1192 + Interview Questions in MATRICES IN GENERAL AWARENESS Page 6 InterviewSolution

251.	Find X and Y, if `X+Y=[[5, 2],[ 0, 9]]`and `X-Y=[[3 ,6],[ 0,-1]]`.
Answer» Adding the given matrices, we get `(X+Y)+(X-Y)=[{:(5,2),(0,9):}]+[{:(3," "6),(0,-1):}]` `implies" "2X=[{:(5+3,2+6),(0+0,9+(-1)):}]` `implies" "2X=[{:(8,8),(0,8):}]impliesX=(1)/(2).[{:(8,8),(0,8):}]=[{:(4,4),(0,4):}].` On subtracting the given matrices, we get `(X+Y)-(X-Y)=[{:(5,2),(0,9):}]-[{:(3,6),(0,-1):}]` `implies" "2Y=[{:(5-3,2-6),(0-0,9-(-1)):}]=[{:(2,-4),(0,10):}]` `implies" "Y=(1)/(2)[{:(2,-4),(0,10):}]=[{:(1,-2),(0," "5):}].` Hence, `X=[{:(4,4),(0,4):}]" and "Y=[{:(1,-2),(0,5):}].`

Discussion

252.	If `2[{:(3,4),(5,x):}]+[{:(1,y),(0,1):}]=[{:(7,0),(10,5):}]`, find (x-y).
Answer» We have `2[{:(3,4),(5,x):}]+[{:(1,y),(0,1):}]=[{:(7,0),(10,5):}]` `implies" "[{:(6,8),(10,2x):}]+[{:(1,y),(0,1):}]=[{:(7,0),(10,5):}]` `implies" "[{:(6+1,8+y),(10+0,2x+1):}]=[{:(7,0),(10,5):}]` `implies" "2x+1=5" and "8+y=0` [comparing the corresponding elements] `implies" "x=2" and "y=-8.` `implies" "(x-y)=2-(-8)=10.`

Discussion

253.	Find a matrix X such that 2A + B + X =0where` A = [[-1,2],[3,4]] and B = [[3,-2],[1,5]]`
Answer» We have `2A+B+X=OimpliesX=-(2A+B).` Now, `(2A+B)=2.[{:(-1,2),(3,4):}]+[{:(3,-2),(1,5):}]` `=[{:(-2,4),(6,8):}]+[{:(3,-2),(1," "5):}]` `=[{:(-2+3,4+(-2)),(" "6+1," "8+5):}]=[{:(1,2),(7,13):}].` `:." "X=-(2A+B)=[{:(-1,-2),(-7,-13):}].`

Discussion

254.	Find a martix X, if `X+[{:(4,6),(-3,7):}]=[{:(3,-6),(5,-8):}].`
Answer» Let `A=[{:(4,6),(-3,7):}]" and "B=[{:(3,-6),(5,-8):}].` Then, the given matrix equation is `X+A=B.` `:." "X+A=BimpliesX=B+(-A)` `=[{:(3,-6),(5,-8):}]+[{:(-4,-6),(3,-7):}]` `=[{:(3+(-4),-6+(-6)),(5+3,-8+(-7)):}]=[{:(-1,-12),(8,-15):}].` Hence, `X=[{:(-1,-12),(8,-15):}].`

Discussion

255.	If A=[0 3 -2 5] then find 'k' so that kA2=5A-6I2
Answer» A \(=\begin{bmatrix}0 &3\\-2&5\end{bmatrix}\) A² \(=\begin{bmatrix}0 &3\\-2&5\end{bmatrix}\)\(\begin{bmatrix}0 &3\\-2&5\end{bmatrix}\) \(=\begin{bmatrix}-6 & 15\\-10 & 19\end{bmatrix}\) Now, 5A – 6I₂ \(=5\begin{bmatrix}0 & 3\\-2 & 5\end{bmatrix}-6\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\) \(=\begin{bmatrix}0 & 15\\-10 & 25\end{bmatrix}\)\(-\begin{bmatrix}6& 0\\0 & 6\end{bmatrix}\) \(=\begin{bmatrix}-6 & 15\\-10 & 19\end{bmatrix}\) = A² Thus k = 1.

255.

If A=[0 3 -2 5] then find 'k' so that kA2=5A-6I2

Answer»

A \(=\begin{bmatrix}0 &3\\-2&5\end{bmatrix}\)

A² \(=\begin{bmatrix}0 &3\\-2&5\end{bmatrix}\)\(\begin{bmatrix}0 &3\\-2&5\end{bmatrix}\)

\(=\begin{bmatrix}-6 & 15\\-10 & 19\end{bmatrix}\)

Now, 5A – 6I₂ \(=5\begin{bmatrix}0 & 3\\-2 & 5\end{bmatrix}-6\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\)

\(=\begin{bmatrix}0 & 15\\-10 & 25\end{bmatrix}\)\(-\begin{bmatrix}6& 0\\0 & 6\end{bmatrix}\)

\(=\begin{bmatrix}-6 & 15\\-10 & 19\end{bmatrix}\)

= A²

Thus k = 1.

Discussion

256.	Using elementary transformation, find the inverse of the matrix `A=[(a,b),(c,((1+bc)/a))]`.
Answer» Correct Answer - `[((1+bc)/a,-b),(-c,a)]` `A=[(a,b),(c,((1+bc)/a))]` We write, `[(a,b),(c,((1+bc)/a))]=[(1,0),(0,1)]A` `implies [(1,b/a),(c,((1+bc)/a))]=[(1/a,0),(0,1)]A" "(R_(1) rarr R_(1)/a)` or `[(1,b/a),(0,1/a)]=[(1/a,0),((-c)/a,1)]A" "(R_(2) rarr R_(2)-cR_(1))` or `[(1,b/a),(0,1)]=[(1/a,0),(-c,a)]A" "(R_(2)=aR_(2))` or `[(1,0),(0,1)]=[((1+bc)/a,-b),(-c,a)]A" "(R_(1) rarr R_(1)- b/a R_(2))` `implies A^(-1)=[((1+bc)/a,-b),(-c,a)]`

Discussion

257.	If a matrix has 5 elements, then write all possible orders it can have.
Answer» Given, a matrix has 5 elements. So, possible order of this matrix are 5 x 1 and 1 x 5.

Discussion

258.	Write the values of x – y + z from the following equation: [(x + y + z), (x + y), (y + z)] = [9, 5, 7]
Answer» By definition of equality of matrices, we have x + y + z = 9 ... (i) x + z =5 .... (ii) y + z =7 ... (iii) (i) – (ii) ⇒ x + y + z – x – z = 9 – 5 ⇒ y = 4 ... (iv) (ii) – (iv) ⇒ x – y + z = 5 – 4 ⇒x – y + z = 1

258.

