1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 1 InterviewSolution

1.	Let `A=" diag "[3,-5,7]" and "B=" diag "[-1,2,4].` ltbr. Find (i) (A+B) (ii) (A-B) (iii) -5A (iv) (2A+3B).`
Answer» We have `A=[{:(3,0,0),(0,-5,0),(0,0,7):}]" and "B=[{:(-1,0,0),(0,2,0),(0,0,4):}].` `:." "(i)A+B=[{:(3,0,0),(0,-5,0),(0,0,7):}]+[{:(-1,0,0),(0,2,0),(0,0,4):}]=[{:(2," "0,0),(0,-3,0),(0," "0,11):}].` (ii) `(A-B)=A+(-B)` `=[{:(3,0,0),(0,-5,0),(0,0,7):}]+[{:(1,0,0),(0,-2,0),(0,0,-4):}]=[{:(4," "0,0),(0,-7,0),(0," "0,3):}].` (iii) `-5A=(-5).A=(-5).[{:(3,0,0),(0,-5,0),(0,0,7):}]=[{:(-15,0," "0),(0,25," "0),(0,0,-35):}].` (iv)`2A+3B=[{:(6,0,0),(0,-10,0),(0,0,14):}]+[{:(-3,0,0),(0,6,0),(0,0,12):}]=[{:(3,0,0),(0,-4,0),(0,0,26):}].`

Discussion

2.	A2=A THEN FIND VALUE OF (I+A)2 -3A
Answer» (I+A)²−3A=I²+2IA+A²−3A =I+2A+A−3A ( since A²=A and IA=A and I²=I ) =I

2.

A2=A THEN FIND VALUE OF (I+A)2 -3A

Answer»

(I+A)²−3A=I²+2IA+A²−3A

=I+2A+A−3A ( since A²=A and IA=A and I²=I )

=I

Discussion

3.	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
Answer» Correct Answer - C `f(x)=(1+x)/(1-x)` or `(1-x) f(x)=1+x` `implies (I-A)f(A)=(I+A)` `implies 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)])` `=([(0,2),(2,0)][(2,2),(2,2)])/(-4)` `=([(4,4),(4,4)])/(-4)` `=[(-1,-1),(-1,-1)]`

Discussion

4.	Id `[1//25 0x1//25]=[5 0-a5]^(-2)`, then the value of `x`is`a//125`b. `2a//125`c. `2a//25`d. none of theseA. `a//125`B. `2a//125`C. `2a//25`D. none of these
Answer» Correct Answer - B Let `A=[(5,0),(-a,5)]` `implies` adj `(A)= [(5,0),(a,5)]` `implies A^(-1) =1/(\|A\|)[(5,0),(a,5)]=1/25 [(5,0),(a,5)]` `implies A^(-2)=(A^(-1))^(2)=1/25 [(5,0),(a,5)]1/25 [(5,0),(a,5)]` `=1/625 [(25,0),(10a, 25)]` `=[(1/25,0),((2a)/125,1/25)]` Now, `[(1//25,0),(x,1//25)]=[(1/25,0),((2a)/125,1/25)]` `implies x=2a//125`

Discussion

5.	If `A=[(a+ib,c+id),(-c+id,a-ib)]` and `a^(2)+b^(2)+c^(2)+d^(2)=1`, then `A^(-1)` is equal toA. `[(a-ib,-c-id),(c-id,a+ib)]`B. `[(a+ib,-c+id),(-c+id,a-ib)]`C. `[(a-ib,-c-id),(-c-id,a+ib)]`D. none of these
Answer» Correct Answer - A We have, `\|A\|=(a+ib) (a-ib)-(-c+id) (c+id)` `=a^(2)+b^(2)+c^(2)+d^(2)=1` and adj `(A)=[(a-ib,-c-id),(c-id,a+ib)]` Then `A^(-1)=[(a-ib,-c-id),(c-id,a+ib)]`

Discussion

6.	If the matrix A is such that `({:(1,3),(0,1):})A=({:(1,1),(0,-1):})`, then what is A equal to ?A. `{:[(1,4),(-1,0)]:}`B. `{:[(1,-4),(1,0)]:}`C. `{:[(1,-4),(0,-1)]:}`D. none of these
Answer» Correct Answer - C

Discussion

7.	If `A=[[cos alpha, -sin alpha] , [sin alpha, cos alpha]], B=[[cos2beta, sin 2beta] , [sin 2 beta, -cos2beta]]` where `0 lt beta lt pi/2` then prove that `BAB=A^(-1)` Also find the least positive value of `alpha` for which `BA^4B= A^(-1)`A.B.C.D.
Answer» Correct Answer - `alpha =(2pi)/3` `because BAB=A^(-1)` `rArr ABAB= I` `rArr (AB)^(2) = I` Now, `AB= [[cos (alpha+2beta),sin (alpha+2beta)],[sin(alpha+2beta),-cos(alpha + 2beta)]]` and `(AB)^(2) = (AB) (AB) = [[1,0],[0,1]]= I [because (AB)(AB)=I]` Also, `BA^(4)B=A^(-1)` or `A^(4) B= B^(-1) A^(-1) =(AB)^(-1) = AB` or`A^(4) = A " "...(i)` Now, `A^(2) = [[cos alpha,-sin alpha],[sin alpha,cos alpha]][[cos alpha,-sin alpha],[sin alpha,cos alpha]]` `=[[cos 2alpha,-sin 2alpha],[sin 2alpha,cos 2alpha]]` Similarly, `A^(4)=[[cos 4alpha,-sin 4alpha],[sin 4alpha,cos4alpha]]` Hence, from Eq. (i) `[[cos 4alpha,-sin 4alpha],[sin 4alpha,cos4alpha]]=[[cos alpha,-sin alpha],[sin alpha,cos alpha]]` or `4 alpha = 2pi + alpha` ` therefore alpha = (2pi)/3`

