24 + Interview Questions in DETERMINANT IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Let `\|(a,l,m),(l,b,n),(m,n,c)\|\|(bc-n^2,mn-lc,ln-bm),(mn-lc,ac-m^2,ml-an),(ln-bm,lm-an,ab-l^2)\|=64.` If the value of `\|(2a+3l,3l+5m,5m+4a),(2l+3b,3b+5n,5n+4l),(2m+3n,3n+5c,5c+4m)\|=lambda` then `[lambda/2]` equalsA. `120`B. `240`C. `360`D. `480`
Answer» Correct Answer - C `(c )` `DeltaDelta^(2)=64` `impliesDelta^(3)=64impliesDelta=4` `\|{:(2a+3l,3l+5m,5m+4a),(2l+3b,3b+5n,5n+4l),(2m+3n,3n+5c,5c+4m):}\|` `=[(2xx3xx5)+(3xx5xx4)]Delta` `=(30+60)Delta` `=90(4)` `=360`

Discussion

2.	Given A = [(4,2,5)(2,0,3)(-1,1,0)], write the value of get.(2AA-1).
Answer» \|2AA^-1\| = \|2\|³[:. AA^-1 = 1] ^{= 8}

2.

Given A = [(4,2,5)(2,0,3)(-1,1,0)], write the value of get.(2AA-1).

Answer»

|2AA^-1| = |2|³[:. AA^-1 = 1]

^{= 8}

Discussion

3.	Let A be a square matrix of order 3 × 3. Write the value of \|2A\|, where \|A\| = 4.
Answer» Since \|2A\| = 2ⁿ \|A\| where n is order of matrix A. Here \|A\| = 4 and n = 3 ∴ \|2A\| = 2³ ×4 = 32

3.

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

Answer»

Since |2A| = 2ⁿ |A| where n is order of matrix A.

Here |A| = 4 and n = 3

∴ |2A| = 2³ ×4 = 32

Discussion

4.	Is A is square matrix of 3 and \|2A\| = k\|A\|, then the value of k.
Answer» \|2A\| = k\|A\|, k = 2³ = 8

4.

Is A is square matrix of 3 and |2A| = k|A|, then the value of k.

Answer»

|2A| = k|A|,

k = 2³ = 8

Discussion

5.	If A and B are square matrices of order 3 such that \|A\| = -1, \|B\| = 3,then find the value of \|2AB\|.
Answer» \|2AB\| = 2³ X \|A\| X \|B\| =8 X (-1) X 3 = -24

5.

If A and B are square matrices of order 3 such that |A| = -1, |B| = 3,then find the value of |2AB|.

Answer»

|2AB| = 2³ X |A| X |B|

=8 X (-1) X 3 = -24

Discussion

6.	If Aij is the cofactor o the element aij of the determinant [(2, - 3, 5), (6, 0, 4), (1, 5, - 7)], then write the value of (i) a32A32(ii) M21
Answer» (i) a₃₂ A₃₂= 5 (-) (-22) = 110 (ii) M₂₁ = 21 - 25 = - 4

6.

If Aij is the cofactor o the element aij of the determinant [(2, - 3, 5), (6, 0, 4), (1, 5, - 7)], then write the value of (i) a32A32(ii) M21

Answer»

(i) a₃₂ A₃₂= 5 (-) (-22) = 110

(ii) M₂₁ = 21 - 25 = - 4

Discussion

7.	If A is a square matrix and \|A\| = 2, then write the value of \| AA'\| , where A' is the transpose of matrix A.
Answer» \|AA\|' = \|A\|. \|A'\| = \|A\|. \|A\|= \|A\|² = 2 x 2 = 4. [since, \| AB\|=\|A\|.\|B\| and\| A\|=\| A'\|, where A and B are square matrices.]

7.

If A is a square matrix and |A| = 2, then write the value of | AA'| , where A' is the transpose of matrix A.

Answer»

|AA|' = |A|. |A'| = |A|. |A|= |A|² = 2 x 2 = 4.

[since, | AB|=|A|.|B| and| A|=| A'|, where A and B are square matrices.]

