1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 23 InterviewSolution

1101.	A square matrix can always be expressed asA. the sum of a symmetric and a skew-symmetric matrix.B. the sum of a diagonal matrix and a symmetric matrixC. a skew-symmetric matrixD. a skew-matrix
Answer» Correct Answer - A see illustration 3 on page 5.

Discussion

1102.	If A and B are two square matrices such that `B=-A^(-1)BA`, then `(A+B)^(2)` is equal toA. OB. `A^2+B^2`C. `A^2+2AB+B^2`D. A + B
Answer» Correct Answer - B We have, `B=-A^(-1)BA` `rArr AB=-A(A^(-1)BA)` `rArr AB=-((A A^(-1))(BA))` `rArr AB=-(I(BA))` `rArr AB=-BA` `rArr AB+BA=O` Now, `(A+B)^2=(A+B)(A+B)` `rArr (A+B)^2=A^2+AB+BA+B^2` `rArr (A+B)^2=A^2+O+B^2` `rArr (A+B)^2=A^2+B^2`

Discussion

1103.	The element in the first row and third column of the inverse of the matrix `{:[(1,2,-3),(0,1,2),(0,0,1)]:}`, isA. -2B. 0C. 1D. none of these
Answer» Correct Answer - C Clearly, the element in the first row and third column of the inverse of the given matrix is cofactor of the element in third row and first column of the given matrix. Hence, required element =`{:abs((2,-3),(1,2))=7:}`

Discussion

1104.	If `A=[(3,-3,4),(2,-3,4),(0,-1,1)]`, then `A^(-1)=`A. AB. `A^(2)`C. `A^(3)`D. `A^(4)`
Answer» Correct Answer - c

Discussion

1105.	Construct a `2xx3` matrix whose elements are given by `a_(ij)=(1)/(2)(i-2j)^(2).`
Answer» `[{:((1)/(2),(9)/(2)),(0,2),((1)/(2),(1)/(2)):}]`

Discussion

1106.	The element of second row and third column in the inverse of `[[1, 2, 1], [2, 1, 0], [-1, 0, 1]]` isA. `-2`B. `-1`C. 1D. 2
Answer» Correct Answer - b

Discussion

1107.	if `A=[a_(ij)]_(2*2)` where `a_(ij)={i+j , i!=j` and `a_(ij)=i^2-2j ,i=j` then `A^-1` is equal toA. `(1)/(9)[(4,1),(-1,2)]`B. `(1)/(9)[(0,-3),(-3,-1)]`C. `(1)/(9)[(0,3),(3,1)]`D. None of these
Answer» Correct Answer - c

Discussion

1108.	Multiplicative inverse of the matrix `[[2,1],[7,4]]` is(i) `[[4,-1],[-7,-2]]`(ii) `[[-4,-1],[7,-2]]`(iii) `[[4,-1],[7,2]]`(iv) `[[4,-1],[-7,2]]`A. `[(4,-1),(-7,-2)]`B. `[(-4,-1),(7,-2)]`C. `[(4,-7),(7,2)]`D. `[(4,-1),(-7,2)]`
Answer» Correct Answer - D

Discussion

1109.	If a matrix has 5 elements, write all possible orders it can have.
Answer» Possible orders are 1x5 and 5x1.

Discussion

1110.	If there are three square matrix A, B, C of same order satisfying the equation `A^2=A^-1 and B=A^(2^n) and C=A^(2^((n-2))`, then prove that `det .(B-C) = 0, n in N`.
Answer» `because B = ^(2^(n)) = A^(2.2^(n-1)) = (A^(2))^(2^(n-1)) = (A^(-1)) ^(2^(n-1) ) [ because A^(2) = A^(-1)]` `= (A^(2^(n-1)))^(-1) = (A^(2.2^(n-2)))^(-1) = [(A^(2))^(2^(n-2))]^(-1)` `= [(A^(-1))^(2^(n-2))]^(-1)= ((A^(-1))^(-1))^(2^(n-2)) = A^(2^(n-2))=C` `rArr B-C=0 rArr det (B-C) = 0`

Discussion

1111.	If A and B are two matrices such that rank of A = m and rank of B = n, thenA. rank (AB)= mnB. `"rank "(AB)gt" rank " (A)`C. `" rank "(AB)gt " rank "(B)`D. `"rank "(AB) lt min ("rank A, rank B")`
Answer» Correct Answer - D