Discussion

8.	If `A=[[cos alpha, -sin alpha] , [sin alpha, cos alpha]], B=[[cos2beta, sin 2beta] , [sin 2 beta, -cos2beta]]` where `0 lt beta lt pi/2` then prove that `BAB=A^(-1)` Also find the least positive value of `alpha` for which `BA^4B= A^(-1)`
Answer» Correct Answer - `alpha=(2pi)/3` `BAB=A^(-1)` `implies ABAB=I` `implies (AB)^(2)=I` Now, `AB=[(cos alpha,-sin alpha),(sin alpha,cos alpha)][(cos 2 beta,sin 2 beta),(sin 2 beta,-cos 2 beta)]` `=[(cos(alpha+2 beta),sin (alpha+2beta)),(sin (alpha+2beta),- cos (alpha +2beta))]` and `(AB)^(2)=[(cos (alpha+2beta),sin (alpha+2beta)),(sin (alpha+2beta),-cos (alpha+2beta))]xx[(cos(alpha+2beta),sin (alpha+2beta)),(sin (alpha+2beta),-cos (alpha+2beta))]` `=[(cos^(2)(alpha+2beta)+sin^(2) (alpha+2beta),0),(0,cos^(2) (alpha+2beta)+sin^(2) (alpha+2beta))]` `=[(1,0),(0,1)]` `=I` `BA^(4)B=A^(-1)` or `A^(4)B=B^(-1)A^(-1)=(AB)^(-1)=AB` or `A^(4)=A` (1) Now, `A^(2)=[(cos alpha,-sin alpha),(sin alpha,cos alpha)][(cos alpha,-sin alpha),(sin alpha,cos alpha)]` `=[(cos 2 alpha,-sin 2 alpha),(sin 2 alpha,cos 2 alpha)]` and `A^(4)=[(cos 4 alpha,-sin 4 alpha),(sin 4 alpha,cos 4 alpha)]` Hence, from Eq. (1), `[(cos 4 alpha,-sin 4 alpha),(sin 4 alpha,cos 4 alpha)]=[(cos alpha,-sin alpha),(sin alpha,cos alpha)]` or `4alpha=2pi+alpha` or `alpha=(2pi)/3`

Discussion

9.	If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal toA. `A(-alpha, -beta)`B. `A(-alpha, beta)`C. `A(alpha, -beta)`D. `A(alpha, beta)`
Answer» Correct Answer - A We have, `A(alpha, beta)^(-1)=1/e^(beta) [(e^(beta) cos alpha,-e^(beta) sin alpha,0),(e^(beta) sin alpha,e^(beta) cos alpha,0),(0,0,1)]` `=A(-alpha, -beta)`

Discussion

10.	`{:A=[(cos^2 alpha,cosalphasinalpha),(cosalphasinalpha,sin^2alpha)]:}` `{:B=[(cos^2 beta,cosbetasinbeta),(cosbetasinbeta,sin^2beta)]:}` are two matrices such that the product AB is the null matrix, then `(alpha-beta)` is
Answer» Correct Answer - C

Discussion

11.	If `E(theta)=[[cos theta, sin theta] , [-sin theta, cos theta]]` then `E(alpha) E(beta)=`A. `E(0^@)`B. `E(alphabeta)`C. `E(alpha+beta)`D. `E(alpha-beta)`
Answer» Correct Answer - C

Discussion

12.	`A=[{:(a,b),(b,-a):}]` and `MA=A^(2m)`, `m in N` for some matrix `M`, then which one of the following is correct ?A. `M=[{:(a^(2m),b^(2m)),(b^(2m),-a^(2m)):}]`B. `M=(a^(2)+b^(2))^(m)[{:(1,0),(0,1):}]`C. `M=(a^(m)+b^(m))[{:(1,0),(0,1):}]`D. `M=(a^(2)+b^(2))^(m-1)[{:(a,b),(b,-a):}]`
Answer» Correct Answer - D `(d)` Clearly option `(d)` satisfies the given conditions.

Discussion

13.	If `A=[(a,b),(b,a)] and A^2=[(alpha, beta),(beta, alpha)]` thenA. `alpha=a^(2)-b^(2),beta=2ab`B. `alpha=2ab,beta=a^(2)+b^(2)`C. `alpha=a^(2)+b^(2),beta=2ab`D. `alpha=2ab,beta=a^(2)-b^(2) `
Answer» Correct Answer - C N/a

Discussion

14.	If `{:E(theta)=[(cos^2 theta,costhetasintheta),(costhetasintheta,sin^2theta)]:},and thetaand phi` differ by an odd multiple of `pi//2," then "E(theta)E(phi)` is aA. null matrixB. unit matrixC. diagonal matrixD. none of these
Answer» Correct Answer - A

Discussion

15.	Let `A=[{:(-5,-8,-7),(3,5,4),(2,3,3):}]` and `B=[{:(x),(y),(1):}]`. If `AB` is a scalar multiple of `B`, then the value of `x+y` isA. `-1`B. `-2`C. `1`D. `2`
Answer» Correct Answer - B `(b)` `AB=lambdaB`, where `lambda` is non-zero scalar. `[{:(-5x-8y-7),(3x+5y+4),(2x+3y+3):}]=lambda[{:(x),(y),(2):}]` i.e., `{:(-5x-8y-7=lambdax),(3x+5y+4=lambday),(2x+3y+3=2lambda):}` Adding `0=lambda(x+y+2)` `lambda ne 0impliesx+y+2=0` `x+y=-2`

Discussion

16.	If `F(x)=[(cos^(2)x,cosxsinx),(cosxsinx,sin^(2)x)]` and the difference of `x` and `y` is the odd Multiple of `(pi)/(2),`then `F(x)F(y) `is :A. Zero matrixB. unit matrixC. diagonal matrixD. None of these
Answer» Correct Answer - B N/a

Discussion

17.	In the matrix\(A = \begin{bmatrix}2 & 5 & 19&-7 \\[0.3em]35 & -2 & \frac{5}{2}&12 \\[0.3em]\sqrt3 &1&-5 & 17\end{bmatrix}\)(i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a13, a21, a33, a24 ,a23,
Answer» (i) The order of the matrix is 3 x 4 (ii) Number of elements = 12 (iii) a₁₃= 19 a₂₁= 35 a₃₃= -5 a₂₄= 12 a₂₃= \(\frac{5}{2}\)

17.

In the matrix\(A = \begin{bmatrix}2 & 5 & 19&-7 \\[0.3em]35 & -2 & \frac{5}{2}&12 \\[0.3em]\sqrt3 &1&-5 & 17\end{bmatrix}\)(i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a13, a21, a33, a24 ,a23,

Answer»

(i) The order of the matrix is 3 x 4

(ii) Number of elements = 12

(iii) a₁₃= 19

a₂₁= 35

a₃₃= -5

a₂₄= 12

a₂₃= \(\frac{5}{2}\)

Discussion

18.	Let `A{:[(5,10),(10,20)]:},{:[(10,5),(15,10)]:}and={:[(-10,35),(25,-5)]:}`. Prove that AB = AC but `B!=C`
Answer» `AB={:[(5,10),(10,20)]:}{:[(10,5),(15,10)]:}` `={:[((5)10+10(15),(5)5+(10)10),((10)10+(20)15,(10)5(20)10)]:}` `={:[(50+150,25+100),(100+300,50+200)]:}={:[(200,125),(400,250)]:}` `AC={:[(5,10),(10,20)]:}{:[(-10,35),(25,-5)]:}` `={:[(5(-10)+10(25),5(35)+10(-5)),(10(-10)+20(25),10(35)+20(-5))]:}` `={:[(-50+250,175-50),(-100+500,350-100)]:}` `={:[(200,125),(400,250)]:}`. Here AB=AC but `B!=C`. Hence Proved.

