1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 19 InterviewSolution

901.	If A and B are square matrices of order `n ,`then prove that `Aa n dB`will commute iff `A-lambdaIa n dB-lambdaI`commute for every scalar `lambdadot`A. `AB=BA`B. `AB+BA=O`C. `A=-B`D. none of these
Answer» Correct Answer - A `(A-lambda I) (B-lambda I)=(B-lambda I) (A-lambda I)` or `AB-lambda(A+B)I+lambda^(2)I^(2)=BA-lambda(B+A)I+lambda^(2)I^(2)` or `AB=BA`

Discussion

902.	If `Aa n dB`are two non-singular matrices of the same order such that `B^r=I ,`for some positive integer `r >1,t h e nA^(-1)B^(r-1)A=A^(-1)B^(-1)A=``I`b. `2I`c. `O`d. -IA. `I`B. `2I`C. 0D. `-I`
Answer» Correct Answer - C Given, `B^(r) = I rArr B^(r) B^(-1) = IB^(-1)` `rArr B^(r-1) = B^(-1)` `therefore A^(-1) B^(r-1) A= A^(-1) B^(-1) A` ` rArr A^(-1) B^(r-1) A- A^(-1) B^(-1) A = 0 `

Discussion

903.	The equation `[1 x y][(1,3,1),(0,2,-1),(0,2,-1)] [(1),(x),(y)]=[0]` hasA. `{:("(i)",(ii)),((p),"(r)"):}`B. `{:("(i)",(ii)),((q),"(p)"):}`C. `{:("(i)",(ii)),((p),"(q)"):}`D. `{:("(i)",(ii)),((r),"(p)"):}`
Answer» Correct Answer - C `[(1,x,y)][(1,3,1),(0,2,-1),(0,0,1)]=[(1),(x),(y)]=10` or `[(1,3+2x,1-x+y)] [(1),(x),(y)]=[0]` or `[1+3x+2x^(2)+y-xy+y^(2)]=[0]` or `2x^(2)+y^(2)+y+3x-xy+1=0` if `y=0, 2x^(2)+3x+1=0` or `(2x+1) (x+1)=0` or `x=-1//2, -1` (rational roots) If `y=-1, 2x^(2)+4x+1=0` or `x=(-4 pm sqrt(12))/4 =(-2 pm sqrt(3))/2` (irrational roots)

Discussion

904.	The number of diagonal matrix, `A`or order`n`which `A^3=A`is a. is a a. 1 b. 0 c. `2^n`d. `3^n`A. 1B. 0C. `2^(n)`D. `3^(n)`
Answer» Correct Answer - D A= diag `(d_(1), d_(2), .... ,d_(n))` Given, `A^(3)=A` `implies` diag `(d_(1)^(3), d_(2)^(3), ...., b_(n)^(3))=` diag `(d_(1), d_(2), ... , d_(n))` `implies d_(1)^(3)=d_(1), d_(2)^(3)=d_(2),..., d_(n)^(3)=d_(n)` Hence, all `d_(1), d_(2), d_(3), ..., d_(n)` have three possible values `pm 1, 0`. Each diagonal element can be selected in three ways. Hence, the number of different matrices is `3^(n)`.

Discussion

905.	Let `A`, `B` are square matrices of same order satisfying `AB=A` and `BA=B` then `(A^(2010)+B^(2010))^(2011)` equals.A. `A+B`B. `2010(A+B)`C. `2011(A+B)`D. `2^(2011)(A+B)`
Answer» Correct Answer - D `(d)` Given `AB=A` and `BA=B` `implies{:(A^(2)=A),(B^(2)=B):}` `implies{:(A^(n)=A),(B^(n)=B):}` `implies(A^(2010)+B^(2010))^(2011)=(A+B)^(2011)` Now `(A+B)^(2)=A^(2)+B^(2)+AB+BA` `=2(A+B)` `implies(A+B)^(k)=2^(k)(A+B)` `implies(A^(2010)+B^(2010))^(2011)=(A+B)^(2011)=2^(2011)(A+B)`

Discussion

906.	If the orthogonal square matrices `A` and `B` of same size satisfy `detA+detB=0` then the value of `det(A+B)`A. `-1`B. `1`C. `0`D. none of these
Answer» Correct Answer - C `(c )` Since `A` and `B` are orthogonal, det `A=-1`, det `B=1`, or the other way round Now `A^(T)(A+B)B^(T)=(A+B)^(T)` We have on taking determinants on both sides `(detA)det(A+B)detB=det(A+B)` `implies-det(A+B)=det(A+B)` `impliesdet(A+B)=0`