Write the values of x – y + z from the following equation: [(x + y + z), (x + y), (y + z)] = [9, 5, 7]

Answer»

By definition of equality of matrices, we have

x + y + z = 9 ... (i)

x + z =5 .... (ii)

y + z =7 ... (iii)

(i) – (ii) ⇒ x + y + z – x – z = 9 – 5

⇒ y = 4 ... (iv)

(ii) – (iv) ⇒ x – y + z = 5 – 4

⇒x – y + z = 1

Discussion

259.	If A is a square matrix such that A2 = A, then write the value of (I + A)2 – 3A.
Answer» (I + A)² – 3A = I² + A² + 2A – 3A = I² + A² - A = I² + A - A [∵ A² = A] = I² = I . I = I

259.

If A is a square matrix such that A2 = A, then write the value of (I + A)2 – 3A.

Answer»

(I + A)² – 3A = I² + A² + 2A – 3A

= I² + A² - A

= I² + A - A [∵ A² = A]

= I² = I . I = I

Discussion

260.	If A is a square matrix such that A2 = A, then write the value of 7A – (I + A)3, where I is an identity matrix.
Answer» 7A – (I + A)³ = 7A - {I³ + 3I²A + 3I.A² + A³} = 7A – {I + 3A + 3A + A²A} [∵ I³ = I² = I, A² = A] = 7A – {I + 6A + A²} = 7A – {I + 6A + A} = 7A – {I + 7A} = 7A – I – 7A = –I

260.

If A is a square matrix such that A2 = A, then write the value of 7A – (I + A)3, where I is an identity matrix.

Answer»

7A – (I + A)³ = 7A - {I³ + 3I²A + 3I.A² + A³}

= 7A – {I + 3A + 3A + A²A} [∵ I³ = I² = I, A² = A]

= 7A – {I + 6A + A²}

= 7A – {I + 6A + A} = 7A – {I + 7A}

= 7A – I – 7A = –I

Discussion

261.	_________ matrix is both symmetric and skew symmetric matrix.
Answer» Null matrix is both symmetric and skew symmetric matrix.

Discussion

262.	Write the values of x - y + z from the following equation :[(x+y+z,x+z,y+z)]=[(9,5,7)]
Answer» we have [(x+y+z,x+z,y+z)]=[(9,5,7)] By definition of equality of matrices, we have x + y + z = 9 ... (i) x + z =5 .... (ii) y + z =7 ... (iii) (i) – (ii) => x + y + z -x - z = 9 -5 => y = 4 ... (iv) (ii) – (iv) => x - y + z =5 -4 => x -y + z =1

262.

Write the values of x - y + z from the following equation :[(x+y+z,x+z,y+z)]=[(9,5,7)]

Answer»

we have

[(x+y+z,x+z,y+z)]=[(9,5,7)]

By definition of equality of matrices, we have
x + y + z = 9 ... (i)
x + z =5 .... (ii)
y + z =7 ... (iii)
(i) – (ii) => x + y + z -x - z = 9 -5
=> y = 4 ... (iv)
(ii) – (iv) => x - y + z =5 -4
=> x -y + z =1

Discussion

263.	Write the order of the product matrix :[(1),(2),(3)] [2,3,4]
Answer» Order is 3 x 3 because it is product of two matrices having order 3 x 1 and 1 x 3.

Discussion

264.	If [(1,0,0)(0,y,0)(0,0,1)] [(x)(-1)(z)] = [(1)(2)(1)], find x + y + z.
Answer» x = 1, y = -2, z = 1 or x + y + z = 0

264.

If [(1,0,0)(0,y,0)(0,0,1)] [(x)(-1)(z)] = [(1)(2)(1)], find x + y + z.

Answer»

x = 1, y = -2, z = 1

or x + y + z = 0

Discussion

265.	Find the matrix A such that `[{:(2,-1), (1,0),(-3,4):}]A=[{:(-1,-8,-10),(1,-2,-5),(9,22,15):}]`
Answer» we have, `[{:(2,-1),(1,0),(-3,4):}]_(3xx2)A=[{:(-1,-8,-10),(1,-2,-5),(9,22,15):}]_(3xx3)` From the given equation , it is clear that order of A should be `2xx3` Let `A=[{:(a,b,c),(d,e,f):}]` `[{:(2,-1),(1,0),(-3,4):}][{:(a,b,c),(d,e,f):}]=[{:(-1,-8,-10),(1,-2,-5),(9,22,15):}]` `rArr[{:(2a-d,2b-e,2c-f),(a+0d,b+0.e,c+0.f),(-3a+4d,3b+4e,-3c+4l):}]=[{:(-1,-8,-10),(1,-2,-5),( 9,22,15):}]` `rArr [{:(2a-d,2b-e,2c-f),(a,b,c),(-3a+4d,-3b+4e,-3c+4f):}]=[{:(-1,-8,-10),(1,-2,-5),(9,22,15):}]` By equality of matrices, we get `a=1,b=-2,c=-5` and `2a-d=-1rArrd=2a+1=3` `rArr 2b-e=-8rArre=2(-2)+8=4` 2c-f=-10`rArr`f=2c+10=0 `therefore A=[{:(1,-2,-5),(3,4,0):}]`

Discussion

266.	If `theta-phi=pi/2,` prove that, `[(cos^2 theta,cos theta sin theta),(cos theta sin theta,sin^2 theta)] [(cos^2 phi,cos phi sin phi),(cos phi sin phi,sin^2 phi)]=0`A. `2npi, in Z`B. `n pi/2, n in Z`C. `(2n+1) pi/2, n in X`D. `npi, n in Z`
Answer» Correct Answer - C `AB=[(cos^(2) theta,),(cos theta sin theta,)][(cos^(2) phi,cos phi sin phi),(cos phi sin phi,sin^(2) phi)]` `=[(cos^(2) theta cos^(2) phi+cos theta cos phi sin theta sin phi ,cos^(2)theta cos phi sin phi+cos theta sin theta sin^(2) phi),(cos theta sin thete cos^(2) phi+sin^(2) theta cos phi sin phi,cos theta cos phi sin theta sin phi+sin^(2) theta sin^(2) phi)]` `=[(cos theta cos phi (cos (theta-phi)),cos theta sin phi (cos(theta-phi))),(sin theta cos phi (cos (theta-phi)),sin theta sin phi (cos (theta-phi)))]` `=(cos (theta-phi)) [(cos theta cos phi,cos theta sin phi),(sin theta cos phi,sin theta sin phi)]` Now, `AB=O implies cos (theta-phi)=0` `implies theta-phi=(2n+1) pi//2, n in Z`.

Discussion

267.	Solve for x and y, given that `[{:(x,y),(3y,x):}][{:(1),(2):}]=[{:(3),(5):}].`
Answer» We have `[{:(x,y),(3y,x):}][{:(1),(2):}]=[{:(3),(5):}]` `implies" "[{:(x+2y),(3y+2x):}]=[{:(3),(5):}]` `implies" "{{:(x+2y=3" "...(i)),(2x+3y=5" "...(ii)):}` Multiplying (i) by 2 and subtracting (ii)from it, we get y=1. Putting y=1 in (i), we get x=1. Hence, x=1 and y=1.

