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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.
| 451. |
If `A=[{:(2,-1),(3,4),(1,5):}]" and "B=[{:(-1,3),(2,1):}]`, find AB. Does BA exist? |
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Answer» Here A is a `3xx2` matrix and B is a `2xx2` matrix. Clearly, the number of columns in A equals the number of rows in B. `:.` AB exists and it is a `3xx2` matrix. Now, `AB=[{:(2,-1),(3,4),(1,5):}][{:(-1,3),(2,1):}]` `=[{:(2.(-1)+(-1).2,2.3+(-1).1),(3.(-1)+4.2,3.3+4.1),(1.(-1)+5.2,1.3+5.1):}]` `=[{:(-4,5),(5,13),(9,8):}].` Further, B is a `2xx2` matrix and A is a `3xx2` matrix. So, the number of columns in B is not equal to the number of rows in A. So, BA does not exist. |
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| 452. |
Solve the equation by inversion method :2x + 6y = 8, x + 3y = 5 |
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Answer» Given, 2x + 6y = 8, x + 3y = 5 The given equations can be written in the matrix form as : \( \begin{bmatrix} 2 & 6 \\[0.3em]1 &3 \\[0.3em] \end{bmatrix} \) \( \begin{bmatrix} x \\[0.3em] y\\[0.3em] \end{bmatrix} \) = \( \begin{bmatrix} 8 \\[0.3em] 5\\[0.3em] \end{bmatrix} \) This is of the form AX = B, where A = \( \begin{bmatrix} 2 & 6 \\[0.3em]1 &3 \\[0.3em] \end{bmatrix} \), X = \( \begin{bmatrix} x \\[0.3em] y\\[0.3em] \end{bmatrix} \) and B = \( \begin{bmatrix} 8 \\[0.3em] 5\\[0.3em] \end{bmatrix} \) Let us find A-1. |A| = \( \begin{bmatrix} 2 & 6 \\[0.3em]1 &3 \\[0.3em] \end{bmatrix} \)= 6 – 6 = 0 ∴ A-1 does not exist. Hence, x and y do not exist. |
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| 453. |
Solve the equation by reduction method :x + 3y = 2, 3x + 5y = 4. |
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Answer» Given, x + 3y = 2, 3x + 5y = 4. The given equations can be written in the matrix form as : \( \begin{bmatrix} 1 & 3 \\[0.3em]3 &5 \\[0.3em] \end{bmatrix} \) \( \begin{bmatrix} x \\[0.3em] y\\[0.3em] \end{bmatrix} \) = \( \begin{bmatrix} 2 \\[0.3em] 4\\[0.3em] \end{bmatrix} \) By R2 – 3R1, we get \( \begin{bmatrix} 1 & 3 \\[0.3em]0 &-4 \\[0.3em] \end{bmatrix} \)\( \begin{bmatrix} x \\[0.3em] y\\[0.3em] \end{bmatrix} \) = \( \begin{pmatrix} 2 \\[0.3em] -2\\[0.3em] \end{pmatrix} \). ∴ \( \begin{bmatrix} x+3 \\[0.3em] 0-4y\\[0.3em] \end{bmatrix} \) = \( \begin{bmatrix} 2 \\[0.3em] -2\\[0.3em] \end{bmatrix} \) By equality of matrices, x + 3y = 2 …(1) - 4y = - 2 From (1), y = \(\frac{1}{2}\) Substituting y = \(\frac{1}{2}\) in (1), we get, x + \(\frac{3}{2}\) = 2 ∴ x = 2 – \(\frac{3}{2}\) = \(\frac{1}{2}\) Hence, x = \(\frac{1}{2}\), y = \(\frac{1}{2}\) is the required solution. |
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| 454. |
If `A=[{:(5,4),(2,3):}]" and "B=[{:(3,5,1),(6,8,4):}]`, find AB and BA whichever exists. |
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Answer» Here, A is a `2xx2` matrix and B is a `2xx3` matrix. Clearly, the number of column in A =number of rows in B. `:.` AB exists and it is a `2xx3` matrix. `AB=[{:(5,4),(2,3):}][{:(3,5,1),(6,8,4):}]` `=[{:(5.3+4.6,5.5+4.8,5.1+4.4),(2.3+3.6,2.5+3.8,2.1+3.4):}]` `=[{:(15+24,25+32,5+16),(6+18,10+24,2+12):}]=[{:(39,57,21),(24,34,14):}].` Again, B is a `2xx3` matrix and A is a `2xx2` matrix. `:.` number of columns in `Bne` number of rows in A. So, BA does not exist. |
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| 455. |
