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InterviewSolution
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.
| 601. |
Statement -1 (Assertion) and Statement - 2 (Reason) Each of these examples also has four alternative choices, ONLY ONE of which is the correct answer. You have to select the correct choice as given below Statement-1 If A and B are two matrices such that AB = B, BA = A, then ` A^(2) + B^(2) = A+B.` Statement-2 A and B are idempotent motrices, then `A^(2) = A, B^(2) = B`.A. Statement - 1 is true, Statement - 2 is true , Statement - 2 is correct explanaction for Statement -2B. Statement -1 is true, Statement - 2 is true, Statement - 2 is not a correct explanation for Statement-2C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is ttrue |
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Answer» Correct Answer - B `because AB = B` `rArr B(AB) = B cdot B` `rArr (BA) B = B^(2) ` [ "by associative law"] `rArr AB = b^(2) [ because BA=A]` `rArr B= B^(2) [because AB=B]` and Ba = A `rArr A(BA) = A cdot A` `rArr (AB) A + A^(2) ` [by associative law] `rArr BA= A^(2) " " [because AB=B]` `rArr A=A^(2) [ because BA=A]` Hence, `therefore A^(2) + B^(2) = A + B` Here, both statments are true and Statement - 2 is not a correct explanation for Statement-1. |
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| 602. |
Statement -1 (Assertion) and Statement - 2 (Reason) Each of these examples also has four alternative choices, ONLY ONE of which is the correct answer. You have to select the correct choice as given below Statement-1 A is singular matrox pf order `nxxn,` then adj A is singular. Statement -2 `abs(adj A) = abs(A)^(n-1)`A. Statement - 1 is true, Statement - 2 is true , Statement - 2 is correct explanaction 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 ttrue |
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Answer» Correct Answer - D If A is non-singular matrix of order `nxxn,` then `abs(adj A) = abs(A)^(n-1)` Hence, Statement- 1 is false and Stement- 2 is true. |
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| 603. |
Let `A= [[a,b,c],[b,c,a],[c,a,b]]` is an orthogonal matrix and `abc = lambda (lt0).` The value ` a^(2) b^(2) + b^(2) c^(2) + c^(2) a^(2)`, isA. `2lambda`B. `-2lambda`C. `lambda^(2)`D. `-lambda` |
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Answer» Correct Answer - B `becauseA` is an orthogonal matrix `therefore A A^(T) =I` `[[a,b,c],[b,c,a],[c,a,b]] [[a,b,c],[b,c,a],[c,a,b]] =1 [[1,0,0],[0,1,0],[0,0,1]]` `[[a^(2)+b^(2)+c^(2),ab + bc+ca,ab + bc+ ca],[ab + bc + ca,a^(2) +b^(2)+c^(2) , ab+ bc+ ca ],[ab+ bc+ca,ab+bc+ca,a^(2) + b^(2) + c^(2)]] =1 [[1,0,0],[0,1,0],[0,0,1]]` By equality of matrices, we get `a^(2) + b^(2) +c^(2) = 1 ` ...(i) `ab + bc + ca= 0` ...(ii) ` (a+b+c)^(2) + a^(2)= b^(2) +c^(2)+ 2 (ab + bc + ca)` `= 1 + 0 = 1` ` therefore a+ b + c = pm 1` ...(iii) `because a^(2) b^(2) + b^(2) a^(2) + c^(2) a^(2) = (ab + bc+ ca) ^(2) - 2abc (a + b + c)` `= 0- 2abc (pm 1) pm 2 lambda [ because abc = lambda ]` ` = - 2 lambda ` `[because lambda lt 0]` |