Discussion

1112.	If `B ,C`are square matrices of order `na n difA=B+C ,B C=C B ,C^2=O ,`then without using mathematical induction, show that for any positiveinteger `p ,A^(p-1)=B^p[B+(p+1)C]`.
Answer» `because A= B+CrArr A^(p+1) = (B+C)^(p+1) ` `= ""^(p+1) C_(0) B^(p+1) + ""^(p=1) C_(1) B^(p) C + ""^(P+1) C_(2) B^(p-1) C^(2) +... + ""^(p+1)C_(p+1) C^(p+1)` `= B^(p+1)+ ""^(p+1) C_(1) B^(p) C + 0 + 0 +... [because C^(2) = 0 rArr C^(2) = C^(3) =... = 0]` `= B^(P) [B+(p+1) C]` Hence, `A^(p+1) = B^(p)[B+(p+1)C]`

Discussion

1113.	If `{:A=[(3,4),(2,4)],B=[(-2,-2),(0,-1)]:}," then " (A+B)^(-1)=`A. is a skew-symmetric matrixB. `A^(-1)+B^(-1)`C. does not existD. none of these
Answer» Correct Answer - D

Discussion

1114.	Let `A = [[3,a,-1],[2,5,c],[b,8,2]]` is symmetric and `B = [[d, 3, a],[b-a, e, -2b-c ],[-2, 6, -f]]` is skew- symmetric, find AB. If AB is symmetric or skew symmetric or neither of them. Justify your answer.
Answer» `because` A is symmetric `therefore c = 5, b=-1 and a =2` …(i) and B is skew- symmetrci `therefore d =e =f=0 and 2b + c = 6, a=b, b-a=-3` …(ii) From Eqs. (i) and (ii), we get `a= 2, b= -1 , c=8, d=0, e=0, f=0` `therefore A= [[3,2,-1],[2,5,8],[-1,8,2]]and B= [[0,3,2],[-3,0,-6],[-2,6,0]]` `rArr AB= [[-4,3,-6],[-31,54,-26],[-28,9, -50]]` which neither symmetric nor skew-symmetric.

Discussion

1115.	Let `{:A=[(a,0,0),(0,a,0),(0,0,a)]:}`, then `A^n` is equal toA. `{:[(a^n,0,0),(0,a^n,0),(0,0,a)]:}`B. `{:[(a^n,0,0),(0,a,0),(0,0,a)]:}`C. `{:[(a^n,0,0),(0,a^n,0),(0,0,a^n)]:}`D. `{:[(na,0,0),(0,na,0),(0,0,na)]:}`
Answer» Correct Answer - C

Discussion

1116.	`A=[(3,a,-1),(2,5,c),(b,8,2)]` is symmetric and `B=[(d,3,a),(b-a,e,-2b-c),(-2,6,-f)]` is skew-symmetric, then find AB.
Answer» Correct Answer - `AB=[(-4,3,-6),(-31,54,-26),(-28,9,-50)]` A is symmetric `implies A^(T)=A` `implies [(3,2,b),(a,5,8),(-1,c,2)]=[(3,a,-1),(2,5,c),(b,8,2)]` `implies a=2, b=-1, c=8` B is skew-symmetric `implies B^(T)=-B` `implies [(d,b-a,-2),(3,e,6),(a,-2b-c,-f)]=[(-d,-3,-a),(a-b,-e,2b+c),(2,-6,f)]` `implies d=-d, f=-f` and `e=-e` `implies d=f=0` So `A=[(3,2,-1),(2,5,8),(-1,8,2)]` and `B=[(0,3,2),(-3,0,-6),(-2,6,0)]` `implies AB=[(3,2,-1),(2,5,8),(-1,8,2)][(0,3,2),(-3,0,-6),(-2,6,0)]` `=[(-4,3,-6),(-31,54,-26),(-28,9,-50)]`

Discussion

1117.	If `{:A=[(costheta,sintheta),(-sintheta,costheta)]:}, " then "lim_(ntooo) 1/nA^n` isA. a null matrixB. an identity matrixC. `{:[(0,1),(-1,0)]:}`D. none of these
Answer» Correct Answer - A

