1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 24 InterviewSolution

1151.	Which of the following is a `2xx3` matrix such that every element in the matrix is zero ?A. `[{:(0,0,0),(0,0,0):}]`B. `[{:(0,0),(0,0),(0,0):}]`C. `[{:(0,0),(0,0)]`D. `[{:(0,0,0,0),(0,0,0,0),(0,0,0,0):}]`
Answer» Correct Answer - A (i) Zero matrix of order `2xx3`. (ii) The matrix with 2 rows and 3 columns with all the elements as zeroes is the required matrix.

Discussion

1152.	`A` and `B` be `3xx3` matrices such that `AB+A=0`, thenA. `(A+B)^(2)=A^(2)+2AB+B^(2)`B. `\|A\|=\|B\|`C. `A^(2)=B^(2)`D. none of these
Answer» Correct Answer - A `(a)` Given `AB+A+B=0` `impliesAB+A+B+I=I` `impliesA(B+I)+(B+I)=(` `implies(A+I)(B+I)=I` `implies (A+I)` and `(B+I)` are inverse of each other `implies(A+I)(B+I)=(B+I)(A+I)` `impliesAB=BA` Thus `A` and `B` are commutative `implies(A+B)^(2)=A^(2)+2AB+B^(2)`

Discussion

1153.	For a matrix `{:[(4,1,2),(2,4,3),(5,8,6)]:}` what is the second row and third column element ?
Answer» Correct Answer - 3

Discussion

1154.	In the matrix `[{:(2,-1,4),(-3,0,5):}]`, the element in the second row, third column is _______
Answer» Correct Answer - 5

Discussion

1155.	The element in the first row and second column of the matrix `[{:(4,5,6),(7,8,9):}]` is _______A. 7B. 8C. 5D. 6
Answer» Correct Answer - C (i) Draw the first horizontal line. (ii) Draw the second vertical line. (iii) The required element in the element on which the above two line meet.

Discussion

1156.	Let a be a matrix of order `2xx2` such that `A^(2)=O`. `(I+A)^(100) =`A. 100 AB. `100 (I+A)`C. `100 I+A`D. `I+100 A`
Answer» Correct Answer - D `(I+A)^(100)= .^(100)C_(0)I^(100)+.^(100)C_(1)I^(99)A+.^(100)C_(2)I^(98) A^(2)+ +.^(100)C_(100)A^(100)` `=I+100 A+O+O+ +O` `=I+100 A`

Discussion

1157.	If the square matrices A and B are such that `AB = A` and `BA = B`, thenA. A, B are idempotentB. only A is idempotentC. only B is idempotentD. none of these
Answer» Correct Answer - A

Discussion

1158.	The elements in the second column of the matrix `[{:(0,5,3),(-5,0,2),(-3,-2,0):}]` are _______A. 0,0,0B. 5,0,-2C. 3,2,0D. None of these
Answer» Correct Answer - B Second column elements in the given are 5,0,-2.

Discussion

1159.	If A and B are two square matrices of order `3xx3` which satify `AB=A` and `BA=B`, then Which of the following is true ?A. If matrix A is singular, then matrix B is nonsingular.B. If matrix A is nonsingular, then materix B is singular.C. If matrix A is singular, then matrix B is also singular.D. Cannot say anything.
Answer» Correct Answer - C `AB=A implies \|AB\|=\|A\|` (1) `implies \|A\|=0` or `\|B\|=1` `BA=B implies \|BA\|=\|B\|` (2) `implies \|A\|=1` or `\|B\|=0` If `\|A\|=0`, then from Eq. (2) `\|B\|=0` If `\|B\|=0`, then from Eq. (1), `\|A\|=0`

Discussion

1160.	Let `{:A=[(1,2),(-5,1)]and A^(-1)=xA+yI:}`, then the values of x and y areA. `x=-1/11,y=2/11`B. `x=-1/11,y=-2/11`C. `x=1/11,y=2/11`D. `x=1/11,y=-2/11`
Answer» Correct Answer - A