Discussion

8.	if A, B are square matrices of the same order, then prove that adj (AB) = (adj B) (adj A).
Answer» We know that (AB) adj(AB) = \|AB\| = adj(AB)(AB) ..(i) (AB)(adj B adj A) = A.B adj B.adj A = A(B adj B) adj A =A(\|B\|)adj A [ :. B adj B = \|B\|I] = \|B\|(A.adj A) = \|B\| \|A\| I [:. A adj A = \|A\|I] = \|A\| \|B\| I = \|AB\|I ....(ii) From (i) and (ii), we get AB(adj AB) = AB(adj B.adj A) Pre-multiplying both sides by(AB)^-1,we get (AB)^-1[(AB)adj AB] = (AB)^-1 [(AB)adj B.adj A] or adj AB = adj B.adj A

8.

if A, B are square matrices of the same order, then prove that adj (AB) = (adj B) (adj A).

Answer»

We know that

(AB) adj(AB) = |AB| = adj(AB)(AB) ..(i)

(AB)(adj B adj A)

= A.B adj B.adj A = A(B adj B) adj A

=A(|B|)adj A [ :. B adj B = |B|I]

= |B|(A.adj A)

= |B| |A| I [:. A adj A = |A|I]

= |A| |B| I

= |AB|I ....(ii)

From (i) and (ii), we get

AB(adj AB) = AB(adj B.adj A)

Pre-multiplying both sides by(AB)^-1,we get (AB)^-1[(AB)adj AB]

= (AB)^-1 [(AB)adj B.adj A]

or adj AB = adj B.adj A

Discussion

9.	If A is square matrix of order 3 such that \|adj A\| = 64, find \|A\|.
Answer» \(\|A\| = \pm8\) Detailed Answer : \|adj A\|= \|A\|^n-1, where n is the order of the matrix \|A\|²=64 Or, \(\|A\| = \pm8\)

9.

If A is square matrix of order 3 such that |adj A| = 64, find |A|.

Answer»

\(|A| = \pm8\)

Detailed Answer :

|adj A|= |A|^n-1, where n is the order of the matrix

|A|²=64

Or, \(|A| = \pm8\)

Discussion

10.	If A is square matrix order 2 and \|adj A\| = 9, find \|A\|.
Answer» \|adj A\| = \|A\|^2-1 \|A\| = 9

10.

If A is square matrix order 2 and |adj A| = 9, find |A|.

Answer»

|adj A| = |A|^2-1

|A| = 9

Discussion

11.	If Δ=\|(5,3,8),(2,0,1),(1,2,3)\| , write the minor of the element a23 .
Answer» Minor of a₂₃=\|(5,3),(1,2)\|=10-3=7.

Discussion

12.	If `alpha, beta. gamma` are the roots of `x^3 + px^2 + q = 0,` where `q=0,` ther `Delta=[(1/alpha,1/beta,1/gamma),(1/beta,1/gamma,1/alpha),(1/gamma,1/alpha,1/beta)]` equalsA. `-p//q`B. `1//q`C. `p^(2)//q`D. None of these
Answer» Correct Answer - D `(d)` We have `betaalpha+gammaalpha+alphabeta=0` `Delta=(1)/(alpha^(3)beta^(3)gamma^(3))\|{:(betagamma,gammaalpha,alphabeta),(gammaalpha,alphabeta,betagamma),(alphabeta,betagamma,gammaalpha):}\|` `=(1)/(alpha^(3)beta^(3)gamma^(3))\|{:(betagamma+gammaalpha+alphabeta,gammaalpha,alphabeta),(gammaalpha+alphabeta+betagamma,alphabeta,betagamma),(alphabeta+betagamma+gammaalpha,betagamma,gammaalpha):}\|` [using `C_(1)toC_(1)+C_(2)+C_(3)`] `=(1)/(alpha^(3)beta^(3)gamma^(3))\|{:(0,gammaalpha,alphabeta),(0,alphabeta,betagamma),(0,betagamma,gammaalpha):}\|=0` [all zero property].