Let `A=[{:(1,-2,3),(-4,2,5):}]" and "B=[{:(2,3),(4,5),(-2,1):}].` Find AB and BA, and show that `ABneBA.` |
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Answer» Here A is a `2xx3` matrix and B is a `3xx2` matrix. So, AB exists and it is a `2xx2` matrix. Now, `AB=[{:(1,-2,3),(-4," "2,5):}][{:(2,3),(4,5),(-2,1):}]` `=[{:(1.2+(-2).4+3.(-2),1.3+(-2).5+3.1),((-4).2+2.4+5.(-2),(-4).3+2.5+5.1):}]` `=[{:(-12,-4),(-10," "3):}].` Again, B is a `3xx2` matrix and A is a `2xx3` matrix. So, BA exists and it is a `3xx3` matrix. Now, `BA=[{:(2,3),(4,5),(-2,1):}][{:(1,-2,3),(-4," "2,5):}]` `=[{:(2.1+3.(-4),2.(-2)+3.2,2.3+3.5),(4.1+5.(-4),4.(-2)+5.2,4.3+5.5),((-2).1+1.(-4),(-2).(-2)+1.2,(-2).3+1.5):}]` `=[{:(-10,2,21),(-16,2,37),(-6,6,-1):}].` Hence, `ABneBA.` |
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| 456. |
If `A=[{:(1,2,3), (0,4,2),(0,0,6):}]`, then the minor of the element `a_(31)` is |
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Answer» Correct Answer - B |
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| 457. |
A finance company has offices located in ewery division, every didtrict and every taluka in a certain state in India. Assume that there are five divisions, thirty districts and 200 talukas in the state. Each office has one head clerk, one cashier, one clerk and one peon. A divisional office has, in addition, one office superntendent, two clerks, one typist and one poen. A district office, has in addition, one clerk and one peon. The basic monthly salaries are as follows : Office superintendernt Rs 500, Head clerk Rs 200, cashier Rs 175, clerks and typist Rs 150 and peon Rs 100. Using matrix motation find the total basic monthly salary bill of each kind of officeA.B.C.D. |
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Answer» Correct Answer - Total basic monthly salary bill of each division of district and taluka offices an Rs 1675, Rs 675 and Rs 625, respectively. Lets us use the symbols Div, Dis, Tal for division, district, taluka respectively and O, H, C, Cl , T and P for office superintendent, Head clerk, Cashier, Clerk, Typist and Peon respectively. Then the number of offices can be arranged as elements of a row matrix A and the composition of staffin various offices can be areanged in a `3xx6` matrix B (say). `{:(Div, Dis, Tal):}` `therefore A = [[5,30,200]]` and `B=[[1,1,1,2+1,1,1+1],[0,1,1,1+1,0,1+1],[0,1,1,1,0,1]]` or `B= [[1,1,1,3,1,2],[0,1,1,2,0,2],[0,1,1,1,0,1]]` The basic monthly salarise of various types of employees of these offies correspond to the elements of the column matrix C. `thereforeC={:[O],[H],[C],[Cl],[T],[P]:}[[500],[200],[175],[150],[150],[100]]` The total basic monthly salary bill of each kind of office = BC `{:[(O,H,C,Cl,T,P),(1,1,1,3,1,2),(0,1,1,2,0,2),(0,1,1,1,0,1)]:}xx[[500],[200],[175],[150],[150],[100]]{:[O],[H],[C],[Cl],[T],[P]:}` `{:[(500+ 200 + 175+ 3xx 150 + 1xx 150 + 2 xx100),(0+1xx200+1xx175+2xx150+0+2xx100),(0+1xx200+1xx175+1xx150+0+1xx100)]:}` `=[[1675],[875],[625]]` i.e. The total basic monthly salary bill of each divisional, distriict and taluka offices are `Rs 1675, Rs 875 and Rs 625, respectively. |
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| 458. |
