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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 551. |
`[{:(3,0,-1),(2,3,0),(0,4,1):}]` |
| Answer» `[{:(3,-4," "3),(-2," "3,-2),(8,-12," "9):}]` | |
| 552. |
If `[(1,-tan theta),(tan theta,1)][(1,tan theta),(-tan theta,1)]^(-1)=[(a,-b),(b,a)],` thenA. `a=cos 2 theta, b = sin 2 theta `B. `a=1, b=1`C. `a=sin 2 theta, b=cos 2 theta `D. `a=sec 2 theta, b=tan 2 theta ` |
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Answer» Correct Answer - a |
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| 553. |
If `[(1,2,3),(2,4,1),(3,2,9)][(x),(y),(z)]=[(6),(7),(14)]`, then the values of `x, y, z` respectively areA. `1, 2, 3`B. `1, 1, 1`C. `2, 2, 2`D. `3, 2, 1` |
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Answer» Correct Answer - b |
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| 554. |
The inverse of `[(1,sin alpha),(-sin alpha,-1)]` isA. `[(1,-sin alpha),(-sin alpha,-1)]`B. ` -sec^(2)[(1,-sin alpha),(sin alpha,-1)]`C. ` sec^(2) alpha[(1,sin alpha),(-sin alpha,-1)]`D. ` cos^(2) alpha[(1,sin alpha),(-sin alpha,-1)]` |
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Answer» Correct Answer - c |
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| 555. |
`A=[(2,-2),(-2,2)], B=[(1,1),(1,1)]` thenA. `A^(-1)=B`B. `B^(-1)` does not existC. `A^(-1)` does not existD. Both (B) and (C) |
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Answer» Correct Answer - d |
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| 556. |
The element in the first row and third column of the inverse of the matrix `[(1,2,-3),(0,1,2),(0,0,1)]` isA. `-2`B. 0C. 1D. 7 |
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Answer» Correct Answer - d |
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| 557. |
The inverse of the matrix `[(1,0,0),(a,1,0),(b,c,1)]` is (A) `[(1,0,0),(-a,1,0),(b,c,1)]` (B) `[(1,0,0),(-a,1,0),(ac,b,1)]` (C) `[(1,-a,ac-b),(-0,1,-c),(0,0,1)]` (D) `[(1,0,0),(-a,1,0),(ac-b,-c,1)]`A. `[(1,0,0),(-a,1,0),(b,-c,1)]`B. `[(1,0,0),(-a,1,0),(ac,b,1)]`C. `[(1,0,0),(-a,1,0),(ac-b,-c,1)]`D. `[(1,-a,ac-b),(0,1,-c),(0,0,1)]` |
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Answer» Correct Answer - c |
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| 558. |
The inverse matrix of `A=[(0,1,2),(1,2,3),(3,1,1)]` isA. `[((1)/(2),-(1)/(2),(1)/(2)),(-4,3,-1),((5)/(2),(-3)/(2),(1)/(2))]`B. `[((1)/(2),-4,(5)/(2)),(1,-6,3),(1,2,-1)]`C. `(1)/(2)[(1,2,3),(3,2,1),(4,2,3)]`D. `(1)/(2)[(1,-1,-1),(-8,6,-2),(5,-3,1)]` |
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Answer» Correct Answer - a |
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| 559. |
If `{:A=[(5,2),(3,1)]:}," then "A^(-1)=`A. `{:[(1,-2),(-3,5)]:}`B. `{:[(-1,2),(3,-5)]:}`C. `{:[(-1,-2),(-3,-5)]:}`D. `{:[(1,2),(3,5)]:}` |
| Answer» Correct Answer - b | |
| 560. |
If `A=[{:(1,-1),(2,-1):}],B=[{:(a,-1),(b,-1):}]" and "(A+B)^(2)=(A^(2)+B^(2))` then find the values of a and b.A. `a=4,b=1`B. `a=1b=4`C. `a=0,b=4`D. `a=2,b=4` |
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Answer» Correct Answer - B We have, `{:A+B=[(a+1,0),(b+2,-2)]:}` `{:A^2=[(1,-1),(2,-1)][(1,-1),(2,-1)]=[(-1,0),(0,-1)]:}` `{:B^2=[(a,1),(b,-1)][(a,1),(b,-1)]=[(a^2+b,a-1),(ab-b,b+1)]:}` `{:(A+B)^2=[(a+1,0),(b+2,-2)][(a+1,0),(b+2,-2)]=[((a+1)^2,0),((a-1)(b+2),4)]:}` `:. (A+B)^2=A^2+B^2` `rArr{:[((a+1)^2,0),((a-1)(b+2),4)]=[(-1,0),(0,-1)]+[(a^2+b,a-1),(ab-b,b+1)]:}` `rArr{:[((a+1)^2,0),((a-1)(b+2),4)]=[(a^2+b-1,a-1),(ab-b,b)]:}` `rArr a-1=0,b=4,(a+1)^2=a^2+b-1,(a-1)(b+2)=a-b` `rArr a=1and b=4` |
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| 561. |