Discussion

19.	If `{:((-2,-1),(3a,b),(4,-6+x)):}={:((-2,-1),(9,-1),(4,2)):}," then "(x)/(a+b)` =A. `(1)/(4)`B. `(-1)/(11)`C. -11D. 4
Answer» Correct Answer - D Find a,b and x by equating corresponding elements of the matrices.

Discussion

20.	Prove that AI=IA=A, for `A{:[(2,-1),(4,3)]:},I={:[(1,0),(0,1)]:}`
Answer» `AI{:[(2,-1),(4,3)]:}{:[(1,0),(0,1)]:}={:[(2,-1),(4,3)]:}=A` `IA={:[(1,0),(0,1)]:}{:[(2,-1),(4,3)]:}={:[(2,-1),(4,3)]:}=A` `:.AI=IA=A`. Here I is called multiplicative identity matrix.

Discussion

21.	If `{:A=[(0,1),(1,0)]:}`,I is the unit matrix of order 2 and a, b are arbitray constants, then `(aI +bA)^2` is equal toA. `a^2I-abA`B. `a^2I+2abA`C. `a^2I+b^2A`D. none of these
Answer» Correct Answer - b

Discussion

22.	If A is a square matrix of order 3 with \|AI= 2, then the value of `\|(A-A^T)^5\|+\|(A-A^T)^3\|` is
Answer» `A`is skew symmetric matrix `\|A-A^T\|=\|A^T-A\|=0` `\|(A-A^T)^5\|+\|(A-A^T)^3\|` `=\|0^5\|+\|0^3\|` `=0`

Discussion

23.	If `A=[{:(cosx,sinx,0),(-sinx,cosx,0),(0,0,1):}]` =f(x), then `A^(-1)` is equal toA. `f(-x)`B. `f(x)`C. `-f(x)`D. `-f(-x)`
Answer» Correct Answer - A

Discussion

24.	`F(x)=[[cosx,-sinx,0],[sinx,cosx,0],[0,0,1]]` and `G(x)=[[cosx,0,sinx],[0,1,0],[-sinx,0,cosx]]`, then `[F(x)G(y)]^(-1)` is equal to (A) `F(-x)G(-y)` (B) `F(x-1)G(y-1)` (C) `G(-y)F(-x)` (D) `G(y^(-1))F(x^(-1))`A. `F(-x)G(-y)`B. `F(x^(-1))_G(y^(-1))`C. `G(-y)F(-x)`D. `G(y^(-1))F(x^(-1))`
Answer» Correct Answer - C

Discussion

25.	If `P(x)=[(cosx, sinx),(-sinx, cosx)]`, then show that `P(x).P(y)=P(x+y)=P(y).P(x)`.
Answer» We have. `P(x)=[{:(cos x,sinx ),(-sinx,cosx):}]` `therefore P(y)=[{:(cos y, sin y),(-siny,cosy):}]` Now, `P(x).P(y)=[{:(cosx,sinx),(-sin x,cosx):}][{:(cosy,siny),(-siny,cosy):}]` `=[{:(cosx.cosy-cos x.siny,-sin x.siny+cosx.cosy),(-sinx.cosy-cosx.siny,-sinx.siny+cosx.cosy):}]` `=[{:(cos(x+y),sin(x+y)),(-sin(x+y),cos(x+y)):}]` `[{:(because cos(x+y)=,cosx.cosy-sinx.siny),("and" sin(x+y)=,sinx.cosy+cosx.siny):}]` and `P(x+y)=[{:(cos(x+y)),(-sin(x+y),cos(x+y )):}]` Also `P(y).P(x)=[{:( cosy,siny),(-siny,cosy):}][{:(cosx,sinx),(-sinx,cosx):}]` `= [{:(cosy.cosx-siny.sinx,cosy.sinx+siny.cosx),(-siny.cosx-sin x.cos y,-sin y.sin x+cos y.cos x):}]` `=[{:(cos(x+y),sin(x+y)),(-sin(x+y),cos(x+y)):}]` Thus, we see from the Eqs. (i), (ii) and (iii) that `P(x).P(y)=P(x+y)=P(y).P(x)` Hence proved.

Discussion

26.	Let P and Q be `3xx3` matrices with `P!=Q` . If `P^3=""Q^3a nd""P^2Q""=""Q^2P`, then determinant of `(P^2+""Q^2)` is equal to (1) 2(2) 1 (3)0 (4) 1
Answer» `P^3 = Q^3 `eqn(1) `P^2Q = Q^2P`eqn(2) `(1)- (2)`equation `P^3 - P^2Q - Q^3 - Q^2P` `P^2(P-Q) = Q^2(Q-P)` `P^2(P-Q) = -Q^2(P-Q)` `(P^2 + Q^2)(P-Q) = 0` `P-Q= 0 or P^2 + Q^2 = 0` `P-Q= 0 ` NOT POSSIBLEso`P^2 + Q^2 = 0` Option 3 is correct

Discussion

27.	LetA be a `2xx2`matrix with real entries. Let I be the `2xx2`identity matrix. Denote by tr (A), the sumof diagonal entries of A. Assume that `A^2=""I`.Statement1: If `A!=I`and `A!=""-I`, then det `A""=-1`.Statement2: If `A!=I`and `A!=""-I`,then `t r(A)!=0`.
Answer» `A= [(a,b),(c,d)]` `A^2 = [(a,b),(c,d)][(a,b),(c,d)]` `= [(a^2+ bc, ab+bd),(ac+cd,bc+d^2)] = I` `a^2 + bc = bc + d^2 = 1` `ac+cd = ab+bd = 0` `c(a+d) = 0` `b(a+d) = 0` `c=0 or a=-d` not possible for c `b= 0 or a=-d` not possible for b `\|(a,b),(c,d)\| = ad - bc = -d^2 - bc` `= -(d^2 + bc) = -1` `tr (A) = a+d= a-a = 0` so, option 4 is correct