Discussion

907.	The number of `2xx2` matrices `A`, that are there with the elements as real numbers satisfying `A+A^(T)=I` and `A A^(T)=I` isA. zeroB. oneC. twoD. infinite
Answer» Correct Answer - C `(c )` `A=({:(a,b),(c,d):})` `A+A^(T)=I` `implies({:(a,b),(c,d):})+({:(a,c),(b,d):})=({:(1,0),(0,1):})` `implies2a=1`, `b+c=0`, `2d=1` `impliesa=(1)/(2)`, `c=-b`, `d=(1)/(2)` `impliesA=({:((1)/(2),b),(-b,(1)/(2)):})` Now `AA^(T)=I` `implies({:((1)/(2),b),(-b,(1)/(2)):})({:((1)/(2),-b),(b,(1)/(2)):})=I` `implies({:((1)/(4)+b^(2),0),(0,b^(2)+(1)/(4)):})=({:(1,0),(0,1):})` `impliesb^(2)+(1)/(4)=1impliesb=+-(sqrt(3))/(2)` `:.A=({:((1)/(2),(sqrt(3))/(2)),(-(sqrt(3))/(2),(1)/(2)):})and({:((1)/(2),-(sqrt(3))/(2)),((sqrt(3))/(2),(1)/(2)):})` No. of matrices `=2`

Discussion

908.	Let `Aa n dB`be two `2xx2`matrices. Consider the statements`A B=O A+OorB=O``A B=I_2 A=B^(-1)``(A+B)^2=A^2+2A B+B^2`(i) and (ii) are false, (iii) is true(ii) and (iii) are false, (i) is true(i) is false (ii) and, (iii) are true(i) and (iii) are false, (ii) is trueA. (i) and (ii) are false, (iii) is trueB. (ii) and (iii) are false, (i) is trueC. (i) is false, (ii) and (iii) are trueD. (i) and (iii) are false, (ii) is true
Answer» Correct Answer - D (i) is false. If `A=[(0,1),(0,-1)]` and `B=[(1,1),(0,0)]`, then `AB=[(0,0),(0,0)]=O` (ii) is true as the product AB is an identity matrix, if and only if B is inverse of the matrix A. (iii) is false since matrix multiplication in not commutative.

Discussion

909.	Give an example of twonon-zero `2xx2`matrices `A`and `B`such that `A B=O`.
Answer» Let `A=[{:(1,1),(1,1):}]" and "B=[{:(" "1,-1),(-1," "1):}]`. Then, `AB=[{:(1,1),(1,1):}][{:(" "1,-1),(-1," "1):}]` `=[{:(1-1,-1+1),(1-1,-1+1):}]=[{:(0,0),(0,0):}]` and `BA=[{:(" "1,-1),(-1," "1):}][{:(1,1),(1,1):}]` `=[{:(" "1-1," "1-1),(-1+1,-1+1):}]=[{:(0,0),(0,0):}].` Thus, `AneO, BneO.` But, AB=BA=O.

Discussion

910.	The number of 3 3 non-singular matrices, with four entries as 1 andall other entries as 0, is(1) 5(2) 6 (3) at least 7 (4) less than 4A. 5B. 6C. atleast 7D. less then 4
Answer» Correct Answer - C First row with exactly one zero `therefore ` Total number of cases = 6 First row 2 zeroes, we get more cases. `therefore ` Total we get more then 7.

Discussion

911.	If `X`and `Y`are `2xx2`matrices, then solvethe following matrix equations for `X`and `Ydot``2X+3Y=[2 3 4 0]`, `3X+2Y=[-2 2 1-5]`
Answer» We have, ` 2X+3Y=[{:(2,3),(4,0):}]` and `3X+2Y=[{:(-2,2),(1,-5):}]` On subtracting Eq. i from Eq. ii we get `therefore (3X+2Y)-(2X_3Y)=[{:(-2,-2,2-3),(1-4,-5,0):}]` `(X-Y)=[{:(-4,-1),(-3,-5):}]` On adding Eqs. I and ii, we get `( 5X+5Y)=[{:(0,5),(5,-5):}]` `rArr (X+Y)=(1)/(5)[{:(-2,0),(-2,-6):}]` `therefore X=[{:(-2,0),(-1 ,-3):}]` From Eq. iv, `[{:(-2,0),(-1,-3):}]+y=[{:(0,1),(1,-1):}]` `therefore Y=[{:(2,1),(2,2):}]` and ` X=[{:(-2,0),(-1, -3):}]`

Discussion

912.	Let `K`be a positive real number and`A=[2k-1 2sqrt(k)2sqrt(k)2sqrt(k)1-2k-2sqrt(k)2k-1]a n dB=[0 2k-1sqrt(k)1-2k0 2-sqrt(k)-2sqrt(k)0]`.If det `(a d jA)+det(a d jB)=10^6,t h e n[k]`is equal to.[Note: `a d jM`denotes the adjoint of a square matix `M`and `[k]`denotes the largest integer less than or equal to `K`].
Answer» Correct Answer - D `abs(A) = (2k-1)(-1+4k^(2))+ 2sqrt(k) (2sqrt(k)+4ksqrt(k))` `+ 2 sqrt(k)(4ksqrt(k)+2sqrt(k)) (2k-1) (4k^(2)-1)` `+ 4k + 8k^(2) + 8k^(2) + 4k` `=(2k-1)(4k^(2)-1)+8k+16k^(2)` `= 8k^(3) - 4k^(2)-2k + 1 + 8k + 16k^(2)` `=8k^(3) + 12k^(2) + 6k +1` `abs(B) = 0 ` is skew-symmetric matrix of odd order. `rArr (8k^(3) + 12k^(2) + 6k+1)^(2) = (10^(3))^(2)` `rArr (2k+1)^(3) = 10^(3)` `rArr 2k +1 =10` `rArr k = 4.5` `rArr [k] = 4`