Discussion

13.	Let `x gt 0`, `y gt 0`, `z gt 0` are respectively the `2^(nd)`, `3^(rd)`, `4^(th)` terms of a `G.P.`and `Delta=\|{:(x^(k),x^(k+1),x^(k+2)),(y^(k),y^(k+1),y^(k+2)),(z^(k),z^(k+1),z^(k+2)):}\|=(r-1)^(2)(1-(1)/(r^(2)))` (where `r` is the common ratio), thenA. `k=-1`B. `k=1`C. `k=0`D. None of these
Answer» Correct Answer - A `(a)` `Delta=x^(k)y^(k)z^(k)\|{:(1,ar,a^(2)r^(2)),(1,ar^(2),a^(2)r^(4)),(1,ar^(3),a^(2)r^(6)):}\|` `=a^(3k)r^(6k)a^(3)r^(3)\|{:(1,1,1),(1,r,r^(2)),(1,r^(2),r^(4)):}\|` `=a^(3(k+1))r^(6k+3)(1-r)(r-r^(2))(r^(2)-1)` Clearly, `k=-1` `:. Delta=r^(-2)(1-r)^(2)(r^(2)-1)` `=(r-1)^(2)(1-(1)/(r^(2)))`

Discussion

14.	If `w ne 1` is a cube root of unity and `Delta=\|{:(x+w^(2),w,1),(w,w^(2),1+x),(1,x+w,w^(2)):}\|=0`, then value of `x` isA. `0`B. `2`C. `-1`D. None of these
Answer» Correct Answer - A `(a)` Applying `C_(1)toC_(1)+C_(2)+C_(3)`. We get `Delta=\|{:(x+w^(2)+w+1,w,1),(w+w^(2)+1+x,w^(2),1+x),(1+x+w+w^(2),x+w,w^(2)):}\|` `=\|{:(x,w,1),(x,w^(2),1+x),(x,x+w,w^(2)):}\|` [using `1+w+w^(2)=0`] `Delta` is clearly equal to `0` for `x=0`.

Discussion

15.	if `omega!=1` is cube root of unity and x+y+z`!=`0 then `\|[x/(1+omega),y/(omega+omega^2),z/(omega^2+1)],[y/(omega+omega^2),z/(omega^2+1),x/(1+omega)],[z/(omega^2+1),x/(1+omega),y/(omega+omega^2)]\|`=0 ifA. `x^(2)+y^(2)+z^(2)=0`B. `x+yomega+zomega^(2)=0` or `x=y=z`C. `x ne y ne z ne 0`D. x=2y=3z
Answer» Correct Answer - B `(b)` As `1+omega+omega^(2)=0` `D=\|{:(-(x)/(omega^(2)),-y,-(z)/(omega)),(-y,-(z)/(omega),-(x)/(omega^(2))),(-(z)/(omega),-(x)/(omega^(2)),-y):}\|=x^(3_+y^(3)+z^(3)-3xyz` `=(1)/(2)(x+y+z){(x-y)^(2)+(y-z)^(2)+(z-x)^(2)}` `=(x+y+z)(x+yomega+zomega^(2))(x+yomega^(2)+zomega)` The determinant varnishes if `x=y=z` or `x+yomega+zomega^(2)=0`

Discussion

16.	If `alpha` is a root of `x^4 = 1` with negative principal argument then the principal argument of `Delta(alpha) = \|(1,1,1), (alpha^n, alpha^(n+1), alpha^(n+3)), (1/alpha^(n+1), 1/alpha^n, 0)\|` isA. `(5pi)/(14)`B. `-(3pi)/(4)`C. `(pi)/(4)`D. `-(pi)/(4)`
Answer» Correct Answer - B `(b)` Clearly `alpha=-i` where `i^(2)=-1` So `Delta(alpha)=0+alpha^(2)+1-1-0-alpha^(3)` `=(-i)^(2)+1-1-(-i)^(3)` =-1+1-1-i=-1-i` So, principal argument of `Delta(alpha)` is `-(3pi)/(4)`.

Discussion

17.	Evaluate:\|(cos15º,sin15º),(sin75º,cos75º)\|
Answer» Expanding the determinant, we get cos 15° . cos 75° - sin 15° . sin 75° = cos (15° + 75° ) = cos 90° = 0 [Note : cos (A + B) = cos A. cos B - sin . sin B]

17.

