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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.
| 351. |
If the order of a matrix is `3xx4`, then the number of elements in the matrix is _______ |
| Answer» Correct Answer - 12 | |
| 352. |
If A is a square matrix such that A2 = I, then (A – I)3 + (A + I)3 – 7A is equal toA. AB. I – AC. I + AD. 3A |
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Answer» As, (A – I)3 + (A + I)3 Use a3 + b3 = (a + b)(a2 + ab + b2) Also A2 = I ∴ then (A – I)3 + (A + I)3 – 7A = 2A + 6A – 7A = A Clearly our answer matches with option (A) ∴ Option (A) is the correct answer. |
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| 353. |
Anitha, Nikita and Ankitha have purchased some books, pencils and pens, This can be respresented in the following matrix. `{:("Anitha"),("Nikitha"),("Ankitha"):}[{:("Books","Pens","Pencils"),(5,7,8),(4,3,2),(7,6,0):}]` The total number of books purchased by Anitha, Nikitha and Ankitha isA. 18B. 10C. 15D. 16 |
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Answer» Correct Answer - D (i) The sum of element of first column. (ii) The sum of the element of first column represent the number of books purchased by Anitha, Nikitha and Ankitha. |
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| 354. |
For any two matrices A and B, we have A. AB = BAB. AB ≠ BA C. AB = O D. None of the above |
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Answer» For any two matrix: Not always option A , B and C are true. ∴ Option (D) is the only suitable answer |
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| 355. |
If `A={:[(0,y),(y,0)]:}` which of the following is true ?A. `A^(3)={:[(0,y^(3)),(y^(3),0)]:}`B. `A^(4)={:[(y^(4),0),(0,y^(4))]:}`C. Both (a) and (b)D. Neither (a) nor (b) |
| Answer» Correct Answer - C | |
| 356. |
If `A={:[(a-3,-5),(c+1,b-2)]:}` is a skew =-symmetric matrix, then a+b-c = _______ .A. 2B. 1C. -1D. 0 |
| Answer» Correct Answer - B | |
| 357. |
Anitha, Nikita and Ankitha have purchased some books, pencils and pens, This can be respresented in the following matrix. `{:("Anitha"),("Nikitha"),("Ankitha"):}[{:("Books","Pens","Pencils"),(5,7,8),(4,3,2),(7,6,0):}]` The number of pencils purchased by Anitha isA. 8B. 2C. 7D. 15 |
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Answer» Correct Answer - A (i) Element of the first row third column. (ii) The element in the first row and third column is the required answer. |
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| 358. |
For any two matrices A and B, we haveA. AB = BAB. AB ≠ BAC. AB = OD. None of the above |
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Answer» For any two matrix: Not always option A , B and C are true. ∴ Option (D) is the only suitable answer |
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| 359. |
`{:(A),(B),(C):}overset(A" " B" " C)({:[(0,3,4),(3,0,5),(4,5,0)]:})` The above matrix represent the number of routes by which we can travel from one place to another. How many ways can a person travel from C to A or B ?A. 3B. 7C. 8D. 9 |