Discussion

1118.	`If A=[[costheta,sintheta],[-sintheta,costheta]],then Lim_(x_>oo)1/nA^n` is
Answer» Correct Answer - Zero Matrix `A=[(cos theta,sin theta),(-sin theta,cos theta)]` `A^(2)=[(cos theta,sin theta),(-sin theta,cos theta)][(cos theta,sin theta),(-sin theta,cos theta)]` `=[(cos^(2) theta-sin^(2) theta,2 sin theta cos theta),(-2 sin theta cos theta,cos^(2) theta - sin^(2) theta)]` `=[(cos 2 theta,sin 2 theta),(-sin 2 theta,cos 2 theta)]` `A^(3)=A^(2) A=[(cos 2 theta,sin 2 theta),(-sin 2 theta,cos 2 theta)][(cos theta,sin theta),(-sin theta,cos theta)]` `=[(cos 3 theta,sin 3 theta),(-sin 3 theta,cos 3 theta)]` Hence, `A^(n)=[(cos n theta,sin n theta),(-sin n theta,cos n theta)]` or `A^(n)/n = [((cos n theta)/n,(sin n theta)/n),((-sin n theta)/n,(cos n theta)/n)]` or `lim_(n rarr oo) A^(n)/n =[(lim_(n rarr oo) (cos n theta)/n,lim_(n rarr oo) (sin n theta)/n),(-lim_(n rarr oo) (sin n theta)/n,lim_(n rarr oo) (cos n theta)/n)]` `=[(0,0),(0,0)]=` Zero matrix

Discussion

1119.	Find the matrix X so that `X[1 2 3 4 5 6]=[-7-8-9""""""2""""""4""""6]`
Answer» Correct Answer - `X=[(1,-2),(2,0)]` It is given that `X[(1,2,3),(4,5,6)]=[(-7,-8,-9),(2,4,6)]` The matrix given on the R.H.S. of the equation is a `2xx3` matrix and the one given on the L.H.S is a `2xx3` matrix. Therefore, X has to be a `2xx2` matrix. Now, let `X=[(a,c),(b,d)]` therefore, we have `[(a,c),(b,d)][(1,2,3),(4,5,6)]=[(-7,-8,-9),(2,4,6)]` or `[(a+4c,2a+5c,3a+6c),(b+4d,2b+5d,3b+6d)]=[(-7,-8,-9),(2,4,6)]` Equating the corresponding elements of the two matrices, we have `{:(a+4c=-7",",2a+5c=-8",",3a+6c=-9),(b+4d=2",",2b+5d=4",",3b+6d=6):}` Now, `a+4c=-7 implies a=-7-4c` `:. 2a+5c=-8 implies -14-8c+5c=-8` or `-3c=6` or `c=-2` `:. a=-7-4(-2)=-7+8=1` Now, `b+4d=2` or `b=2-4d` `:. 2b+2d=4implies 4-8d+5d=4` `implies -3d=0` `d=0` `:. b=2-4(0)=2` Thus, `a=1, b=2, c=-2, d=0` Hence, the required matrix X is `[(1,-2),(2,0)]`

Discussion

1120.	for what values of x: `[1" "2" "1][{:(1,2,0),(2,0,1),(1,0,2):}][{:(0),(2),(x):}]=0? `
Answer» Correct Answer - x=-1 We have `[(1,2,1)][(1,2,0),(2,0,1),(1,0,2)][(0),(2),(x)]=0` or `[(1+4+1,2+0+0,0+2+2)][(0),(2),(x)]=0` or `[6(0)+2(2)+4(x)]=0` or `[4+4x]=[0]` or `4+4x=0` or `x=-1`

Discussion

1121.	If `A = [[1,2],[3,-5]]` and `B=[[1,0],[0,2]]` and X is a matrix such that `A=BX`, then X=
Answer» Correct Answer - `X=1/2 [(2,4),(3,-5)]` Let `X=[(a,b),(c,d)]` then `[(1,2),(3,-5)]=[(1,0),(0,2)][(a,b),(c,d)]=[(a,b),(2c,2d)]` `:. a=1, b=2, 2c=3, 2d=-5` `:. X=[(1,2),(3//2,-5//2)]=1/2 [(2,4),(3,-5)]`

Discussion

1122.	If `A=[(3,-4),(1,-1)]` prove that `A^k=[(1+2k,-4k),(k,1-2k)]` where `k` is any positive integer.
Answer» We have. `A^(2)=[(3,-4),(1,-1)][(3,-4),(1,-1)]=[(5,-8),(2,-3)]=[(1+2xx2,-4xx2),(2,1-2xx2)]` and `A^(3)=[(5,-8),(2,-3)][(3,-4),(1,-1)]=[(7,-12),(3,-5)]=[(1+2xx3,-4xx3),(3,1-2xx3)]` Thus, it is true for indices 2 and 3. Now, assume `A^(k)=[(1+2k,-4k),(k,1-2k)]` to be true. Then, `A^(k+1)=[(1+2k,-4k),(k,1-2k)][(3,-4),(1,-1)]` `=[(3+2k,-4(k+1)),(k+1,-1-2k)]` `=[(1+2(k+1),-4(k+1)),(k+1,1-2(k+1))]` thus, if the law is true for `A^(k)`, it is also true for `A^(k+1)`. but it is true for `k=2, 3` etc. Hence, by induction, the required result follows.