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| 604. |
Suppose A and B be two ono-singular matrices such that `AB= BA^(m), B^(n) = I and A^(p) = I `, where `I` is an identity matrix. Which of the following orderd triplet `(m, n, p)` is false?A. `(3, 2, 80)`B. `(6, 3, 215)`C. `(8, 3, 510)`D. `(2, 8, 255)` |
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Answer» Correct Answer - C `because AB = BA^(m) ` ` rArr B = A^(-1) BA^(m)` `therefore B^(n) = underset("n times")(underbrace((A^(-1) BA^(m))(A^(-1)BA^(m))... (A^(-1) BA^(m))))` `=A^(-1) underset("n times")(underbrace(BA^(m-1)BA^(m-1)... BA^(m-1)BA^(m-1)))A` ...(i) Given, ` AB = BA^(m)` ` rArr A AB = ABA^(m) = BA^(2m) rArr A A AB = BA^(3m)` Similarly, `A^(x) B = BA^(mx) AA m in N` From Eq. (i) we get `B^(n)=A^(-1) BA^(m-1) underset("(n-1) times")(underbrace(BA^(m-1)BA^(m-1)... BA^(m-1)BA^(m-1)))A` `=A^(-1) B(A^(m-1) B)A^(m-1) underset("(n-2) times")(underbrace(BA^(m-1)... BA^(m-1)BA^(m-1)))A` `=A^(-1) BBA^((m-1)m) A^(m-1) underset("(n-2) times")(underbrace(BA^(m-1)... BA^(m-1)BA^(m-1)))A` `=A^(-1) B^(2)A^((m^(2)-1)) underset("(n-2) times")(underbrace(BA^(m-1)... BA^(m-1)BA^(m-1)))A` `..." " ..." "... " "... ` `=A^(-1) B^(m) (A) ^(m^(n)-1)A` `I = A^(-1) I A^(m^(n)-1) A [because B^(n) = I]` `I = A^(-1) A^(m^(n)-1) A= A^(-1) A^(m^(n))` `rArr I = A^(m^(n)-1)` `therefore p= m^(n)-1 " "...(ii) [because A^(p) = I]` From Eq. (ii), we get `510 ne 8^(3) -1` |
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| 605. |
Suppose `A` and `B` are two non singular matrices such that `B != I, A^6 = I` and `AB^2 = BA`. Find the least value of `k` for `B^k = 1`A. `31`B. `32`C. `64`D. `63` |
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Answer» Correct Answer - D `(d)` `A^(6)=IimpliesBA^(6)=B` `implies(BA)A^(5)=B` `impliesAB^(2)A^(5)=B` `impliesAB(AB^(2))A^(4)=B` `impliesA^(2)B^(4)A^(4)=B` Proceeding like this we get `A^(6)B^(64)=BimpliesB^(64)=B` `impliesB^(63)=I` `impliesk=63` |
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| 606. |
Suppose `A` and `B` are two non singular matrices such that `B != I, A^6 = I` and `AB^2 = BA`. Find the least value of `k` for `B^k = 1`A.B.C.D. |
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Answer» Given, `AB = BA^(2) rArr B= A^(-1) BA^(2) rArr B^(3) = I` `rArr (A^(-1) BA A) ( A^(-1) BA A ) (A^(-1) BA A) = I` `rArr (A^(-1) BA A) (BA ) ( BA A) = I [because A^(-1) A=I]` `rArr A^(-1) B (BA^(2)) (BA^(2)) A A=I` `rArr A^(-1) B(BA^(2)) (BA^(2)) A A = I [ because AB =BA^(2) ] ` `rArr A^(-1) BBA (AB) A^(4) = I` `rArr A^(-1) BBA (BA^(2) )A^(4) = I [because AB = BA^(2)]` `rArr A^(-1) BB (AB) A^(6) = I` `rArr A^(-1) BB(BA^(2)) A^(6) = I [because AB = BA^(2)]` `rArr A^(-1) B^(3) A^(8) = I` ` rArr (A^(-1)I) A^(8) = I [ B^(3) = I]` `rArr A^(-1) A^(8) = I` `rArr A^(7) = I = A^(k) [ because A^(k) = I]` `rArr A^(k) = A^(7)` `therefore` Least value of `k` is 7. |