Discussion

913.	Let P be an odd prime number and `T_(p)` be the following set of `2xx2` matrices : The number of A in `T_(p)` such that the trace of a is not divisible by p but det (A) divisible by p is [Note : The trace of matrix is the sum of its diaginal entries].A. `(p-1) (p^(2)-p+1)`B. `p^(3)-(p-1)^(2)`C. `(p-1)^(2)`D. `(p-1) (p^(2)-2)`
Answer» Correct Answer - C

Discussion

914.	The number of `3xx3`matrices `A`whose entries are either `0or1`and for which the system `A[x y z]=[1 0 0]`has exactly two distinct solution isa. 0 b. `2^9-1`c. `168`d. `2`A. oB. `2^(9) -1`C. 168D. 2
Answer» Correct Answer - A Three planes cannot meet only at two distinect points. Hence. Number of matrices = 0

Discussion

915.	Let A be the set of all `3xx3` symmetric matrices all of whose either 0 or 1. Five of these entries are 1 and four of them are 0. The number of matrices A in A for which the system of linear equations `A[(x),(y),(z)]=[(1),(0),(0)]` has a unique solution isA. less than 4B. at least 4 but less than 7C. at least 7 but less than 10D. at leat 10
Answer» Correct Answer - B `[(0,a,b),(a,0,c),(b,c,1)]` Either `b=0` or `c=0implies \|A\| ne 0` `implies` two matrices `A=[(0,a,b),(a,1,c),(b,c,0)]` Either `a=0` or `c=0 implies \|A\| ne 0` `implies` two matrices `A=[(1,a,b),(a,0,c),(b,c,0)]` Either `a=0` or `b=0implies \|A\| ne 0` `implies` two matrices `A=[(1,a,b),(a,1,c),(b,c,1)]` `a=b=0 implies \|A\|=0` `a=c=0 implies \|A\|=0` `b=c=0 implies \|A\| =0` Therefore, there will be only six matrices.

Discussion

916.	Let p be an odd prime number and `T_p` be the following set of 2 x 2 matrices `T_p={A=[(a,b),(c,a)]} , a,b,c in ` {0,1,2,…, p -1} The number of A in `T_p` such that A is either symmetric or skew-symmetric or both and det(A) is divisible by p is: [Note: the trace of a matrix is the sum of its diagonal entries.]A. `2P^(2)`B. `p^(3)-5p`C. `p^(3) 3p`D. `P^(3) = p^(2)`
Answer» Correct Answer - B If A is symmetric matrix, then b = c `therefore det (A) = abs((a,b),(b,a))= a^(2) - b^(2) = (a+b) (a-b)` `a, b, c, in {0, 1, 2, 3,..., P-1}` Number of numbers of type `np=1` `np+1=1` `np + 2 =1` ` ………` `……….` `np_(p-1) = 1 AA n in I` Total number of `A= pxx pxxp=p^(3)` Number of A such that det `(A)` divisible by `p` `= (p-1)^(2)+` numbre of A in which `a = 0` `= (p-1)^(2)+ p + p -1 =p^(2)` Required number `= p^(3) - p^(2)`

Discussion

917.	Let p be an odd prime number and `T_p`, be the following set of `2 xx 2` matrices `T_p={A=[(a,b),(c,a)]:a,b,c in {0,1,2,.........p-1}}` The number of A in `T_p`, such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p isA. `(p-1)(p^(2)-p+1)`B. `p^(3)-(p-1)^(2)`C. `(p-1)^(2)`D. `(p-1) (p^(2) - 2)`
Answer» Correct Answer - A If A is symmetric matrix, then b = c `therefore det (A) = abs((a,b),(b,a))= a^(2) - b^(2) = (a+b) (a-b)` `a, b, c, in {0, 1, 2, 3,..., P-1}` Number of numbers of type `np=1` `np+1=1` `np + 2 =1` ` ………` `……….` `np_(p-1) = 1 AA n in I` As Tr (A) not divisible by `p rArr a ne 0 ` det `(A)` is divisible by `p rArr a^(2) - bc` divisible by `p` Number of ways of selection of a, b, c `= (p-1) [(p-1) xx1] = (p-1)^(2)`

Discussion

918.	Let p be an odd prime number and `T_p` be the following set of 2 x 2 matrices `T_p={A=[(a,b),(c,a)]} , a,b,c in ` {0,1,2,…, p -1} The number of A in `T_p` such that A is either symmetric or skew-symmetric or both and det(A) is divisible by p is: [Note: the trace of a matrix is the sum of its diagonal entries.]A. `(p-1)^(2)`B. `2(p-1)`C. `(p-1)^(2)+1`D. `2p-1`
Answer» Correct Answer - D If A is symmetric matrix, then b = c `therefore det (A) = abs((a,b),(b,a))= a^(2) - b^(2) = (a+b) (a-b)` `a, b, c, in {0, 1, 2, 3,..., P-1}` Number of numbers of type `np=1` `np+1=1` `np + 2 =1` ` ………` `……….` `np_(p-1) = 1 AA n in I` as det (A) is divisible by `p rArr` either `a+b` divisible by `p` corresponding number of ways `= (p -1)` [excluding zero] or `(a-b)` is divisible by `p` corresponding number of ways `= p` Total Number of ways ` = 2p -1`