Discussion

1123.	Consider, `A=[(a,2,1),(0,b,0),(0,-3,c)]`, where a, b and c are the roots of the equation `x^(3)-3x^(2)+2x-1=0`. If matric B is such that `AB=BA, A+B-2I ne O` and `A^(2)-B^(2)=4I-4B`, then find the value of det. (B)
Answer» Given that a, b and c are roots of the equation `x^(3)-3x^(2)+2x-1=0`. Let `x^(3)-3x^(2)+2x-1=(x-a) (x-b) (x-c)` (1) Now, `A^(2)-B^(2)=4I-4B` `:. A^(2)=B^(2)-4B+4I` `implies A^(2)=(B-2I)^(2)` `implies A^(2)-(B-2I)^(2)=O` `implies (A+B-2I) (A-B+2I)=O" "("as "AB=BA)` Now, `A+B-2I ne O` `:. A-B+2I=O` [as otherwise `(A-B+2I)^(-1)` exists and `A+B-2I=O`] `:. B=A+2I =[(a+2,2,1),(0,b+2,0),(0,-3,c+2)]` `:. \|B\|=(a+2) (b+2) (c+2)` (2) Putting `x=-2` in (1), we get `-8-12-4-1=(-2-a) (-2-b) (-2-c)` `:. 25=(a+2) (b+2) (c+2)` `:. \|B\|=25`

Discussion

1124.	if `A=[[i,0] , [0,i]]` where `i=sqrt(-1)` and `x epsilon N` then `A^(4x)` equals to:
Answer» `A^(2)=[(i,0),(0,i)][(i,0),(0,i)]=[(i^(2),0),(0,i^(2))]=[(-1,0),(0,-1)]` `A^(4)=[(-1,0),(0,-1)][(-1,0),(0,-1)]=[(1,0),(0,1)]=I` `implies A^(4n)=I^(n)=I`

Discussion

1125.	Let `A=[("tan"pi/3,"sec" (2pi)/3),(cot (2013 pi/3),cos (2012 pi))]` and P be a `2 xx 2` matrix such that `P P^(T)=I`, where I is an identity matrix of order 2. If `Q=PAP^(T)` and `R=[r_("ij")]_(2xx2)=P^(T) Q^(8) P`, then find `r_(11)`.
Answer» `R=P^(T)Q^(8)P` `=P^(T)(PAP^(T))^(8)P` `=P^(T)PAP^(T) (PAP^(T))^(7)P` `=IAP^(T) (PAP^(T))^(7) P` `=AP^(T) PAP^(T) (PAP^(T))^(6)P` `=A^(2)P^(T) (PAP^(T))^(6)P` ... ... `=A^(8) P^(T)P` `=A^(8)` Now, `A^(2)=A A=[(sqrt(3),-2),(0,1)][(sqrt(3),-2),(0,1)]` `=[((sqrt(3))^(2),-2sqrt(3)-2),(0,1)]` `A^(3)=A^(2)A=[(3,-2sqrt(3)-2),(0,1)][(sqrt(3),-2),(0,1)]` `=[((sqrt(3))^(3),-6-2sqrt(3)-2),(0,1)]` `:. R=[r_("ij")]_(2xx2)=P^(T) Q^(8) P=A^(8)=[((sqrt(3))^(8),-),(-,-)]` `implies r_(11)=81`

Discussion

1126.	The matrix R(t) is defined by `R(t)=[(cos t,sin t),(-sin t,cos t)]`. Show that `R(s)R(t)=R(s+t)`.
Answer» `R(s)R(t)=[(cos s ,sin s),(cos t,sin t)][(cos t,sin t),(-sin t,cos t)]` `=[(cos s cos t-sin s sin t,cos s sin t + sin s cos t),(-sin s cos t-cos s sin t,cos s cos t-sin s sin t)]` `=[(cos(s+t),sin (s+t)),(-sin (s+t),cos (s+t))]` `=R(s+t)`