Discussion

1161.	Write the order of matrix `[{:(-3,2),(4,-1),(0,2):}]`
Answer» In the given matric, there are three rows and two columns. `:.` The order of the matrix = Number of rows `xx` number of columns=`3xx2`

Discussion

1162.	If A is a `3xx3` matrix and B is a matrix such that `A^TB and BA^(T)` are both defined, then order of B isA. `3xx4`B. `3xx3`C. `4xx4`D. `4xx3`
Answer» Correct Answer - A

Discussion

1163.	Write a matrix of order `3xx3` in which every element is equal to 3.
Answer» The given matrix is of order `3xx3` and each element is equal to 3. `:.` The required matrix `=({:(3,3,3),(3,3,3),(3,3,3):})`.

Discussion

1164.	The number of rows of the matrix `[{:(3,4),(5,6):}]` is ______A. 3B. 4C. 2D. 1
Answer» Correct Answer - C Cound the number of horizontal arrays of elements.

Discussion

1165.	If `A` is a square matrix of order `3` such that `\|A\|=2`, then `\|(adjA^(-1))^(-1)\|` isA. `1`B. `2`C. `4`D. `8`
Answer» Correct Answer - C `(c )` `\|adjA^(-1)\|=\|A^(-1)\|^(2)=(1)/(\|A\|^(2))` `\|(adjA^(-1))^(-1)\|=(1)/(\|adjA^(-1)\|)=\|A\|^(2)=2^(2)=4`

Discussion

1166.	Write all the possible orders of the matrix containing 6 elements.
Answer» The possible orders for the matrix that contain six element are `1xx6,2xx3,3xx2 and 6xx1`

Discussion

1167.	If the matrix `A` and `B` are of `3xx3` and `(I-AB)` is invertible, then which of the following statement is/are correct ?A. `I-BA` is not invertibleB. `I-BA` is invertibleC. `I-BA` has for its inverse `I+B(I-AB)^(-1)A`D. `I-BA` has for its inverse `I+A(I-BA)^(-1)B`
Answer» Correct Answer - B::C `(b,c)` Let `(I-AB)^(-1)=P` `impliesP(I-AB)=I` `impliesP-PAB=I` `impliesPB^(-1)-PA=B^(-1)` `impliesBPB^(-1)-BPA=I` `impliesBPB^(-1)=I+BPA` Now `BPB^(-1)=B(I-AB)^(-1)B^(-1)` `=B(B(I-AB))^(-1)` `=(B^(-1))^(-1)(B(I-AB))^(-1)` `=(B(I-AB)B^(-1))^(-1)` `=((B-BAB)B^(-1))^(-1)` `=(I-BA)^(-1)`

Discussion

1168.	If a matrix contain 6 elements then the order of the matrix can be ______A. `3xx2`B. `2xx3`C. `1xx6`D. All the above.
Answer» Correct Answer - D Number of rows `xx` Number of coumns =6

Discussion

1169.	If `A` is a square matrix such that `A*(AdjA)=[{:(4,0,0),(0,4,0),(0,0,4):}]`, thenA. `\|A\|=4`B. `\|adjA\|=16`C. `(\|adj(adjA)\|)/(\|adjA\|)=16`D. `\|adj2A\|=128`
Answer» Correct Answer - A::B::C `(a,b,c)` `A(adjA)=\|A\|I=4I` `:.\|A\|=4` `\|adjA\|=\|A\|^(3-1)=16` `adjKA=K^(3-1)adjA` `\|adj(adjA)\|=\|A\|^((3-1)^(2))=\|A\|^(4)=256` `:. \|adjKA\|=(K^(3-1))^(3)\|adjA\|` `:.\|adj2A\|=2^(6)*16`

Discussion

1170.	If a matrix contain 5 rows and 3 columns, then the number of elements of the matrix is _____A. 6B. 8C. 10D. 15
Answer» Correct Answer - D Number of element=(Number of rows ) `xx` (Number of columns )