Discussion

268.	Find the value of x, if `[1x1][{:(1,3,2),(2,5,1),(15,3,2):}][{:(1),(2),(x):}]=O.`
Answer» We have `[1x1]_(1xx3)[{:(1,3,2),(2,5,1),(15,3,2):}]_(3xx3)[{:(1),(2),(x):}]_(3xx1)=O` `implies" "[1+2x+15" "3+5x+3" "2+x+2][{:(1),(2),(x):}]=O` `implies" "[16+2x" "6+5x" "4+x][{:(1),(2),(x):}]=O` `implies" "[(16+2x).1+(6+5x).2+(4+x).x]=O` `implies" "(16+2x)+(12+10x)+(4x+x^(2))=0` `implies" "x^(2)+16x+28=0` `implies" "(x+14)(x+2)=0` `implies" "x+14=0" or "x+2=0` `implies" "x=-14" or "x=-2.` Hence, `x=-14" or "x=-2.`

Discussion

269.	If `A^(2) - A + I = 0` then `A^(-1) ` is equal toA. `A^(-2)`B. `A+1`C. `I-A`D. `A-I`
Answer» Correct Answer - c

Discussion

270.	If `A[{:(3,-4),(1,1),(2,0):}]` and `B=[{:(2,1,2),(1,2,4):}]` and `B=[{:(4,1),(2,3),(1,2):}]`
Answer» We have, `A=[{:(2,1,2),(1,2,4):}]_(2xx3) "and" B=[{:(4,1),(2,3),(1,2):}]_(3xx2)` So, AB and BA both are possible [since, in both A-B and B-A, the number of columns of first is equal to the number of rows of second] `therefore AB=[{:(2,1,2),(1,2,4):}]_(2xx3)[{:(4,1),(2,3),(1,2):}]_(3xx2)` `=[{:(8+2+2,2+3+4),(4+4+4,1+ 6+8):}]=[{:(12,9),(12,15):}]` and `BA=[{:(4,1),(2,3),(1,2):}]_(3xx2)[{:(2,1,2),(1,2,4):}]_(2xx3)` `= [{:(4xx2+1,4+2,8+4),(4+3,2+ 6,4+12),(2+2,1+4,2+8):}]=[{:(9,6,12),(7,8,16),(4,5,10):}]`

Discussion

271.	If `A=[a b0a]`is nth root of `I_2,`then choose the correct statements:If `n`is odd, `a=1,b=0`If `n`is odd, `a=-1,b=0`If `n`is even, `a=1,b=0`If `n`is even, `a=-1,b=0`a. i, ii, iii, iv b.ii, iii, ivc. i, ii, iii, iv d. i,iii, ivA. i, ii, iiiB. ii, iii, ivC. i, ii, iii, ivD. i, iii, iv
Answer» Correct Answer - D If A is nth root of `I_(2)`, then `A^(n)=I_(2)`. Now, `A^(2)=[(a,b),(0,a)][(a,b),(0,a)]=[(a^(2),2ab),(0,a^(2))]` `A^(3)=A^(2) A=[(a^(2),2ab),(0,a^(2))][(a,b),(0,a)]=[(a^(3),3a^(2)b),(0,a^(3))]` Thus, `A^(n)=[(n^(n),na^(n-1)b),(0,a^(n))]` Now, `A^(n)=I implies [(a^(n),na^(n-1)b),(0,a^(n))]=[(1,0),(0,1)]` `implies a^(n)=1, b=0`

Discussion

272.	If for the matrix `A , A^3=I`, then `A^(-1)=``A^2`(b) `A^3`(c) `A`(d) none of theseA. `A^(2)`B. `A^(3)`C. AD. None of these
Answer» Correct Answer - a

Discussion

273.	If `[alphabetagamma-alpha]`is to be square root of two-rowed unit matrix, then `alpha,betaa n dgamma`should satisfy therelation.`1-alpha^2+betagamma=0`b. `alpha^2+betagamma=0`c. `1+alpha^2+betagamma=0`d. `1-alpha^2-betagamma=0`A. `1-alpha^(2)+beta gamma=0`B. `alpha^(2)+beta gamma-1=0`C. `1+alpha^(2)+beta gamma=0`D. `1-alpha^(2)-beta gamma=0`
Answer» Correct Answer - B We have `[(alpha,beta),(gamma,-alpha)][(alpha,beta),(gamma,-alpha)]=[(1,0),(0,1)]` or `[(alpha^(2)+beta gamma,0),(0,alpha^(2)+beta gamma)]=[(1,0),(0,1)]` or `alpha^(2)+beta gamma-1=0`

Discussion

274.	If `A=[i-i-i i]a n dB=[1-1-1 1],t h e nA^8`equals`4B`b. `128 B`c. `-128 B`d. `-64 B`A. 4BB. 128BC. `-128` BD. `-64`B
Answer» Correct Answer - B We have, `A=iB` or `A^(2)=(iB)^(2)=i^(2)B^(2)=-B^(2)=- [(2,-2),(-2,2)]=-2B` or `A^(4)=(-2B)^(2)=4B^(2)=4(2B)=8B` or `(A^(4))^(2)=(8B)^(2)` or `A^(8)=64 B^(2)=128 B`

Discussion

275.	Prove that matrix `[((b^(2)-a^(2))/(a^(2)+b^(2)),(-2ab)/(a^(2)+b^(2))),((-2ab)/(a^(2)+b^(2)),(a^(2)-b^(2))/(a^(2)+b^(2)))]` is orthogonal.
Answer» `A=[((b^(2)-a^(2))/(a^(2)+b^(2)),(-2ab)/(a^(2)+b^(2))),((-2ab)/(a^(2)+b^(2)),(a^(2)-b^(2))/(a^(2)+b^(2)))]=[((1-a^(2)/b^(2))/(1+a^(2)/b^(2)),(-2 a/b)/(1+a^(2)/b^(2))),((-2a/b)/(1+a^(2)/b^(2)),-(1-a^(2)/(2))/(1+a^(2)/b^(2)))]` `=[(cos 2 theta,-sin 2 theta),(-sin 2 theta,-cos 2 theta)]`, where `a/b= tan theta` `:. A A^(T)=[(cos 2 theta,-sin 2 theta),(-sin 2 theta,-cos 2 theta)][(cos 2 theta,-sin 2 theta),(- sin 2 theta,-cos 2 theta)]` `=[(1,0),(0,1)]` Thus, A is orthogonal.