Discussion

1127.	If matrix `A=[(0,1,-1),(4,-3,4),(3,-3,4)]=B+C`, where B is symmetric matrix and C is skew-symmetric matrix, then find matrices B and C.
Answer» Correct Answer - `B=1/2 [(0,5,2),(5,-6,1),(2,1,8)], C=1/2[(0,-3,-4),(3,0,7),(4,-7,0)]` Here matrix A is expressed as the sum of symmetric and skew-symmetric matrix. Then `B=1/2(A+A^(T))` and `C=1/2 (A-A^(T))` Now `A=[(0,1,-1),(4,-3,4),(3,-3,4)]` `implies A^(T)=[(0,4,3),(1,-3,-3),(-1,4,4)]` `implies B=1/2 ([(0,1,-1),(4,-3,4),(3,-3,4)]+[(0,4,3),(1,-3,-3),(-1,4,4)])` `=1/2 [(0,5,2),(5,-6,1),(2,1,8)]` and `C=1/2 ([(0,1,-1),(4,-3,4),(3,-3,4)]-[(0,4,3),(1,-3,-3),(-1,4,4)])` `=1/2 [(0,-3,-4),(3,0,7),(4,-7,0)]`

Discussion

1128.	Let `A=B B^(T)+C C^(T)`, where `B=[(cos theta),(sin theta)], C=[(sin theta),(-cos theta)], theta in R`. Then prove that a is unit matrix.
Answer» `A=B B^(T)+C C^(T)` `=[(cos theta),(sin theta)][(cos theta),(sin theta)]+[(sin theta),(-cos theta)][(sin theta, -cos theta)]` `=[(cos^(2) theta,cos theta . sin theta),(sin theta.costheta,sin^(2) theta)]+[(sin^(2) theta,0sintheta.cos theta),(-sin theta. cos theta,cos^(2) theta)]` `=[(1,0),(0,1)]`

Discussion

1129.	Consider the matrices `A=[(4,6,-1),(3,0,2),(1,-2,5)], B=[(2,4),(0,1),(-1,2)], C=[(3),(1),(2)]` Out of the given matrix products, which one is not defined ?A. `(AB)^(T)C`B. `C^(T)C (AB)^(T)`C. `C^(T)AB`D. `A^(T)AB B^(T)C`
Answer» Correct Answer - B `A rarr 3xx3, B rarr 3xx2, C rarr 3xx1` `AB rarr 3xx2 implies (AB)^(T)=2xx3 implies (AB)^(T)C` is defined `C^(T) rarr 1xx3, implies C^(T) C rarr 1xx1` Hence `C^(T)C(AB)^(T)` is not defined. Now, `C^(T) AB` is also defined. `A^(T) rarr 3xx3, B^(T) rarr 2xx3` `A^(T)A rarr 3xx3` `B B^(T) rarr 3xx3` `implies A^(T) AB B^(T) rarr 3xx3` `implies A^(T) AB B^(T) C` is defined

Discussion

1130.	If one of the eigenvalues of a square matrix a order `3xx3` is zero, then prove that det `A=0`.
Answer» Let `A=[(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))]` `implies A-lambdaI=[(a_(1)-lambda,b_(1),c_(1)),(a_(2),b_(2)-lambda,c_(2)),(a_(3),b_(3),c_(3)-lambda)]` `implies` det. `(A-lambdaI)=(a_(1)-lambda)[(b_(2)-lambda)(c_(3)-lambda)-b_(3)c_(2)]` `-b_(1) [a_(2) (c_(3)-lambda)-c_(2)a_(3)]+c_(1)[a_(2)b_(3)-a_(3)(b_(2)-lambda)]` Now if one of the eigenvalues is zero, one root of `lambda` should be zero. Therefore, constant term in the above polynominal is zero. `implies a_(1) b_(2)c_(3)-a_(1)b_(3)c_(2)-b_(1)a_(2)c_(3)+b_(1)c_(2)a_(3)+a_(1)a_(2)a_(3)-c_(1)a_(3)b_(2)=0` But in above equation L.H.S. is the value of determinant of A. Hence, det. `A=0`

Discussion

1131.	Show that if `lambda_(1), lambda_(2), ...., lamnda_(n)` are `n` eigenvalues of a square matrix a of order n, then the eigenvalues of the matric `A^(2)` are `lambda_(1)^(2), lambda_(2)^(2),..., lambda_(n)^(2)`.
Answer» `AX=lambdaX` or `A(AX)=A(lambdaX)` or `A^(2)X=lambda(AX)=lambda(lambdaX)=lambda^(2)X` i.e., `A^(2)X=lambda^(2)X` Hence, eigenvalue of `A^(2)` is `lambda^(2)`. Thus, if `lambda_(1), lambda_(2), ..., lambda_(n)` are eigenvalue of A, then `lambda_(1)^(2), lambda_(2)^(2), ..., lambda_(n)^(2)` are eigenvalues of `A^(2)`.