Evaluate:|(cos15º,sin15º),(sin75º,cos75º)|

Answer»

Expanding the determinant, we get

cos 15° . cos 75° - sin 15° . sin 75°
= cos (15° + 75° ) = cos 90° = 0

[Note : cos (A + B) = cos A. cos B - sin . sin B]

Discussion

18.	Evaluate \(\begin{vmatrix} \sqrt 6 & \sqrt5\\[0.3em] \sqrt{20} &\sqrt{24} \\[0.3em] \end{vmatrix}\) Evaluate \|(√6 √5),(√20,√21)\|
Answer» Find determinant √6 × √ 24-√ 20 × √ 5 √ 144-√ 100. =12-10 =2.

18.

Evaluate \(\begin{vmatrix} \sqrt 6 & \sqrt5\\[0.3em] \sqrt{20} &\sqrt{24} \\[0.3em] \end{vmatrix}\) Evaluate |(√6 √5),(√20,√21)|

Answer»

Find determinant

√6 × √ 24-√ 20 × √ 5

√ 144-√ 100. =12-10

=2.

Discussion

19.	Evaluate: \|(cos15°, sin15°), (sin 75°, cos 75°)\|
Answer» Expanding the determinant, we get cos 15° . cos 75° - sin 15° . sin 75° = cos (15° + 75° ) = cos 90° = 0 [since cos (A + B) = cos A. cos B – sinA . sin B]

19.

Evaluate: |(cos15°, sin15°), (sin 75°, cos 75°)|

Answer»

Expanding the determinant, we get

cos 15° . cos 75° - sin 15° . sin 75°

= cos (15° + 75° ) = cos 90° = 0

[since cos (A + B) = cos A. cos B – sinA . sin B]

Discussion

20.	Evaluate x if : \|(2,4)(5,1)(2x,4)(2x,4)(6,x)\|.
Answer» 2-20 = 2x² - 24 x = ±√3

20.

Evaluate x if : |(2,4)(5,1)(2x,4)(2x,4)(6,x)|.

Answer»

2-20 = 2x² - 24

x = ±√3

Discussion

21.	In a triangle `ABC`, if `a,b,c` are the sides opposite to angles `A`, `B`, `C` respectively, then the value of `\|{:(bcosC,a,c cosB),(c cosA,b,acosC),(acosB,c,bcosA):}\|` isA. `1`B. `-1`C. `0`D. `acosA+bcosB+c cosC`
Answer» Correct Answer - C `(c )` `\|{:(bcosC,a,c cosB),(c cosA,b,acosC),(acosB,c,bcosA):}\|` Apply `C_(1)toC_(1)+C_(3)` and using projection rule. `D=\|{:(bcosC,a,c cosB+bcosC),(c cosA,b,acosC+c cosA),(acosB,c,bcosA+acosB):}\|` `=\|{:(bcosC,a,a),(c cosA,b,b),(acosB,c,c):}\|=0`

Discussion

22.	`\|[1/c,1/c,-(a+b)/c^2],[-(b+c)/c^2,1/a,1/a],[(-b(b+c))/(a^2c),(a+2b+c)/(ac),(-b(a+b))/(ac^2)]\|` isA. dependent on `a,b,c`B. dependent on `a`C. dependent on `b`D. independent on `a,b` and `c`
Answer» Correct Answer - A `(a)` Multiplying `C_(1)` by `a`, `C_(2)` by `b` and `C_(3)` by `c`, we obtain `Delta=(1)/(abc)\|{:((a)/(c),(b)/(c),-(a+b)/(c)),(-(b+c)/(a),(b)/(a),(c)/(a)),(-(b(b+c))/(ac),(b(a+2b+c))/(ac),-(b(a+b))/(ac)):}\|` Applying `C_(1)toC_(1)+C_(2)+C_(3)` we get `Delta=(1)/(abc)\|{:(0,(b)/(c),-(a+b)/(c)),(0,(b)/(a),(c)/(a)),(0,(b(a+2b+c))/(ac),-(b(a+b))/(ac)):}\|` This shows that `Delta` is independent of `a`, `b` and `c`.