Discussion

276.	If `e^(A)` is defined as `e^(A)=I+A+A^(2)/(2!)+A^(3)/(3!)+...=1/2 [(f(x),g(x)),(g(x),f(x))]`, where `A=[(x,x),(x,x)], 0 lt x lt 1` and I is identity matrix, then find the functions f(x) and g(x).
Answer» `A=[(x,x),(x,x)]` `implies A^(2)=[(2x^(2),2x^(2)),(2x^(2),2x^(2))], A^(3)=[(2^(2)x^(3),2^(2)x^(3)),(2^(2)x^(3),2^(2)x^(3))]` and so on Then `e^(A)=I+A+A^(2)/(2!)+A^(3)/(3!)+...` `=[(1+x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+...,x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+...),(x+(2x^(2))/(2!)+(2^(2)x^(3))/(3!)+...,1+x+ (2x^(2))/(2!)+(2^(2)x^(3))/(3!)+...)]` `=[(1/2(1+2x+(2^(2)x^(2))/(2!)+...)+1/2,1/2(1+2x+(2^(2)x^(2))/(2!)+...)-1/2),(1/2(1+2x+(2^(2)x^(2))/(2!)+...)-1/2,1/2(1+2x+(2^(2)x^(2))/(2!)+...)+1/2)]` `=1/2 [(e^(2x)+1,e^(2x)-1),(e^(2x)-1,e^(2x)+1)]` `implies f(x)=e^(2x)+1` and `g(x)=e^(2x)-1`

Discussion

277.	FindA, if `[{:(4),(1),(3):}]A=[{:(-4,8,4),(-1,2,1),(-3,6,3):}]`
Answer» We have, `[{:(4),(1),(3):}]_(3xx1)A=[{:(-4,8,4),(-1,2,1),(-3,6,3):}]_(3xx3)` `rArr [{:(4x,4y,4z),(x,y,z),(3x,3y,3z):}]=[{:(-4,8,4),(-1,2,1),(-3,6,3):}]` lt

Discussion

278.	If `I_(3)` is identity matrix of order 3, then `I_(3)^(-1)=`
Answer» Correct Answer - c

Discussion

279.	`if[{:(2,1),(3,2):}]A[{:(-3,2),(5,-3):}]=[{:(1,0),(0,1):}],"then" A=?`
Answer» We have, `[{:(2,1),(3,2):}]_(2xx2)A.[{:(-3,2),(5,-3):}]_(2xx2)=[{:(1,0),(0,1):}]_(2xx2)` Let `A=[{:(a,b),(c,d):}]_(2xx2)` `therefore [{:(2,1),(3,2):}] [{:(a,b),(c,d):}][{:(-3,2),(5,-3):}]=[{:(1,0),(0,1):}]` `rArr [{:( 2a+c,2b+d),(3a+2c,3b+2d):}][{:(-3,2),(5,-3):}]=[{:( 1,0),(0,1):}]` `rArr [{:(-6a-3c_10b+5d,a+2c-6b-3d),(-9a-6c+15b+10d,6a+4c-9b-6d):}]=[{:(1,0),(0,1):}]` `rArr -6a-3c+10b+5d=1` `rArr 4a+2c-6b-3d=0` `rArr -9a-6c+15b+10d=0` `rArr 6a+4c-9b-6d=1` On edding Eqs. i and iv, we get `c+b-d=2rArrd=c+b-2` On adding Eqs. v and vii `v+b-2=1-arArra+b+c=3` `rArr a=3-b-c` Now, using values of a and d in eq. ii, we get `4(3-b-c)+2c-6b-3(b+c-2)=0` `rArr 12-4b-4c+2c-6b-3b-3c+6=0` `rArr -13b-5c=-18` On multiplying Eq. ix by 5 and Eq. x by 13, then adding we get `=169b-65c=-234` `170b+65c=235` b=1 `rArr -13xx1-5c=-18` `rArr -5c =-18+13=-5rArrc=1` `therefore a=3-1-1=1` and d=1-1=0 `therefore A=[{:(1,1),(1,0):}]`

Discussion

280.	If A, B, and C are three square matrices of the same order, then `AB=AC implies B=C`. ThenA. `\|A\| ne 0`B. A is invertibleC. A may be orthogonalD. A is symmetric
Answer» If `\|A\| ne 0`, then `AB=AC` or `A^(-1) AB=A^(-1) AC` or `B=C` Also, if A is orthogonal matrix, then `A A^(T)=I` `implies \|A A^(T)\|=1 implies \|A\|^(2)=1implies` is invertible

Discussion

281.	Let `A=[0-tan(alpha//2)tan(alpha//2)0]`and `I`be the identity matrixof order 2. Show that `I+A=(I-A)[cosalpha-sinalphasinalphacosalpha]`.A. `-I+A`B. `I-A`C. `-I-A`D. none of these
Answer» Correct Answer - B Since `I=[(1,0),(0,1)]` and given `A=[(0,tan alpha//2),(-tan alpha//2,0)]` `:. I-A=[(1,-tan alpha//2),(tan alpha//2,1)]` Now, `(I-A) [(cos alpha,-sin alpha),(sin alpha,cos alpha)]` `=[(1,tan alpha//2),(-tan alpha//2,1)][(cos alpha,-sin alpha),(sin alpha,cos alpha)]` `=[(1,tan alpha//2),(-tan alpha//2,1)][((1-tan^(2) alpha//2)/(1+tan^(2)alpha//2),-(2 tan alpha//2)/(1+tan^(2) alpha//2)),((2 tan alpha//2)/(1+tan^(2) alpha//2),(1-tan^(2) alpha//2)/(1+tan^(2) alpha//2))]` `=[((1-tan^(2) alpha//2)/(1+tan^(2)alpha//2)+(2 tan^(2) alpha//2)/(1+tan^(2) alpha//2),-(2 tan alpha//2)/(1+tan^(2) alpha//2)),((-tan alpha//2 (1-tan^(2) alpha//2))/(1+tan^(2) alpha//2)+(2 tan alpha//2)/(1+tan^(2) alpha//2),(2 tan^(2) alpha//2)/(1+tan^(2) alpha//2)+(1-tan^(2) alpha//2)/(1+tan^(2) alpha//2))]` `=[(((1+tan^(2) alpha//2))/((1+tan^(2) alpha//2)),-(tan alpha//2(1+tan^(2) alpha//2))/((1+tan^(2) alpha//2))),((tan alpha//2(1+tan^(2) alpha//2))/((1+tan^(2) alpha//2)),((1+tan^(2) alpha//2))/((1+tan^(2) alpha//2)))]` `=[(1,-tan alpha//2),(tan alpha//2,a)]` `=I-A` [Using (1)]