Discussion

28.	LetA be a `2xx2`matrix with real entries. Let I be the `2xx2`identity matrix. Denote by tr (A), the sumof diagonal entries of A. Assume that `A^2=""I`.Statement1: If `A!=I`and `A!=""-I`, then det `A""=-1`.Statement2: If `A!=I`and `A!=""-I`,then `t r(A)!=0`.(1)Statement 1is false, Statement `( 2) (3)-2( 4)`is true(6)Statement 1is true, Statement `( 7) (8)-2( 9)`(10)is true, Statement `( 11) (12)-2( 13)`is a correct explanation forStatement 1(15)Statement 1is true, Statement `( 16) (17)-2( 18)`(19)is true; Statement `( 20) (21)-2( 22)`is not a correct explanationfor Statement 1.(24)Statement 1is true, Statement `( 25) (26)-2( 27)`is false.A. Statement -1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1B. Statement -1 is true, Statement - 2 is true, Statement -2 is not a correct explanation for Statement-1C. Statement-1 is true, Statement-2 is falseD. Statement-1 is false, Statement-2 is true
Answer» Correct Answer - C Let `A = [[1,0],[0,-1]]or [[-1,0],[0,1]]` Then `A^(2) = I` `therefore A = abs((1,0),(0,-1))= - 1 and tr (A)=0`

Discussion

29.	Let A and B be two symmetric matrices of order 3.Statement-1 : A(BA) and (AB)A are symmetric matrices.Statement-2 : AB is symmetric matrix if matrix multiplication of A with B iscommutative. Statement-1 is true, Statement-2 is true; Statement-2 is a correctexplanation for Statement-1. Statement-1 is true, Statement-2 is true; Statement-2 is true;Statement-2 is not a correctexplanation for Statement-1. Statement-1 is true, Statement-2 is false. Statement-1 is false, Statement-2 is true.
Answer» `A^T = A & B^T = B` 1) `[A(BA)]^T = [(BA)^T * A^T] = A^T B^TA^T= ABA ` `[(AB)A]^T = A^T(AB)^T = A^TB^TA^T = ABA` option 2 is correct

Discussion

30.	If f(x) = `[(cos x , - sinx,0),(sinx,cosx,0),(0,0,1)]` then show f(x) . f(y) = f(x+y)
Answer» Here, `f(x) = [[cosx,-sinx,0],[sinx,cosx,0],[0,0,1]]` `:. f(y) = [[cosy,-siny,0],[siny,cosy,0],[0,0,1]]` `:.f(x+y) = [[cos(x+y),-sin(x+y),0],[sin(x+y),cos(x+y),0],[0,0,1]]` Now,`L.H.S. = f(x)*f(y)` `= [[cosx,-sinx,0],[sinx,cosx,0],[0,0,1]] [[cosy,-siny,0],[siny,cosy,0],[0,0,1]]` `=[[cosxcosy-sinxsiny,-cosxsiny-sinxsiny,0],[sinxcosy+cosxsiny,-sinxsiny+cosxcosy,0],[0,0,1]]` `=[[cos(x+y),-sin(x+y),0],[sin(x+y),cos(x+y),0],[0,0,1]]` `=f(x+y) = R.H.S.`

Discussion

31.	Let A be a `2xx2`matrix with non-zero entriesand let `A^2=""I`, where I is `2xx2`identity matrix. Define Tr(A) =sum of diagonal elements of A and \|A\| = determinant of matrix A.Statement-1:`T r(A)""=""0`Statement-2:`\|A\|""=""1`(1)Statement-1 istrue, Statement-2 is true; Statement-2 is not the correct explanation forStatement-1(2)Statement-1 istrue, Statement-2 is false(3)Statement-1 isfalse, Statement-2 is true(4)Statement-1 istrue, Statement-2 is true; Statement-2 is the correct explanation forStatement-1
Answer» `A= [(a,b),(c,d)]` `A^2 = [(a,b),(c,d)][(a,b),(c,d)] = [ (a^2 + bc,ab+bd),(ac+cd, bc+d^2)]= I` `I= [(1,0),(0,1)]` now,`a^2 + bc = bc+d^2 = 1` `ab + bd = ac + cd = 0` `b(a+d) = c(a+d) = 0` `a+d = 0` 1)`T_r = (A) = a+d = 0` 2) `det(A) = ad*bc = a(-a) - bc= -a^2 - bc` `= -(a^2 +bc)` option 2 is correct

Discussion

32.	Which of the following statements is true? (a) Every zero matrix is a square matrix (b) A unit matrix is a diagonal matrix(c) \(\begin{bmatrix}1&0\\[0.3em]0&1\end{bmatrix}\) is the identity matrix for addition of 2 × 2 matrix(d) \(\begin{bmatrix}4&x\\[0.3em]1&3\end{bmatrix}\) = \(\begin{bmatrix}4&5\\[0.3em]1&0\end{bmatrix}\), if x = 5
Answer» (b) • Every zero matrix is not necessarily a square matrix. • A unit matrix is a diagonal matrix whose diagonal elements are all 1. • The null matrix is the identity matrix for addition. If x = 5, then the matrix on LHS = \(\begin{bmatrix}4&5\\[0.3em]1&3\end{bmatrix}\) ≠ \(\begin{bmatrix}4&5\\[0.3em]1&0\end{bmatrix}\)

32.

Which of the following statements is true? (a) Every zero matrix is a square matrix (b) A unit matrix is a diagonal matrix(c) \(\begin{bmatrix}1&0\\[0.3em]0&1\end{bmatrix}\) is the identity matrix for addition of 2 × 2 matrix(d) \(\begin{bmatrix}4&x\\[0.3em]1&3\end{bmatrix}\) = \(\begin{bmatrix}4&5\\[0.3em]1&0\end{bmatrix}\), if x = 5

Answer»

(b) • Every zero matrix is not necessarily a square matrix.

• A unit matrix is a diagonal matrix whose diagonal elements are all 1.

• The null matrix is the identity matrix for addition.

If x = 5, then the matrix on

LHS = \(\begin{bmatrix}4&5\\[0.3em]1&3\end{bmatrix}\) ≠ \(\begin{bmatrix}4&5\\[0.3em]1&0\end{bmatrix}\)

Discussion

33.	If \(\begin{bmatrix}2x&3\\[0.3em]0&y-1\end{bmatrix} \) = \(\begin{bmatrix}x-3&3\\[0.3em]0&2\end{bmatrix},\) then the values of x and y respectively are :(a) 3, –3 (b) –3, 3 (c) –3, –3 (d) 3, 3
Answer» (b) Equating corresponding elements, we get 2x = x – 3 ⇒ x = –3 and y – 1 = 2 ⇒ y = 3

33.

If \(\begin{bmatrix}2x&3\\[0.3em]0&y-1\end{bmatrix} \) = \(\begin{bmatrix}x-3&3\\[0.3em]0&2\end{bmatrix},\) then the values of x and y respectively are :(a) 3, –3 (b) –3, 3 (c) –3, –3 (d) 3, 3

Answer»

(b) Equating corresponding elements, we get

2x = x – 3 ⇒ x = –3

and y – 1 = 2 ⇒ y = 3

Discussion

34.	A=[aij] mxn is a square matrix, if(A) m < n (B) m > n (C) m = n (D) None of these
Answer» The correct answer is C. It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns. Therefore, A=[_ij]_mxnis a square matrix, if m = n.