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Answer» Correct Answer - D The sum of elements in the third row. |
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| 360. |
Find the order of the matrix `[{:(1,2,3,4,5,6,7),(5,6,7,8,4,5,6),(9,10,11,0,10,9,11):}]` |
| Answer» Correct Answer - `3xx7` | |
| 361. |
Which of the following is a `2xx1` matrix ?A. [a,b]B. `[{:(a),(b):}]`C. `[{:(a,b),(c,a):}]`D. None of these |
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Answer» Correct Answer - B The order of `[{:(a),(b):}]` is `2xx1` |
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| 362. |
The order of a matrix is `4xx1` and every element in the matrix is 5. Find the matrix. |
| Answer» Correct Answer - `[{:(5),(5),(5),(5):}]_(4xx1)` | |
| 363. |
If a matrix has 11 elements, then the order of the matrix can be _____A. `5xx6`B. `6xx5`C. `10xx1`D. None of these |
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Answer» Correct Answer - D Since the matrix contain 11 elements, `1xx11 or 11 xx 1` are possible orders. |
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| 364. |
Fill in the blanks:______ matrix is both symmetric and skew symmetric matrix. |
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Answer» A Zero matrix ∴ Let A be the symmetric and skew symmetric matrix. ⇒ A’=A (Symmetric) ⇒ A’=-A (Skew-Symmetric) Considering the above two equations, ⇒ A=-A ⇒ 2A=0 ⇒ A=0 (A Zero Matrix) Hence Zero matrix is both symmetric and skew symmetric matrix. |
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| 365. |
Fill in the blanks:Sum of two skew symmetric matrices is always _______ matrix. |
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Answer» A skew symmetric matrix ∴ Let A and B are two skew symmetric matrices. ⇒ A’=-A ..(1) ⇒ B’=-B ..(2) Now Let A+B=C ..(3) ⇒ C’=(A+B)’=A’+B’ ⇒ A’+B’=(-A)+(-B) ⇒ (-A)+(-B)=-(A+B)=-C ⇒ C’=-C (Skew Symmetric matrix) |
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| 366. |
If `({:(11, -4), (8, -3):})({:(-x, 4), (-8, y):}) = -({:(2, 3),(6, 9):})`, then find 2x - y.A. `-5`B. 5C. 0D. 14 |
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Answer» Correct Answer - A Apply matrix multiplication concept. |
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| 367. |
If a matrix contains 8 elements, then the order of the matrix can be _____A. `2xx4`B. `4xx2`C. `1xx8`D. All of these |
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Answer» Correct Answer - D Since a matrix contains 8 elements, `2xx4 or xx2 or 1xx8 or 8xx1` are possible orders. |
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| 368. |
If `[(a^x),(a^(-x))][(1,2)]=[(p,a^(- 2)),(q,log_2 2)],(a >0)` then `a^(p-q)=`A. `2^((3)/(2))`B. `2^((-3)/(2))`C. 1D. `4^((3)/(2))` |
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Answer» Correct Answer - D Multiphy the left side matrices and equate and equate the corresponding elements. |
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| 369. |
If `A = [{:(4, p), (3, -4):}] " and " A-A^(-1) = 0`, then p = ____.A. 4B. 3C. `-5`D. 5 |
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Answer» Correct Answer - C `A^(-1) = (1)/(ad-bc)[{:(d, -b), (-c, a):}]` |