The matrix `{:A=[(1,-3,-4),(-1,3,4),(1,-3,-4)]:}`is nilpotent of indexA. 2B. 3C. 4D. 6 |
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Answer» Correct Answer - A We know that a square matrix A is nilpotent of index n, if n is the least positive such that `A^n=O`(null matrix). For the give matrix, we have `A^2=O` Hence,it is a nilpotent matrix of index 2. |
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| 562. |
Let `P=[[1,0,0],[4,1,0],[16,4,1]]`and `I` be the identity matrix of order `3`. If `Q = [q_()ij ]` is a matrix, such that `P^(50)-Q=I`, then `(q_(31)+q_(32))/q_(21)` equalsA. 52B. 103C. 201D. 205 |
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Answer» Correct Answer - B We have, `{:P=[(1,0,0),(4,1,0),(16,4,1)]:}`. `:. {:P^2=PP=[(1,0,0),(4,1,0),(16,4,1)][(1,0,0),(4,1,0),(16,4,1)]=[(1,0,0),(8,1,0),(16+32,8,1)]:}` `:. {:P^3=P^2P=[(1,0,0),(8,1,0),(16+32,8,1)][(1,0,0),(4,1,0),(16,4,1)]=[(1,0,0),(12,1,0),(16+32+48,12,1)]:}` By observing the symmetry, we obtain `{:P^50=[(1,0,0),(4xx50,1,0),(16+32+48+...50"terms",4xx50,1)]:}` `rArr{:P^50[(1,0,0),(200,1,0),((16xx50xx51)/2,200,1)]:}` `:.P^50-Q=I` `rArr{:Q=P^50-I[(0,0,0),(200,0,0),(20400,200,0)]:}` `rArr q_21=200,q_31=20400and q_32=200` `:. (q_31+q_32)/q_21=(20400+200)/200=20600/200=103` |
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| 563. |
Find x, y, a and b if \(\begin{bmatrix}3x+4y & 2 & x-2y \\[0.3em]a+b & 2a-b & -1 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}2 & 2 & 4 \\[0.3em]5 &-5 & -1 \\[0.3em]\end{bmatrix}\) |
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Answer» Given two matrices are equal, \(\begin{bmatrix}3x+4y & 2 & x-2y \\[0.3em]a+b & 2a-b & -1 \\[0.3em]\end{bmatrix}\)= \(\begin{bmatrix}2 & 2 & 4 \\[0.3em]5 &-5 & -1 \\[0.3em]\end{bmatrix}\) We know that if two matrices are equal then the elements of each matrices are also equal. ∴ 3x + 4y = 2 …… (1) and x – 2y = 4 …… (2) and a + b = 5 ……(3) 2a – b = – 5 ……(4) Multiplying equation (2) by 2 and adding to equation (1). 3x + 4y + 2x – 4y = 2 + 8 ⇒ 5x = 10 ⇒ x = 2 Now, Putting the value of x in equation (1) 3 × 2 + 4y = 2 ⇒ 6 + 4y = 2 ⇒ 4y = 2 – 6 ⇒ 4y = – 4 ⇒ y = – 1 Adding equation (3) and (4), a + b + 2a – b = 5 + (–5) ⇒ 3a = 5 – 5 = 0 ⇒ a = 0 Now, Putting the value of a in equation (3) 0 + b = 5 ⇒ b = 5 ∴ a = 0, b = 5, x = 2 and y = – 1 |
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| 564. |
Let `A=[{:(2,4),(3,2):}],B=[{:(1,3),(-2,5):}]" and "C=[{:(-2,5),(3,4):}].` Find: (i) `A+2B` (ii) `B-4C` (iii) `A-2B+3C` |
| Answer» (i) `[{:(4,10),(-1,12):}]" "(ii)[{:(9,-17),(-14,-11):}]" "(iii) [{:(-6,13),(16,4):}]` | |
| 565. |
Construct a 3 × 2 matrix whose elements are given by aij = (2i – j). |
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Answer» Given: aij = (2i – j) Now, a11 = (2 × 1 – 1) = 2 – 1 = 1 a12 = 2 × 1 – 2 = 2 – 2 = 0 a21 = 2 × 2 – 1 = 4 – 1 = 3 a22 = 2 × 2 – 2 = 4 – 2 = 2 a31 = 2 × 3 – 1 = 6 – 1 = 5 a32 = 2 × 3 – 2 = 6 – 2 = 4 Therefore, A = \(\begin{bmatrix}1 &0 \\[0.3em]3& 2 \\[0.3em]5 & 4\end{bmatrix}\) |
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| 566. |
Find x, y, a and b if \(\begin{bmatrix} 2x -3y & a - b & 3 \\[0.3em] 1 & x + 4y & 3a + 4b \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 1& -2 &3 \\[0.3em] 1 & 6 & 29 \\[0.3em] \end{bmatrix}\) |