Discussion

1132.	Find x, y and z so that A = B, where \(A=\begin{bmatrix}x-2 & 3 & 2z \\[0.3em]18z & y+2 &6z \\[0.3em]\end{bmatrix},\)\(B=\begin{bmatrix}y & z & 6 \\[0.3em]6y & z &2y \\[0.3em]\end{bmatrix}\)
Answer» Given two matrices are equal as A = B. \(\begin{bmatrix} x-2 & 3 & 2z \\[0.3em] 18z & y+2 &6z \\[0.3em] \end{bmatrix} \)\(=\begin{bmatrix} y & z & 6 \\[0.3em] 6y & z &2y \\[0.3em] \end{bmatrix} \) We know that if two matrices are equal then the elements of each matrices are also equal. ∴x – 2 = y …(1) z = 3 And y + 2 = z … (2) 2y = 6z ⇒ y = 3z …(3) Putting the value of z in equation (3), ∴ y = 3z = 3 × 3 = 9 Putting the value of y in equation (1), x – 2 = 9 ⇒ x – 2 = 9 ⇒ x = 9 + 2 ⇒ x = 11 ∴ x = 11, y = 9, z = 3.

1132.

Find x, y and z so that A = B, where \(A=\begin{bmatrix}x-2 & 3 & 2z \\[0.3em]18z & y+2 &6z \\[0.3em]\end{bmatrix},\)\(B=\begin{bmatrix}y & z & 6 \\[0.3em]6y & z &2y \\[0.3em]\end{bmatrix}\)

Answer»

Given two matrices are equal as A = B.

\(\begin{bmatrix} x-2 & 3 & 2z \\[0.3em] 18z & y+2 &6z \\[0.3em] \end{bmatrix} \)\(=\begin{bmatrix} y & z & 6 \\[0.3em] 6y & z &2y \\[0.3em] \end{bmatrix} \)

We know that if two matrices are equal then the elements of each matrices are also equal.

∴x – 2 = y …(1)

z = 3 And

y + 2 = z … (2)

2y = 6z

⇒ y = 3z …(3)

Putting the value of z in equation (3),

∴ y = 3z

= 3 × 3 = 9

Putting the value of y in equation (1),

x – 2 = 9

⇒ x – 2 = 9

⇒ x = 9 + 2

⇒ x = 11

∴ x = 11,

y = 9,

z = 3.

Discussion

1133.	The inventor of input-output analysis is: (a) Sir Francis Galton (b) Fisher (c) Prof. Wassily W. Leontief (d) Arthur Cayley
Answer» (c) Prof. Wassily W. Leontief

Discussion

1134.	Which of the following matrix has no inverse?(a) \( \begin{pmatrix} -1&1\\1&-4 \end{pmatrix}\)(b) \( \begin{pmatrix} 2&-1\\-4&2 \end{pmatrix}\)(c) \( \begin{pmatrix} cos\,a&sin\,a\\-sin\,a&cos\,a \end{pmatrix}\)(d) \( \begin{pmatrix} sin\,a&sin\,a\\-cos\,a&cos\,a \end{pmatrix}\)
Answer» (b) \( \begin{pmatrix} 2&-1\\-4&2 \end{pmatrix}\) So \( \begin{pmatrix} 2&-1\\-4&2 \end{pmatrix}\) has no inverse.

1134.

Which of the following matrix has no inverse?(a) \( \begin{pmatrix} -1&1\\1&-4 \end{pmatrix}\)(b) \( \begin{pmatrix} 2&-1\\-4&2 \end{pmatrix}\)(c) \( \begin{pmatrix} cos\,a&sin\,a\\-sin\,a&cos\,a \end{pmatrix}\)(d) \( \begin{pmatrix} sin\,a&sin\,a\\-cos\,a&cos\,a \end{pmatrix}\)

Answer»

(b) \( \begin{pmatrix} 2&-1\\-4&2 \end{pmatrix}\)

So \( \begin{pmatrix} 2&-1\\-4&2 \end{pmatrix}\) has no inverse.