Discussion

23.	`{:("If",a = 1 + 2 + 4 + ..."to n terms"),(,b = 1 + 3 + 9 + ..."to n terms"),(,c =1 + 5 + 25+ ..."to n terms"):}` then `\|(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n))\|=`A. `(30)^(n)`B. `(10)^(n)`C. `0`D. `2^(n)+3^(n)+5^(n)`
Answer» Correct Answer - C `(c )` a=2^(n)-1`, `b=(3^(n-1))/(2)`, `c=(5^(n)-1)/(4)` `:.Delta=\|{:(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n)):}\|` `Delta=2*\|{:(2^(n)-1,3^(n)-1,5^(n)-1),(1,1,1),(2^(n),3^(n),5^(n)):}\|=0`

Discussion

24.	If `a_1, a_2, a_3,54,a_6,a_7, a_8, a_9`are in H.P., and `D=\|a_1a_2a_3 5 4a_6a_7a_8a_9\|`, then the value of `[D]i sw h e r e[dot]`represents the greatest integer functionA. `4`B. `5`C. `6`D. `7`
Answer» Correct Answer - B `(b)` We have `D=\|{:(a_(1),a_(2),a_(3)),(5,4,a_(6)),(a_(7),a_(8),a_(9)):}\|` Since `a_(n)=(20)/(n)`, `d=(1)/(20)` Hence `D=\|{:(20,(20)/(2),(20)/(3)),((20)/(4),(20)/(5),(20)/(6)),((20)/(7),(20)/(8),(20)/(9)):}\|=((20)^(3))/(4xx7)\|{:(1,(1)/(2),(1)/(3)),(1,(4)/(5),(2)/(3)),(1,(7)/(8),(7)/(9)):}\|` Applying `R_(1)toR_(1)-R_(2)` and `R_(2)toR_(2)-R_(3)` `=((20)^(3))/(4xx7)\|{:(0,(-3)/(10),(-1)/(3)),(0,(-3)/(40),(-1)/(9)),(1,(7)/(8),(7)/(9)):}\|` `=(50)/(21)`

Discussion

Explore topic-wise InterviewSolutions in .

Let `|(a,l,m),(l,b,n),(m,n,c)||(bc-n^2,mn-lc,ln-bm),(mn-lc,ac-m^2,ml-an),(ln-bm,lm-an,ab-l^2)|=64.` If the value of `|(2a+3l,3l+5m,5m+4a),(2l+3b,3b+5n,5n+4l),(2m+3n,3n+5c,5c+4m)|=lambda` then `[lambda/2]` equalsA. `120`B. `240`C. `360`D. `480`

Given A = [(4,2,5)(2,0,3)(-1,1,0)], write the value of get.(2AA-1).

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

Is A is square matrix of 3 and |2A| = k|A|, then the value of k.

If A and B are square matrices of order 3 such that |A| = -1, |B| = 3,then find the value of |2AB|.

If Aij is the cofactor o the element aij of the determinant [(2, - 3, 5), (6, 0, 4), (1, 5, - 7)], then write the value of (i) a32A32(ii) M21

If A is a square matrix and |A| = 2, then write the value of | AA'| , where A' is the transpose of matrix A.

if A, B are square matrices of the same order, then prove that adj (AB) = (adj B) (adj A).

If A is square matrix of order 3 such that |adj A| = 64, find |A|.

If A is square matrix order 2 and |adj A| = 9, find |A|.

If Δ=|(5,3,8),(2,0,1),(1,2,3)| , write the minor of the element a23 .

If `alpha, beta. gamma` are the roots of `x^3 + px^2 + q = 0,` where `q=0,` ther `Delta=[(1/alpha,1/beta,1/gamma),(1/beta,1/gamma,1/alpha),(1/gamma,1/alpha,1/beta)]` equalsA. `-p//q`B. `1//q`C. `p^(2)//q`D. None of these

Let `x gt 0`, `y gt 0`, `z gt 0` are respectively the `2^(nd)`, `3^(rd)`, `4^(th)` terms of a `G.P.`and `Delta=|{:(x^(k),x^(k+1),x^(k+2)),(y^(k),y^(k+1),y^(k+2)),(z^(k),z^(k+1),z^(k+2)):}|=(r-1)^(2)(1-(1)/(r^(2)))` (where `r` is the common ratio), thenA. `k=-1`B. `k=1`C. `k=0`D. None of these