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| 370. |
If `A =[(7,6),(-8,-7)]` then find `(A^12345)^-1`.A. `A^(T)`B. AC. ID. Cannot be determined |
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Answer» Correct Answer - B (i) Calculate `A^(2)`. (ii) Find `A^(2), A^(3)`,….. And observe. |
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| 371. |
Fill in the blanks:If A is a skew symmetric matrix, then A2 is a _________. |
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Answer» A2 is a symmetric matrix. Given: A’=-A ⇒ (A2)’=(AA)’=A’A’ ⇒ A’A’=(-A)(-A)=A2 ⇒ (A2)’=A2 (symmetric matrix) |
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| 372. |
If `A =((4,22),(-1,-6))`, then find `A +A^-1.`A. `[{:(8, -11), (-1, -6):}]`B. `[{:(7, 33), ((1)/(2), -4):}]`C. `[{:(7, 33), (-(3)/(2), -8):}]`D. `[{:(7, 33), (-(3)/(2), -4):}]` |
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Answer» Correct Answer - C `A^(-1) = (1)/(ad-bc)[{:(d, -b), (-c, a):}]` |
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| 373. |
If A is a `2 xx 2 scalar matrix and 7 is the one of the elements in its principal diagonal, then the inverse of A is_____.A. `[{:((-1)/(7), 0), (0, (-1)/(7)):}]`B. `[{:(0, -7), (-7, 0):}]`C. `[{:(7, 0), (0, 7):}]`D. `[{:((1)/(7), 0), (0, (1)/(7)):}]` |
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Answer» Correct Answer - D (i) Refer the definition of scalar matrix and write A. (ii) `A^(-1) = (1)/(ad-bc)[{:(d, -b), (-c, a):}]` |
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| 374. |
The inverse of a diagonal matrix, whose principal diagonal elements are `l, m` isA. `[{:((1)/(l), 0), (0, (1)/(m)):}]`B. `[{:(l, 0), (0, m):}]`C. `[{:(l^(2), 0), (0, m^(2)):}]`D. `[{:(2l, 0), (0, 2m):}]` |
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Answer» Correct Answer - A The inverse of diagonal matrix `[{:(a, 0), (0, b):}] " is " [{:((1)/(a), 0), (0, (1)/(b)):}]` |
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| 375. |
`A_(1), A_(2), A_(3), ……, A_(n) " and " B_(1), B_(2), B_(3), …., B_(n)` are non-singular square matrices of order n such that `A_(1)B_(1) = I_(n), A_(2)B_(2) = I_(n), A_(3)B_(3) = I_(n),……A_(n)B_(n) = I_(n) " then"(A_(1) A_(2)A_(3)….. A_(n))^(-1)` = ______.A. `B_(1)B_(2)B_(3)… B_(n)`B. `B_(1)^(-1)B_(2)^(-1)B_(3)^(-1)… B_(n)^(-1)`C. `B_(n)B_(n)-_(1)B_(n)-_(2)….B_(1)`D. `B_(n-1)B_(n-1^(-1)) B_(n-2^(-1))...B_(1)^(-1)` |
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Answer» Correct Answer - C (i) `AB = I rArr A^(-1) = B`. (ii) `(AB)^(-1) = B^(-1)A^(-1)` |
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| 376. |
Fill in the blanks:If A is a symmetric matrix, then A3 is a ______ matrix. |
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Answer» A3 is Also a symmetric matrix. Given: A’=A ..(1) ⇒ (A2)’=(AA)’=A’A’ ⇒ A’A’=(A)(A)=A2 ⇒ (A2)’=A2 (symmetric matrix) ..(2) ⇒ (A3)’=(A(A2))’=(A2)’A’ ⇒ (A2)’A’=A2A= A3 (Using (1) and (2) ) ⇒ (A3)’=A3 (symmetric matrix) |
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| 377. |