A finance company has offices located in ewery division, every didtrict and every taluka in a certain state in India. Assume that there are five divisions, thirty districts and 200 talukas in the state. Each office has one head clerk, one cashier, one clerk and one peon. A divisional office has, in addition, one office superntendent, two clerks, one typist and one poen. A district office, has in addition, one clerk and one peon. The basic monthly salaries are as follows : Office superintendernt Rs 500, Head clerk Rs 200, cashier Rs 175, clerks and typist Rs 150 and peon Rs 100. Using matrix motation find the total unmber of posts of each kind in all the offices taken together,A.B.C.D. |
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Answer» Correct Answer - Number of posts in all the offices taken together are 5 office superintendents; 235 had clerks; 235 cashiers; 275 clerks; 5 typisit and 270 Lets us use the symbols Div, Dis, Tal for division, district, taluka respectively and O, H, C, Cl , T and P for office superintendent, Head clerk, Cashier, Clerk, Typist and Peon respectively. Then the number of offices can be arranged as elements of a row matrix A and the composition of staffin various offices can be areanged in a `3xx6` matrix B (say). `{:(Div, Dis, Tal):}` `therefore A = [[5,30,200]]` and `B=[[1,1,1,2+1,1,1+1],[0,1,1,1+1,0,1+1],[0,1,1,1,0,1]]` or `B= [[1,1,1,3,1,2],[0,1,1,2,0,2],[0,1,1,1,0,1]]` The basic monthly salarise of various types of employees of these offies correspond to the elements of the column matrix C. `thereforeC={:[O],[H],[C],[Cl],[T],[P]:}[[500],[200],[175],[150],[150],[100]]` Total number of Posts = AB `{:(O,H,C,Cl,T,P):}` `{:(,"Div","Dis","Tal"),("[",5,30,200" ]"):}xx =[{:(,1,1,1,3,1,2),(,0,1,1,2,0,2),(,0,1,1,1,0,1):}]` `{:( ,O,H,C,Cl,T,P),(=,"["5,235,235,275,5,270"]"):}` i.e. Required number of posta in all the offices taken together are 5 office Suprintendents, 235 Head Clareks, 235 Cashiers, 275 Clerks, 5 Typists and 270 Peons. |
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| 459. |
Find the inverse of the following matrices by the adjoint method :\( \begin{bmatrix}2&-2 \\[0.3em]4 & 3 \\[0.3em] \end{bmatrix} \)[2,-2,4,3] |
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Answer» Let A = [2,-2,4,3] \( \begin{bmatrix}2&-2 \\[0.3em]4 & 3 \\[0.3em] \end{bmatrix} \) |A| = = 6 + 8 = 14 ≠ 0 ∴ A-1 exist First we have to find the co-factor matrix = [Aij]2x2 where Aij = (-1)i+jMij Now, A11 = (-1)1+1 M11 = 3 A12 = (-1)1+2M = -4 A21 = (-2)2+1M21 = (-2) = 2 A22 = (-1)2+2M22 = 2 Hence the co-factor matrix = \( \begin{bmatrix}A_{11}&A_{12} \\[0.3em]A_{21} & A_{22} \\[0.3em] \end{bmatrix} \) = \( \begin{bmatrix}3&-4 \\[0.3em]2 & 2 \\[0.3em] \end{bmatrix} \) ∴ adj A = \( \begin{bmatrix}3&2 \\[0.3em]-4& 2 \\[0.3em] \end{bmatrix} \) ∴ A-1 = \(\frac{1}{|A|}\) (adj A) = \(\frac{1}{14}\)\( \begin{pmatrix}3&2 \\[0.3em]-4& 2 \\[0.3em] \end{pmatrix} \) |
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| 460. |
If `A=[(1,-1),(2,3)]` , then `M_(21)`=A. 1B. `-1`C. 2D. 3 |
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Answer» Correct Answer - B |
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| 461. |
Show that the matris `[[l_(1),m_(1),n_(1)],[l_(2),m_(2),n_(2)],[l_(3),m_(3),n_(3)]]` is orthogonal, `if l_(1)^(2) + m_(1)^(2) + n_(1)^(2) = Sigmal_(1)^(2) = 1 = Sigma l_(2)^(2) = Sigma_(3) ^(2) and` `l_(1) l_(2) + m_(1)m_(2) + n_(1) n_(2) = Sigma l_(1)l_(2) =0 = Sigma l_(2)l_(3) = Sigma l_(3) l_(1).`A.B.C.D. |