Discussion

919.	Let p be an odd prime number and `T_p`, be the following set of `2 xx 2` matrices `T_p={A=[(a,b),(c,a)]:a,b,c in {0,1,2,.........p-1}}` The number of A in `T_p`, such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p isA. `2p^(2)`B. `p^(3)-5p`C. `p^(3)-3p`D. `p^(3)-p^(2)`
Answer» Correct Answer - D

Discussion

920.	Let p be an odd prime number and `T_p`, be the following set of `2 xx 2` matrices `T_p={A=[(a,b),(c,a)]:a,b,c in {0,1,2,.........p-1}}` The number of A in `T_p`, such that A is either symmetric or skew-symmetric or both, and det (A) divisible by p isA. `(p-1)^(2)`B. `2(p-1)`C. `(p-1)^(2)+1`D. `2p-1`
Answer» Correct Answer - D We must have `a^(2)-b^(2)=kp` `implies (a+b) (a-b)=kp` `implies` either `a-b=0` or `a+b` is multiple of p when a=b, number of matrices is p and when `a+b=` multiple of `p implies a, b` has `p-1` `:.` total number of matrices `=p+p-1=2p-1`.

Discussion

921.	Let a,b, and c be three real numbers satistying `[a,b,c][(1,9,7),(8,2,7),(7,3,7)]=[0,0,0]` If the point `P(a,b,c)` with reference to (E), lies on the plane `2x+y+z=1`, the the value of `7a+b+c` is (A) 0 (B) 12 (C) 7 (D) 6
Answer» Correct Answer - D `a+ 8b +7c = 0, 9a + 2b 3c=0` `7a + 7b +7c=0` Solving these equations, we get `b=6a` `rArr c= -7a` Now, `2x + y + z = 0` `rArr 2a + 6 a + (-7a) = 1` `rArr a= 1, b = 6, c=-7` `therefore 7a + b+ c= + 6 - = 6`

Discussion

922.	Let a,b, and c be three real numbers satistying `[a,b,c][(1,9,7),(8,2,7),(7,3,7)]=[0,0,0]` Let `omega` be a solution of `x^3-1=0` with `Im(omega)gt0. I fa=2` with b nd c satisfying (E) then the vlaue of `3/omega^a+1/omega^b+3/omega^c` is equa to (A) -2 (B) 2 (C) 3 (D) -3A. -2B. 2C. 3D. -3
Answer» Correct Answer - A `because a = 2 ` with b and c satisfying ( E) `therefore 2+ 8b+ 7c = 0, 18 + 2b + 3c = 0` and ` 2+ b + c= 0` we get ` b = 12 and c= -14` Hence, `3/omega^(a) + 1/omega^(b) + 3/omega^(c) = 3/omega^(2) + 1/omega^(12) + 3/omega^(-14) ` `= (3omega)/omega^(3) + 1/1 + 3omega^(14) ` `= 3 omega + 1 + 3omega^(2)` `= 1 + 3(omega+omega^(2))` `= 1+ 3(-1)=-2`

Discussion

923.	Let a,b, and c be three real numbers satistying `[a,b,c][(1,9,7),(8,2,7),(7,3,7)]=[0,0,0]`Let b=6, with a and c satisfying (E). If alpha and beta are the roots of the quadratic equation `ax^2+bx+c=0 then sum_(n=0)^oo (1/alpha+1/beta)^n` is (A) 6 (B) 7 (C) `6/7` (D) ooA. 6B. 3C. `6/7`D. `infty`
Answer» Correct Answer - B `because b= 6, ` with a and c satisgying ( E) `therefore a + 48 + 7 c = 0, 9a + 12 + 3c= 0, a + 6 +c=0` we get `a = 1, c -7` Given, `alpha ,beta` are the roots of `ax^(2) + bx+c=0` `therefore alpha + beta = -b/a = -6, ` `alpha beta = c/a = -7` Now, `1/alpha + 1/beta=(alpha + beta)/(alpha beta) = (-6)/(-7) = 6/7` `therefore sum _(n=0) ^(infty) (1/alpha +1/beta)^(n) = sum _(n=0)^(infty)(6/7)^(n) ` `= 1+ ( 6/7) + (6/7) ^(2) +...infty` `1/(1-6//7)=7`

Discussion

924.	Let `M`be a `3xx3`matrix satisfying`M[0 1 0]=M[1-1 0]=[1 1-1],a n dM[1 1 1]=[0 0 12]`Then the sum of the diagonal entries of `M`is _________.A.B.C.D.
Answer» Let `M =[[a ,b,c],[d,e,f],[g,h,i]]` `M= [[0],[1],[0]]= [[-1],[2],[3]]rArr b =-1, e=2. h = 3` `M [[1],[-1],[0]] = [[1],[1],[-1]] rArr a = 0, d= 3, g= 2` `M [[1],[1],[1]] = [[0],[0],[12]] rArr g+h+i= 12rArri=7` `therefore ` Sum of diagonal elements `= a + e + i = 0 + 2+7=9`