Discussion

282.	A matrix denotes a number
Answer» False A matrix is an ordered rectangular array of numbers or functions.

Discussion

283.	If `A=[{:(0,1),(1,1):}] "and" B=[{:(0,-1),(1,0):}]`, then show that `(A+B)(A-B)neA^(2)-B^(2)`
Answer» We have, `A=[{:(0,1),(1,1):}]"and" B=[{:(0,-1),(1,0):}]` `(A+ B)=[{:(0+0,1 -1),(1+1,1+0):}]=[{:( 0,0),(2,1):}]` `therefore (A+B)=[{:(0+0,1-1),(1+1,1+0):}]=[{:(0,0),(2,1):}]_(2xx2)` and ` (A-B)=[{:(0-0,1+1),( 1-1,1-0):}]=[{:(0,2),(0,1):}]_(2xx2)` Since (A+B),(A-B) is defined, if the number of columns of (A+ B) is equal to the number of rows of (A-B) so here multiplication of matrices (A+B ),(A-B) is possible Now, `(A+B)_(2xx2).(A-B)_(2x2)=[{:(0+0,0+0),(0+0,4+1):}]=[{:(0,0),(0,5):}]` Also `A^(2)=A,A` `=[{:(0,1),(1,1):}],[{:(0,1),(1,1):}]` `=[{:(0+1,0+1) ,(0+1,1+1):}]=[{:(1,1) ,(1,2):}] ` and `B^(2)=B,B =[{:(0,-1),(1,0):}][{:(0,-1),(1,0)]` `=[{:(0-1, 0+0),(0+0,-1+0):}] =[{:(-1,0),( 0,-1):}]` `rArr (A+B),(A-B)neA^(2)-B^(2)` `rArr [{:( 0,0),(0,5):}] ne[{:(2,1),(1,3):}]`

Discussion

284.	show that `[(1,1,3),(5,2,6),(-2,-1,-3)]=A` is nipotent matrix of order 3.
Answer» Let `A=[(1,1,3),(5,2,6),(-2,-1,-3)]` `therefore A^(2)=A.A=[(1,1,3),(5,2,6),(-2,-1,-3)]xx[(1,1,3),(5,2,6),(-2,-1,-3)]` `[(1+5-6,1+2-3,3+6-9),(5+10-12,5+4-6,15+12-18),(-2-5+6,-2-2+3,-6-6+9)]` `=[(0,0,0),(3,3,9),(-1,-1,-3)]` `thereforeA^(3)=A^(2).=[(0,0,0),(3,3,9),(-1,-1,-3)]xx[(1,1,3),(5,2,6),(-2,-1,-3)]` `[(0+0+0,0+0+0,0+0+0),(3+15-18,3+6-9,9+18-27),(-1-5+16,-1-2+3,-3-6+9)]=[(0,0,0),(0,0,0),(0,0,0)]=0` `therefore" " A^(3)=Oi.e.,A^(k)=O` Here `k=3`

Discussion

285.	Show that `A =[{:(5,3),(-1,-2):}]` satisfies the equation `A^(2)-3A-7I=0` and hence find the value of `A^(-1)`
Answer» We have. `A=[{:(5,3),(-1,-2):}]` `therefore A^(2)=A.A=[{:(5,3) ,(-1,-1) :}][{:(5,3),(-1,-2):}]` `=[{:(25-3,15-6),(-5+2,-3+4):}]=[{:(22,9),(-3,1):}]` `3A=3[{:(5,3),(-1,-2):}]=[{:(15,9),(-3,-6):}]` and `7I=7[{:(22,9),(-3,1):}]-[{:(15,9),(-3,-6):}]-[{:(7,0),(0,7):}]` `=[{:(22-15 -7,9-9-0),( -3 +3-0,1+ 6-7):}]` ` =[{:(0,0) ,(0,0):}]` =0 Since, `A^(2)-3A-7I=0` `rArr A^(-1)[A(A^(2))-3A-7I]=A^(-1)0` `rArr A^(-1),A-3A^(-1)-7A^(-1)I=0` `rArr IA-3I-7A^(-1)=0 [because A^(-1)A=I]` `rArr A-3I-7A^(-1)=0 [because A^(-1)I=A^(-1)]` `rArr -7A^(-1)=-A+3I` `= [{:(-5,-3),(1,2):}]+[{:(3,0),(0,3):]=[{:(-2,-3),(1,5):}]` `therefore A^(-1)=(-1)/(7)[{:(-2,-3),(1,5):}]`

Discussion

286.	If A is a square matrix, using mathematical induction prove that (AT)n = (An)T for all n ϵ N.
Answer» Given, A is a square matrix. We need to prove that (A^T)ⁿ = (Aⁿ)^T. We will prove this result using the principle of mathematical induction. Step 1 : When n = 1, we have ∴ (A^T)¹ = A^T Hence, The equation is true for n = 1. Step 2 : Let us assume the equation true for some n = k, Where k is a positive integer. ⇒ (A^T)^k = (A^k)^T To prove the given equation using mathematical induction, we have to show that, (A^T)^k+1 = (A^k+1)^T. We know, (A^T)^k+1 = (A^k+1)^T ⇒ (A^T)^k+1 = (A^k)^T × A^T We have, (AB)^T = B^TA^T. ⇒ (A^T)^k+1 = (A^k)^T ⇒ (A^T)^k+1 = (A^1+k)^T ∴ (A^T)k+1 = (A^k+1)^T Hence, The equation is true for n = k + 1 under the assumption that it is true for n = k. Therefore, By the principle of mathematical induction, the equation is true for all positive integer values of n. Thus, (A^T)ⁿ = (Aⁿ)^T for all n ϵ N.

286.

If A is a square matrix, using mathematical induction prove that (AT)n = (An)T for all n ϵ N.

Answer»

Given,

A is a square matrix.

We need to prove that (A^T)ⁿ = (Aⁿ)^T.

We will prove this result using the principle of mathematical induction.

Step 1 :

When n = 1, we have

∴ (A^T)¹ = A^T

Hence,

The equation is true for n = 1.

Step 2 :

Let us assume the equation true for some n = k, Where k is a positive integer.

⇒ (A^T)^k = (A^k)^T

To prove the given equation using mathematical induction, we have to show that,

(A^T)^k+1 = (A^k+1)^T.

We know,

(A^T)^k+1 = (A^k+1)^T

⇒ (A^T)^k+1 = (A^k)^T × A^T

We have,

(AB)^T = B^TA^T.

⇒ (A^T)^k+1 = (A^k)^T

⇒ (A^T)^k+1 = (A^1+k)^T

∴ (A^T)k+1 = (A^k+1)^T

Hence,

The equation is true for n = k + 1 under the assumption that it is true for n = k.

Therefore,

By the principle of mathematical induction, the equation is true for all positive integer values of n.

Thus,

(A^T)ⁿ = (Aⁿ)^T for all n ϵ N.