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Answer» `Let A =[[l_(1) , m_(1), n_(1)],[l_(2), m_(2), n_(2) ],[l_(3),m_(3),n_(3)]]` `therefore A^(T) = [[l_(1),l_(2), 1_(3)],[m_(1) ,m_(2),m_(3) ],[n_(1),n_(2),n_(3)]]` Now, `A A^(T) =[[l_(1) , m_(1), n_(1)],[l_(2), m_(2), n_(2) ],[l_(3),m_(3),n_(3)]]xx [[l_(1),l_(2), 1_(3)],[m_(1) ,m_(2),m_(3) ],[n_(1),n_(2),n_(3)]]` `= [[Sigmal_(1)^(2),Sigmal_(1) l_(2) ,Sigmal_(3)l_(1)],[Sigmal_(1)l_(2),Sigmal_(2)^(2), Sigmal_(2)l_(3) ],[Sigmal_(3)l_(1), Sigmal_(2)l_(3),Sigmal_(3)^(2)]]= [[1,0,0],[0,1,0],[0,0,1]]=I` Hence, matrix A is orthogonal. |
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| 462. |
Find the adjoint of the matrices :\( \begin{bmatrix}2&-3 \\[0.3em]3 & 5 \\[0.3em]\end{bmatrix}\)[2,-3,3,5] |
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Answer» Let A = [2,-3,3,5] \( \begin{bmatrix}2&-3 \\[0.3em]3 & 5 \\[0.3em]\end{bmatrix}\) Here, a11 = 2, M11 = 5 ∴ A11 = (-1)1+1(5) = 5 a12 = -3, M12 = 3 ∴ A12 = (-1)1+2(3) = -3 a21 = 3, M21 = -3 ∴ A21 = (-1) (-3) = 3 a22 = 5, M22 = 2 ∴ A22 = (-1)2+1 = 2 ∴The co-factor matrix = \( \begin{bmatrix}A_{11}&A_{12} \\[0.3em]A_{21} & A_{22} \\[0.3em]\end{bmatrix}\) = \( \begin{bmatrix}5&-3 \\[0.3em]3 & 2 \\[0.3em]\end{bmatrix}\) ∴ adj A = \( \begin{pmatrix}5&3 \\[0.3em]-3 & 2 \\[0.3em]\end{pmatrix}\) |
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| 463. |
If `A=[{:(" 1",1," 1"), (" 2",1,-3),(-1,2," 3"):}]` , then `M_(23)=`A. 3B. 4C. 5D. `-1` |
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Answer» Correct Answer - A |
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| 464. |
The co-factor `A_(12)` for the matrix `A=[(-2,3),(-3,5)]` isA. 3B. `-3`C. 2D. `-5` |
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Answer» Correct Answer - A |
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| 465. |
Find the matrix of co-factors for the matrices :\( \begin{bmatrix}1& 3 \\[0.3em]4 & -1 \\[0.3em]\end{bmatrix}\)[1,3,4,-1] |
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Answer» Let A = [1,3,4,-1] \( \begin{bmatrix}1& 3 \\[0.3em]4 & -1 \\[0.3em]\end{bmatrix}\) Here, a11 = 1, M11 = -1 ∴ A11 = (-1)1+1(-1) = -1 a12 = 3, M12 = 4 ∴ A12 = (-1)1+2(4) = -4 a21 = 4, M21 = 3 ∴ A21 = (-1)2+1(3) = -3 a22 = -1, M22 = 1 ∴ A22 = (-1)2+1(1) = 1 ∴ The co-factor matrix = \( \begin{bmatrix}A_{11}& A_{12} \\[0.3em]A_{21} &A_{22} \\[0.3em]\end{bmatrix}\) = \( \begin{pmatrix}-1& -4 \\[0.3em]-3 &1 \\[0.3em]\end{pmatrix}\) |
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| 466. |
Find the co-factors of the elements of matrices :\( \begin{bmatrix}-1&2 \\[0.3em]-3 & 4 \\[0.3em]\end{bmatrix}\)[-1,2,-3,4] |
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Answer» Let A = [-1,2,-3,4] \( \begin{bmatrix} -1&2 \\[0.3em] -3 & 4 \\[0.3em] \end{bmatrix} \) Here, a11 = -11,M11 = 4 ∴ A11 = (-1)1+1(4) = 4 a12 = 2, M12 = -3 ∴ A12 = (-1)1+2(- 3) = 3 a21 = – 3, M21 = -2 ∴ A21 = (- 1)2+1(2) = -2 a22 = 4, M22 = -1 ∴ A22 = (-1)2+2(-1) = -1. |
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| 467. |
if `A=[[2,-3],[-4,1]]` then `(3A^2+12A)=?`A. `[[72,-63],[-84,51]]`B. `[[72,-84],[-63,51]]`C. `[[51,63],[84,72]]`D. `[[51,84],[63,72]]` |
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Answer» Correct Answer - C `because A^(2) = [[2,-3],[-4,1]][[2, -3],[-4, 1]]= [[16,-9],[-12,13]]` |
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| 468. |
If `A=[[1, 1, 1],[ 1, 1, 1],[ 1, 1, 1]]`, then prove that `A^n=[[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)],[3^(n-1),3^(n-1),3^(n-1)]]`for every positive integer `n`. |