Discussion

925.	Let `P=[a_("ij")]` be a `3xx3` matrix and let `Q=[b_("ij")]`, where `b_("ij")=2^(i+j) a_("ij")` for `1 le i, j le 3`. If the determinant of P is 2, then the determinant of the matrix Q isA. `2^(10)`B. `2^(11)`C. `2^(12)`D. `2^(13)`
Answer» Correct Answer - D `\|Q\|=\|(2^(2)a_(11),2^(3)a_(12),2^(2)a_(13)),(2^(3)a_(21),2^(4) a_(22),2^(5) a_(23)),(2^(4) a_(31),2^(5) a_(32),2^(6) a_(33))\|` `=2^(2). 2^(3). 2^(4)\|(a_(11),a_(12),a_(13)),(2a_(21),2a_(22),2a_(23)),(2^(2)a_(31),2^(2)a_(32),2^(2) a_(33))\|` `=2^(9). 2. 2^(2) \|(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(31),a_(32),a_(33))\|` `=2^(12) \|P\|` `=2^(13)`

Discussion

926.	State, true or false. If false, give a reason. The matrices A2 × 3 and B2 × 3 are conformable for subtraction.
Answer» Answer is True

Discussion

927.	State whether the statements is true or false: Matrix subtraction is associative.
Answer» Answer is False

Discussion

928.	Let `omega!=1`be cube root of unity and `S`be the set of all non-singular matrices of the form `[1a bomega1comega^2theta1],w h e r e`each of `a ,b ,a n dc`is either `omegaoromega^2dot`Then the number of distinct matrices in the set `S`isa. 2 b. `6`c. `4`d. `8`A. 2B. 6C. 4D. 8
Answer» Correct Answer - A For being nonsingular, `\|(1,a,b),(omega,1,c),(omega^(2),omega,1)\| ne 0` or `ac omega^(2)-(a+c) omega+1 ne 0` Hence, number of possible triplets of (a, b, c) is 2. i.e., `(omega, omega^(2), omega)` and `(omega, omega, omega)`.

Discussion

929.	A rectangular arrangement of number in rows and columns is called _______
Answer» Correct Answer - a matrix

Discussion

930.	If A is a matrix of order 3 x 4 and B is a matrix of order 4 x 3, then find order of matrix (AB).
Answer» Order of matrix AB is 3 x 3

Discussion

931.	If [(2x+y, 3y)(0,4)] = [(6,0)(0,4)] then find the value of x.
Answer» 3y = 0 y = 0 2x+y = 6 2x = 6 x = 3

931.

If [(2x+y, 3y)(0,4)] = [(6,0)(0,4)] then find the value of x.

Answer»

3y = 0

y = 0

2x+y = 6

2x = 6

x = 3

Discussion

932.	Let m and N be two 3x3 matrices such that MN=NM. Further if `M!=N^2` and `M^2=N^4` then which of the following are correct.A. determinant of `(M^(2)+ MN^(2))` is 0B. there is a `3xx3` non-zero matrix `U` such that `(M^(2) + MN^(2)) U` is the zero matrixC. determinant of`(m^(2) + MN^(2)) ge 1`D. for a `3xx3` matrix `U` if `(M^(2)+MN^(2))U` equals the zero matrix, then `U` is the zero matrix
Answer» Correct Answer - A::B Given, `MN = NM, M ne N^(2) and M^(2) = N^(4) ` Then, `M^(2) = N^(2)` lt brgt `rArr (M+N^(2)) (M-N^2) = 0` `therefore M+N^(2) = 0 " " [because M ne N^(2)]` `rArr abs( M+ N^(2)) = 0` (a) `abs( M^(2)+ MN^(2)) = abs(M) abs( M+ N^(2)) = 0` `therefore` Option (a) is correct. `( M^(2)+ MN^(2))U = M( M+ N^(2))U = 0` `therefore` Option (b) is correct. (c) `becauseabs( M^(2)+ MN^(2)) =0` from opttion (a) `therefore abs(M^(2) +MN^(2))cancel(ge) 1` `therefore` OPtion (c ) is incorrect. (d)If ` AX = 0 and abs(A) = 0` then X can be non-zero.

Discussion

933.	If A is `2 xx 2` square matrix, such that det A = 9, then det (9A) =______.A. `(1)/(3^(30))`B. 9C. 81D. 729
Answer» Correct Answer - D If A is an `n xx n` matrix, then `\|kA\| = k^(n)\|A\|.`

Discussion

934.	If K`in R_(0)`, then det `(K I_(0))` is equal toA. `k^(n-1)`B. `k^(n(n-1))`C. `k^n)`D. k
Answer» Correct Answer - B

Discussion

935.	If det (A) = 5, then find det (15A) where A is of order `2 xx 2`A. 225B. 75C. 375D. 1125
Answer» Correct Answer - D `"det"(Ka) = k^(2)\|A\| ("where A is " 2 xx 2 " matrix")`.

Discussion

936.	There are 25 software engineers and 10 testers in Infosys and 15 software engineers and 8 testers in Wipro. In both the companies, a software engineer is paid Rs. 5000 per month and a tester is paid Rs 3000 per month. Find the total amount paid by each of the companies per month by representing the data in matrix form.A. `((155000), (99000))`B. `((23000), (24000))`C. `((50000), (30000))`D. `((155000), (100000))`
Answer» Correct Answer - A Write the given data in matrix form.