If `w ne 1` is a cube root of unity and `Delta=|{:(x+w^(2),w,1),(w,w^(2),1+x),(1,x+w,w^(2)):}|=0`, then value of `x` isA. `0`B. `2`C. `-1`D. None of these

if `omega!=1` is cube root of unity and x+y+z`!=`0 then `|[x/(1+omega),y/(omega+omega^2),z/(omega^2+1)],[y/(omega+omega^2),z/(omega^2+1),x/(1+omega)],[z/(omega^2+1),x/(1+omega),y/(omega+omega^2)]|`=0 ifA. `x^(2)+y^(2)+z^(2)=0`B. `x+yomega+zomega^(2)=0` or `x=y=z`C. `x ne y ne z ne 0`D. x=2y=3z

If `alpha` is a root of `x^4 = 1` with negative principal argument then the principal argument of `Delta(alpha) = |(1,1,1), (alpha^n, alpha^(n+1), alpha^(n+3)), (1/alpha^(n+1), 1/alpha^n, 0)|` isA. `(5pi)/(14)`B. `-(3pi)/(4)`C. `(pi)/(4)`D. `-(pi)/(4)`

Evaluate:|(cos15º,sin15º),(sin75º,cos75º)|

Evaluate \(\begin{vmatrix} \sqrt 6 & \sqrt5\\[0.3em] \sqrt{20} &\sqrt{24} \\[0.3em] \end{vmatrix}\) Evaluate |(√6 √5),(√20,√21)|

Evaluate: |(cos15°, sin15°), (sin 75°, cos 75°)|

Evaluate x if : |(2,4)(5,1)(2x,4)(2x,4)(6,x)|.

In a triangle `ABC`, if `a,b,c` are the sides opposite to angles `A`, `B`, `C` respectively, then the value of `|{:(bcosC,a,c cosB),(c cosA,b,acosC),(acosB,c,bcosA):}|` isA. `1`B. `-1`C. `0`D. `acosA+bcosB+c cosC`

`|[1/c,1/c,-(a+b)/c^2],[-(b+c)/c^2,1/a,1/a],[(-b(b+c))/(a^2c),(a+2b+c)/(ac),(-b(a+b))/(ac^2)]|` isA. dependent on `a,b,c`B. dependent on `a`C. dependent on `b`D. independent on `a,b` and `c`

`{:("If",a = 1 + 2 + 4 + ..."to n terms"),(,b = 1 + 3 + 9 + ..."to n terms"),(,c =1 + 5 + 25+ ..."to n terms"):}` then `|(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n))|=`A. `(30)^(n)`B. `(10)^(n)`C. `0`D. `2^(n)+3^(n)+5^(n)`

If `a_1, a_2, a_3,54,a_6,a_7, a_8, a_9`are in H.P., and `D=|a_1a_2a_3 5 4a_6a_7a_8a_9|`, then the value of `[D]i sw h e r e[dot]`represents the greatest integer functionA. `4`B. `5`C. `6`D. `7`

Explore topic-wise InterviewSolutions in .

Let `|(a,l,m),(l,b,n),(m,n,c)||(bc-n^2,mn-lc,ln-bm),(mn-lc,ac-m^2,ml-an),(ln-bm,lm-an,ab-l^2)|=64.` If the value of `|(2a+3l,3l+5m,5m+4a),(2l+3b,3b+5n,5n+4l),(2m+3n,3n+5c,5c+4m)|=lambda` then `[lambda/2]` equalsA. `120`B. `240`C. `360`D. `480`

Given A = [(4,2,5)(2,0,3)(-1,1,0)], write the value of get.(2AA-1).

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

Is A is square matrix of 3 and |2A| = k|A|, then the value of k.

If A and B are square matrices of order 3 such that |A| = -1, |B| = 3,then find the value of |2AB|.

If Aij is the cofactor o the element aij of the determinant [(2, - 3, 5), (6, 0, 4), (1, 5, - 7)], then write the value of (i) a32A32(ii) M21

If A is a square matrix and |A| = 2, then write the value of | AA'| , where A' is the transpose of matrix A.

if A, B are square matrices of the same order, then prove that adj (AB) = (adj B) (adj A).

If A is square matrix of order 3 such that |adj A| = 64, find |A|.

If A is square matrix order 2 and |adj A| = 9, find |A|.