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Answer» Given \(\begin{bmatrix} 2x -3y & a - b & 3 \\[0.3em] 1 & x + 4y & 3a + 4b \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} 1& -2 &3 \\[0.3em] 1 & 6 & 29 \\[0.3em] \end{bmatrix}\) As we know that if two matrices are equal then the elements of each matrices are also equal. Given as two matrices are equal. So by equating them 2a + b = 4 …… (1) And a – 2b = – 3 …… (2) And 5c – d = 11 …… (3) 4c + 3d = 24 …… (4) On multiplying equation (1) by 2 and adding to equation (2) 4a + 2b + a – 2b = 8 – 3 ⇒ 5a = 5 ⇒ a = 1 Substitute this value of a in equation (1) 2 × 1 + b = 4 ⇒ 2 + b = 4 ⇒ b = 4 – 2 ⇒ b = 2 On multiplying equation (3) by 3 and adding to equation (4) 15c – 3d + 4c + 3d = 33 + 24 ⇒ 19c = 57 ⇒ c = 3 Substitute this value of c in equation (4) 4 × 3 + 3d = 24 ⇒ 12 + 3d = 24 ⇒ 3d = 24 – 12 ⇒ 3d = 12 ⇒ d = 4 So a = 1, b = 2, c = 3 and d = 4 |
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| 567. |
Find x, y, a and b if \(\begin{bmatrix}3x + 4y & 2 & x - 2y \\[0.3em]a + b & 2a - b & -1 \\[0.3em]\end{bmatrix}\) = \(\begin{bmatrix}2& 2 &4 \\[0.3em]5 & -5 & -1 \\[0.3em]\end{bmatrix}\) |
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Answer» Given as \( \begin{bmatrix} 3x + 4y & 2 & x - 2y \\[0.3em] a + b & 2a - b & -1 \\[0.3em] \end{bmatrix}\) = \( \begin{bmatrix} 2& 2 &4 \\[0.3em] 5 & -5 & -1 \\[0.3em] \end{bmatrix}\) Given as two matrices are equal. As we know that if two matrices are equal then the elements of each matrices are also equal. So, by equating them 3x + 4y = 2 …… (1) x – 2y = 4 …… (2) a + b = 5 …… (3) 2a – b = – 5 …… (4) On multiplying equation (2) by 2 and adding to equation (1), 3x + 4y + 2x – 4y = 2 + 8 ⇒ 5x = 10 ⇒ x = 2 Substitute this value of x in equation (1) 3 × 2 + 4y = 2 ⇒ 6 + 4y = 2 ⇒ 4y = 2 – 6 ⇒ 4y = – 4 ⇒ y = – 1 On adding equation (3) and (4) a + b + 2a – b = 5 + (– 5) ⇒ 3a = 5 – 5 = 0 ⇒ a = 0 Again substitute this value of a in equation (3), 0 + b = 5 ⇒ b = 5 So, a = 0, b = 5, x = 2 and y = – 1 |
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| 568. |
If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = AnBn for all n ∈ N. |
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Answer» (i) Let P(n) be the statement P(n) ⇒ Abn = BnA P(1) ⇒ AB = BA ∴ P( 1) is true Let P(k) be true P(k) ⇒ ABk = BkA P(k +1) ⇒ ABk+1 = A(BkB) = (ABk)B ∵ matrix multiplication is associative. ⇒ (BkA) B ⇒ Bk (AB) = Bk (BA) = (BkB) A = Bk+1 A Hence P(k + 1) is true. ∴ P(n) is true for all the values of n ∈ N (ii) Let P(n) be the statement P(n) ⇒ (AB)n = AnBn P(1) = (AB)1 = AB ∴ P(1) is true Let P(k) be true P(k) ⇒ (AB)k = AkBkP(k+1) ⇒ (AB)k+1 = (AB)kAB = (AkBk) AB = Ak (BkA)B = Ak (ABk) B (∵ ABn = BnA whenever Ab = BA) = Ak+1 Bk+1 ∴ P (k+1) is true Hence P(n) is true for all the natures of n ∈ N |
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| 569. |
What are the possible order of matrix, having 6 elements ? |
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Answer» 6 x 1, 1 x 6, 2 x 3 and 3 x 2. |
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| 570. |
Find the values of a, b and c if matrices A and B are equal, whereA = \(\begin{bmatrix}a - 2&3&2c\\[0.3em]12c& b + 2 & bc\end{bmatrix}, B = \begin{bmatrix}b&c&6\\[0.3em]6b&a&3b\end{bmatrix}\) |
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Answer» Given, A = B A = \(\begin{bmatrix}a - 2&3&2c\\[0.3em]12c& b + 2 & bc\end{bmatrix}, B = \begin{bmatrix}b&c&6\\[0.3em]6b&a&3b\end{bmatrix}\) On comparing a – 2 = b ⇒ a – b = 2 …..(i) 3 = c 12c = 6b ⇒ b = (12 x 3)/6 = 6 ⇒ b = 6 From b + 2 = a, a – b = 2 ∴ a = 2 + b = 2 + 6 = 8 So, a = 8, b = 6, c = 3. |
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| 571. |
If \(\begin{bmatrix}k + 4&-1\\[0.3em]3&k - 6\end{bmatrix} = \begin{bmatrix}a&-1\\[0.3em]3&-4\end{bmatrix}\), then find the value of a. |