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Answer» We shall prove the result by using the principle of mathematical induction. When n=1, we have `A^(1)=[{:(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)):}]=[{:(3^(0),3^(0),3^(0)),(3^(0),3^(0),3^(0)),(3^(0),3^(0),3^(0)):}]=[{:(1,1,1),(1,1,1),(1,1,1):}].` Thus, the result is true for `n=1`. Let it be true for n=k. Then, `A^(k)=[{:(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)):}].` `:." "A^(k-1)=A.A^(k)` `=[{:(1,1,1),(1,1,1),(1,1,1):}][{:(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)),(3^(k-1),3^(k-1),3^(k-1)):}]` `=[{:(3(3^(k-1)),3(3^(k-1)),3(3^(k-1))),(3(3^(k-1)),3(3^(k-1)),3(3^(k-1))),(3(3^(k-1)),3(3^(k-1)),3(3^(k-1))):}]=[{:(3^(k),3^(k),3^(k)),(3^(k),3^(k),3^(k)),(3^(k),3^(k),3^(k)):}].` Thus, the result is true for `n=(k+1)`, whenever it is true for `n=k.` So, the result is true for all `n in N.` Hence, `A^(n)=[{:(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)),(3^(n-1),3^(n-1),3^(n-1)):}]`, for all values of `n inN.` |
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| 469. |
If A=`[[3,-4] , [1,-1]]` then by the method of mathematical induction prove that `A^n=[[1+2n,-4n] , [n,1-2n]]` |
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Answer» We shall prove the result by using the principle of mathematical induction. When `n=1`, we have `A^(1)=[{:(1+2.1,-4.1),(1,1-2.1):}]=[{:(1+2,-4),(1,1-2):}]=[{:(3,-4),(1,-1):}].` Thus, the result it true for n=1. Let the result be true for n=k. Then, `A^(k)=[{:(1+2k,-4k),(k,1-2k):}].` `:." "A^(k+1)=A.A^(k)` `=[{:(3,-4),(1,-1):}][{:(1+2k,-4k),(k,1-2k):}]` `=[{:(3+6k-4k,-12k-4+8k),(1+2k-k,-4k-1+2k):}]=[{:(3+2k,-4k-4),(k+1,1-2k):}]` `=[{:(1+2(k+1),-4(k+1)),(k+1,1-2(k+1)):}].` Thus, the result is true for `n=(k+1)`, whenever it is true for `n=k.` So, the result is true for all `n inN.` Hence, `A^(n)=[{:(1+2n,-4n),(n,1-2n):}]` for all values of `n inN.` |
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| 470. |
If `[2x4][{:(x),(-8):}]=O`, find the positive value of x. |
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Answer» The given matrix equation is `AB=O`, where A is a `(1xx2)` matrix and B is a `(2xx1)` matrix. So, AB is a `(1xx1)` matrix. So, O is a `(1xx1)` matrix. `:." "[2x4][{:(x),(-8):}]=[0]` `implies" "2x^(2)-32=0implies2x^(2)=32` `implies" "x^(2)=16impliesx=sqrt(16)=4.` Hence, x=4. |
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| 471. |
Show that the matrix B′ AB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric. |
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Answer» We suppose that A is a symmetric matrix, then A′ = A Consider (B′ AB)′ ={B′ (AB)}′ = (AB)′ (B′ )′ [∵ (AB)′ = B ′ A′ ] = B′ A′ (B) [∵ (B′ )′ = B] = B′ (A′ B) = B′ (AB) [∵ A′ = A] ⇒ (B′ AB)′ = B′ AB which shows that B′ AB is a symmetric matrix. Now, we suppose that A is a skew-symmetric matrix. Then, A′ = − A Consider (B′ AB)′ = [B′ (AB)]′ = (AB)′ (B′ )′ [∵(AB)′ = B′ A′ and (A′ )′ = A] = (B′ A′ )B = B′ (−A)B= − B′ AB [∵ A′ = − A] ⇒ (B′ AB)′ = − B′ AB which shows that B′ AB is a skew-symmetric matrix. |
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| 472. |
Matrices A and B will be inverse of each other only ifA.AB = BA B. AB = BA = 0C. AB = 0, BA = I D. AB = BA = I |
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Answer» We know that if A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is said to be the inverse of A. In this case, it is clear that A is the inverse of B. Thus, matrices A and B will be inverses of each other only if AB = BA = I. |