34.

A=[aij] mxn is a square matrix, if(A) m < n (B) m > n (C) m = n (D) None of these

Answer»

The correct answer is C.
It is known that a given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

Therefore, A=[_ij]_mxnis a square matrix, if m = n.

Discussion

35.	If A = [aij]m x n' B = [bij]m x n' then the element C23 of the matrix C=A+B is:(a) a13 + b13 (b) a23 + b32 (c) a23 + b23 (d) a32 + b23
Answer» (c) A = [a_ij]_{m × n'} B = [b_ij]_{m × n} ∴ C = A + B ⇒ [c_ij]_{m × n ,} = [a_ij+ b_ij]_{m ×n} Hence, c₂₃ = a₂₃ + b₂₃

35.

If A = [aij]m x n' B = [bij]m x n' then the element C23 of the matrix C=A+B is:(a) a13 + b13 (b) a23 + b32 (c) a23 + b23 (d) a32 + b23

Answer»

(c) A = [a_ij]_{m × n'} B = [b_ij]_{m × n}

∴ C = A + B ⇒ [c_ij]_{m × n ,} = [a_ij+ b_ij]_{m ×n}

Hence, c₂₃ = a₂₃ + b₂₃

Discussion

36.	If A is a non-singular square matrix such that `A^2-7A+5I=0`, then `A^-1`A. `7A -I`B. `(7)/(5)I - (1)/(5)A`C. `(7)/(5)I + (1)/(5)A`D. `(A)/(5) - (7)/(5)`
Answer» Correct Answer - B Pre-multiply the given equation with `A^(-1)` and use the relation, `A^(-1)A= I, A^(-1)I = A^(-1)`.

Discussion

37.	Construct `A_(2xx3)` matrix whose elements are given by `a_(ij)=(1)/(2)\|5i-3j\|.`
Answer» `A2xx3` matrix has 2 rows and 3 columns. In general,`a2xx3` matrix is given by `A=[{:(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)):}]_(3xx2)` Thus, `a_(ij)=(1)/(2)\|5i-3j\|" where "i=1,2" and "j=1,2,3.` `:." "a_(11)=(1)/(2)\|5xx1-3xx1\|=(1)/(2).\|2\|=(1)/(2)xx2=1,` `a_(12)=(1)/(2)\|5xx1-3xx2\|=(1)/(2).\|5-6\|=(1)/(2).\|-1\|=(1)/(2)xx1=(1)/(2),` `a_(13)=(1)/(2)\|5xx1-3xx3\|=(1)/(2).\|5-9\|=(1)/(2).\|-4\|=(1)/(2)xx4=2,` `a_(21)=(1)/(2)\|5xx2-3xx1\|=(1)/(2).\|10-3\|=(1)/(2).\|7\|=(1)/(2)xx7=(7)/(2),` `a_(22)=(1)/(2)\|5xx2-3xx2\|=(1)/(2).\|10-6\|=(1)/(2).\|4\|=(1)/(2)xx4=2,` `a_(23)=(1)/(2)\|5xx2-3xx3\|=(1)/(2).\|10-9\|=(1)/(2).\|1\|=(1)/(2)xx1=(1)/(2).` Hence, `A=[{:(1,(1)/(2),2),((7)/(2),2,(1)/(2)):}].`

Discussion

38.	Write the order of each of the following matrices:(i) A = [(3,5,4,-2),(0,√3,-1,4/9)](ii) B = [(6,-5),(1/2,3/4),(-2,-1)](iii) C = [7,-√2,5,0](iv) D = [8, -3](v) E = [(-2),(3),(0)](vi) F = [6]
Answer» (i) The order of matrix A is 2 × 4. (ii) The order of matrix B is 3 × 2. (iii) The order of matrix C is 1 × 4. (iv) The order of matrix D is 1 × 2. (v) The order of matrix E is 3 × 1. (vi) The order of matrix F is 1 × 1.

38.

Write the order of each of the following matrices:(i) A = [(3,5,4,-2),(0,√3,-1,4/9)](ii) B = [(6,-5),(1/2,3/4),(-2,-1)](iii) C = [7,-√2,5,0](iv) D = [8, -3](v) E = [(-2),(3),(0)](vi) F = [6]

Answer»

(i) The order of matrix A is 2 × 4.

(ii) The order of matrix B is 3 × 2.

(iii) The order of matrix C is 1 × 4.

(iv) The order of matrix D is 1 × 2.

(v) The order of matrix E is 3 × 1.

(vi) The order of matrix F is 1 × 1.

Discussion

39.	If `A = ({:(4, 3), (-2, 1):})`, then find `A + 10 A^(-1),`
Answer» Correct Answer - 5I

Discussion

40.	If `X=[{:( 3,1,-1) ,(5,-2,-3) :}] "and" Y= [{:( 2,1,-1),( 7,2,4):}]` then find (i) x+y, (ii) 2x-3y. (iii) a matrix Z such that `X+Y+Z` is a zero matrix.
Answer» We have , `X=[{:(3,1 ,-1) ,(5,-2,-3):}]_(2xx3) "and" =[{:(2,1,-1),(7,2,4):}]_(2xx3)` (i) `x+y=[{:(3+2,1+1,-1-1),(5+7,-2+2,-3+4):}]=[ {:(5,2,-2) ,(12,0,1):}]` `because 2x=2[{:( 3,1,-1),(5,-2,-3) ,(5,-2,-3):}]=[{:(6, 2,-2) ,(10,-4,-6):}]` and `3Y=3[{:(2,1,-1),(7,2,4):}] =[{:(6,3,-3),(21,6,12):}]` `therefore 2X-3Y=[{:(6,-6,2-3,-2+3),(10-21,,-4-6,-6-12):}]=[{:(0,-1,1),(-11,-10,- 18) :}]` (iii) `X+Y=[{:(3+2,1+1,-1-1),(5+7,-2+ 2,-3+4):}]=[{:(5,2,-2),(12,0,+1):}]` Also, `X+Y+Z=[{:(0,0,0),(0,0,0):}]` We see that Z is the additive inverse of (X+y) or negative or (X+Y) `Z=[{:(-5,- 2,2),(-12,0,-1):}] [because Z=-(X+Y)]`