If `A = [{:(3, -2),(6, 4):}], " then AA"^(-1)` =_____.A. `[{:(0, 1),(1, 0):}]`B. `[{:(0, 1),(0, 0):}]`C. `[{:(1, 0),(0, 1):}]`D. `[{:(1, 0),(0, 0):}]` |
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Answer» Correct Answer - C `A^(-1) = (1)/(ad-bc)[{:(d, -b), (-c, a):}]` |
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| 378. |
If `A = [{:(5, 6), (9, 9):}], B= [{:(1, 3), (p, 3):}]` and AB = BA, then p=_____.A. `(9)/(2)`B. `(-2)/(9)`C. `(-9)/(2)`D. `(2)/(9)` |
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Answer» Correct Answer - A Find AB and BA. |
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| 379. |
Fill in the blanks:Matrix multiplication is _____ over addition. |
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Answer» Distributive ⇒ Matrix multiplication is distributive over addition. i.e A(B+C)=AB+AC and (A+B)C=AC+BC |
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| 380. |
The inverse of a scalar matrix A of order `2 xx 2`, where one of the principal diagoanl elements is 5, is _____.A. 5IB. IC. `(1)/(5)I`D. `(1)/(25)I` |
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Answer» Correct Answer - C Write matrix A and find `A^(-1)` |
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| 381. |
The order of the matrix `[{:(1,2,3),(4,5,6):}]` isA. `2xx3`B. `3xx2`C. `4xx1`D. `3xx3` |
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Answer» Correct Answer - A Order of a matrix = Number of rows `xx` Number of columns. |
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| 382. |
A matrix which is not a square matrix is called a..........matrix. |
| Answer» A matrix which is not a square matrix is called rectangular matrix. For example a rectangular matrix is `A=[a_(ij)]_(mxxn)`. Where `mnen` | |
| 383. |
The product of any matrix by the scalar ......... Is the null matrix. |
| Answer» The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 | |
| 384. |
The matrix obtained by multiplying each of the given matrix A with -1 is called the ____ of A and is denoted by _____. |
| Answer» Correct Answer - additive inverse, (-A) | |
| 385. |
If A and B are square matrices of the same order such that `A B = B A`, then proveby induction that `A B^n=B^n A`. Further, prove that `(A B)^n=A^n B^n`for all `n in N`. |
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Answer» We are given, `AB = BA` We have to prove, `AB^n = B^nA` When `n = 1`, `AB^1 = B^1A => AB = BA` So, given equation is true for `n =1` Let this equation is true for `n=k`. Then, `AB^k = B^kA` Now, we have to prove it is true for `n = k+1` that is `AB^(k+1) = B^(k+1)A` `L.H.S = AB^(k+1) = AB^k*B =B^kA*B = B^k(AB) = B^k(BA)= B^(k+1)A = R.H.S.` So, given equation is true for `n =k+1` if it is true for `n=k`. `:. AB^n = B^nA` Now, we will prove the second part. `(AB)^n = A^nB^n` When `n = 1`, `L.H.S. = (AB)^1 = AB` `R.H.S. = A^1B^1 = AB` So, given equation is true for `n =1` Let this equation is true for `n=k`. Then, `(AB)^k = A^kB^k` Now, we have to prove it is true for `n = k+1` that is `(AB)^(k+1) = A^(k+1)B^(k+1)` `L.H.S = (AB)^(k+1) = (AB)^(k)AB =(A^kB^k)AB = (A^kA)*(B^kB) = A^(k+1)B^(k+1) = R.H.S.` So, given equation is true for `n =k+1` if it is true for `n=k`. `:. (AB)^n = A^nB^n` |