Discussion

1135.	The inverse matrix of \( \begin{pmatrix} \frac{4}{5}&\frac{5}{12}\\\frac{2}{5}&\frac{1}{2} \end{pmatrix}\) is(a) \(\frac{7}{30}\)\( \begin{pmatrix} \frac{1}{2}&\frac{5}{12}\\\frac{2}{5}&\frac{4}{5} \end{pmatrix}\)(b) \(\frac{7}{30}\)\( \begin{pmatrix} \frac{1}{2}&\frac{-5}{12}\\\frac{-2}{5}&\frac{1}{5} \end{pmatrix}\)(c) \(\frac{30}{7}\)\( \begin{pmatrix} \frac{1}{2}&\frac{5}{12}\\\frac{2}{5}&\frac{4}{5} \end{pmatrix}\)(d) \(\frac{30}{7}\)\( \begin{pmatrix} \frac{1}{2}&\frac{-5}{12}\\\frac{-2}{5}&\frac{4}{5} \end{pmatrix}\)
Answer» (c) \(\frac{30}{7}\)\( \begin{pmatrix} \frac{1}{2}&\frac{5}{12}\\\frac{2}{5}&\frac{4}{5} \end{pmatrix}\)

Discussion

1136.	Find the adjoint of the matrix A = \(\begin{bmatrix} 2&3\\1&4 \end{bmatrix}\)
Answer» A = \(\begin{bmatrix} 2&3\\1&4 \end{bmatrix}\) Adj A = \(\begin{bmatrix}4&-3\\-1&2\end{bmatrix}\)

1136.

Find the adjoint of the matrix A = \(\begin{bmatrix} 2&3\\1&4 \end{bmatrix}\)

Answer»

A = \(\begin{bmatrix} 2&3\\1&4 \end{bmatrix}\)

Adj A = \(\begin{bmatrix}4&-3\\-1&2\end{bmatrix}\)

Discussion

1137.	adj (AB) is equal to: (a) adj A adj B (b) adj AT adj BT (c) adj B adj A (d) adj BT adj AT
Answer» (c) adj B adj A

Discussion

1138.	If A = \( \begin{pmatrix}-1&2\\1&-4\end{pmatrix}\) then A (adj A) is: (a) \( \begin{pmatrix}-4&-2\\-1&-1\end{pmatrix}\)(b) \( \begin{pmatrix}4&-2\\-1&1\end{pmatrix}\)(c) \( \begin{pmatrix}2&0\\0&2\end{pmatrix}\)(d) \( \begin{pmatrix}0&2\\2&0\end{pmatrix}\)
Answer» (c) \( \begin{pmatrix} 2&0\\0&2 \end{pmatrix}\) A = \( \begin{pmatrix} -1&2\\1&-4 \end{pmatrix}\) \|A\| = 4 – 2 = 2 We know that A (adj A) = \|A\| ⇒ 2\( \begin{pmatrix} 1&0\\0&1 \end{pmatrix}\) = \( \begin{pmatrix} 2&0\\0&2 \end{pmatrix}\)

1138.

If A = \( \begin{pmatrix}-1&2\\1&-4\end{pmatrix}\) then A (adj A) is: (a) \( \begin{pmatrix}-4&-2\\-1&-1\end{pmatrix}\)(b) \( \begin{pmatrix}4&-2\\-1&1\end{pmatrix}\)(c) \( \begin{pmatrix}2&0\\0&2\end{pmatrix}\)(d) \( \begin{pmatrix}0&2\\2&0\end{pmatrix}\)

Answer»

(c) \( \begin{pmatrix} 2&0\\0&2 \end{pmatrix}\)

A = \( \begin{pmatrix} -1&2\\1&-4 \end{pmatrix}\)

|A| = 4 – 2 = 2

We know that A (adj A) = |A|

⇒ 2\( \begin{pmatrix} 1&0\\0&1 \end{pmatrix}\) = \( \begin{pmatrix} 2&0\\0&2 \end{pmatrix}\)

Discussion

1139.	......... Matrix is both symmetric and skew-symmetric matrix.
Answer» Null matrix is both symmetric and skew-symmetric matrix.

Discussion

1140.	The negative of a matrix is obtained b y multiplying it by ...........
Answer» Let A is a given matrix, `therefore -A=-1[A]` So, the negative of a matrix is obtained by multiplying it by -1 .

Discussion

1141.	Sum of two skew-symmetric matrices is always ......... Matrix.
Answer» Let A is a given matrix, then (-A) is a skew symmetric matrix Similarly, for a given matrix -B is a skew -symmetric matrix. Hence -A-B=-(A+B)`rArr` sum of two skew symmetric matrices is always skew symmetric m atrix.

Discussion

1142.	`{:(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 B to C?A. 3B. 5C. 0D. 4
Answer» Correct Answer - B Element in the second row third column.

Discussion

1143.	A matrix of order `mxxn` contains 7 elements, then how many different order pairs (m,n) can take ?A. 2B. 1C. 3D. 7
Answer» Correct Answer - A Since the matrix contain 7 elements `implies 1xx7 or 7xx1` `:.` Two different order pairs are possible.