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Answer» Given, \(\begin{bmatrix}k + 4&-1\\[0.3em]3&k - 6\end{bmatrix} = \begin{bmatrix}a&-1\\[0.3em]3&-4\end{bmatrix}\) ∴ On comparing ∴ k +4 = a …….(i) k – 6 = -4 …….(ii) From (i), k = -4 + 6 = 2 From (ii), a = k + 4 = 6 ∴ a = 6 |
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| 572. |
If \(\begin{bmatrix}2x&3x + y\\[0.3em]-x + z& 3y - 2p\end{bmatrix} = \begin{bmatrix}4&5\\[0.3em]-4&-3\end{bmatrix}\), then find the value of x, y, z and p. |
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Answer» \(\begin{bmatrix}2x&3x + y\\[0.3em]-x + z& 3y - 2p\end{bmatrix} = \begin{bmatrix}4&5\\[0.3em]-4&-3\end{bmatrix}\) On comparing 2x = 4 ⇒ x = 2 ⇒ 3x + y = 5 y = 5 – 3 x 2 = 5 – 6 = -1 -x + z = -4 z = -4 + x = -4 + 2 = – 2 ⇒ 3y – 2p = -3 ⇒ 2p = 3y + 3 = 3 x -1 + 3 = 0 So, p = 0 x = 2, y = -1, z = -2, p = 0. |
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| 573. |
The rank of the matrix `A={:[(1,2,3),(3,6,9),(1,2,3)]:}`, isA. 1B. 2C. 3D. none of these |
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Answer» Correct Answer - A We observe that `absA=0` and every square sub-matrix of order 2 is singular. But, A is a non-null matrix. Therefore, r(A) =1. |
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| 574. |
The rank of the matrix `A={:[(1,2,3,4),(4,3,2,1)]:}`, isA. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B We find that `{:[(1,2),(3,4)]:}` is a non-singular square submatrix of A of order 2. Therefore, r(A) = 2` |
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| 575. |
The rank of the matrix `{:A=[(1,2,3),(4,5,6),(3,4,5)]:}` isA. 1B. 2C. 3D. none of these |
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Answer» Correct Answer - B We have, `{:absA=abs((1,2,3),(4,5,6),(3,4,5)):}=1(25-24)-2(20-18)+3(16-15)=0` Therefore, r(A) is less than 3. We observe that `{:[(5,6),(4,5)]:}` is a non-singular square submatrix of order 2. Hence, r (A) =2. |
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| 576. |
If `A=[(1,2,2),(2,1,2),(2,2,1)],` then the value of `|A^4-18A^2-32A|` isA. 1B. 2C. 3D. none of these |
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Answer» Correct Answer - B We have, `{:absA=[(1,2,2),(2,1,2),(2,2,1)]:}` `rArrabsA=1(1-4)-2(2-4)+2(4-2)=5ne0` Thus, A is non-singular matrix of order 3 and hence r (A) =3. |
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| 577. |
if `A=[(cos0,sin0),(-sin0,cos0)],then A^(2)=l`is true forA. `theta=0`B. `theta=(pi)/(4)`C. `theta=(pi)/(2)`D. None of these |
| Answer» Correct Answer - A | |
| 578. |
if matrix `A=(1)/sqrt2[(1,i),(-i,a)], i=sqrt-1` is unitary matrix, a is equal toA. 2B. `-1`C. 0D. 1 |
| Answer» Correct Answer - B | |
| 579. |
If `{:A=alpha[(1,1+i),(1-i,-1)]:}a in R`, is a unitary matrix then `alpha^2` isA. `1/2`B. `1/3`C. `1/4`D. `2/9` |
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Answer» Correct Answer - B It is given that A is a unitary matrix. `:. A(overset-A)^T=I` `rArr{:alpha[(1,1+i),(1-i,-1)]alpha[(1,1+i),(1-i,-1)]=[(1,0),(0,1)]:}` `{:alpha^2[(3,0),(0,3)]=[(1,0),(0,1)]:}` `rArr 3alpha^2=1rArr alpha^2=1/3` |
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| 580. |
If `[(0,2b,c),(a,b,-c),(a,-b,c)]` is orthogonal matrix, then the value of `|abc|` is equal to (where `|*|` represents modulus function)A. `1/2`B. `1/3`C. `1/6`D. 1 |
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Answer» Correct Answer - C If A is an orthogonal matrix, then `A A^T=I=A^TA` `rArr{:[(0,2b,c),(a,b,-c),(a,-b,c)][(0,a,a),(2b,b,-b),(c,-c,c)]=[(1,0,0),(0,1,0),(0,0,1)]:}` `rArr{:[(4b^2+c^2,2b^2-c^2,-2b^2+c^2),(2b^2-c^2,a^2+b^2+c^2,a^2-b^2-c^2),(-2b^2+c^2,a^2-b^2-c^2,a^2+b^2+c^2)]=[(1,0,0),(0,1,0),(0,0,1)]:}` `rArr 4b^2+c^2=1,2b^2-c^2=0 and a^2-b^2-c^2=0` `rArra^2=1/2,b^2=1/6and c^2=1/3rArrabs(abc)=1/6` |