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| 473. |
If A and B are symmetric matrices, prove that AB − BA is a skew-symmetric matrix. |
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Answer» Here, A and B are symmetric matrices, then A′ = A and B′ = B Now, (AB − BA)′ = (AB)′ − (BA)′ (∵(A − B)′ = A′ − B′ and (AB)′ = (B′A′) = B ′A′ − A′B′ = BA − AB (∵ B′ = B and A′ = A) = −(AB − BA) ⇒(AB − BA)′ = −(AB − BA) Thus, (AB − BA) is a skew-symmetric matrix. |
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| 474. |
Transform `[(1,3,3),(2,4,10),(3,8,4)]` into a unit matrix. |
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Answer» Let `A=[(1,3,3),(2,4,10),(3,8,4)]` Applying `R^(2)to R^(2)-2R_(1) and R_(3) to R_(3)-3R_(1), ` we get `A~[(1,3,3),(0,-2,4),(0,-1,-5)]` Applying `R_(2) to ((-1)/(2))R_(2) and R_(2)to (-1)R_(2),` we get `A~[(1,3,3),(0,1,-2),(0,1,5)]` Applying `R^(1)to R^(1)-3R_(2) and R_(3) to R_(3)-R_(1),` we get `A~[(1,0,9),(0,1,-2),(0,0,7)]` Applying `R_(3) to ((1)/(7))R_(3)` we get `A~[(1,0,9),(0,1,-2),(0,0,1)]` Applying` R_(1) to R_(1)-9R_(3) and R_(2) to R_(2) +2R_(3),` we get `A~[(1,0,0),(0,1,0),(0,0,1)]` Hence `A~I` |
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| 475. |
Given `A=[(1,1,1),(2,4,1),(2,3,1)], B=[(2,3),(3,4)].` Find P such that BPA= `[(1,0,1),(0,1,0)]` |
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Answer» Given, BPA=`[(1,0,1),(0,1,0)]` ` therefore " " p=b^(-1)[(1,0,1),(0,1,0)]A^(-1)` `therefore " " B=[(2,3),(3,4)] rArr B^(-1)=(1)/((-1))[(4,-3),(-3,2)]=[(-4,3),(3,-2)]` `therefore" " B^(-1)=[(-4,3),(3,-2)]` and ` A= [(1,1,1),(2,4,1),(2,3,1)` `therefore" " |A|=1(4-3)-1(2-2)+1(6-8)=-1!=0` Now, adj `A=[(1,2,-3),(0,-1,1),(-2,-1,2)]` `therefore" " A^(-1)=(adjA)/(|A|)=[(-1,-2,3),(0,1,-1),(2,1,-2)]` substiruting the values of `A^(-1)` and `B^(-1)` from Eqs. (ii) and (iii) in Eq. (i), then `P=[(-4,3),(3,-2)][(1,0,1),(0,1,0)][(-1,-2,3),(0,1,-1),(2,1,-2)]` `P=[(-4,3),(3,-2)][(1,-1,1),(0,1,-1)][(-4,7,-7),(3,-5,5)]` |
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| 476. |
find the invese of the matraix `[(1,2,5),(2,3,1),(-1,1,1)],` using elementary row operaations. |
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Answer» Let `A=[(1,2,5),(2,3,1),(-1,1,1)],` `therefore" " |A|=[(1,2,5),(2,3,1),(-1,1,1)],=1(3-1)-2(2+1)+5(2+3)=21!=0` `therefore" " A^(-1)` exists. We write A=IA `rArr" " [(1,2,5),(2,3,1),(-1,1,1)]=[(1,0,0),(0,1,0),(-0,0,1)]` Applying `R_(2) to R_(2)` and `R_(3) to R_(3) +R_(1),` we get ` [(1,2,5),(0,-1,-9),(0,3,6)]=[(1,0,0),(-2,1,0),(1,0,1)]A` Applying `R_(2) to (-1)R_(2)` and `R_(3) to ((1)/(3))R_(3),` we get ` [(1,2,5),(0,1,9),(0,1,2)]=[(1,0,0),(2,-1,0),((1)/(3),0,(1)/(3))]A` Applying `R_(1) to R_(1)-2R_(2)` and `R_(3)to R_(3)-R_(2),` we get ` [(1,0,-13),(0,1,9),(0,0,-7)]=[(-3,2,0),(2,-1,0),((-5)/(3),1,(1)/(3))]A` Applying `R_(3) to (-(1)/(7))` we get ` [(1,0,-13),(0,1,9),(0,0,1)]=[(-3,2,0),(2,-1,0),((-5)/(21),-(1)/(7),(1)/(21))]A` Applying `R_(2) to R_(2)-9R_(3)` and `R_(1) to R_(1) +13R_(3),` we get ` [(1,0,0),(0,1,0),(0,0,1)]=[((2)/(21),(1)/(7),-(13)/(21)),(-(1)/(7),(2)/(7),(3)/(7)),((5)/(21),-(1)/(7),-(1)/(21))]A` Hence `A^(-1)=[((2)/(21),(1)/(7),-(13)/(21)),(-(1)/(7),(2)/(7),(3)/(7)),((5)/(21),-(1)/(7),-(1)/(21))]` |
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| 477. |
Given `A = ({:(3, P),(2, 5):}) " and B" = ({:(1, 0), (5, 6):}).` If AB= BA, then find p. |