Discussion

937.	Check whether the following matrices are in vertible or not :\( \begin{bmatrix} 1 & 2 \\[0.3em] 3 & 3 \\[0.3em] \end{bmatrix}\)[1,2,3,3]
Answer» Given, [1,2,3,3] Let A =\( \begin{bmatrix} 1 & 2 \\[0.3em] 3 & 3 \\[0.3em] \end{bmatrix}\) Then, \|A\| = \( \begin{vmatrix} 1 & 2 \\[0.3em] 3 & 3 \\[0.3em] \end{vmatrix}\) = 3 – 6 = -3 ≠ 0. ∴ A is a non-singular matrix. Hence, A^-1 exist.

937.

Check whether the following matrices are in vertible or not :\( \begin{bmatrix} 1 & 2 \\[0.3em] 3 & 3 \\[0.3em] \end{bmatrix}\)[1,2,3,3]

Answer»

Given,

[1,2,3,3]

Let A =\( \begin{bmatrix} 1 & 2 \\[0.3em] 3 & 3 \\[0.3em] \end{bmatrix}\)

Then,

|A| = \( \begin{vmatrix} 1 & 2 \\[0.3em] 3 & 3 \\[0.3em] \end{vmatrix}\)

= 3 – 6

= -3 ≠ 0.

∴ A is a non-singular matrix.

Hence,

A^-1 exist.

Discussion

938.	Check whether the following matrices are in vertible or not :\( \begin{bmatrix} 2 & 3 \\[0.3em] 10 & 15 \\[0.3em] \end{bmatrix}\)[2,3,10,15]
Answer» Given, [2,3,10,15] Let A =\( \begin{bmatrix} 2 & 3 \\[0.3em] 10 & 15 \\[0.3em] \end{bmatrix}\) Then, \|A\| = \( \begin{vmatrix} 2 & 3 \\[0.3em] 10 & 15 \\[0.3em] \end{vmatrix}\) = 30 – 30 = 0. ∴ A is a singular matrix. Hence, A^-1 does not exist.

938.

Check whether the following matrices are in vertible or not :\( \begin{bmatrix} 2 & 3 \\[0.3em] 10 & 15 \\[0.3em] \end{bmatrix}\)[2,3,10,15]

Answer»

Given,

[2,3,10,15]

Let A =\( \begin{bmatrix} 2 & 3 \\[0.3em] 10 & 15 \\[0.3em] \end{bmatrix}\)

Then,

|A| = \( \begin{vmatrix} 2 & 3 \\[0.3em] 10 & 15 \\[0.3em] \end{vmatrix}\)

= 30 – 30

= 0.

∴ A is a singular matrix.

Hence,

A^-1 does not exist.

Discussion

939.	Check whether the following matrices are in vertible or not :\( \begin{bmatrix} 1 & 1 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}\)[1,1,1,1]
Answer» Given, [1,1,1,1] Let A = \( \begin{bmatrix} 1 & 1 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}\) Then, \|A\| = \( \begin{vmatrix} 1 & 1 \\[0.3em] 1 & 1 \\[0.3em] \end{vmatrix}\) = 1 – 1 = 0. ∴ A is a singular matrix. Hence, A^-1 does not exist.

939.

Check whether the following matrices are in vertible or not :\( \begin{bmatrix} 1 & 1 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}\)[1,1,1,1]

Answer»

Given,

[1,1,1,1]

Let A = \( \begin{bmatrix} 1 & 1 \\[0.3em] 1 & 1 \\[0.3em] \end{bmatrix}\)

Then,

|A| = \( \begin{vmatrix} 1 & 1 \\[0.3em] 1 & 1 \\[0.3em] \end{vmatrix}\)

= 1 – 1 = 0.

∴ A is a singular matrix.

Hence,

A^-1 does not exist.

Discussion

940.	Convert [1,-1,2,3] into an identity matrix by suitable row transformations.
Answer» Let A = [1,-1,2,3] = \( \begin{bmatrix}1 & -1 \\[0.3em]2 &3 \\[0.3em]\end{bmatrix}\) By R₂ – 2R₁, we get, A ~ \( \begin{bmatrix}1 & -1 \\[0.3em]0 &5 \\[0.3em]\end{bmatrix}\) By (\(\frac{1}{5}\)) R₂, we get, A ~ \( \begin{bmatrix}1 & -1 \\[0.3em]0 &1 \\[0.3em]\end{bmatrix}\) By R₁ + R₂, we get, A ~ \( \begin{bmatrix}1 & 0 \\[0.3em]0 &1 \\[0.3em]\end{bmatrix}\) This is an identity matrix.

940.

Convert [1,-1,2,3] into an identity matrix by suitable row transformations.