Discussion

287.	If `A=[[1,2,-3] , [5,0,2] , [1,-1,1]]` and `B=[[3,-1,2] , [4,2,5] , [2,0,3]]` then find matrix C such that `A+2C=B`
Answer» `2c=B-A=[{:(3,-1,2),(4,2,5),(2,0,3):}]-[{:(1,2,-3),(5,0,2),(1,-1,1):}]` `=[{:(3-1,-1-2,2+3),(4-5,2-0,5-2),(2-1,0+1,3-1):}]` `2C=[{:(2,-3,5),(-1,2,3),(1,1,2):}] rArrC=(1)/(2)[{:(2,-3,5),(-1,2,3),(1,1,1):}]` `=[{:(1,-3//2,5//2),(-1//2,1,3//2),(1//2,1//2,1):}]`

Discussion

288.	The statement are True or False:If A and B are any two matrices of the same order, then (AB)’ = A’B’.
Answer» False ∵If A and B are any two matrices for which AB is defined, then (AB)’=B’A’.

288.

The statement are True or False:If A and B are any two matrices of the same order, then (AB)’ = A’B’.

Answer»

False

∵If A and B are any two matrices for which AB is defined, then

(AB)’=B’A’.

Discussion

289.	The statement are True or False:If A, B and C are square matrices of same order, then AB = AC always implies that B = C.
Answer» False ∵ If AB = AC => B=C The above condition is only possible if matrix A is invertible (i.e \|A\|≠0). ⇒ If A is invertible, then ⇒ A^-1(AB)= A^-1(AC) ⇒ (A^-1A)B = (A^-1A)C ⇒ IB=IC ⇒ B=C

289.

The statement are True or False:If A, B and C are square matrices of same order, then AB = AC always implies that B = C.

Answer»

False

∵ If AB = AC => B=C

The above condition is only possible if matrix A is invertible

(i.e |A|≠0).

⇒ If A is invertible, then

⇒ A^-1(AB)= A^-1(AC)

⇒ (A^-1A)B = (A^-1A)C

⇒ IB=IC

⇒ B=C

Discussion

290.	If `A`is a square matrix,using mathematical induction prove that `(A^T)^n=(A^n)^T`for all `n in N`.
Answer» Let `P(n): (A)^(n)=(A^(n))` `therefore P(1):(A)^(1)=(A)` `rArr A=ArArrP(1)` is true Now, `P(k):(A)^(n)=(A^(k))` where `k in N` and `P(k+1) .(A)^(k+1)=(A^(k+1))` Where `P(k+1)` is true whenever P (k) is true. `therefore P(k+1)(A)^(k),(A)^(1)= [A^(k+1)]` `(A^(k)) ,(A)=[A^(k+1)]` `(A.A^(k))=[A^(k+1)]` `(A^(k+1))=[A^(k+1)]`

Discussion

291.	Matrices of different order cannot be subtracted.
Answer» True Two matrices of same order can be subtracted.

Discussion

292.	show that the matrix `A=[(2,-2,-4),(-1,3,4),(1,-2,-3)]` is idempotent.
Answer» `A^(2)=A.A=[(2,-2,-4),(-1,3,4),(1,-2,-3)]xx[(2,-2,-4),(-1,3,4),(1,-2,-3)]` `=[{:(2.2+(-2).(-1)(-4).1),((-1).2+3.(-1)+4.1),(12+(-2).(-1)+(-3).1),(2.(-2)+(-2).3+(-4).(-2)),((-1).(-2)+3.3+4.(-2)),(1.(-2)+(-2).3+(-3).(-2)),(2.(-4)+(-2).4+(-4).(-3)),((-1).(-4)+3.4.+4.(-3)),(1.(-4)+(-2).4+(-3).(-3)):}` `[(2,-2,-4),(-1,3,4),(1,-2,-3)]=A` Hence the matrix A is idempotent.

Discussion

293.	Two matrices are equal. If they have same number of rows and same number columns.
Answer» False If two matrices have same number of rows and same number of columns, then they are said to be square matrix and it two square matrices have same elements in both the matrices only then they are called equal.

Discussion

294.	Solve the following equations for X and Y : `2X-Y=[(3,-3,0),(3,3,2)], 2Y+X=[(4,1,5),(-1,4,-4)]`
Answer» Given `2X-Y=[{:(3,-3,0),(3,3,2):}]` on multiplying both sides by 2, we get `4x-2y=2[{:(3,-3,0),(3,3,2):}], 4x-2y=[{:(6,-6,0),(6,6,4):}]` also given `x+2y= [{:(4,4,4),(-1,4,-4):}]` Adding Eqs. (i) and (ii) , we get `5X=[{:(6,-6,0),(6,6,4):}]+[{:(4,1,5),(-1,4,-4):}]` `= [{:(6+4,-6+1,0+5),(6-1,6+4,4-4):}]=[{:(10,-5,5),(5,10,0):}]` `X=(1)/(5)[{:(10,-5,5),(5,10,0):}]rArr X=[{:(2,-1,1),(1,2,0):}]` putting the value of X in Eq. (ii), we get `[{:(2,-1,1),(1,2,0):}]+2Y=[{:(4,1,5),(-1,4,-4):}]` `rArr 2Y = [{:(4,1,5),(-1,4,-4):}]-[{:(2,-1,1),(1,2,0):}]` `=[{:(4-2,1+1,5-1),(-1-1,4-2,-4-0):}]=[{:(2,2,4),(-2,2,-2):}]` `Y = [{:(1,1,2),(-1,1,-2):}]` Hence, `X = [{:(2,-1,1),(1,2,0):}] and Y= [{:(2,-1,1),(-1,1,-2):}]`

Discussion

295.	If A = \(\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 0 \\[0.3em] \end{bmatrix}\), find A16.
Answer» Given : A = \(\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 0 \\[0.3em] \end{bmatrix}\) We will find A², A² = A x A = \(\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 0 \\[0.3em] \end{bmatrix}\)\(\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 0 \\[0.3em] \end{bmatrix}\) ⇒ A² = \(\begin{bmatrix} 0+0 & 0+0 \\[0.3em] 0+0 & 0+0 \\[0.3em] \end{bmatrix}\) ⇒ A² = 0 Hence, A¹⁶= (A²)⁸ = (0)⁸ = 0 Hence A¹⁶ is a nill matrix.

295.

If A = \(\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 0 \\[0.3em] \end{bmatrix}\), find A16.

Answer»

Given : A = \(\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 0 \\[0.3em] \end{bmatrix}\)

We will find A²,

A² = A x A

= \(\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 0 \\[0.3em] \end{bmatrix}\)\(\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 0 \\[0.3em] \end{bmatrix}\)

⇒ A² = \(\begin{bmatrix} 0+0 & 0+0 \\[0.3em] 0+0 & 0+0 \\[0.3em] \end{bmatrix}\)

⇒ A² = 0

Hence,

A¹⁶= (A²)⁸ = (0)⁸ = 0

Hence A¹⁶ is a nill matrix.