Discussion

1171.	If a matrix contians 3 columns and 5 rows, find the order of the matrix.
Answer» Correct Answer - `5xx3`

Discussion

1172.	If `a_j=0(i!=j)anda_(i j)=4(i=j)`,,then the matrix `A =[a_(i j)]_(n xx n)` is a matrix
Answer» Correct Answer - scalar

Discussion

1173.	The order of column matrix containing n rows is _____.
Answer» Correct Answer - `n xx 1

Discussion

1174.	If A `= [{:(-1, 0, 0),(0, x, 0),(0, 0, m):}]` is a scalar matrix, then x + m = ____.
Answer» Correct Answer - -2

Discussion

1175.	Find x, y satisfying the matrix equations. [x y + 2 z - 3] + [y 4 5] = [4 9, 12]
Answer» [X+Y Y+2+4 Z- 3+ 5] = [4 9, 12] We know that, corresponding entries of equal matrices are equal. = X + Y = 4 ……(i) = Y + 6=9 ……(ii) = Z + 2=12 ……(iii) On solving equation(i),(ii) and equation(iii) we get, = Y = 9 - 6 = Y = 3 = Z = 12 - 2 = Z = 10 Put the value of Y in equation(i)…we get, = X + 3 = 4 = X = 4 - 3 = X - 1 Hence, X = 1,Y = 3 and Z = 10

1175.

Find x, y satisfying the matrix equations. [x y + 2 z - 3] + [y 4 5] = [4 9, 12]

Answer»

[X+Y Y+2+4 Z- 3+ 5] = [4 9, 12]

We know that, corresponding entries of equal matrices are equal.

= X + Y = 4 ……(i)

= Y + 6=9 ……(ii)

= Z + 2=12 ……(iii)

On solving equation(i),(ii) and equation(iii) we get,

= Y = 9 - 6

= Y = 3

= Z = 12 - 2

= Z = 10

Put the value of Y in equation(i)…we get,

= X + 3 = 4

= X = 4 - 3

= X - 1

Hence, X = 1,Y = 3 and Z = 10

Discussion

1176.	If P ` = [{:(3, 0), (0, lambda):}]` is scalar matrix then `lambda` = ____.
Answer» Correct Answer - `lambda= 3`

Discussion

1177.	Let matrix `A=[(4,6,6),(1,3,2),(-1,-4,-3)],` Find the non-zero column vector X such that `AX= lambdaX` for some scalar `lambda.`
Answer» The characteristic equation is `\|A-lambdaI\|=0` `rArr" " [(4-lambda,6,6),(1,3-lambda,2),(-1,-4,-3-lambda)]=0` `rArr " " lambda^(3)-4lambda^(2)-lambda+4=0` or `(lambda+1)(lambda-1)(lambda-4)=0` The eigen values are `lambda=-1,1,4` If `lambda=-1`, we get `5x+6y+6z=0,x+4y+2z=0` and `-x-4y-2z=0` Giving `(x)/(6)=(y)/(2)=(z)/(-7),X=[(,6),(,2),(,-7)]` If `lambda =4`, we get `0.x+6y+6z=0,x-y+2z=0` and `-x-4y-7=0` Giving `(x)/(3)=(y)/(1)=(3)/(-1),X=[(,0),(,1),(,-1)]` Hence, vector are X `X=[(,6),(,2),(,-7)][(,0),(,1),(,-1)][(,3),(,1),(,-1)]`

Discussion

1178.	If \(2\begin{bmatrix}3& 4 \\[0.3em]5 & x\end{bmatrix}+\)\(\begin{bmatrix}1& y \\[0.3em]0 & 1\end{bmatrix}=\)\(\begin{bmatrix}7& 0 \\[0.3em]10 & 5\end{bmatrix},\)find x and y.
Answer» \(2\begin{bmatrix}3& 4 \\[0.3em]5 & x\end{bmatrix}+\)\(\begin{bmatrix}1& y \\[0.3em]0 & 1\end{bmatrix}=\)\(\begin{bmatrix}7& 0 \\[0.3em]10 & 5\end{bmatrix}\) \(=\begin{bmatrix}6& 8 \\[0.3em]10 & 2x\end{bmatrix}+\)\(\begin{bmatrix}1& y \\[0.3em]0 & 1\end{bmatrix}=\)\(\begin{bmatrix}7& 0 \\[0.3em]10 & 5\end{bmatrix}\) We know that, corresponding entries of equal matrices are equal. = Y + 8=0 2X + 1=5 = Y = -8 2X = 5 - 1 = Y= - 8 X = \(\frac{4}{2}\) = Y= - 8 X = 2 Hence, X = 2 Y = -8

1178.