Answer»

Let A = [1,-1,2,3]

= \( \begin{bmatrix}1 & -1 \\[0.3em]2 &3 \\[0.3em]\end{bmatrix}\)

By R₂ – 2R₁, we get,

A ~ \( \begin{bmatrix}1 & -1 \\[0.3em]0 &5 \\[0.3em]\end{bmatrix}\)

By (\(\frac{1}{5}\)) R₂, we get,

A ~ \( \begin{bmatrix}1 & -1 \\[0.3em]0 &1 \\[0.3em]\end{bmatrix}\)

By R₁ + R₂, we get,

A ~ \( \begin{bmatrix}1 & 0 \\[0.3em]0 &1 \\[0.3em]\end{bmatrix}\)

This is an identity matrix.

Discussion

941.	Check whether the following matrices are in vertible or not :\( \begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)[1,0,0,1]
Answer» Given, [1,0,0,1] Let A =\( \begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\) Then, \|A\| = \( \begin{vmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em]\end{vmatrix}\) = 1 – 0 = 1 ≠ 0. ∴ A is a non-singular matrix. Hence, A^-1 exists.

941.

Check whether the following matrices are in vertible or not :\( \begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)[1,0,0,1]

Answer»

Given,

[1,0,0,1]

Let A =\( \begin{bmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em]\end{bmatrix}\)

Then,

|A| = \( \begin{vmatrix}1 & 0 \\[0.3em]0 & 1 \\[0.3em]\end{vmatrix}\)

= 1 – 0 = 1 ≠ 0.

∴ A is a non-singular matrix.

Hence,

A^-1 exists.

Discussion

942.	If A =\( \begin{bmatrix} 1 & 0 & 0 \\[0.3em] 2 & 1 & 0 \\[0.3em] 3 &3 &1 \end{bmatrix}\)then reduce it to I3 by using column transformations.[1,0,0,2,1,0,3,3,1]
Answer» Given, [1,0,0,2,1,0,3,3,1] \|A\| =\( \begin{vmatrix}1 & 0 & 0 \\[0.3em]2 & 1 & 0 \\[0.3em]3 &3 &1\end{vmatrix}\) = 1(1 – 0) – 0 + 0 = 1 ≠ 0 ∴ A is a non-singular matrix. Hence, The required transformation is possible. Now, A = \( \begin{bmatrix}1 & 0 & 0 \\[0.3em]2 & 1 & 0 \\[0.3em]3 &3 &1\end{bmatrix}\) By C₁ – 2C₂, we get, A ~\( \begin{bmatrix}1 & 0 & 0 \\[0.3em]0 & 1 & 0 \\[0.3em]-3 &3 &1\end{bmatrix}\) By C₁ + 3C₃ and C₂ – 3C₃, we get, A ~\( \begin{bmatrix}1 & 0 & 0 \\[0.3em]0 & 1 & 0 \\[0.3em]0 &0 &1\end{bmatrix}\) = I₃.

942.

If A =\( \begin{bmatrix} 1 & 0 & 0 \\[0.3em] 2 & 1 & 0 \\[0.3em] 3 &3 &1 \end{bmatrix}\)then reduce it to I3 by using column transformations.[1,0,0,2,1,0,3,3,1]

Answer»

Given,

[1,0,0,2,1,0,3,3,1]

|A| =\( \begin{vmatrix}1 & 0 & 0 \\[0.3em]2 & 1 & 0 \\[0.3em]3 &3 &1\end{vmatrix}\)

= 1(1 – 0) – 0 + 0 = 1 ≠ 0

∴ A is a non-singular matrix.

Hence,

The required transformation is possible.

Now,

A = \( \begin{bmatrix}1 & 0 & 0 \\[0.3em]2 & 1 & 0 \\[0.3em]3 &3 &1\end{bmatrix}\)

By C₁ – 2C₂, we get,

A ~\( \begin{bmatrix}1 & 0 & 0 \\[0.3em]0 & 1 & 0 \\[0.3em]-3 &3 &1\end{bmatrix}\)

By C₁ + 3C₃ and C₂ – 3C₃, we get,

A ~\( \begin{bmatrix}1 & 0 & 0 \\[0.3em]0 & 1 & 0 \\[0.3em]0 &0 &1\end{bmatrix}\) = I₃.

Discussion

943.	Show that the two matrices A, `P^(-1) AP` have the same characteristic roots.
Answer» Let `P^(-1) AP=B` `:. B-lambdaI=P^(-1) AP-lambdaI` `=P^(-1) AP-P^(-1) lambdaIP` `=P^(-1) (A-lambdaI)P` `implies \|B-lambda I\|=\|P^(-1)\|\|A-lambdaI\|\|P\|` `=\|A-lambdaI\|\|P^(-1)\|\|P\|` `=\|A-lambdaI\|\|P^(-1) P\|` `=\|A-lambdaI\|\|I\|=\|A-lambdaI\|` Thus, the two matrices A and B have the same characteristic determinants and hence the same characteristic equations and the same characteristic roots. the same thing may also be seen in another way. Now, `AX=lambdaX` `implies P^(-1) AX=lambdaP^(-1) X` `implies (P^(-1) AP) (P^(-1)X)=lambda(P^(-1) X)`

Discussion

944.	If A is symmetric as well as skew-symmetric matrix, then A isA. diagonal matrixB. null matrixC. triangular materixD. none of these
Answer» Correct Answer - B Let `A=[a_("ij")]`. Since a is skew-symmetric, we have `a_(ii)=0` and `a_("ij")=-a_("ij") (i ne j)` A is symmetric as well, so `a_("ij")=a_("ji")` for all `i` and `j`. `:. A_("ij")=0` for all `i ne j` Hence, `a_("ij")=0` for all `i` and `j`, i.e., A is a null matrix.