Discussion

296.	The value of x so that `[[1,x,1]][[1,3,2] , [0,5,1] , [0,3,2]][[1], [1], [x]]=0` is
Answer» we have `[1" " x" "1][(1,3,2),(0,5,1),(0,3,2)][(,1),(,1),(,x)]=0` `rArr" "[1" "5x+6" "x+4][(,1),(,1),(,x)]=0` `rArr" "[1+5x+6+x^(2)+4x]=0` `or" "x^(2)+9x+7=0` `therefore" " x= (-9pm sqrt((81-28)))/(2) implies x = (-9 pm sqrt(53))/(2)`

Discussion

297.	If A `=[{:(0,1,0),(0,0,1),(p,q,r):}]`, show that ltbargt `A^(3)= pI+qA+rA^(2)`
Answer» we have, `A(2)=A.A` `[(0,1,0),(0,0,1),(p,q,r)]xx[(0,1,0),(0,0,1),(p,q,r)]` `[(0,0,1),(p,q,r),(pr,p+qr,q+r^(2))]` `therefore (3)=A^(2)=[(0,1,1),(p,q,r),(pr,p+qr,q+r^(2))]xx[(0,1,0),(0,0,1),(p,q,r)]` `=[(p,q,r),(pr,p+qr,q+r^(2)),(pq+r^(2)p,pr+q^(2)+qr^(2),p+2qr+r^(3))]" "therefore(i)` and `pI+qA+rA^(2) =p[(1,0,0),(0,1,0),(0,0,1)]+q[(0,1,0),(0,0,1),(p,q,r)]+r[(0,0,1),(p,q,r),(pr,p+qr,q+r^(2))]` `[(p+0+0,0+q+0,0+0+r),(0+0+pr,p+0+qr,0+q+r^(2)),(0+pq+pr^(2),0+q^(2)+pr+qr^(2),p+2qr+r^(3))]` `[(p,q,r),(pr,p+qr,q+r^(3)),(pq+pr^(2),q^(2)+pr+qr^(2),p+2qr+r^(3))]" "therefore(ii)` thus, from Eqs. (i) and (ii), we get `A = pI+qA+rA^(2)`

Discussion

298.	If `A = [{:(1,2),(-2,3):}], B= [{:(2,1),(2,3):}] and C = [{:(-3,1),(2,0):}]`,verify that (AB)C = A(BC) and A(B+C)=AB+AC.
Answer» we have, AB `AB = [{:(1,2),(-2,3):}],xx [{:(2,1),(2,3):}]` `=[{:(1.2+2.2,1.1+2.3),((-2).2+3.2,(-2).1+3.3):}]=[{:(6,7),(2,7):}]` `BC =[{:(2,1),(2,3):}]xx[{:(-3,1),(2,0):}]=[{:(2(-3)+1.2,2.1+1.0),(2(-3)+3.2,2.1+3.0):}]` `=[{:(-6+2,2+0),(-6+6,2+0):}]=[{:(-4,2),(0,2):}]` `BC =[{:(2,1),(-2,3):}]xx[{:(-3,1),(2,0):}]=[{:(1(-3)+2.2,,1.1+2.0),((-2)(-3)+3.2,,(-2)1+3.0):}]` `=[{:(-3+4,1+0),(6+6,-2+0):}]=[{:(1,1),(12,-2):}]` ltbegt `B+C =[{:(2,1),(2,3):}]+[{:(-3,1),(2,0):}]=[{:(2-3,1+1),(2+2,3+0):}]=[{:(-1,2),(4,3):}]` Now, (AB)C `[{:(6,7),(2,3):}]xx[{:(-3,1),(2,0):}]=[{:(-18-3,6+0),(-6+14,2+0):}]` `=[{:(-4,6),(8,2):}]` A(BC)` =[{:(1,2),(-2,3):}]xx[{:(-4,2),(0,2):}]` `=[{:(-4+0,2+4),(8+0,-4+6):}]=[{:(-4,2),(8,2):}]` Thus, from Eqs. (i) and (ii), we get, (AB)C =A(BC) Now, A(B+C) `=[{:(1,2),(-2,3):}]xx[{:(-1,2),(4,3):}]=[{:(-1+8,2+6),(2+12,-4+9):}]` `=[{:(7,8),(14,5):}]`and AB+AC `=[{:(6,7),(2,7):}]xx[{:(1,1),(12,-2):}]=[{:(6+1,7+1),(2+12,7-2):}]` `=[{:(7,8),(14,5):}]` Thus, from Eqs, (iii) and (iv), we get

Discussion

299.	If `A=[{:(" "1,-1," "2),(" "3," "2," "0),(-2," "0," "1):}],B=[{:(3,1),(0,2),(-2,5):}]" and "C=[{:(2,1,-3),(3,0,-1):}` then verify that `(AB)C=A(BC).`
Answer» We have `AB=[{:(1,-1,2),(3,2,0),(-2,0,1):}][{:(3,1),(0,2),(-2,5):}]` `=[{:(3-0-4,1-2+10),(9+0-0,3+4+0),(-6+0-2,-2+0+5):}]=[{:(-1,9),(" "9,7),(-8,3):}]` `implies" "(AB)C=[{:(-1,9),(" "9,7),(-8,3):}][{:(2,1,-3),(3,0,-1):}]` `=[{:(-2+27,-1+0,3-9),(18+21,9+0,-27-7),(-16+9,-8+0,24-3):}]=[{:(25,-1,-6),(39,9,-34),(-7,-8,21):}]` Also, `BC=[{:(3,1),(0,2),(-2,5):}][{:(2,1,-3),(3,0,-1):}]` `=[{:(6+3,3+0,-9-1),(0+6,0+0,0-2),(-4+15,-2+0,6-5):}]=[{:(9,3,-10),(6,0,-2),(11,-2,1):}]` `implies" "A(BC)=[{:(1,-1,2),(3,2,0),(-2,0,1):}][{:(9,3,-10),(6,0,-2),(11,-2," "1):}]` `=[{:(9-6+22,3-0-4,-10+2+2),(27+12+0,9+0-0,-30-4+0),(-18+0+11,-6+0-2,20-0+1):}]` `=[{:(25,-1,-6),(39,9,-34),(-7,-8,21):}].` Hence, `(AB)C=A(BC).`

Discussion

300.	If A = \(\begin{bmatrix}4& 1 \\[0.3em]5& 8\\[0.3em]\end{bmatrix}\), show that (A + A’) is symmetric.A = [(4,1)(5,8)]
Answer» Given A = \(\begin{bmatrix}4 &1 \\[0.3em]5 & 8 \\[0.3em]\end{bmatrix}\). To Prove: A + A’ is symmetric.(Note:A matrix P is symmetric if P’ = P) Proof: We will find A’, A' = \(\begin{bmatrix}4 &1 \\[0.3em]5 & 8 \\[0.3em]\end{bmatrix}\) Now let us take P = A + A’ \(\Rightarrow P'=P\) Hence A + A’ is a symmetric matrix.

300.