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| 386. |
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:Market Products I 10.000 2.000 18.000II 6.000 20.000 8.000(a) If unit sale prices of x, y and z are Rs 2.50, Rs 1.50 and Rs 1.00, respectively, find the total revenue in each market with the help of matrix algebra.(b) If the unit costs of the above three commodities are Rs 2.00, Rs 1.00 and 50paise respectively. Find the gross profit. |
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Answer» Let `A` is the matrix that represents annual sales for both markets. `A = [[10000,2000,18000],[6000,20000,8000]]` (a) Let `B` is the matrix that represents unit sale prices of the products. Then, `B = [[2.5],[1.5],[1]]` `:.` Revenue ` = AB = [[10000,2000,18000],[6000,20000,8000]] [[2.5],[1.5],[1]]` `=>AB = [[10000**2.5+2000**1.5+18000**1],[6000**2.5+20000**1.5+8000**1]] = [[46000],[53000]]` (b) Let `C` is the matrix that represents unit cost prices of the products. Then, `C = [[2],[1],[0.5]]` So, the cost price `= AC = [[10000,2000,18000],[6000,20000,8000]] [[2],[1],[0.5]]` `=>AC = [[10000**2+2000**1+18000**0.5],[6000**2+20000**1+8000**0.5]]= [[31000],[36000]]` `:.` Gross Profit `= AB -AC = [[46000],[53000]] - [[31000],[36000]] = [[15000],[17000]]` |
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| 387. |
Find the matrix X so that `X[[1, 2, 3],[ 4 ,5, 6]]`= `[[-7,-8,-9],[2,4,6]]` |
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Answer» With the given equation, `X` should be a `2xx2` matrix. Let `X = [[a,b],[c,d]]` Then, `[[a,b],[c,d]] [[1,2,3],[4,5,6]] = [[-7,-8,-9],[2,4,6]]` `=>[[a+4b,2a+5b,3a+6b],[c+4d,2c+5d,3c+6d]] = [[-7,-8,-9],[2,4,6]]` `:. a+4b = -7->(1)` `2a+5b = -8->(2)` `3a+6b = -9=> a+2b = -3->(3)` `c+4d = 2->(4)` `2c+5d = 4->(5)` `3c+6d = 6 => c+2d = 2->(6)` Subtracting (3) from (1), `2b = -4=> b = -2` `=>a+4(-2) =-7 => a = 1` Similarly, subtracting (6) from (4), `2d = 0=> d = 0` `c+4(0) = 2 => c = 2` So, `X = [[1,-2],[2,0]]` |
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| 388. |
Using elementary transformations, find the inverse of each of the matrices`[[3 ,10],[ 2, 7]]` |
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Answer» Here, `A = [[3,10],[2,7]]` We know, `A = IA` `=>[[3,10],[2,7]]= [[1,0],[0,1]]A` Applying `R_1->R_1-R_2` `=>[[1,3],[2,7]] = [[1,-1],[0,1]]A` Applying `R_2->R_2-2R_1` `=>[[1,3],[0,1]] = [[1,-1],[-2,3]]A` Applying `R_1->R_1-3R_2` `=>[[1,0],[0,1]] = [[7,-10],[-2,3]]A` `=>I = [[7,-10],[-2,3]]A` Comparing it with , `I = A^-1A` So, `A^-1 = [[7,-10],[-2,3]]A` |
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| 389. |
For a matrix `A=[[1,2r-1] , [0,1]]` then `prod_(r=1)^(60) [[1,2r-1] , [0,1]]=`A. ` [[1, 100],[0,1]]`B. ` [[1, 4950],[0,1]]`C. ` [[1, 5050],[0,1]]`D. ` [[1, 2500],[0,1]]` |
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Answer» Correct Answer - D `prod_(r=1) ^(50) [[1, 2r-1],[0,1]]= [[1, 1+3+5+...+99],[0,1]]` `=[[1 ,(50)^(2)],[0,1]]=[[1,2500],[0,1]]` |
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| 390. |