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| 607. |
Let `A=[{:(3,5," "4),(1,2,-3):}]" and "O=[{:(0,0,0),(0,0,0):}]`, then verify that `A+O=O+A=A.` |
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Answer» Clearly, each one of A and O is a matrix of order `(2xx3).` So, `(A+O)` and `(O+A)` are both defined. Now, `A+O=[{:(3,5,4),(1,2,-3):}]+[{:(0,0,0),(0,0,0):}]` `=[{:(3+0,5+0,4+0),(1+0,2+0,-3+0):}]+[{:(3,5,4),(1,2,-3):}]=A.` And, `O+A=[{:(0,0,0),(0,0,0):}]+[{:(3,5,4),(1,2,-3):}]` `=[{:(0+3,0+5,0+4),(0+1,0+2,0+(-3)):}]=[{:(3,5,4),(1,2,-3):}]=A.` Hence, `A+O=O+A=A.` |
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| 608. |
If `A=[{:(3,-2," "0),(-5," "7,sqrt(2)):}],` find (-A) and verify that `A+(-A)=(-A)+A=0.` |
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Answer» Clearly, we have: `(-A)=[{:(-3," "2," "0),(5,-7,-sqrt(2)):}]` `A+(-A)=[{:(3,-2,0),(-5," "7,sqrt(2)):}]+[{:(-3," "2," "0),(5,-7,-sqrt(2)):}]` `=[{:(3+(-3),-2+2," "0+0),(-5+5,7+(-7),sqrt(2)+"("-sqrt(2)")"):}]=[{:(0,0,0),(0,0,0):}]=O.` And, `(-A)+A=[{:(-3," "2," "0),(5,-7,-sqrt(2)):}]+[{:(3,-2," "0),(-5," "7,sqrt(2)):}]` `=[{:(-3+3,2+(-2)," "0+0),(5+(-5),-7+7,-sqrt(2)+sqrt(2)):}]=[{:(0,0,0),(0,0,0):}]` Hence, `A+(-A)=(-A)+A=O.` |
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| 609. |
Let `A=[{:(" "2,3,5),(-1,0,4):}]" and "B=[{:(4,-2,3),(2," "6,-1):}].` Verify that `A+B=B+A.` |
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Answer» Here, A is a `2xx3` matrix and B is a `2xx3` matrix. So, A and B are comparable. Therefore, `(A+B)` and `(B+A)` both exist and each is a `2xx3` matrix. Now, `A+B=[{:(" "2,3,5),(-1,0,4):}]+[{:(4,-2," "3),(2," "6,-1):}]` `=[{:(2+4,3+(-2),5+3),(-1+2,0+6,4+(-1)):}]=[{:(6,1,8),(1,6,3):}].` And, `B+A=[{:(4,-2," "3),(2," "6,-1):}]+[{:(" "2,3,5),(-1,0,4):}]` `=[{:(4+2,-2+3,3+5),(2+(-1)," "6+0,(-1)+4):}]=[{:(6,1,8),(1,6,3):}].` Hence, `A+B=B+A.` |
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| 610. |
Solve for x and y, when `[{:(3,-4),(1," "2):}][{:(x),(y):}]=[{:(3),(11):}].` |
| Answer» Correct Answer - `x=5,y=3` | |
| 611. |
Let `A=[{:(1,-2),(5," "4),(3," "0):}],B=[{:(" "3," "1),(" "0," "2),(-3," "5):}]" and "C=[{:(" "4," "3),(-2," "2),(" "1," "6):}].` Verify that `(A+B)+C=A+(B+C).` |
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Answer» Clearly, each one of the martices A, B, C is a `(3xx2)` matrix. So, `(A+B)+C` and `A+(B+C)` are both defined and each one is a `3xx2` matrix. Now, `(A+B)=[{:(1,-2),(5," "4),(3," "0):}]+[{:(3,1),(0,2),(-3,5):}]` `=[{:(1+3,-2+1),(5+0," "4+2),(3+(-3)," "0+5):}]+[{:(4,-1),(5," "6),(0," "5):}].` `:." "(A+B)+C=[{:(4,-1),(5," "6),(0," "5):}]+[{:(" "4," "3),(-2," "2),(" "1," "6):}]` `=[{:(4+4,-1+3),(5+(-2)," "6+2),(0+1," "5+6):}]=[{:(8,2),(3,8),(1,11):}].` Also, `(B+C)=[{:(3,1),(0,2),(-3,5):}]+[{:(4,3),(-2,2),(1,6):}]` `=[{:(3+4,1+3),(0+(-2),2+2),(-3+1,5+6):}]=[{:(7,4),(-2,4),(-2,11):}]` `:." "A+(B+C)=[{:(1,-2),(5," "4),(3," "0):}]+[{:(7,4),(-2,4),(-2,11):}]` `=[{:(1+7,-2+4),(5+(-2)," "4+4),(3+(-2)," "0+11):}]=[{:(8,2),(3,8),(1,11):}].` Hence, `(A+B)+C=A+(B+C).` |