If A = \(\begin{bmatrix}4& 1 \\[0.3em]5& 8\\[0.3em]\end{bmatrix}\), show that (A + A’) is symmetric.A = [(4,1)(5,8)]

Answer»

Given A = \(\begin{bmatrix}4 &1 \\[0.3em]5 & 8 \\[0.3em]\end{bmatrix}\).

To Prove: A + A’ is symmetric.(Note:A matrix P is symmetric if P’ = P)

Proof: We will find A’,

A' = \(\begin{bmatrix}4 &1 \\[0.3em]5 & 8 \\[0.3em]\end{bmatrix}\)

Now let us take P = A + A’

\(\Rightarrow P'=P\)

Hence A + A’ is a symmetric matrix.

Discussion

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If `theta-phi=pi/2,` prove that, `[(cos^2 theta,cos theta sin theta),(cos theta sin theta,sin^2 theta)] [(cos^2 phi,cos phi sin phi),(cos phi sin phi,sin^2 phi)]=0`A. `2npi, in Z`B. `n pi/2, n in Z`C. `(2n+1) pi/2, n in X`D. `npi, n in Z`

Solve for x and y, given that `[{:(x,y),(3y,x):}][{:(1),(2):}]=[{:(3),(5):}].`

Find the value of x, if `[1x1][{:(1,3,2),(2,5,1),(15,3,2):}][{:(1),(2),(x):}]=O.`

If `A^(2) - A + I = 0` then `A^(-1) ` is equal toA. `A^(-2)`B. `A+1`C. `I-A`D. `A-I`

If `A[{:(3,-4),(1,1),(2,0):}]` and `B=[{:(2,1,2),(1,2,4):}]` and `B=[{:(4,1),(2,3),(1,2):}]`

If `A=[a b0a]`is nth root of `I_2,`then choose the correct statements:If `n`is odd, `a=1,b=0`If `n`is odd, `a=-1,b=0`If `n`is even, `a=1,b=0`If `n`is even, `a=-1,b=0`a. i, ii, iii, iv b.ii, iii, ivc. i, ii, iii, iv d. i,iii, ivA. i, ii, iiiB. ii, iii, ivC. i, ii, iii, ivD. i, iii, iv

If for the matrix `A , A^3=I`, then `A^(-1)=``A^2`(b) `A^3`(c) `A`(d) none of theseA. `A^(2)`B. `A^(3)`C. AD. None of these

If `A=[i-i-i i]a n dB=[1-1-1 1],t h e nA^8`equals`4B`b. `128 B`c. `-128 B`d. `-64 B`A. 4BB. 128BC. `-128` BD. `-64`B

Prove that matrix `[((b^(2)-a^(2))/(a^(2)+b^(2)),(-2ab)/(a^(2)+b^(2))),((-2ab)/(a^(2)+b^(2)),(a^(2)-b^(2))/(a^(2)+b^(2)))]` is orthogonal.

If `e^(A)` is defined as `e^(A)=I+A+A^(2)/(2!)+A^(3)/(3!)+...=1/2 [(f(x),g(x)),(g(x),f(x))]`, where `A=[(x,x),(x,x)], 0 lt x lt 1` and I is identity matrix, then find the functions f(x) and g(x).

FindA, if `[{:(4),(1),(3):}]A=[{:(-4,8,4),(-1,2,1),(-3,6,3):}]`

If `I_(3)` is identity matrix of order 3, then `I_(3)^(-1)=`

`if[{:(2,1),(3,2):}]A[{:(-3,2),(5,-3):}]=[{:(1,0),(0,1):}],"then" A=?`

If A, B, and C are three square matrices of the same order, then `AB=AC implies B=C`. ThenA. `|A| ne 0`B. A is invertibleC. A may be orthogonalD. A is symmetric

Let `A=[0-tan(alpha//2)tan(alpha//2)0]`and `I`be the identity matrixof order 2. Show that `I+A=(I-A)[cosalpha-sinalphasinalphacosalpha]`.A. `-I+A`B. `I-A`C. `-I-A`D. none of these

A matrix denotes a number

If `A=[{:(0,1),(1,1):}] "and" B=[{:(0,-1),(1,0):}]`, then show that `(A+B)(A-B)neA^(2)-B^(2)`

show that `[(1,1,3),(5,2,6),(-2,-1,-3)]=A` is nipotent matrix of order 3.

Show that `A =[{:(5,3),(-1,-2):}]` satisfies the equation `A^(2)-3A-7I=0` and hence find the value of `A^(-1)`

If A is a square matrix, using mathematical induction prove that (AT)n = (An)T for all n ϵ N.

If `A=[[1,2,-3] , [5,0,2] , [1,-1,1]]` and `B=[[3,-1,2] , [4,2,5] , [2,0,3]]` then find matrix C such that `A+2C=B`

The statement are True or False:If A and B are any two matrices of the same order, then (AB)’ = A’B’.

The statement are True or False:If A, B and C are square matrices of same order, then AB = AC always implies that B = C.

If `A`is a square matrix,using mathematical induction prove that `(A^T)^n=(A^n)^T`for all `n in N`.

Matrices of different order cannot be subtracted.

show that the matrix `A=[(2,-2,-4),(-1,3,4),(1,-2,-3)]` is idempotent.

Two matrices are equal. If they have same number of rows and same number columns.

Solve the following equations for X and Y : `2X-Y=[(3,-3,0),(3,3,2)], 2Y+X=[(4,1,5),(-1,4,-4)]`

If A = \(\begin{bmatrix} 0 &amp; 0 \\[0.3em] 4 &amp; 0 \\[0.3em] \end{bmatrix}\), find A16.

The value of x so that `[[1,x,1]][[1,3,2] , [0,5,1] , [0,3,2]][[1], [1], [x]]=0` is

If A `=[{:(0,1,0),(0,0,1),(p,q,r):}]`, show that ltbargt `A^(3)= pI+qA+rA^(2)`

If `A = [{:(1,2),(-2,3):}], B= [{:(2,1),(2,3):}] and C = [{:(-3,1),(2,0):}]`,verify that (AB)C = A(BC) and A(B+C)=AB+AC.

If `A=[{:(" "1,-1," "2),(" "3," "2," "0),(-2," "0," "1):}],B=[{:(3,1),(0,2),(-2,5):}]" and "C=[{:(2,1,-3),(3,0,-1):}` then verify that `(AB)C=A(BC).`

If A = \(\begin{bmatrix}4&amp; 1 \\[0.3em]5&amp; 8\\[0.3em]\end{bmatrix}\), show that (A + A’) is symmetric.A = [(4,1)(5,8)]

If A = \(\begin{bmatrix} 0 & 0 \\[0.3em] 4 & 0 \\[0.3em] \end{bmatrix}\), find A16.

If A = \(\begin{bmatrix}4& 1 \\[0.3em]5& 8\\[0.3em]\end{bmatrix}\), show that (A + A’) is symmetric.A = [(4,1)(5,8)]