Discussion

945.	If A and B are square matrices of the same order such that `A B=B A`, then show that `(A+B)^2=a^2+2A B+B^2`
Answer» False For two square matrices of same order it is not always true that AB=BA.

Discussion

946.	If `A`and `B`are square matrices ofthe same order such that `A B=B A`, then show that `(A+B)^2=A^2+2A B+B^2`.A. `AB = I`B. `BA=I`C. `AB=BA`D. none of these
Answer» Correct Answer - C

Discussion

947.	If `A`and `B`are symmetric matricesof the same order, write whether `A B-B A`is symmetric orskew-symmetric or neither of the two.A. symmetric matrixB. skew-symmetric matrixC. null matrixD. unit matrix
Answer» Correct Answer - B

Discussion

948.	If A and B are symmetric matrices of the same order, write whether AB – BA is symmetric or skew-symmetric or neither of the two.
Answer» A and B are symmetric matrices, ∴ A’ = A and B’ = B …(i) Consider, (AB – BA)’ = (AB)’ – (BA)’ [(a – b)’ = a’ – b’] = B’A’ – A’B’ [(AB)’ = B’A’] = BA – AB [from (i)] = - (AB – BA) ∴ (AB – BA)’ = - (AB – BA) Hence, (AB – BA) is a skew symmetric matrix.

948.

If A and B are symmetric matrices of the same order, write whether AB – BA is symmetric or skew-symmetric or neither of the two.

Answer»

A and B are symmetric matrices,

∴ A’ = A and B’ = B …(i)

Consider,

(AB – BA)’ = (AB)’ – (BA)’ [(a – b)’ = a’ – b’]

= B’A’ – A’B’ [(AB)’ = B’A’]

= BA – AB [from (i)]

= - (AB – BA)

∴ (AB – BA)’ = - (AB – BA)

Hence,

(AB – BA) is a skew symmetric matrix.

Discussion

949.	Find AT , if A = \(\begin{vmatrix}1&3\\-4&5\end{vmatrix}\)A = [(1, 3) (-4, 5)]
Answer» A = \(\begin{vmatrix}1&3\\-4&5\end{vmatrix}\) A^T= \(\begin{vmatrix}1&-4\\3&5\end{vmatrix}\)

949.

Find AT , if A = \(\begin{vmatrix}1&3\\-4&5\end{vmatrix}\)A = [(1, 3) (-4, 5)]

Answer»

A = \(\begin{vmatrix}1&3\\-4&5\end{vmatrix}\)

A^T= \(\begin{vmatrix}1&-4\\3&5\end{vmatrix}\)

Discussion

950.	Find AT , if A = \(\begin{vmatrix}2&-6&1\\-4&0&5\end{vmatrix}\)A = [(2, -6, 1) (-4, 0, 5)]
Answer» A= \(\begin{vmatrix}2&-6&1\\-4&0&5\end{vmatrix}\) A^T=\(\begin{vmatrix}2&-4\\-6&0\\1&5\end{vmatrix}\)

950.

Find AT , if A = \(\begin{vmatrix}2&-6&1\\-4&0&5\end{vmatrix}\)A = [(2, -6, 1) (-4, 0, 5)]

Answer»

A= \(\begin{vmatrix}2&-6&1\\-4&0&5\end{vmatrix}\)

A^T=\(\begin{vmatrix}2&-4\\-6&0\\1&5\end{vmatrix}\)

Discussion

Explore topic-wise InterviewSolutions in .

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Check whether the following matrices are in vertible or not :\( \begin{bmatrix} 2 &amp; 3 \\[0.3em] 10 &amp; 15 \\[0.3em] \end{bmatrix}\)[2,3,10,15]

Check whether the following matrices are in vertible or not :\( \begin{bmatrix} 1 &amp; 1 \\[0.3em] 1 &amp; 1 \\[0.3em] \end{bmatrix}\)[1,1,1,1]

Convert [1,-1,2,3] into an identity matrix by suitable row transformations.

Check whether the following matrices are in vertible or not :\( \begin{bmatrix}1 &amp; 0 \\[0.3em]0 &amp; 1 \\[0.3em]\end{bmatrix}\)[1,0,0,1]

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If A and B are symmetric matrices of the same order, write whether AB – BA is symmetric or skew-symmetric or neither of the two.

Find AT , if A = \(\begin{vmatrix}1&amp;3\\-4&amp;5\end{vmatrix}\)A = [(1, 3) (-4, 5)]

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Check whether the following matrices are in vertible or not :\( \begin{bmatrix} 1 & 2 \\[0.3em] 3 & 3 \\[0.3em] \end{bmatrix}\)[1,2,3,3]

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Find AT , if A = \(\begin{vmatrix}1&3\\-4&5\end{vmatrix}\)A = [(1, 3) (-4, 5)]

Find AT , if A = \(\begin{vmatrix}2&-6&1\\-4&0&5\end{vmatrix}\)A = [(2, -6, 1) (-4, 0, 5)]