Discussion

41.	Find the possible orders for matrices A and B if they have 18 and 19 elements respectively.
Answer» Correct Answer - The orders possible for A are ` 1 xx 18, 2 xx 9, 3 xx 6, 6 xx3, 9 xx 2, 18 xx 1` the orders possible for B are `1 xx 19 " and " 19 xx 1`

Discussion

42.	Find values of a and b if A = B, whereA = \(\begin{bmatrix} a + 4& 3b \\[0.3em] 8&-6 \end{bmatrix}\) and B = \(\begin{bmatrix} 2a + 2 & b^2 + 2 \\[0.3em] 8 & b^2 - 5b \end{bmatrix}\)
Answer» Given, matrix A = matrix B Then their corresponding elements are equal. So, we have a₁₁ = b₁₁; a + 4 = 2a + 2 ⇒ a = 2 a₁₂ = b₁₂; 3b = b² + 2 ⇒ b² – 3b + 2 = 0 ⇒ b = 1, 2 a₂₂ = b₂₂; -6 = b² – 5b ⇒ b² – 5b + 6 = 0 ⇒ b = 2, 3 Hence, a = 2 and b = 2 (common value)

42.

Find values of a and b if A = B, whereA = \(\begin{bmatrix} a + 4& 3b \\[0.3em] 8&-6 \end{bmatrix}\) and B = \(\begin{bmatrix} 2a + 2 & b^2 + 2 \\[0.3em] 8 & b^2 - 5b \end{bmatrix}\)

Answer»

Given, matrix A = matrix B

Then their corresponding elements are equal.

So, we have

a₁₁ = b₁₁; a + 4 = 2a + 2 ⇒ a = 2

a₁₂ = b₁₂; 3b = b² + 2 ⇒ b² – 3b + 2 = 0 ⇒ b = 1, 2

a₂₂ = b₂₂; -6 = b² – 5b ⇒ b² – 5b + 6 = 0 ⇒ b = 2, 3

Hence, a = 2 and b = 2 (common value)

Discussion

43.	If possible find the sum of the matrics A and B, where `A=[{:( sqrt(3),1),(2,3):}]"and" B=[{:(x,y,z) ,(a,b,c):}]`
Answer» We have, `A=[{:(sqrt(3),1),(2,3):}]_(2xx2)"and" B=[{:(x,y,x),(a,b, 6):}]_(2xx3)` Here A and B are of different orders, Also we know that the addition of two matrices A and B is possible only if order of both the matricles A and B should be same. Hence , the sum of matrices A and B is not possible.

Discussion

44.	If A and B are square matrices of the same order, then (A + B)(A – B) is equal to :A. A2 – B2 B. A2 – BA – AB – B2 C. A2 – B2 + BA – AB D. A2 – BA + B2 + AB
Answer» (C). A² – B² + BA – AB (A + B)(A – B) = A (A – B) + B (A – B) = A.A – A.B + B.A – B.B =A² – A.B + B.A – B.B = A² – AB + BA – BB Matrix multiplication does not have a commutative property i.e., A.B ≠ B.A Hence, Option (C) is the answer.

44.

If A and B are square matrices of the same order, then (A + B)(A – B) is equal to :A. A2 – B2 B. A2 – BA – AB – B2 C. A2 – B2 + BA – AB D. A2 – BA + B2 + AB

Answer»

(C). A² – B² + BA – AB

(A + B)(A – B) = A (A – B) + B (A – B)

= A.A – A.B + B.A – B.B

=A² – A.B + B.A – B.B

= A² – AB + BA – BB

Matrix multiplication does not have a commutative property

i.e.,

A.B ≠ B.A

Hence,

Option (C) is the answer.

Discussion

45.	If `A = [{:(3, 2),(1, 4):}], B = [{:(-5, 9), (3, 4):}]," and C" = [{:(-3, 6),(2, 1):}],` then find 2A + 3B - 4C.
Answer» Correct Answer - `[{:(3, 7),(3, 16):}]`

Discussion

46.	If `[{:(a+4,3b),(8,-6):}]=[{:(2a+2,b+2),(" "8,a-8b):}]`, then find the value of `(a-2b)`.
Answer» Comparing the corresponding elements of given equal matrices, we have `a+4=2a+2,3b=b+2` and `a-8b=-6.` From these equations, we get `a=2` and `b=1.` `:." "(a-2b)=(2-2xx1)=(2-2)=0.`

Discussion

47.	Find the values of a and b, if A=B,where `A=[{:(a+4,3b),(8,-6):}] "and" B=[{:(2a +2,b^(2)+2),(8,b^(2)-5b):}]`
Answer» We h ave `A=[{:( a+4,3b) ,( 8,-6):}]_(2xx2)"and" B=[{:(2a+2,b^(2)+2),( 8,b^(2)-5b):}]_(2xx2)` Also, A=B By equality of matrices we know that each element of A is equal to the corresponding element of B, that is `a_(ij)=b_(ij)` for all i and j. `a_(11)=b_(11)rArra+4=2a+2rArra=2` `a_(12)=b_(12)rArr3bj=b^(2)+2rArrb^(2)=3b-2` and `a_(22)=b_(22)rArr-6=b^(2)-5b [because b^(2)=3b-2]` `rArr -6=3b-2-5b` `rArr 2b=4rArrb=2` `therefore a=2` and b=2

Discussion

48.	If A and B are square matrices of the same order, explain, why in general (i) (A + B)2 ≠ A2 + 2AB + B2 (ii) (A – B)2 ≠ A2 – 2AB + B2 (iii) (A + B)(A – B) = A2 – B2
Answer» (i) Given that A and B are square matrices of the same order. We know, (A + B)² = (A + B)(A + B) ⇒ (A + B)² = A(A + B) + B(A + B) ∴ (A + B)² = A² + AB + BA + B² For the equation, (A + B)² = A²+ 2AB + B² to be valid, we need AB = BA. As the multiplication of two matrices does not satisfy the commutative property in general, AB ≠ BA. Thus, (A + B)² ≠ A² + 2AB + B². (ii) Given that A and B are square matrices of the same order. We know, (A – B)² = (A – B)(A – B) ⇒ (A – B)² = A(A – B) – B(A – B) ∴ (A – B)² = A² – AB – BA + B² For the equation, (A – B)² = A² – 2AB + B² to be valid, we need AB = BA. As the multiplication of two matrices does not satisfy the commutative property in general, AB ≠ BA. Thus, (A – B)² ≠ A² – 2AB + B². (iii) Given that A and B are square matrices of the same order. We have, (A + B)(A – B) = A(A – B) + B(A – B) ∴ (A + B)(A – B) = A² – AB + BA – B² For the equation, (A + B)(A – B) = A² – B² to be valid, we need AB = BA. As the multiplication of two matrices does not satisfy the commutative property in general, AB ≠ BA. Thus, (A + B)(A – B) ≠ A² – B².

