz^(2),,yz^(2),,z^(3)+1):}|=11`
Multiplying `R_(1) " by " x, R_(2)` by y and `R_(3) ` by z we bet
`(1)/(xyz) |{:(x^(4)+x,,x^(2)y,,x^(3)z),(xy^(3),,y^(4)+y,,y^(3)z),(xz^(3),,yz^(3),,z^(4)+z):}|=11`
Taking x,y,z common from `C_(1) ,C_(2),C_(3)` respectively we get
`|{:(x^(4)+x,,x^(2)y,,x^(3)z),(xy^(3),,y^(4)+y,,y^(3)z),(xz^(3),,yz^(3),,z^(4)+z):}|=11`
Using `R_(1) to R_(1) +R_(2)+R_(3)` we get
`(x^(3)+y^(3)+z^(3)+1) |{:(1,,1,,1),(y^(2),,y^(3)+1,,y^(3)),(z^(3),,z^(3),,z^(3)+1):}|=11`
Using `C_(2) to C_(2)-C_(1) " and " C_(3) toC_(3)-C_(1)` we get
`(x^(3) +y^(3) +z^(3) +1) |{:(1,,0,,0),(y^(3),,1,,0),(z^(2),,0,,1):}|=11`
Hence `,x^(3) +y^(3)+z^(3)=10`
Therefore the ordered triplets are (2,1,1) ,(1,2,1),(1,1,2)
Discussion
633.

if x=31,y=32 and z=33 then the value of `|{:((x^(2)+1)^(2),,(xy+1)^(2),,(xz+1)^(2)),((xy+1)^(2),,(y^(2)+1)^(2),,(yz+1)^(2)),((xz+1)^(2),,(yz+1)^(2),,(z^(2)+1)^(2)):}|" is " "____"`

Answer» Correct Answer - 8
` |{:((x^(2)+1)^(2),,(xy+1)^(2),,(xz+1)^(2)),((xy+1)^(2),,(y^(2)+1)^(2),,(yz+1)^(2)),((xz+1)^(2),,(yz+1)^(2),,(z^(2)+1)^(2)):}|`
` = |{:(1,,x,,x^(2)),(1,,y,,y^(2)),(1,,z,,z^(2)):}||{:(1,,1,,1),(2x,,2y,,2z),(x^(2),,y^(2),,z^(2)):}|`
`=2 (x-y)^(2)(y-z)^(2)(z-x)^(2)`
Discussion
634.

Find the value of the `"determinant" |{:(1,x,y+z),(1,y,z+x),(1,z,x+y):}|`

Answer» `|{:(1,x,y+z),(1,y,z+x),(1,z,x+y):}|`
`=|{:(1,x,x+y+z),(1,y,x+y+z),(1,z,x+y+z):}|`
`(C_(3)toC_(3)+C_(2))`
`=(x+y+z)|{:(1,x,1),(1,y,1),(1,z,1):}|`
`=(x+y+z)xx0(therefore C_(1)" and "C_(3)" are idential")`
=0.
Discussion
635.

`|(x,x^2,y2),(y,y^2,2x),(z,z^2,xy)| = (x-y)(y-z)(z-x)(xy+yz+2x)`

Answer» `L.H.S. = |[x,x^2 ,yz] ,[y , y^2, zx],[z,z^2,xy]|`
Operating `R_1->R_1- R_2 and R_2->R_2-R_3`
`=|[x-y,x^2-y^2,yz-zx],[y-z,y^2-z^2,zx-xy],[z,z^2,xy]|`
`=|[x-y,(x-y)(x+y),yz-zx],[y-z,(y-z)(y+z),zx-xy],[z,z^2,xy]|`
`=(x-y)(y-z)|[1,x+y,-z],[1,y+z,-x],[z ,z^2,xy]|`
Now, operating `R_1->R_1-R_2`
`=(x-y)(y-z)|[0,x-z,x-z],[1,y+z,-x],[z ,z^2,xy]|`
`=(x-y)(y-z)(z-x)|[0,-1,-1],[1,y+z,-x],[z ,z^2,xy]|`
`=(x-y)(y-z)(z-x)[1(xy+zx)-1(z^2-yz-z^2)]`
`=(x-y)(y-z)(z-x)(xy+yz+zx)= R.H.S.`
Discussion
636.