If \(2\begin{bmatrix}3& 4 \\[0.3em]5 & x\end{bmatrix}+\)\(\begin{bmatrix}1& y \\[0.3em]0 & 1\end{bmatrix}=\)\(\begin{bmatrix}7& 0 \\[0.3em]10 & 5\end{bmatrix},\)find x and y.

Answer»

\(2\begin{bmatrix}3& 4 \\[0.3em]5 & x\end{bmatrix}+\)\(\begin{bmatrix}1& y \\[0.3em]0 & 1\end{bmatrix}=\)\(\begin{bmatrix}7& 0 \\[0.3em]10 & 5\end{bmatrix}\)

\(=\begin{bmatrix}6& 8 \\[0.3em]10 & 2x\end{bmatrix}+\)\(\begin{bmatrix}1& y \\[0.3em]0 & 1\end{bmatrix}=\)\(\begin{bmatrix}7& 0 \\[0.3em]10 & 5\end{bmatrix}\)

We know that, corresponding entries of equal matrices are equal.

= Y + 8=0

2X + 1=5

= Y = -8

2X = 5 - 1

= Y= - 8

X = \(\frac{4}{2}\)

= Y= - 8

X = 2

Hence,

X = 2

Y = -8

Discussion

1179.	Find the value of λ, a non-zero scalar, if \(\lambda\begin{bmatrix}1 & 0 & 2 \\[0.3em]1 & 4 & 5 \\[0.3em]\end{bmatrix}+\)\(2\begin{bmatrix}1 & 2 & 3 \\[0.3em]-1 & -3 & 2 \\[0.3em]\end{bmatrix}=\)\(\begin{bmatrix}4 & 4 & 10 \\[0.3em]4 & 2 & 14 \\[0.3em]\end{bmatrix}.\)
Answer» = \(\lambda\begin{bmatrix}1 & 0 & 2 \\[0.3em]1 & 4 & 5 \\[0.3em]\end{bmatrix}+\)\(\begin{bmatrix}2 & 4 &6 \\[0.3em]-2 & -6 & 4 \\[0.3em]\end{bmatrix}=\)\(\begin{bmatrix}4 & 4 & 10 \\[0.3em]4 & 2 & 14 \\[0.3em]\end{bmatrix}\) We know that, corresponding entries of equal matrices are equal. = \(\begin{bmatrix}\lambda+2 & 0+4 & 2\lambda+6 \\[0.3em]3\lambda-2 & 4\lambda-6 & 5\lambda+4 \\[0.3em]\end{bmatrix}=\)\(\begin{bmatrix}4 & 4 & 10 \\[0.3em]4 & 2 & 14 \\[0.3em]\end{bmatrix}\) = λ + 2 = 4 = λ = 4 - 2 = λ = 2 Hence, λ = 2 = λ[110425]+λ[102145]+[2−24−664]=[246−2−64]=[44421014][44104214] We know that, corresponding entries of equal matrices are equal. = [λ+23λ−20+44λ−62λ+65λ+4] =[λ+20+42λ+63λ−24λ−65λ+4] =[44421014][44104214] = λ + 2 = 4 = λ = 4 - 2 = λ = 2 Hence, λ = 2

1179.