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| 581. |
Let `A= [[a,b,c],[b,c,a],[c,a,b]]` is an orthogonal matrix and `abc = lambda (lt0).` The value of ` a^(3) + b^(3)+c^(3)` isA. `lambda`B. `2lambda`C. `3lambda`D. None of these |
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Answer» Correct Answer - D `becauseA` is an orthogonal matrix `therefore A A^(T) =I` `[[a,b,c],[b,c,a],[c,a,b]] [[a,b,c],[b,c,a],[c,a,b]] =1 [[1,0,0],[0,1,0],[0,0,1]]` `[[a^(2)+b^(2)+c^(2),ab + bc+ca,ab + bc+ ca],[ab + bc + ca,a^(2) +b^(2)+c^(2) , ab+ bc+ ca ],[ab+ bc+ca,ab+bc+ca,a^(2) + b^(2) + c^(2)]] =1 [[1,0,0],[0,1,0],[0,0,1]]` By equality of matrices, we get `a^(2) + b^(2) +c^(2) = 1 ` ...(i) `ab + bc + ca= 0` ...(ii) ` (a+b+c)^(2) + a^(2)= b^(2) +c^(2)+ 2 (ab + bc + ca)` `= 1 + 0 = 1` ` therefore a+ b + c = pm 1` ...(iii) ` because a^(3) + b^(3) +c^(3) - 3abc = (a+b+c) ` `(a^(2) + b^(2) +c^(2) - ab - bc - ca)` `rArr a^(3) + b^(3) + c^(3) - 3lambda = (pm 1) (1-0) ` [from Eqs.(i), (ii) and (iii) and abc` = lambda`] `rArr a^(3) + b^(3)+ c^(3) = 3lambda pm 1` |
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| 582. |
Let `A= [[a,b,c],[b,c,a],[c,a,b]]` is an orthogonal matrix and `abc = lambda (lt0).` The equation whose roots are `a, b, c, ` isA. `x^(3)- 2x^(2) + lambda = 0`B. `x^(3) -lambda x^(2) + lambda x + lambda = 0 `C. `x^(3) - 2 x^(2) + 2 lambda x + lambda = 0 `D. `x^(3) pm x^(2) - lambda = 0 ` |
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Answer» Correct Answer - D `becauseA` is an orthogonal matrix `therefore A A^(T) =I` `[[a,b,c],[b,c,a],[c,a,b]] [[a,b,c],[b,c,a],[c,a,b]] =1 [[1,0,0],[0,1,0],[0,0,1]]` `[[a^(2)+b^(2)+c^(2),ab + bc+ca,ab + bc+ ca],[ab + bc + ca,a^(2) +b^(2)+c^(2) , ab+ bc+ ca ],[ab+ bc+ca,ab+bc+ca,a^(2) + b^(2) + c^(2)]] =1 [[1,0,0],[0,1,0],[0,0,1]]` By equality of matrices, we get `a^(2) + b^(2) +c^(2) = 1 ` ...(i) `ab + bc + ca= 0` ...(ii) ` (a+b+c)^(2) + a^(2)= b^(2) +c^(2)+ 2 (ab + bc + ca)` `= 1 + 0 = 1` ` therefore a+ b + c = pm 1` ...(iii) Equation whose roots are a, b,c is `x^(3) - (a+b+c) x^(2) + (ab+ bc+ca) x - abc = 0` `rArr x^(3) - (pm1) = 0 -lambda = 0` `therefore x^(3) pm x^(2) - lambda = 0` |
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| 583. |
The number of Hawkins-Simon conditions for the viability of input-output analysis is: (a) 1 (b) 3 (c) 4 (d) 2 |
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Answer» Answer is (d) 2 |
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| 584. |
If \(\begin{bmatrix}x+3 & z+4 & 2y-7 \\[0.3em]4x+6 & a-1 & 0 \\[0.3em]b-3 & 3b &z+2c\end{bmatrix}\)= \(\begin{bmatrix}0 & 6 & 3y-2 \\[0.3em]2x & -3 & 2c+2 \\[0.3em]2b+4 & -21 &0\end{bmatrix}\)Obtain the values of a, b, c, x, y and z. |
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Answer» Given two matrices are equal. \(\begin{bmatrix}x+3 & z+4 & 2y-7 \\[0.3em]4x+6 & a-1 & 0 \\[0.3em]b-3 & 3b &z+2c\end{bmatrix}\)= \(\begin{bmatrix}0 & 6 & 3y-2 \\[0.3em]2x & -3 & 2c+2 \\[0.3em]2b+4 & -21 &0\end{bmatrix}\) We know that if two matrices are equal then the elements of each matrices are also equal. ∴x + 3 = 0 ⇒ x = 0 – 3 = – 3 …(1) And z + 4 = 6 ⇒ z = 6 – 4 = 2 …(2) And 2y – 7 = 3y – 2 ⇒ 2y – 3y = – 2 + 7 ⇒ – y = 5 ⇒ y = – 5 …(3) 4x + 6 = 2x …(4) a – 1 = – 3 ⇒ a = – 3 + 1 = – 2 …(5) 2c + 2 = 0 ⇒ 2c = – 2 ⇒ c = – 1 … (6) b – 3 = 2b + 4 ⇒ b – 2b = 4 + 3 ⇒ – b = 7 ⇒ b = – 7 … (7) ∴ x = – 3, y = – 5, z = 2 and a = – 2, b = – 7, c = – 1 |
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| 585. |