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| 612. |
`(AB)^(-1)` =____. |
| Answer» Correct Answer - `B^(-1)A^(-1)` | |
| 613. |
Find the values of x and y for which `[{:(2,-3),(1," "1):}][{:(x),(y):}]=[{:(1),(3):}].` |
| Answer» Correct Answer - `x=2,y=1 | |
| 614. |
The inverse of matrix A, if `A^(2)` = 1, is ____. |
| Answer» Correct Answer - A | |
| 615. |
Find the additive inverse of the matrix `A=[{:(2,-5," "0),(4," "3,-1):}].` |
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Answer» Clearly, the additive inverse of the given matrix A is the matrix -A, given by `-A=[{:(-2,-(-5)," "0),(-4," "-3,-(-1)):}]=[{:(-2," "5,0),(-4,-3,1):}].` |
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| 616. |
`adj AB -(adj B)(adj A) =`A. `adj A-adj B`B. 1C. 0D. non of these |
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Answer» Correct Answer - c |
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| 617. |
If A is unimidular, then which of the following is unimodular ?A. `-A`B. `A^(-1)`C. adj AD. `omegaA`, where `omega` is cube root of unity |
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Answer» Correct Answer - B::C det `(-A)=(-1)^(n)` det (A) det `(A^(-1))=1/("det (A)")=1` det (adj A) `=|A|^(n-1)=1` `|omega A|=omega^(n)|A|=1` only when `n=3k, k in Z`. |
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| 618. |
Is A `=[{:(4, 6), (2, 3):}]` singular? |
| Answer» Correct Answer - Given matrix is singular matrix | |
| 619. |
There are two possible values of A in the solution of the matrix equation `[[2A+1,-5],[-4,A]]^(-1) [[A-5,B],[2A-2,C]]= [[14,D],[E,F]]`, where A, B, C, D, E, F are real numbers. The absolute value of the difference of these two solutions, isA. `8/3`B. `11/3`C. `1/3`D. `19/3` |
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Answer» Correct Answer - D `because [[A-5,B],[2A-2,C]]= [[2A+1,-5],[-4,A]][[14,D],[E,F]]` `rArr A-5 = 28 A + 14 - 5E` `rArr 5e = 27 A + 19` …(i) `2A - 2 = -56 + AE` ` rArr AE = 2A +54 ` (ii) From eq. (i), we get `5AE = 27A^(2) + 19A` `rArr 5 (2A+54)=27 A^(2) + 19A ` [from Eq. (ii) ] `rArr 27A^(2) + 9 A - 270 = 0` `rArr 9 (A-3) (3A+10)=0` `therefore A= 3, A= -10/3` `therefore` Absolute value of difference `=abs(3+10/3 ) = 19/3` |
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| 620. |
The additive inverse of `[{:(-1, 3, 4), (5, -7, 8):}]` is ____. |
| Answer» Correct Answer - `[{:(1, -3, -4), (-5, 7, -8):}]` | |
| 621. |
If `A=[{:(1,2),(4,-3):}]` and `f(x)=2x^(3)+4x+5`, find f(A). |
| Answer» `f(A)=[{:(-5," "68),(136,-141):}]` | |