Find the value of λ, a non-zero scalar, if \(\lambda\begin{bmatrix}1 & 0 & 2 \\[0.3em]1 & 4 & 5 \\[0.3em]\end{bmatrix}+\)\(2\begin{bmatrix}1 & 2 & 3 \\[0.3em]-1 & -3 & 2 \\[0.3em]\end{bmatrix}=\)\(\begin{bmatrix}4 & 4 & 10 \\[0.3em]4 & 2 & 14 \\[0.3em]\end{bmatrix}.\)

Answer»

= \(\lambda\begin{bmatrix}1 & 0 & 2 \\[0.3em]1 & 4 & 5 \\[0.3em]\end{bmatrix}+\)\(\begin{bmatrix}2 & 4 &6 \\[0.3em]-2 & -6 & 4 \\[0.3em]\end{bmatrix}=\)\(\begin{bmatrix}4 & 4 & 10 \\[0.3em]4 & 2 & 14 \\[0.3em]\end{bmatrix}\)

We know that, corresponding entries of equal matrices are equal.

= \(\begin{bmatrix}\lambda+2 & 0+4 & 2\lambda+6 \\[0.3em]3\lambda-2 & 4\lambda-6 & 5\lambda+4 \\[0.3em]\end{bmatrix}=\)\(\begin{bmatrix}4 & 4 & 10 \\[0.3em]4 & 2 & 14 \\[0.3em]\end{bmatrix}\)

= λ + 2 = 4

= λ = 4 - 2

= λ = 2

Hence, λ = 2

= λ[110425]+λ[102145]+[2−24−664]=[246−2−64]=[44421014][44104214]

We know that, corresponding entries of equal matrices are equal. = [λ+23λ−20+44λ−62λ+65λ+4]

=[λ+20+42λ+63λ−24λ−65λ+4]

=[44421014][44104214] = λ + 2 = 4 = λ = 4 - 2 = λ = 2

Hence, λ = 2

Discussion

1180.	If x1, x2, x3 as well as y1, y2, y3 are in geometric progression with the same common ratio, then the points (x1, y1), (x2, y2), (x3, y3) are (a) vertices of an equilateral triangle (b) vertices of a right angled triangle (c) vertices of a right angled isosceles triangle (d) collinear
Answer» Answer is (d) collinear

Discussion

1181.	If `A= [[3,2,2],[2,4,1],[-2,-4,-1]] ` and X,Y are two non-zero column vectors such that `AX=lambda X, AY=muY , lambdanemu, ` find angle between X and Y.
Answer» `because AX = lambda X rArr (A-lambdaI) X = 0` `because = X ne 0` `therefore det (A-lambdaI)=0` `rArr [[3-lambda,2,2],[2, 4-lambda,1],[-2, -4,-1-lambda]]=0` Applying `R-(3) rarr R_(3) + R_(2)`, then ` [[3-lambda,2,2],[2, 4-lambda,1],[0, -lambda,-lambda]]=0` Applying `C_(2 ) rarr C_(2) C_(3),` then `rArr [[3-lambda,0,2],[2, 3-lambda,1],[0, 0,-lambda]]=0` `rArr -lambda (3-lambda )^(2) = 0` ` rArr lambda = 0,3` It is clear that `lambda = 0 mu = 3` for `lambda = 0, AX = 0 rArr [[3,2,2],[2,4,1],[-2,-4,-1]][[x],[y],[z]] = [[0],[0],[0]]` `rArr 3x+ 2y+ 2z = 0 and 2x + 4y + z=0` `therefore x/-5= y/1 = z/8` So, `X= [[-6],[1],[8]]` For` mu= 3, (A-3I) Y=0` ` rArr [[0,2,2],[2,1,1],[-2,-4,-4]][[alpha],[beta],[gamma]] = [[0],[0],[0]]` `rArr 0.alpha +2beta + 2gamma = 0and 2 alpha + beta + gamma = 0` `therefore alpha/0 = beta / 4 = gamma/(-4)` `rArr alpha/0 = beta / -1= gamma/1` So, `Y=[[0],[-1],[1]]` If `theta` angle between X and Y, then `cos theta = (0.(-6)+(-1)cdot 1 + 1cdot8)/(sqrt((0+1+1))sqrt(36+1 +64) ) = 7/sqrt(202)`

Discussion

1182.	Use matrix method to solve the equations `5x - 7y = 2` and `7x - 5y = 3`A. `x=(11)/(24), y=(1)/(24)`B. `x=(10)/(24, y=(5)/(24)`C. `x=-6, y=-5`D. `x=2, y=1`
Answer» Correct Answer - a