Find the value of the `"determinant" |{:(3,4,7),(-1,6,5),(2,8,10):}|`

Answer» `|{:(3,4,7),(-1,6,5),(2,8,10):}|=|{:(7,4,7),(5,6,5),(10,8,10):}|`
`(C_(1)toC_(1)+C_(1))`
= 0
(`therefore C_(1)" and " C_(3)" are identical ") `
Discussion
637.

If there are two values of a which makes determinant, `Delta=|(1,-2,5),(2,a,-1),(0,4,2a)|=86` then the sum of these number isA. `4`B. `5`C. `-4`D. `9`

Answer» Correct Answer - C
We have
`Delta=|(1,-2,5),(2,a,-1),(0,4,2a)|=86`
`implies 1(2a^(2)+4)-2(-4a-20)+0=86` [expanding along first column]
`implies 2a^(2)+4+8a+40=86`
`implies 2a^(2)+8a+44-86=0`
`implies a^(2)+4a-21=0`
`implies a^(2)+7a-3a-21=0`
`= (a+7)(a-3)=0`
`a=-7` and `3`
`:.` Required sum `=-7+3=-4`
Discussion
638.

If `A` and `B` are invertible matrices then which of the following is not correct?A. `adjA=|A|.A^(-1)`B. `det(A)^(-1)=["det"(A)]^(-1)`C. `(AB)^(-1)=B^(-1)A^(-1)`D. `(A+B)^(-1)=B^(-1)+A^(-1)`

Answer» Correct Answer - D
Since A and B are invertible matrices So we can say that
`(AB)^(-1)=B^(-1)A^(-1)`…………..i
Also `A^(-1)=1/(|A|)(adjA)`
`implies adjA=|A|.A^(-1)`………….ii
Also `det(A)^(-1)=[det(A)]^(-1)`
`implies det(A)^(-1)=1/([det(A)])`
`implies det(A).det(A)^(-1)=1`...............iii
which is true
Again `(A+B)^(-1)=1/(|(A+B)|)(adj(A+B)`
`implies (A+B)^(-1)!=B^(-1)+A^(-1)`..........iv
So only option (d) is incorrect
Discussion
639.

Prove that: `|(a,b-c,c-b),(a-c,b,c-a),(a-b,b-a,c)|=(a+b-c)(b+c-a)(c+a-b)`

Answer» `[[a+b+c-c-b,b-c,c-b],[a-c+b+c-a,b,c-a],[a-b+b-a+c,b-a,c]]`
`C_2->C_2+C_3`
`[[a,b-c+c-b,c-b],[b,b+c-a,c-a],[c,b-a+c,c]]`
`[[a,0,c-b],[b,b+c-a,c-a],[c,b+c-a,c]]`
`(b+c-a)[[a,0,c-b],[b,1,c-a],[c,1,c]]`
`(b+c-a)[a(c-c+a)+(-b)(b-c)-(b-c)(b-c)]`
`(b+c-a)(a-(b-c))(a+(b-c))`
`(a+b-c)(b+c-a)(a+c-b)`.
Discussion
640.

If A+B+C=0, then prove that `Det[[1,cosC,cosB],[cosC,1,cosA],[cosB,cosA,1]]=0`

Answer» `=1[1-cos^2A]-cosc[cosc-cosAcosB]+cosB[cosAcosB-cosB]`
`=1-cos^2A-cos^2C+cosAcosBcosC+cosAcosBcosC-cos^2B`
`=1-[cos^A+cos^2B+cos^2C]+2coaAcosBcosC`
`=1-[cos^2A+cos^2B+(cos(A+B)^2)+2cosAcosB[A+B]]`
`=1-cos^2A-cos^2B-cos^2Acos^2B-acos^2b-2cosAcosBsin^2Asin^2B+2cosAcosBsinAsinB+2cos^2Acos^2B-2cosAcosBsinAsinB`
`=1-cos^2A-cos^2B-sin^2Asin^2B+cos^2Acos^2B`
`=sin^2A-cos^2B-sin^2Asin^2B+cos^2Acos^2B`
`=sin^2Acos^2B-cos^2Bsin^2A`
`LHS=0=RHS`
Discussion
641.