The number of solutions of the matrix equation `X^2=[1 1 2 3]`isa. more than2 b. `2`c. `0`d. `1`A. more then 2B. 2C. 0D. 1 |
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Answer» Correct Answer - A Let `X^(2)=[[a,b],[c,d]]` `therefore X^(2)=[[a^(2)+bc,b(a+d)],[c(a+d),bc+d^(2)]]=[[1,1],[2,3]]` `rArr a^(2) + bc = 1, b(a+d) = 1, ` ` c(a+d) = 2 and bc + d^(2) = 3 ` `rArr d^(2) - a^(2) = 2` `rArr d-a = 2/(d+a) = 2b and d+a = 1/b` `therefore 2d = 2b + 1/b and 2a = 1/b - 2b` Also, c= 2b Now, from `bc+ d^(2) = 3` `rArr 2b^(2) + ( b+ 1/(2b))^(2) = 3 rArr 3b^(2) + 1/ 4b^(2) - 2 = 0 ` `rArr 12 b^(4) - 8 b^(2) + 1 = 0 ` or `(6b^(2)-1) (2b^(2) -1) = 0` `rArr b=pm 1/sqrt(6) or b = pm 1/sqrt2` Therefoer, matrices are `[[0,1/sqrt(2)],[sqrt(2),sqrt(2)]],[[0 ,-1/sqrt(2)],[-sqrt(2) ,-sqrt(2)]],[[2/sqrt(6),1/sqrt(6)],[2/sqrt(6),4/sqrt(6)]]and [[-2/sqrt(6),-1/sqrt(6)],[-2/sqrt(6),-4/sqrt(6)]]` |
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| 391. |
If `A=[(1,2),(2,1)]` and `f(x)=(1+x)/(1-x)`, then f(A) isA. `[[1 ,1],[1,1]]`B. `[[2 ,2],[2,2]]`C. `[[-1 ,-1],[-1,-1]]`D. None of these |
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Answer» Correct Answer - C `because f(x) = (1+x)/(1-x)` `rArr (1-x) f(x) = 1 +x` `rArr (I-A)f(A) = (I+A)` ` rArr f(A) = (I-A)^(-1) (I+A)` `([[1,0],[0,1]]-[[1,2],[2,1]])^(-1)([[1,0],[0,1]]+[[1,2],[2,1]])` `[[0,-2],[-2,0]]^(-1)[[2,2],[2,2]]=-1/4[[0,2],[2,0]][[2,2],[2,2]]` `=-1/4 [[4,4],[4,4]]=[[-1,-1],[-1,-1]]` |
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| 392. |
If `{:A=[(a,b),(c,d)]:}` such that `ad - bc ne 0`, then `A^(-1)`, isA. `1/(ad-bc){:[(a,-b),(-c,a)]:}`B. `1/(ad-bc){:[(a,-b),(-c,a)]:}`C. `{:[(d,b),(-c,a)]:}`D. none of these |
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Answer» Correct Answer - A We have, `absA=ad-bcne0` Cofactor of `a_11=d," Cofactor of "a_12=-c` Cofactor of `a_21=-b," Cofactor of "a_22 =a` `:. A^(-1) =1/absAadjA =1/(ad-bc){:[(d,-b),(-c,a)]:}` |
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| 393. |
Let a be a matrix of order `2xx2` such that `A^(2)=O`. `A^(2)-(a+d)A+(ad-bc)I` is equal toA. `I`B. `O`C. `-I`D. none of these |
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Answer» Correct Answer - B Let `A=[(a,b),(c,d)]` `implies A^(2)-(a+d)A+(ad-bc)I` `=[(a,b),(c,d)][(a,b),(c,d)]-(a+d)[(a,b),(c,d)]+(ad-bc)[(1,0),(0,1)]` `=[(a^(2)+bc,ab+bd),(ac+cd,bc+d^(2))]-[(a^(2)+ad,ab+bd),(ac+cd,bc+d^(2))]+[(a^(2)+ad,ab+bd),(ac+cd,ad+d^(2))]` `=O` |
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| 394. |
If `A = [[1 ,1],[1,1]]` and det `(A^(n) - 1) = 1 -lambda ^(n), n in N,` then the value of `lambda` isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B `because " "A = [[1 ,1],[1,1]]` `therefore " "A^(2) = [[2 ,2],[2,2]] = 2[[1 ,1],[1,1]] =2A` ` rArr A^(3) = A^(2) cdot A = 2A^(2) = 2^(2) A` Similarly, `A_(n) = 2^(n-1) A ` `therefore " "A^(n) -I= [[2^(n-1) ,2^(n-1)],[2^(n-1),2^(n-1)]] -[[1 ,0],[0,1]]` `= [[2^(n-1)-1 ,2^(n-1)],[2^(n-1),2^(n-1)-1]] ` `rArr "det" (A^(n)-I) = (2^(n-1)-1)^(2) - (2^(n-1))^(2)` `= 1- 2^(n) = 1 - lambda ^(n)` [given] `therefore lambda = 2` |
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| 395. |