Discussion

1183.	If for a matrix A, `absA=6and adj A=`{:[(1,-2,4),(4,1,1),(-1,k,0)]:}`, then k is equal toA. -1B. 0C. 1D. 2
Answer» Correct Answer - B If A is a square matrix of order `nxxn` then `abs(adjA)=abs(A)^(n-1)`. Here, n =3,`absA=6and adjA={:[(1,-2,4),(4,1,1),(-1,k,0)]:}`. `:. A={:abs((1,-2,4),(4,1,1),(-1,k,0))=6^2:}` `-1,(-2-4)-k(1-36)rArr6+15k=36rArrk=2`

Discussion

1184.	If A and B are square matrices of order 3 such that `absA=-1,absB=3," then "abs(3AB)` equalsA. -9B. -81C. -27D. 81
Answer» Correct Answer - a

Discussion

1185.	Let A be a square matrix of order n. Statement - 1 : `abs(adj(adj A))=absA^(n-1)^2` Statement -2 : `adj(adj A)=absA^(n-2)A`A. Statement -1 is True, Statement -2 is true, Statement -2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement -2 is True, Statement -2 is not a correct explanation for Statement -1.C. Statement -1 is True, Statement -2 is False.D. Statement -1 is False, Statement -2 is True.
Answer» Correct Answer - A Statement -1 is ture (see Theorem 8 on page 16.7) `:. abs(adj(adj A))=abs(absA^(n-2)A)=absA^((n-2)n)absA` `rArr abs(adj(adjA))=abs A^(n^2-2n+1)=absA^((n-1))^2`

Discussion

1186.	The system of equations:`x+y+z=5``x+2y+3z=9``x+3y+lambdaz=mu`has a unique solution, if`lambda=5,mu=13`(b) `lambda!=5``lambda=5,mu!=13`(d) `mu!=13`A. `lambda=5,mu=13`B. `lambda ne 5`C. `lambda=5, mu ne 13`D. `mu ne 13`
Answer» Correct Answer - B The system oif equations will have unique solution if `{:[(1,1,1),(1,2,3),(1,3,lambda)]:} ne 0 rArrlambda -5 ne 0 rArr lambda ne 5`

Discussion

1187.	. For what values of `lambda` and `mu` the system of equations `x+y+z=6, x+2y+3z=10, x+2y+lambdaz=mu` has (i) Unique solution (ii) No solution (iii) Infinite number of solutions
Answer» we can write the above system of equations in the matrix form `[(1,1,1),(1,2,3),(1,2,lambda)][(,x),(,y),(,z)]=[(,6),(,10),(,mu)]` `rArr AX=B` where A`A=[(1,1,1),(1,2,3),(1,2,lambda)],X=[(,x),(,y),(,z)]andB=[(,6),(,10),(,mu)]` `therefore` augmented matrix `C=[A:B]=[(1,1,1,vdots,6),(1,2,3,vdots,10),(1,2,lambda,vdots,mu)]` Appying `R_(2) to R_(2) and R_(3) to R_(3)-R_(1),` we get `C=[(1,1,1,vdots,6),(0,1,2,vdots,4),(0,1,lambda-1,vdots,mu-6)]` Applying `R_(3) to R_(3)-R_(2),` we get (I) No solution `p(A)!=p(c)` `i.e.," " lambda-3!=0 and mu-10!= 0` (ii) A unique solution `p(A) = p(c) =3` `ie.," " lambda-3and mu in R` `therefore" " lambda!=3 and mu in R` (iii) Infinite number of solutions `p(A)=p(c)(angle3)` `ie.," " lambda-3=0 and mu -10=0` `therefore " " lambda=3 and mu =10`