| Answer» Correct Answer - p=0 | |
| 478. |
Let `A=[0 1 0 0]`show that `(a I+b A)^n=a^n I+n a^(n-1)b A`, where I is the identitymatrix of order 2 and `n in N`. |
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Answer» Let `p(n):(aI+ba)^(n)=a^(n)I+na^(n-1)ba` step I for `n=1` `LHS= (aI+ba)^(1) =aI+ba` and RHS `= a^(1)I+1.a^(0) ba=aI+ba` LHS=RHS therefore, `p(1) is true. step II Assume that `p(k) is true , then `p(k): (aI+ba)^(k) I+ka^(k-1)ba` step III for `n=k+1,` we have to prove that `p(k+1):(aI+ba)^(k+1)k=a^(k+1) I+(k+I) a^(k)bA ` LHS `=(aI+bA)^(k+1) = (aI+bA)^(k) (aI+bA)` `=a^(k+1) I^(2) + a^(k)b (IA) + ka^(k)b (AI)+k a^(k-1)b^(2) A^(2)` `=a^(k+1) I+(k+1)a^(k)b A+0` `[therefore AI=A,A^(2)=0and I^(2) = I]` `=a^(k+1)I+(k+1)a^(k)bA=RHS` therefore, `P(k+1)` is true. Hence, by the principal of mathematical12 induction `p(n)` is true for all n `in` N. |
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| 479. |
If `A = [(3,1),(2,1)]` find the value of `|a|+|b|` such that `A^(2)+aA+bl=O.` Hece,find `A^(-1)` |
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Answer» we have, `A=[(3,1),(2,0)]` then `|A|=|(3,1),(2,1)|=3-2=1!=0` `therefore" " A^(-1)`. Exists. Now `A^(2)=A.A=[(3,1),(2,1)][(3,1),(2,1)]=[(11,4),(8,3)]` `rArr" " [(11,4),(8,3)]+a[(3,1),(2,1)]+b[(1,0),(0,1)]=[(0,0),(0,0)]` `rArr" " [(11+3a+b,4+a),(8+2a,3+a+b)]+[(0,0),(0,0)]` Equating the corresponding elements, we get `11a+3a+b=0` `4+a=0` ` 3+a+b=0` from Eqs. (ii) and (iv), we get `a=-4` and `b=1` `therefore" " |a|+|b|=|-4|+|1|=4+1=5` As`" " A^(2)+aA+bI=O` ` rArr " " A^(2)-4A+I+O rArr I=4A-A^(2)` `rArr " " IA^(-1) = 4(A A^(-1))-A(A A^(-1))` `=4I-AI=4I-A` `= 4[(1,0),(0,1)]-[(3,1),(2,1)]=[(4,0),(0,4)]-[(3,1),(2,1)]` `therefore " " A(-1)=[(1,-1),(-2,3)]` |
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| 480. |
If `A=[{:(1,-3,2),(2," "0,2):}]" and "B=[{:(2,-1,-1),(1," "0,-1):}]`, find a matrix C such that `(A+B+C)` is a zero matrix. |
| Answer» `[{:(-3,4,-1),(-3,0,-1):}]` | |
| 481. |
Let M be a `2xx2` symmetric matrix with integer entries. Then , M is invertible, ifA. the first column of M is the transpose of the second row of MB. The second row of M is the transpose of the first column of MC. m is a diagonal matrix with non- zero entries in the main diagonalD. the product of entries in the main diagonal of M is not the square of an integer |
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Answer» Correct Answer - C::D Let `M= [[a,b],[c,d]]`, where `a, b, c, in I` M is invertible if `abs((a,b),(b,c)) ne 0 rArr ac- b^(2) ne 0 ` (a) `[[a],[b]]=[[b],[c]]rArr a = b =c rArr ac-b^(2)=0` `therefore` Option (a) is incorrect (b) `[(b,c)]= [(a,b)] rArr a = b = c rArr ac - b^(2) = 0` `therefore` Option (b) is incorrect (c) `M= [[a,0],[0,c]], ` then` abs(M) = ac ne 0` `therefore` M is invertible `therefore` Potion ( c) is correct. (d) As `acne"Integre """^(2)rArrac ne b^(2)` `therefore ` Option (d)is correct. |
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| 482. |
Let m and N be two 3x3 matrices such that MN=NM. Further if `M!=N^2` and `M^2=N^4` then which of the following are correct.A. determinant of `(M^(2)+Mn^(2))` is 0B. there is a `3xx3` non-zero matrix U such that `(M^(2)+MN^(2))U` is the zero matrixC. determinant of `(M^(2)+MN^(2)) ge 1`D. for a `3xx3` matrix `U`, is the zero matrix |