If Δ=|(5,3,8),(2,0,1),(1,2,3)| , write the minor of the element a23 .

If `alpha, beta. gamma` are the roots of `x^3 + px^2 + q = 0,` where `q=0,` ther `Delta=[(1/alpha,1/beta,1/gamma),(1/beta,1/gamma,1/alpha),(1/gamma,1/alpha,1/beta)]` equalsA. `-p//q`B. `1//q`C. `p^(2)//q`D. None of these

Let `x gt 0`, `y gt 0`, `z gt 0` are respectively the `2^(nd)`, `3^(rd)`, `4^(th)` terms of a `G.P.`and `Delta=|{:(x^(k),x^(k+1),x^(k+2)),(y^(k),y^(k+1),y^(k+2)),(z^(k),z^(k+1),z^(k+2)):}|=(r-1)^(2)(1-(1)/(r^(2)))` (where `r` is the common ratio), thenA. `k=-1`B. `k=1`C. `k=0`D. None of these

If `w ne 1` is a cube root of unity and `Delta=|{:(x+w^(2),w,1),(w,w^(2),1+x),(1,x+w,w^(2)):}|=0`, then value of `x` isA. `0`B. `2`C. `-1`D. None of these

if `omega!=1` is cube root of unity and x+y+z`!=`0 then `|[x/(1+omega),y/(omega+omega^2),z/(omega^2+1)],[y/(omega+omega^2),z/(omega^2+1),x/(1+omega)],[z/(omega^2+1),x/(1+omega),y/(omega+omega^2)]|`=0 ifA. `x^(2)+y^(2)+z^(2)=0`B. `x+yomega+zomega^(2)=0` or `x=y=z`C. `x ne y ne z ne 0`D. x=2y=3z

If `alpha` is a root of `x^4 = 1` with negative principal argument then the principal argument of `Delta(alpha) = |(1,1,1), (alpha^n, alpha^(n+1), alpha^(n+3)), (1/alpha^(n+1), 1/alpha^n, 0)|` isA. `(5pi)/(14)`B. `-(3pi)/(4)`C. `(pi)/(4)`D. `-(pi)/(4)`

Evaluate:|(cos15º,sin15º),(sin75º,cos75º)|

Evaluate \(\begin{vmatrix} \sqrt 6 &amp; \sqrt5\\[0.3em] \sqrt{20} &amp;\sqrt{24} \\[0.3em] \end{vmatrix}\) Evaluate |(√6 √5),(√20,√21)|

Evaluate: |(cos15°, sin15°), (sin 75°, cos 75°)|

Evaluate x if : |(2,4)(5,1)(2x,4)(2x,4)(6,x)|.

In a triangle `ABC`, if `a,b,c` are the sides opposite to angles `A`, `B`, `C` respectively, then the value of `|{:(bcosC,a,c cosB),(c cosA,b,acosC),(acosB,c,bcosA):}|` isA. `1`B. `-1`C. `0`D. `acosA+bcosB+c cosC`

`|[1/c,1/c,-(a+b)/c^2],[-(b+c)/c^2,1/a,1/a],[(-b(b+c))/(a^2c),(a+2b+c)/(ac),(-b(a+b))/(ac^2)]|` isA. dependent on `a,b,c`B. dependent on `a`C. dependent on `b`D. independent on `a,b` and `c`

`{:("If",a = 1 + 2 + 4 + ..."to n terms"),(,b = 1 + 3 + 9 + ..."to n terms"),(,c =1 + 5 + 25+ ..."to n terms"):}` then `|(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n))|=`A. `(30)^(n)`B. `(10)^(n)`C. `0`D. `2^(n)+3^(n)+5^(n)`

If `a_1, a_2, a_3,54,a_6,a_7, a_8, a_9`are in H.P., and `D=|a_1a_2a_3 5 4a_6a_7a_8a_9|`, then the value of `[D]i sw h e r e[dot]`represents the greatest integer functionA. `4`B. `5`C. `6`D. `7`

Evaluate \(\begin{vmatrix} \sqrt 6 & \sqrt5\\[0.3em] \sqrt{20} &\sqrt{24} \\[0.3em] \end{vmatrix}\) Evaluate |(√6 √5),(√20,√21)|