Discussion

1188.	If `{:A=[(-4,-1),(3,1)]:}`, then the determint of the matrix `(A^2016-2A^2015-A^2014)`,isA. 2014B. 2016C. `-175`D. `-25`
Answer» Correct Answer - B Let `B=A^2016-2A^2015-A^2014`. Then, `B=A^2014(A^2-2A-1)` `rArr absB=absA^2014abs(A^2-2A-1)` Now,`A=[(-4,-1),(3,1)]` `:.{:A^2-2A-I=[(-4,-1),(3,1)][(-4,-1),(3,1)]-2[(-4,-1),(3,1)]-[(1,0),(0,1)]:}` `{:= [(13,3),(-9,-2)]+[(8,2),(-6,-2)]+[(-1,0),(0,-1)]=[(20,5),(-15,-5)]:}` `rArrabs(A^2-2A-I)={:[(20,5),(-15,-5)]=-25:}` `and absA={:abs((-4,-1),(3,1))=-1:}` Hence,`absB=(-1)^2014=-25`

Discussion

1189.	Let, `C_(k) = ""^(n)C_(k) " for" 0 le kle n and A_(k) = [[C_(k-1)^(2),0],[0,C_(k)^(2)]]` for `k ge l and ` `A_(1) + A_(2) + A_(3) +...+ A_(n) = [[k_(1),0],[0, k_(2)]]`, thenA. `k_(1) = K_(2)`B. `k_(1) + k_(2) = 2`C. `k_(1) = ""^(2n)C_(n)-1`D. `k_(2) = ""^(2n)C_(n+1)`
Answer» Correct Answer - A::C `A_(1) + A_(2) + A_(3) +... + A_(n) = [[C_(0)^(2),0],[0,C_(1)^(2)]]+ [[C_(1)^(2),0],[0, C_(2)^(2)]]` `+ [[C_(2)^(2),0],[0,C_(3)^(2)]]+...+ [[C_(n-1)^(2),0],[0, C_(n)^(2)]]` ` = [[C_(0)^(2)+C_(1)^(2) + C_(2)^(2)+...+C_(n-1)^(2),0],[0,C_(1)^(2)+C_(2)^(2)+C_(2)^(3)+...+C_(n)^(2)]]` `[[""^(2n)C_(n)-1,0],[0,""^(2n)C_(n)-1]] =[[k_(1),0],[0,k_(2)]]` [given ] `therefore k_(1) = k_(2) =""^(2n) C_(n)-1`

Discussion

1190.	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. The relation between m, n and p, isA. `p = mn^(2)`B. `p=m^(n)-1`C. `p=n^(m) - 1`D. `p = m^(n-1)`
Answer» Correct Answer - B `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 `p = m^(n) - 1`

Discussion

1191.	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. If `m = 2 and n = 5 ` then p equals to
Answer» `AB=BA^2` `B^-1(AB)= B^-1(BA^2)` `= (B^-1B)A^2` `= A^2` `A^2 = B^-1AB` `A^4= A^2A^2= (B^-1AB)(B^-1AB) = B^-1ABB^-1AB` `= B^-1A(BB^-1)AB` `= B^-1A(I)AB` `=B^-1A^2B` `=B^-1(B^-1AB)B` `=B^-2AB^2` `A^8=A^4A^4= (B^-2AB^2)(B^-2AB^2)` `= B^-2A^2B^2` `=B^-2(B^-1AB)B^2` `= B^-3AB^3` `A^16= A^8A^8 = B^-3B^3B^-3AB^3` `= B^-3A^2B^3` `=B^-3(B^-1AB)B^3` `= B^-4AB^4` `A^32= A^16A^16 = B^-4AB^4B^-4AB^4` `= B^-4A^2B^4` `= B^-4(B^-1AB)B^4` `=B^-5AB^5` as, `B^5 = I ` so, `B^-5 = I` putting in eqn `=IAI` `A^32=A` multilpying by `A^-1` to both sides `A^32 = A` `A^32A^-1 = A*A^-1` `A^31 = I`

Discussion

1192.	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. If `m = 2 and n = 5 ` then p equals toA. 30B. 31C. 33D. 81
Answer» Correct Answer - B `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]` Put `m = 2, n = 5 ` in Eq. (ii), we get `p= 2^(5) - 1 = 31`

Discussion

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