| 622. |
The inverse of the matrix `[(1,0,0),(3,3,0),(5,2,-1)]` isA. `-(1)/(3)[(-3,0,0),(3,1,0),(9,2,-3)]`B. `-(1)/(3)[(-3,0,0),(3,-1,0),(-9,-2,3)]`C. `-(1)/(3)[(3,0,0),(-3,-1,0),(-9,-2,3)]`D. `-(1)/(3)[(-3,0,0),(-3,-1,0),(-9,-2,3)]` |
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Answer» Correct Answer - b |
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| 623. |
If `P = [[cos frac(pi)(6), sin frac(pi)(6) ],[-sinfrac(pi)(6),cosfrac(pi)(6)]], A = [[1,1],[0,1]]and Q = PAP^(T)` then `P^(T) Q^(2007) P ` is equal toA. `[[1,sqrt(3)//2],[0,2007]]`B. `[[1,2007],[0,1]]`C. `[[sqrt(3)//2, 2007 ],[0,1]]`D. `[[sqrt(3)//2,-1//2],[1,2007]] |
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Answer» Correct Answer - B We have, `P^(T) = P^(-1) " " [ because PP^(T) =I]` Now, ,,`Q= PAP^(T) = PAP^(-1)` `therefore Q^(2007) = PA^(2007) P^(-1) ` `therefore P^(T) Q^(2007 ) P= P^(-1) (PA^(2007)P^(-1)) P` `= A^(2007 = [[1,2007],[0,1]][becauseA^(2)=[[1,2],[0,1]],A^(3) = [[1,3],[0,1]],...]` |
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| 624. |
If `A xx [{:(1, 2, 3),(4, 5, 6):}] = [{:(1, 2, 3), (3, 2, 1), (3, 1, 2):}]`, then the order of A is ____. |
| Answer» Correct Answer - `3 xx 2` | |
| 625. |
If `A=[{:(-1,2),(3,1):}]`, find f(A), where `f(x)=x^(2)-2x+3.` |
| Answer» `[{:(12,-4),(-6," "8):}]` | |
| 626. |
If `A=[(1,-1,0),(1,0,0),(0,0,-1)]`, then `A^(-1)` isA. `A^(T)`B. `A^(2)`C. AD. I |
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Answer» Correct Answer - a |
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| 627. |
If the order of matrices A, B and C are `3 xx 4, 7 xx 3 " and " 4 xx 7` respectively, then the order of (AC) B is _____. |
| Answer» Correct Answer - `3 xx 3` | |
| 628. |
If A `A=[3-2 4-2]`and `I=[1 0 0 1]`, find k so that `A^2=k A-2I`. |
| Answer» Correct Answer - k=1 | |
| 629. |
Express the equations 2x-y + 6 =0 and 6x + y + 8 = 0, in the matrix equation form. |
| Answer» Correct Answer - `[{:(2, -1),(6, 1):}][(x), (y)] = [(-6), (-8)]` | |
| 630. |
What is the inverse of `A=[{:(0,0,1),(0,1,0),(1,0,0):}]`?A. AB. `A^(T)`C. `[(1,0,0),(0,1,0),(0,0,1)]`D. `[(1,0,0),(1,0,0),(0,1,0)]` |
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Answer» Correct Answer - a |
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| 631. |
If `A+B={:[(01,-11),(9,7)]:}andA-B={:[(-8,9),(9,-5)]:}`, then B=A. `{:[(9,-10),(0,-6)]:}`B. `{:[(9,10),(0,-6)]:}`C. `{:[(9/2,-10),(0,6)]:}`D. `{:[(9,10),(0,6)]:}` |
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Answer» Correct Answer - C 2B=(A+B)-(A-B). |
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| 632. |
If A`=[{:(a, b), (c, d):}] " then " A^(-1)` = ____. |
| Answer» Correct Answer - `(1)/(ad-bc) [{:(d, -b), (-c, a):}]` | |
| 633. |
If `A=[((k)/(2),0,0),(0,(l)/(2),0),(0,0,(m)/(4))] and A^(-1)= [((1)/(2),0,0),(0,(1)/(3),0),(0,0,(1)/(4))]` then `k+l+m=`A. 1B. 9C. 14D. 29 |