If \(\begin{bmatrix} x & 3x-y\\[0.3em] 2x+z & 3y-\omega \\[0.3em] \end{bmatrix} \)\(=\begin{bmatrix} 3 & 2 \\[0.3em] 4 & 7 \\[0.3em] \end{bmatrix},\) find x,y,z,ω. |
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Answer» Given two matrices are equal, \(\begin{bmatrix} x & 3x-y\\[0.3em] 2x+z & 3y-\omega \\[0.3em] \end{bmatrix} \)\(=\begin{bmatrix} 3 & 2 \\[0.3em] 4 & 7 \\[0.3em] \end{bmatrix}\) We know that if two matrices are equal then the elements of each matrices are also equal. ∴x = 3 …(1) And 3x – y = 2 …(2) And 2x + z = 4 …(3) 3y – ω = 7 …(4) Putting the value of x in equation (2), 3 × 3 – y = 2 ⇒ 9 – y = 2 ⇒ y = 9 – 2 ⇒ y = 7 Now, putting the value of y in equation (4), 3×7 – ω = 7 ⇒ 21 – ω = 7 ⇒ ω = 21 – 7 ⇒ ω = 14 Again, Putting the value of x in equation (3), 2 × 3 + z = 4 ⇒ 6 + z = 4 ⇒ z = 4 – 6 ⇒ z = – 2 ∴ x = 3, y = 7, z = – 2 and ω = 14 |
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| 586. |
If \(\begin{bmatrix}2x+1 & 5x \\[0.3em]0 & y^2+1 \\[0.3em]\end{bmatrix}\) = \(\begin{bmatrix}x+3 & 10 \\[0.3em]0 & 26 \\[0.3em]\end{bmatrix},\)find the value of (x + y). |
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Answer» Given two matrices are equal. \(\begin{bmatrix}2x+1 & 5x \\[0.3em]0 & y^2+1 \\[0.3em]\end{bmatrix}\) = \(\begin{bmatrix}x+3 & 10 \\[0.3em]0 & 26 \\[0.3em]\end{bmatrix}\) We know that if two matrices are equal then the elements of each matrices are also equal. ∴2x + 1 = x + 3 … (1) ⇒ 2x – x = 3 – 1 ⇒ x = 2 And 5x = 10 … (2) y2 + 1 = 2 … (3) ⇒ y2 = 26 – 1 ⇒ y2 = 25 ⇒ y = 5 or – 5 ∴ x = 2, y = 5 or – 5 ∴ x + y = 2 + 5 = 7 Or x + y = 2 – 5 = – 3 |
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| 587. |
Give an example of a row matrix which is also a column matrix. |
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Answer» As we know that order of a row matrix = 1 x n and order of a column matrix = m x 1 So, order of a row as well as column matrix = 1 x 1 Therefore, required matrix A = [aij]1 x 1 |
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| 588. |
If `A=[alphabetagamma-alpha]`is such that `A^2=I`, then`1+alpha^2+betagamma=0`(b) `1-alpha^2+betagamma=0`(c) `1-alpha^2-betagamma=0`(d) `1+alpha^2-betagamma=0`A. `1+alpha^(2)+betagamma =0`B. `1-alpha^(2) +beta gamma =0`C. `1-alpha^(2)-beta gamma=0`D. `1+alpha^(2)-beta gamma =0` |
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Answer» Correct Answer - A `A^(2) =I` `implies [{:(alpha,beta),(gamma,-alpha):}][{:(alpha,beta),(gamma,-alpha):}]=I` `implies[{:(alpha^(2)+betagamma,alphabeta-betaalpha),(gammaalpha -alphagamma,betagamma +alpha^(2)):}]=I` `implies [{:(alpha^(2) +beta gamma ,0),(0, alpha^(2) +betagamma ):}]=[{:(1,0),(0,1):}]` `alpha^(2) +beta gamma =1 implies 1-alpha^(2) -betagamma =0` |
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| 589. |
If `A=[1 1-1 2 0 3 3-1 2]`, `B=[1 0-13 2 4]`and `C=[1 2 3-4 2 0-2 1]`, find A(BC), (AB)C and show that `(A B)C = A(B C)`. |
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Answer» Here, `BC = [[1+6,2+0,3-6,-4+3],[0+4,0+0,0-4,0+2],[-1+8,-2+0,-3-8,4+4]]` `= [[7,2,-3,-1],[4,0,-4,2],[7,-2,-11,8]]` Now, `A(BC) = [[1,1,-1],[2,0,3],[3,-1,2]][[7,2,-3,-1],[4,0,-4,2],[7,-2,-11,8]]` `A(BC) = [[7+4-7,2+0+2,-3-4+11,-1+2-8],[14+0+21,4+0-6,-6+0-33,-2+0+24],[21-4+14,6-0-4,-9+4-22,-3-2+16]]` `=[[4,4,4,-7],[35,-2,-39,22],[31,2,-27,11]]` Now, `AB = [[1,1,-1],[2,0,3],[3,-1,2]] [[1,3],[0,2],[-1,4]]` `=[[1+0+1,3+2-4],[2+0-3,6+0+12],[3+0-2,9-2+8]]` `=[[2,1],[-1,18],[1,15]]` `(AB)C = [[2,1],[-1,18],[1,15]][[1,2,3,-4],[2,0,-2,1]]` `=[[2+2,4+0,6-2,-8+1],[-1+36,-2+0,-3-36,4+18],[1+30,2+0,3-30,-4+15]]` `=[[4,4,4,-7],[35,-2,-39,22],[31,2,-27,11]]` `:. A(BC) = (AB)C. ` |