If `x , y , z`are different from zero and `|1+x1 1 1 1+y1 1 1 1+z|=0`then the value of `x^(-1)+y^(-1)+2^(-1)`is`x y z`(b) `x^(-1)y^(-1)z^(-1)`(c) `-x-y-z`(d) `-1`A. `xyz`B. `x^(-1)y^(-1)z^(-1)`C. `-x-y-z`D. `-1`

Answer» Correct Answer - D
We have `|(1+x,1,1),(1,1+y,1),(1,1,1+z)|=0`
Applying `C_(1)toC_(1)-C_(3)` and `C_(2)toC_(2)-C_(3)`
`implies |(x,0,1),(0,y,1),(-z,-z,1+z)|=0`
Expanding along `R_(1)`
`x[y(a+z)+z]-0+1(yz)=0`
`impliesx(y+yz+z)+yz=0`
`implies xy+xyz+xz+yz=0`
`= (xy)/(x y)=(x yz)/(x yz)+(xz)/(x yz)+(yz)/(x yz)=0` [on dividing `(xyz)` from both sides]
`implies 1/x+1/y+1/z+1=0`
`implies 1/x+1/y+1/z=-1`
`:. x^(-1)+y^(-1)+z^(-1)=-1`
Discussion
642.

If `D=|1 1 1 1 1+x1 1 1 1+y|"f o r"x!=0, y!=0`then D is(1)divisible by neither x nor y(2) divisible by both x and y(3) divisible by x but not y (4) divisible by ybut not x

Answer» `D=[[1,1,1],[1,1+x,1],[1,1,1+y]]`
`=(1+x)(1+y)-1-(1+y)+1+1-(1+x)`
`=1+x+y+xy-1-1-y+1+1-1-x`
`=3-3+xy`
`=xy`
option`2`
Discussion
643.

Prove that `det ((yx-x^2,zx-y^2,xy-z^2),(zx-y^2,xy-z^2,yz-x^2),(xy-z^2,yz-x^2,zx-y^2))` is divisible by (x+y+z) and hence find the quotient.

Answer» `Delta = |[yz-x^2,zx-y^2,xy-z^2],[zx-y^2,xy-z^2,yz-x^2],[xy-z^2,yz-x^2,zx-y^2]|`
Applying `C_1->C_1+C_2+C_3`
`Delta = |[xy+yz+zx-x^2-y^2-z^2,zx-y^2,xy-z^2],[xy+yz+zx-x^2-y^2-z^2,xy-z^2,yz-x^2],[xy+yz+zx-x^2-y^2-z^2,yz-x^2,zx-y^2]|`
`= (xy+yz+zx-x^2-y^2-z^2)|[1,zx-y^2,xy-z^2],[1,xy-z^2,yz-x^2],[1,yz-x^2,zx-y^2]|`
Now, applying `R_1->R_1-R_3` and `R_2->R_2-R_3`
`= (xy+yz+zx-x^2-y^2-z^2)|[0,zx-y^2-yz+x^2,xy-z^2-zx+y^2],[0,xy-z^2-yz+x^2,yz-x^2-zx+y^2],[1,yz-x^2,zx-y^2]|`
`= (xy+yz+zx-x^2-y^2-z^2)|[0,(x-y)(x+y+z),(y-z)(x+y+z)],[0,(x-z)(x+y+z),(y-x)(x+y+z)],[1,yz-x^2,zx-y^2]|`
`= (xy+yz+zx-x^2-y^2-z^2)(x+y+z)^2|[0,(x-y),(y-z)],[0,(x-z),(y-x)],[1,yz-x^2,zx-y^2]|`
`= (xy+yz+zx-x^2-y^2-z^2)(x+y+z)^2[(x-y)(y-x) - (x-z)(y-z)]`
`=(xy+yz+zx-x^2-y^2-z^2)(x+y+z)^2[xy+yz+zx-x^2-y^2-z^2]`
`Delta = (xy+yz+zx-x^2-y^2-z^2)^2(x+y+z)^2`
`:. Delta` is divisible by `(x+y+z)` and quotient is `(xy+yz+zx-x^2-y^2-z^2)^2(x+y+z).`
Discussion