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Answer» Correct Answer - d |
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| 634. |
If `A={:((2,-2),(-2,2)):},B={:((5,3),(-7,2)):}and=C={:((4,2),(-8,1)):}` thenA. AB=BCB. AB=ACC. BC=ACD. None of these |
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Answer» Correct Answer - B (i) Find the product AB,BC,AC. (ii) Find the products AB,BC and AC and verify from options. |
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| 635. |
If `A={:((-2,-1),(-5,-3)):}B={:((-3,1),(5,-2)):}and(AB)^(n)=I` then n is (a/an)A. odd number.B. even numberC. `AAninN`.D. None of these |
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Answer» Correct Answer - C (i) Find the product of AB. (ii) Find `AB,(AB)^(2)` and so on till you get I to know the value of n. |
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| 636. |
If `A=[{:(2,-2),(-3," "4):}]` then find `(-A^(2)+6A).` |
| Answer» Correct Answer - `[{:(2,0),(0,2):}]` | |
| 637. |
If `A(theta)=[(sin theta, i cos theta),(i cos theta, sin theta)]`, then which of the following is not true ?A. `A(theta)^(-t)=A(pi-theta)`B. `A(theta)+A(pi+theta)` is a null matrixC. `A(theta)` is invertible for all `theta in R`D. `A(theta)^(-1)=A(-theta)` |
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Answer» Correct Answer - A::B::C We have, `|A(theta)|=1` Hence, A is invertiable. `A(pi+theta)=A(pi + theta)=[(sin(pi+theta),i cos (pi+theta)),(i cos (pi+theta),sin (pi+theta))]` `=[(-sin theta,-i co theta),(-i cos theta,-sin theta)]=-A(theta)` adj `(A(theta))=[(sin theta,-i cos theta),(-i cos theta,sin theta)]` `implies A(theta)^(-1) =[(sin theta,-i cos theta),(-i cos theta,sin theta)]=A(pi-theta)` |
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| 638. |
The invrse of the matrix `[(2,0,0),(0,3,0),(0,0,4)]` isA. `(1)/(24)[(2,0,0),(0,3,0),(0,0,4)]`B. `[(2,0,0),(0,3,0),(0,0,4)]`C. `(1)/(24)[(1,0,0),(0,1,0),(0,0,1)]`D. `[((1)/(2),0,0),(0,(1)/(3),0),(0,0,(1)/(4))]` |
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Answer» Correct Answer - d |
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| 639. |
If `A={:((4,3),(-5,2)):}`, then `A^(2)-6A` =A. 23IB. 24IC. `-23I`D. `24I` |
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Answer» Correct Answer - C (i) Find the square of A,i.e., `(AxxA)`. (ii) `A^(2)-6A=A(A-6I)`. |
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| 640. |
If `{:((2,3),(p,1)):}{:((5,2),(-4,6)):}={:((-2,q),(16,r)):}`, then 2p+r=A. qB. 21C. 2qD. q-r |
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Answer» Correct Answer - A (i) Equate the corresponding elements on both sides. (ii) Find the product of the two matrices which are on LHS. (iii) Equate the corresponding elements and get the values p,q and r, then find 2p+r. |
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| 641. |