If `A=[(a,b),(c,d)]`, where a, b, c and d are real numbers, then prove that `A^(2)-(a+d)A+(ad-bc) I=O`. Hence or therwise, prove that if `A^(3)=O` then `A^(2)=O` |
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Answer» Given, `A=[(a,b),(c,d)]` `implies A^(2)=[(a,b),(c,d)].[(a,b),(c,d)]=[(a^(2)+bc,ab+bd),(ac+cd,bc+d^(2))]` Hence, `A^(2)-(a+d)A+(ad-bc) I` `=[(a^(2)+bc,ab+bd),(ac+cd,bc+d^(2))]-(a+d) [(a,b),(c,d)]+(ad-bc) [(1,0),(0,1)]` `=[(a^(2)+bc-(a^(2)+ad)+(ad-bc),ab+bd-(ab+bd)),(ac+cd-(ac+cd),bc+d^(2)-(ad+d^(2))+(ad-bc))]` `=[(0,0),(0,0)]=O` given `A^(3)=O` `implies |A|=0` or `ad-bc=0` `implies A^(2)-(a+d)A=O` or `A^(2)=(a+d)A` (1) Case I : `a+d=0` From equation (A), `A^(2)=O`. Case II : `a+d ne 0` Given `A^(3)=O` `implies A^(2)A=O` `implies (a+d)A A=O` `implies A^(2)=O` |
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| 396. |
If `A=[(k,l),(m,n)]` and `kn!=lm,` show that `A^(2)-(k+n)A+(kn-lm)l=O.` Hence, find `A^(-1)` |
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Answer» We, have, `A[(k,l),(m,n)]`, then `|A|=|(k,l),(m,n)|` `=kn-ml!=0` `therefore" " A^(-1) exists.` Now, `A^(2)=A.A=[(k,l),(m,n)][(k,l)(m,n)=[(k^(2)+lm,kl+ln),(mk+nm,ml+n^(2))]` `therefore " " A^(2)-(k+n)A+(kn-lm)I` `= [(k^(2)+lm,kl+ln),(mk+nm,ml+n^(2))]-[(k-n)[(k,l),(m,n)]+(kn-lm)[(1,0),(0,1)]` `= [(k^(2)+lm,kl+ln),(mk+nm,ml+n^(2))][(k^(2)+nk,kl+nl),(km+nm,kn+n^(2))] +[(kn-lm,0),(0,kn-lm)]` `[(k^(2)+lm-K^(2)-nk+kn-lm,kl+ln-kl-ln),(mk+nm-km-nm,ml+n^(2)-kn-n^(2)+kn-lm)]` `[(0,0),(0,0)]=O` `AsA^(2)-(k+n)A+(kn-lm)I=O` `rArr" " (kn-lm)I=(k+n)A-A^(2)` `rArr" " (kn-lm)IA^(-1)=(k+n)A-A^(2))A^(-1)` `rArr" " (kn-lm)A^(-1)=(k+n)A A^(-1)-A(A A^(-1))` `=(k+n)I-AI` `=(k+n)I-A` `=(k+n)[(1,0),(0,1)]-[(k,l),(m,n)]` `=[(k+n,0),(0,k+n)]-[(k,l),(m,n)]` ` rArr " " (kn-lm)A^(-1)=[(n,-1),(-m,k)]` Hence `A^(-1)=(1)/((kn-lm))[(n,-1),(-m,k)]` |
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| 397. |
Find the characteristic equation of the matrix` A= [(2,1),(3,2)]` and hence find its inverse using Cayley-hamilton theorem. |
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Answer» Charaacteritic equation is `|A-lambdaI|=0rArr [(2-lambda,1),(3,2-lambda)]=0` `rArr (2-lambda)^(2)-3=0` `rArr lambda^(2)-4lambda+1=0` therefore Cayley-hamiltion theorem, `A^(2)-4A+I=O or I=4A-A^(2)` Multiplying by` A^(-1)`, we get ` A^(-1)=4A^(-1)A-A^(-1)A A` =` 4I-IA=4I-A` =`4[(1,0),(0,1)]-{(2,1),(3,2)]` ` [(2,-1),(-3,2)]` |
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| 398. |
Find the matrix X such that `2A-B+X=O`, where `A=[{:(3,1),(0,2):}]" and "B=[{:(-2,1),(0,3):}].` |
| Answer» `[{:(-8,-1),(0,-1):}]` | |
| 399. |
If `A = ({:(9, -7), (-4, 3):}) " and B"=({:(-3, -7),(-4, -9):})`, then find AB and hence find `A^(-1)`. |
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Answer» Correct Answer - `AB = [{:(1, 0), (0, 1):}]` `A^(-1) = B= [{:(-3, -7), (-4, -9):}]` |
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| 400. |
If `A=[{:(-2," "3),(" "4," "5),(" "1," "-6):}]" and "B=[{:(" "5," "2),(-7," "3),(" "6," "4):}]`, find a matrix C such that `A+B-C=O.` |
| Answer» `[{:(3," "5),(-3," "8),(7,-2):}]` | |