48.

If A and B are square matrices of the same order, explain, why in general (i) (A + B)2 ≠ A2 + 2AB + B2 (ii) (A – B)2 ≠ A2 – 2AB + B2 (iii) (A + B)(A – B) = A2 – B2

Answer»

(i) Given that A and B are square matrices of the same order.

We know,

(A + B)² = (A + B)(A + B)

⇒ (A + B)² = A(A + B) + B(A + B)

∴ (A + B)² = A² + AB + BA + B²

For the equation,

(A + B)² = A²+ 2AB + B² to be valid, we need

AB = BA.

As the multiplication of two matrices does not satisfy the commutative property in general,

AB ≠ BA.

Thus,

(A + B)² ≠ A² + 2AB + B².

(ii) Given that A and B are square matrices of the same order.

We know,

(A – B)² = (A – B)(A – B)

⇒ (A – B)² = A(A – B) – B(A – B)

∴ (A – B)² = A² – AB – BA + B²

For the equation,

(A – B)² = A² – 2AB + B² to be valid, we need

AB = BA.

As the multiplication of two matrices does not satisfy the commutative property in general,

AB ≠ BA.

Thus,

(A – B)² ≠ A² – 2AB + B².

(iii) Given that A and B are square matrices of the same order.

We have,

(A + B)(A – B) = A(A – B) + B(A – B)

∴ (A + B)(A – B) = A² – AB + BA – B²

For the equation,

(A + B)(A – B) = A² – B² to be valid, we need

AB = BA.

As the multiplication of two matrices does not satisfy the commutative property in general,

AB ≠ BA.

Thus,

(A + B)(A – B) ≠ A² – B².

Discussion

49.	The monthly incomes of Aryan and Babban are in the ration 3: 4 and their monthly expenditures are in the ratio 5: 7. If each saves 15000 per month, find their monthly incomes using the matrix method. This problem reflects which value?
Answer» Let us represent the situation through a matrix. We will make two matrices: Income and Expenditure Matrices. We know that Saving = Income – Expenditure. Let the incomes of Aryan and Babban be 3x and 4x respectively and the expenditures be 5y and 7y respectively. Income Matrix = \(\begin{bmatrix} 3x \\[0.3em] 4x\end{bmatrix}\) Expenditure Matrix = \(\begin{bmatrix} 5y \\[0.3em] 7y\end{bmatrix}\) Now, Saving = \(\begin{bmatrix} 3x \\[0.3em] 4x\end{bmatrix}\)- \(\begin{bmatrix} 5y \\[0.3em] 7y\end{bmatrix}\) Given : Saving = 15000 each Therefore, we have, \(\begin{bmatrix} 15000 \\[0.3em] 15000\end{bmatrix}\)= \(\begin{bmatrix} 3x \\[0.3em] 4x\end{bmatrix}\)- \(\begin{bmatrix} 5y \\[0.3em] 7y\end{bmatrix}\) So, 3 x – 5 y = 15000 ….(1) 4 x – 7 y = 15000 …..(2) Solving equations 1 and 2, we get, Multiplying eq(1) by 4 and eq(2) by 3 we get, 12 x – 20 y = 60000 ….(3) 12 x – 21 y = 45000 …..(4) Eq(3) – Eq(4), Y = 15000 Putting this value in eq(1) we get, 3 x – 4 × 15000 = 15000 X = 25000. There monthly incomes are, 3 x = 3 × 15000 = 45000 and 4 x = 4 × 15000 = 60000.

49.

The monthly incomes of Aryan and Babban are in the ration 3: 4 and their monthly expenditures are in the ratio 5: 7. If each saves 15000 per month, find their monthly incomes using the matrix method. This problem reflects which value?

Answer»

Let us represent the situation through a matrix.

We will make two matrices: Income and Expenditure Matrices.

We know that Saving = Income – Expenditure.

Let the incomes of Aryan and Babban be 3x and 4x respectively and the expenditures be 5y and 7y respectively.

Income Matrix = \(\begin{bmatrix} 3x \\[0.3em] 4x\end{bmatrix}\)

Expenditure Matrix = \(\begin{bmatrix} 5y \\[0.3em] 7y\end{bmatrix}\)

Now,

Saving = \(\begin{bmatrix} 3x \\[0.3em] 4x\end{bmatrix}\)- \(\begin{bmatrix} 5y \\[0.3em] 7y\end{bmatrix}\)

Given :

Saving = 15000 each

Therefore, we have,

\(\begin{bmatrix} 15000 \\[0.3em] 15000\end{bmatrix}\)= \(\begin{bmatrix} 3x \\[0.3em] 4x\end{bmatrix}\)- \(\begin{bmatrix} 5y \\[0.3em] 7y\end{bmatrix}\)

So,

3 x – 5 y = 15000 ….(1)

4 x – 7 y = 15000 …..(2)

Solving equations 1 and 2, we get,

Multiplying eq(1) by 4 and eq(2) by 3 we get,

12 x – 20 y = 60000 ….(3)

12 x – 21 y = 45000 …..(4)

Eq(3) – Eq(4),

Y = 15000

Putting this value in eq(1) we get,

3 x – 4 × 15000 = 15000

X = 25000.

There monthly incomes are,

3 x = 3 × 15000 = 45000 and

4 x = 4 × 15000 = 60000.

Discussion

50.	If `A={:[(3,2,7),(1,1,4),(-1,-1,0)]:},B={:[(1,0,3),(2,1,0),(0,-1,-3)]:}andC{:[(1,0,0),(0,1,0),(0,0,1)]:}`, then find 2A+3B-7C.A. `{:[(2,4,23),(8,0,8),(-2,-5,16)]:}`B. `{:[(2,4,23),(8,-2,8),(-2,-5,-16)]:}`C. `{:[(2,4,23),(8,-2,-8),(-2,5,-16)]:}`D. `{:[(2,4,2-3),(8,-2,8),(-2,-5,16)]:}`
Answer» Correct Answer - D (i) Evalute 2A, 3B, 7C and simplify. (ii) 2A means multiply each element of A with 2. (iii) 3B means multiply each element of B with 3. (iv) Similarly find 7C, then find 2A+3B-7C.

Discussion

Explore topic-wise InterviewSolutions in .