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| 590. |
Two farmers Ramkishan and Gurcharan Singh cultivates only three varieties of rice namely Basmati, Permal and Naura. The sale (in Rupees) of these varieties of rice by both the farmers in the month of September and October are given by the following matrices A and B.(i) Find the combined sales in September and October for each farmer in each variety.(ii) Find the decrease in sales from September to October.(iii) if both farmers receive 2% profit on gross sales, compute the profit for each farmer and for each variety sold in October. |
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Answer» combined sale`=A+B` `[[10000+5000,20000+10000,30000+6000],[50000+20000,30000+10000,10000+10000]]` `=[[15000,30000,36000],[70000,40000,20000]]` Decrease in sale`=A+B` `[[10000-5000,20000-10000,30000-6000],[50000-20000,30000-10000,10000-10000]]` `=[[5000,10000,24000],[30000,20000,0]]` Profit`=2%of sales` `2%`of sales in october `=2/(100)xxB` `=0.02xxB` `=0.02[[5000,10000,6000],[20000,10000,10000]]` `=[[0.02xx5000,0.02xx10000,0.02xx6000],[0.02xx20000,0.02xx10000,0.02xx10000]]` `=[[100,200,120],[400,200,200]]` |
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| 591. |
If `n=p`, then the order of the matrix `7X-5Z`is:(A) `p xx2` (B) `2xx n` (C) `n xx3` (D) `p xxn` |
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Answer» it is given that X=2Xn Z=2Xp n=p (7X-5Z)=it either can be (2Xn or 2Xp) option 2 is correct. |
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| 592. |
the restriction on n, k and p so that PY +Wywill be defined are :A. K=3,p=nB. K is arbitrary , p=2C. p is arobitray , k-3D. `k=2,p=3` |
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Answer» Correct Answer - A (A) Order of matrix p=p `xx`k Order of matrix `Y=3xxk` `therfore `PY will be defined if K =3 and the order pf `Py=3xxk=pxx3` Order of matrix `W=nxx3` Order of matrix` Y=3xxK` here the matrix WY is definded and it order `=nxk=nxx3.` Now ,Py +Wy will be definded if p=n `therefore p=n and l=3` |
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| 593. |
Matrices A and B will be inverse of each other onlyifA. AB=BAB. AB=BA=0C. AB=O,BA=ID. AB+BA=I |
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Answer» Correct Answer - A (d) if a is any square matrix and a matrix B of same order is such that AB=1 BA then A and B are inverse of each other. |
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| 594. |
If `n" "=" "p`, then theorder of the matrix `7X" "" "5Z`is:(A) `p xx2`(B) `2xx n`(C) `n xx3`(D) `p xxn`A. `pxx2`B. `2xxn`C. `nxx3`D. `pxxn` |
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Answer» Correct Answer - A (b) Order of mattrix`X=2xxn` Order of matrix `Z=2xxp` `therefore 7X-5Z` will be definded if `p=n` and its order `=2xxn` |
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| 595. |
If A is a square matric of order 5 and `2A^(-1)=A^(T)`, then the remainder when `|"adj. (adj. (adj. A))"|` is divided by 7 isA. 2B. 3C. 4D. 5 |
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Answer» Correct Answer - A We have, `2A^(-1)=A^(T)` `implies |2A^(-1)|=|A^(T)|` `implies 2^(5) 1/(|A|)=|A|` `implies |A|^(2)=2^(5)` `:. |"adj. (adj. (adj. A))"|=|A|^(4^(3))=(|A|^(2))^(32)=2^(160)` `2^(160)/7=(2(2^(3))^(53))/7=(2(7+1)^(53))/7= m+2/7`, where m is integer Thus, required remainder is 2. |
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| 596. |
Let A be a square matric of order 3 × 3. Write the value of |2A|, where |A| = 4. |