Discussion

1144.	Write a matrix which contains 1 row and 3 columns.
Answer» Correct Answer - [a,b,c]

Discussion

1145.	There are 4 routes from city A to city B, 5 routes from city B to city C and 3 routes from city C to city A. Convert the above information into matrix form.
Answer» Correct Answer - `[{:(0,4,0),(0,0,5),(3,0,0):}]`

Discussion

1146.	The order of the matrix formed with the information given in the following table is ______. A. `1xx9`B. `9xx1`C. `3xx3`D. None of these
Answer» Correct Answer - C `3xx3`

Discussion

1147.	The element in the second row and third columns of the matrix `[{:(x,y,z,p),(a,b,c,d):}]` is _______A. xB. pC. dD. c
Answer» Correct Answer - D The element in the second row and third column is c.

Discussion

1148.	In the matrix `[{:(2,-1,4),(-3,0,5):}]`, the element in the second row and third column is __________A. 5B. 0C. 3D. 4
Answer» Correct Answer - A 5

Discussion

1149.	The order of the matrix formed with , the information given in the following tables is _______
Answer» Correct Answer - `3xx3`

Discussion

1150.	The element in the second row and third column of the matrix `[{:(1,2,3),(4,5,6),(7,8,9):}]` is ________A. 7B. 6C. 8D. 9
Answer» Correct Answer - A (i) Draw the second horizontal line. (ii) Draw the third vertical line. (iii) The required element is the element on which the above two lines meet.

Discussion

Explore topic-wise InterviewSolutions in .

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The inventor of input-output analysis is: (a) Sir Francis Galton (b) Fisher (c) Prof. Wassily W. Leontief (d) Arthur Cayley

Find the adjoint of the matrix A = \(\begin{bmatrix} 2&amp;3\\1&amp;4 \end{bmatrix}\)

adj (AB) is equal to: (a) adj A adj B (b) adj AT adj BT (c) adj B adj A (d) adj BT adj AT

If A = \( \begin{pmatrix}-1&amp;2\\1&amp;-4\end{pmatrix}\) then A (adj A) is: (a) \( \begin{pmatrix}-4&amp;-2\\-1&amp;-1\end{pmatrix}\)(b) \( \begin{pmatrix}4&amp;-2\\-1&amp;1\end{pmatrix}\)(c) \( \begin{pmatrix}2&amp;0\\0&amp;2\end{pmatrix}\)(d) \( \begin{pmatrix}0&amp;2\\2&amp;0\end{pmatrix}\)

......... Matrix is both symmetric and skew-symmetric matrix.

The negative of a matrix is obtained b y multiplying it by ...........

Sum of two skew-symmetric matrices is always ......... Matrix.

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Write a matrix which contains 1 row and 3 columns.

There are 4 routes from city A to city B, 5 routes from city B to city C and 3 routes from city C to city A. Convert the above information into matrix form.

The order of the matrix formed with the information given in the following table is ______. A. `1xx9`B. `9xx1`C. `3xx3`D. None of these

The element in the second row and third columns of the matrix `[{:(x,y,z,p),(a,b,c,d):}]` is _______A. xB. pC. dD. c

In the matrix `[{:(2,-1,4),(-3,0,5):}]`, the element in the second row and third column is __________A. 5B. 0C. 3D. 4

The order of the matrix formed with , the information given in the following tables is _______

The element in the second row and third column of the matrix `[{:(1,2,3),(4,5,6),(7,8,9):}]` is ________A. 7B. 6C. 8D. 9

Find x, y and z so that A = B, where \(A=\begin{bmatrix}x-2 & 3 & 2z \\[0.3em]18z & y+2 &6z \\[0.3em]\end{bmatrix},\)\(B=\begin{bmatrix}y & z & 6 \\[0.3em]6y & z &2y \\[0.3em]\end{bmatrix}\)

Find the adjoint of the matrix A = \(\begin{bmatrix} 2&3\\1&4 \end{bmatrix}\)

If A = \( \begin{pmatrix}-1&2\\1&-4\end{pmatrix}\) then A (adj A) is: (a) \( \begin{pmatrix}-4&-2\\-1&-1\end{pmatrix}\)(b) \( \begin{pmatrix}4&-2\\-1&1\end{pmatrix}\)(c) \( \begin{pmatrix}2&0\\0&2\end{pmatrix}\)(d) \( \begin{pmatrix}0&2\\2&0\end{pmatrix}\)