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If A is a square matrix of order 3 with |AI= 2, then the value of `|(A-A^T)^5|+|(A-A^T)^3|` is

If `A=[{:(cosx,sinx,0),(-sinx,cosx,0),(0,0,1):}]` =f(x), then `A^(-1)` is equal toA. `f(-x)`B. `f(x)`C. `-f(x)`D. `-f(-x)`

`F(x)=[[cosx,-sinx,0],[sinx,cosx,0],[0,0,1]]` and `G(x)=[[cosx,0,sinx],[0,1,0],[-sinx,0,cosx]]`, then `[F(x)G(y)]^(-1)` is equal to (A) `F(-x)G(-y)` (B) `F(x-1)G(y-1)` (C) `G(-y)F(-x)` (D) `G(y^(-1))F(x^(-1))`A. `F(-x)G(-y)`B. `F(x^(-1))_G(y^(-1))`C. `G(-y)F(-x)`D. `G(y^(-1))F(x^(-1))`

If `P(x)=[(cosx, sinx),(-sinx, cosx)]`, then show that `P(x).P(y)=P(x+y)=P(y).P(x)`.

Let P and Q be `3xx3` matrices with `P!=Q` . If `P^3=""Q^3a nd""P^2Q""=""Q^2P`, then determinant of `(P^2+""Q^2)` is equal to (1) 2(2) 1 (3)0 (4) 1

LetA be a `2xx2`matrix with real entries. Let I be the `2xx2`identity matrix. Denote by tr (A), the sumof diagonal entries of A. Assume that `A^2=""I`.Statement1: If `A!=I`and `A!=""-I`, then det `A""=-1`.Statement2: If `A!=I`and `A!=""-I`,then `t r(A)!=0`.

If f(x) = `[(cos x , - sinx,0),(sinx,cosx,0),(0,0,1)]` then show f(x) . f(y) = f(x+y)

If \(\begin{bmatrix}2x&amp;3\\[0.3em]0&amp;y-1\end{bmatrix} \) = \(\begin{bmatrix}x-3&amp;3\\[0.3em]0&amp;2\end{bmatrix},\) then the values of x and y respectively are :(a) 3, –3 (b) –3, 3 (c) –3, –3 (d) 3, 3

A=[aij] mxn is a square matrix, if(A) m &lt; n (B) m &gt; n (C) m = n (D) None of these

If A = [aij]m x n' B = [bij]m x n' then the element C23 of the matrix C=A+B is:(a) a13 + b13 (b) a23 + b32 (c) a23 + b23 (d) a32 + b23

If A is a non-singular square matrix such that `A^2-7A+5I=0`, then `A^-1`A. `7A -I`B. `(7)/(5)I - (1)/(5)A`C. `(7)/(5)I + (1)/(5)A`D. `(A)/(5) - (7)/(5)`

Construct `A_(2xx3)` matrix whose elements are given by `a_(ij)=(1)/(2)|5i-3j|.`

Write the order of each of the following matrices:(i) A = [(3,5,4,-2),(0,√3,-1,4/9)](ii) B = [(6,-5),(1/2,3/4),(-2,-1)](iii) C = [7,-√2,5,0](iv) D = [8, -3](v) E = [(-2),(3),(0)](vi) F = [6]

If `A = ({:(4, 3), (-2, 1):})`, then find `A + 10 A^(-1),`

If `X=[{:( 3,1,-1) ,(5,-2,-3) :}] "and" Y= [{:( 2,1,-1),( 7,2,4):}]` then find (i) x+y, (ii) 2x-3y. (iii) a matrix Z such that `X+Y+Z` is a zero matrix.

Find the possible orders for matrices A and B if they have 18 and 19 elements respectively.

Find values of a and b if A = B, whereA = \(\begin{bmatrix} a + 4&amp; 3b \\[0.3em] 8&amp;-6 \end{bmatrix}\) and B = \(\begin{bmatrix} 2a + 2 &amp; b^2 + 2 \\[0.3em] 8 &amp; b^2 - 5b \end{bmatrix}\)

If possible find the sum of the matrics A and B, where `A=[{:( sqrt(3),1),(2,3):}]"and" B=[{:(x,y,z) ,(a,b,c):}]`

If A and B are square matrices of the same order, then (A + B)(A – B) is equal to :A. A2 – B2 B. A2 – BA – AB – B2 C. A2 – B2 + BA – AB D. A2 – BA + B2 + AB

If `A = [{:(3, 2),(1, 4):}], B = [{:(-5, 9), (3, 4):}]," and C" = [{:(-3, 6),(2, 1):}],` then find 2A + 3B - 4C.

If `[{:(a+4,3b),(8,-6):}]=[{:(2a+2,b+2),(" "8,a-8b):}]`, then find the value of `(a-2b)`.

Find the values of a and b, if A=B,where `A=[{:(a+4,3b),(8,-6):}] "and" B=[{:(2a +2,b^(2)+2),(8,b^(2)-5b):}]`

If A and B are square matrices of the same order, explain, why in general (i) (A + B)2 ≠ A2 + 2AB + B2 (ii) (A – B)2 ≠ A2 – 2AB + B2 (iii) (A + B)(A – B) = A2 – B2

The monthly incomes of Aryan and Babban are in the ration 3: 4 and their monthly expenditures are in the ratio 5: 7. If each saves 15000 per month, find their monthly incomes using the matrix method. This problem reflects which value?

If `A={:[(3,2,7),(1,1,4),(-1,-1,0)]:},B={:[(1,0,3),(2,1,0),(0,-1,-3)]:}andC{:[(1,0,0),(0,1,0),(0,0,1)]:}`, then find 2A+3B-7C.A. `{:[(2,4,23),(8,0,8),(-2,-5,16)]:}`B. `{:[(2,4,23),(8,-2,8),(-2,-5,-16)]:}`C. `{:[(2,4,23),(8,-2,-8),(-2,5,-16)]:}`D. `{:[(2,4,2-3),(8,-2,8),(-2,-5,16)]:}`

In the matrix\(A = \begin{bmatrix}2 & 5 & 19&-7 \\[0.3em]35 & -2 & \frac{5}{2}&12 \\[0.3em]\sqrt3 &1&-5 & 17\end{bmatrix}\)(i) The order of the matrix, (ii) The number of elements, (iii) Write the elements a13, a21, a33, a24 ,a23,

If \(\begin{bmatrix}2x&3\\[0.3em]0&y-1\end{bmatrix} \) = \(\begin{bmatrix}x-3&3\\[0.3em]0&2\end{bmatrix},\) then the values of x and y respectively are :(a) 3, –3 (b) –3, 3 (c) –3, –3 (d) 3, 3

A=[aij] mxn is a square matrix, if(A) m < n (B) m > n (C) m = n (D) None of these

Find values of a and b if A = B, whereA = \(\begin{bmatrix} a + 4& 3b \\[0.3em] 8&-6 \end{bmatrix}\) and B = \(\begin{bmatrix} 2a + 2 & b^2 + 2 \\[0.3em] 8 & b^2 - 5b \end{bmatrix}\)