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Answer» |2A| = 2n |A| Where n is order of matrix A. |
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| 597. |
Find the characteristic roots of the two-rowed orthogonal matrix `[(cos theta,-sin theta),(sin theta,cos theta)]` and verify that they are of unit modulus. |
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Answer» We have, `|A-lambdaI|=|(cos theta - lambda,-sin theta),(sin theta,cos theta - lambda)|` `=(cos theta-lambda)^(2)+sin^(2) theta` Therefore, characteristic equation of A is `(cos theta-lambda)^(2)+sin^(2) theta=0` or `cos theta-lambda= pm i sin theta` or `lambda=cos theta pm i sin theta` Which are of unit modulus. |
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| 598. |
Construct a `3 xx 4`matrix, whose elements are given by:(i) `a_(i j)=1/2|-3i+j|` (ii) `a_(i j)=2i-j` |
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Answer» Correct Answer - `[(1,1/2,0,1/2),(5/2,2,3/2,1),(4,7/2,3,5/2)]` A `3xx4` matrix is given by `A=[(a_(11),a_(12),a_(13),a_(14)),(a_(21),a_(22),a_(23),a_(24)),(a_(31),a_(32),a_(33),a_(34))]` `a_("ij")=1/2 |-3i+j|, i=1, 2, 3` and `j=1, 2, 3, 4` `:. a_(11)=1/2 |-3xx1+1|=1/2 |-3+1|=1/2|-2|=2/2=1` `a_(21)=1/2|-3xx2+1|=1/2 |-6+1|=1/2|-5|=5/2` `a_(31)=1/2|-3xx3+1|=1/2|-9+1|=1/2|-8|=8/2=4` `a_(12)=1/2 |-3xx1+2|=1/2|-3+2|=1/2 |-1|=1/2` `a_(22)=1/2 |-3xx2+2|=1/2|-6+2|=1/2 |-4|=4/2=2` `a_(32)=1/2 |-3xx3+2|=1/2 |-9+2|=1/2 |-7|=7/2` `a_(13)=1/2 |-3xx1+3|=1/2 |-3+3|=0` `a_(23)=1/2 |-3xx2+3|=1/2 |-6+3|=1/2|-3|=3/2` `a_(33)=1/2|-3xx3+3|=1/2|-9+3|=1/2 |-6|=6/2=3` `a_(14)=1/2 |-3xx1+4|=1/2 |-3+4|=1/2 |1|=1/2` `a_(24)=1/2 |-3xx2+4|=1/2 |-6+4|=1/2 |-2|=2/2=1` `a_(34)=1/2 |-3xx3+4|=1/2 |-9+4|=1/2 |-5|=5/2` Therefore, the required matrix is `A=[(1,1/2,0,1/2),(5/2,2,3/2,1),(4,7/2,3,5/2)]` |
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| 599. |
If A = \(\begin{bmatrix} 1 &a \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\),then An (where n ∈ N) equals :A. \(\begin{bmatrix} 1 &na \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\) B. \(\begin{bmatrix} 1 &n^2a \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\) C. \(\begin{bmatrix} 1 &na \\[0.3em] 0 &0 \\[0.3em] \end{bmatrix}\)D.\(\begin{bmatrix} n &na \\[0.3em] 0 &n \\[0.3em] \end{bmatrix}\) |
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Answer» (A). \(\begin{bmatrix} 1 &na \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\) A = \(\begin{bmatrix} 1 &a \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\) An = \(\begin{bmatrix} 1 &a \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\)x \(\begin{bmatrix} 1 &a \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\)x\(\begin{bmatrix} 1 &a \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\)x\(\begin{bmatrix} 1 &a \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\) {n times, (where n ∈ N)} An = \(\begin{bmatrix} 1 &na \\[0.3em] 0 &1 \\[0.3em] \end{bmatrix}\) Option (A) is the answer. |
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| 600. |
If A is idempotent matrix, then show that `(A+I)^(n) = I+(2^(n)-1) A, AAn in N,` where I is the identity matrix having the same order of A. |
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Answer» `because ` A is idempotent matrix `therefore A^(2) = A`, similarly `A = A^(2) = A^(3) =A^(4) =... = A^(n)` …(i) Now, `(A+I)^(n) = (I+A)^(n)` `I+""^(n)C_(1)A+""^(n) C_(2) A^(2)+""^(n)C_(3) A^(3) +...+ ""^(n) C_(n) A^(n) ` `I+(""^(n)C_(1)+""^(n) C_(2) +""^(n)C_(3) +...+ ""^(n) C_(n)) A ` [ from Eq.(i)] `I+(2^(n)-1)A` Hence, `(A+I)^(n) = I + (2^(n)-1) A, AA n in N.` |
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