If `{:[(4,-5),(3,6)]:}+2X={:[(8,-1),(-7,2)]:}`, then `X^(T)` =A. `{:[(2,2),(-5,2)]:}`B. `{:[(2,-5),(2,-2)]:}`C. `{:[(1,0),(0,1)]:}`D. `{:[(2,-5),(-2,2)]:}` |
| Answer» Correct Answer - B | |
| 642. |
If AB=A and BA=Bm then which of the following is/are true ?A. A is idempotentB. B is idempotentC. `A^T` is idempotentD. none of these |
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Answer» Correct Answer - A::B::C Given, `AB=A, BA=B` `implies BxxAB=BxxA` or `(BA)B=B` or `B^(2)=B` Also, `AxxBA=AB` `implies (AB)A=A` `implies A^(2)=A` Now `(A^(T))^(2)=(A^(T)xxA^(T))=(AxxA)^(T)=(A^(2))^(T)=A^(T)` Similarly, `(B^(T))^(2)=B^(T)` Hence, `A^(T)` and `B^(T)` are idempotent. |
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| 643. |
If `AB = KI`, where `K in R`, then `A^-1=` |
| Answer» Correct Answer - `(1)/(K)B` | |
| 644. |
If A`=[{:(a, b), (c, d):}] " then " A^(-1)` = ____.A. `(1)/(ab-cd)[(b,-c),(-d,a)]`B. `(1)/(ad-bc)[(b,-c),(-d,a)]`C. `(1)/(ab-cd)[(b,d),(c,a)]`D. None of these |
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Answer» Correct Answer - a |
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| 645. |
If `[{:(2, -4), (9, d-3):}]=4`, then d=____. |
| Answer» Correct Answer - -13 | |
| 646. |
If the matrix `({:(10, -9), (5x + 7, 5):})` is non-singular, then the range of x.A. `(113)/(45)`B. `R -{(-113)/(45)}`C. `R -{(113)/(45)}`D. `(-113)/(45)` |
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Answer» Correct Answer - B Determinant is non-zero. |
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| 647. |
If `A={:((1,5),(3,6)):}and=((34)/(39))`, then find the matrix X such that AX=B.A. `((0),(7))`B. `((7),(-1))`C. `((-1),(7))`D. `((9),(-1))` |
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Answer» Correct Answer - C (i) Find the order of X using product rule. (ii) Order of X is `2xx1`. (iii) Verify from the options. |
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| 648. |
If `A={:[(p,2),(-4,-5)]:},B={:[(q,r-s),(r,-5)]:}" and A=B. Find"{:[(p,q),(r,s)]:}`, if the trace of A+B=-2.A. `{:[(2,6),(-4,-7)]:}`B. `{:[(4,4),(-4,-7)]:}`C. `{:[(3,5),(-4,-7)]:}`D. `{:[(1,0),(0,1)]:}` |
| Answer» Correct Answer - B | |
| 649. |
If `A=[(1,-1),(2,1)], B=[(a,1),(b,-1)]` and `(A+B)^(2)=A^(2)+B^(2)+2AB`, thenA. `a=-1`B. `a=1`C. `b=2`D. `b=-2` |
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Answer» Correct Answer - A::D Given, `(A+B)^(2)=A^(2)+B^(2)+2AB` or `(A+B) (A+B)=A^(2)+B^(2)+2AB` or `A^(2)+AB+BA+B^(2)=A^(2)+B^(2)+2AB` or `BA=AB` or `[(a,1),(b,-1)] [(1,-1),(2,1)]=[(1,-1),(2,1)][(a,1),(b,-1)]` or `[(a+2,-a+1),(b-2,-b-1)]=[(a-b,1+1),(2a+b,2-1)]` The corresponding elements of equal matrices are equal. `a+2=a-b, -a+1=2 implies a=-1` `b-2=2a+b, -b-1=1implies b=-2` `implies a=-1, b=-2` |
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| 650. |
If p `=({:(1, 0), (0, 1):}), " then " P^(-1)` =____. |
| Answer» Correct Answer - `({:(1, 0), (0, 1):})` | |