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Answer» Correct Answer - A::B `M^(2)=N^(4)` `implies M^(2)-N^(4)=O` `implies (M-N^(2)) (M+N^(2))=O" "` (as M, N commute) Also, `M ne N^(2)`. det. `((M-N^(2))(M+N^(2)))=0` as `M ne N^(2) implies` dte. `(M+N^(2))=0` Also, det `(M^(2)+MN^(2))=("det. M") ("det. "(M+N^(2)))=0` Hence, there exist non-null U such that `(M^(2)+MN^(2)) U=O`. |
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| 483. |
If A is a skew-symmetric matrix of order 2 and B, C are matrices `[[1,4],[2,9]],[[9,-4],[-2,1]]` respectively, then `A^(3) (BC) + A^(5) (B^(2)C^(2)) + A^(7) (B^(3) C^(3)) + ... + A^(2n+1) (B^(n) C^(n)),` isA. a symmetric matrixB. a skew-symmetric matrixC. an identity matrixD. None of these |
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Answer» Correct Answer - B Let, `A= [[0,a],[-a,0]], ` `BC= [[1,4],[2,9]][[9,-4],[-2,1]]=[[1,0],[0,1]]=I` `therefore B^(2) C^(2) = (BC^(2)) = I^(2) = I` Similarly, `B^(2) C^(2) = B^(3) C^(3) = ...= B^(n) C^(n) = I` Let, `D= A^(3) (BC) + A^(5) (B^(2)C^(2)) + A^(7) (B^(3) C^(3)) + ...+ A^(2n+1) (B^(n)C^(n))` `= A^(3) + A^(5) + A^(7) +...+A^(2n+1)` `=A(A^(2) + A^(4) +A^(6) +...+ A^(2n)) ` Let, `A=[[0,a],[-a,0]]` ` rArr A^(2) = [[-a^(2),0],[0,-a^(2)]]` `therefore D = IA (-a^(2) + a^(4) - a^(6)+...+ (-1)^(n) a^(2n))[agt0]` `= A (-a^(2) + a^(4) - a^(6)+...+ (-1)^(n) a^(2n)` Hence, D is skew-symmetric. |
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| 484. |
If A and B are any two different square matrices of order n with `A^3=B^3` and `A(AB)=B(BA)` thenA. `A^2+B^2=O`B. `A^2+B^2=I`C. `A^3+B^3=I`D. none of these |
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Answer» Correct Answer - D We have, `(A^2+B^2)(A-B)=A^3-A^2B+B^2A-B^3` `=A^3-A(AB)+B(BA)-B^3=O` It is given that `A-BneO " but " (A^2+B^2)(A-B)=O` may not imply that `A^2+B^2=O`. Because, the product of two non-null matrices can be a null matrix. |
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| 485. |
Let `A =[(1,-1,1),(2,1,-3),(1,1,1)]` and `10B=[(4,2,2),(-5,0,alpha),(1,-2,3)]`. If B is the inverse of A, then `alpha` is :A. 2B. -1C. 3D. 5 |
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Answer» Correct Answer - D It is given that B is the inverse of matrix A. `:. AB=I` `rArr A(10B)=10I` `rArr{:[(1,-1,1),(2,1,-3),(1,1,1)][(4,2,2),(-5,0,alpha),(1,-2,3)]=[(10,0,0),(0,10,0),(0,0,10)]:}` `rArr{:[(10,0,5-alpha),(0,10,alpha-5),(0,0,alpha+5)]=[(10,0,0),(0,10,0),(0,0,10)]rArralpha=5:}` ALITER : We have, `AB=I` `rArr A(10B)=10IrArr (A(10B))_13=0` `rArr{:[(1,-1,1)][(2),(alpha),(3)]=0rArr 5-alpha=0rArr a=5:}` |
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| 486. |
Let `{:A=[(0,0,-1),(0,-1,0),(-1,0,0)]:}`. The only correct statement aboul the matrix A isA. `A^(-1)` does not existB. `A=(-1)I` is a unit matrixC. A is a zero matrixD. `A^2=I` |
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Answer» Correct Answer - D We have, `absA=1 ne 0`. So, `A^(-1)` exists. Clearly, `A ne (-1)I and A` is not a zero or null matrix. options (a),(b) and (c ) are not correct. Now, `{:A^2=[(0,0,-1),(0,-1,0),(-1,0,0)][(0,0,-1),(0,-1,0),(-1,0,0)]=[(1,0,0),(0,1,0),(0,0,1)]=I:}` Hence, option (d) is correct. |
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| 487. |
Let `A =[(1,-1,1),(2,1,-3),(1,1,1)]` and `10B=[(4,2,2),(-5,0,alpha),(1,-2,3)]`. If B is the inverse of A, then `alpha` is :A. 5B. -1C. 2D. -2 |
| Answer» Correct Answer - a | |
| 488. |
Let `A =[(1,-1,1),(2,1,-3),(1,1,1)]` and `10B=[(4,2,2),(-5,0,alpha),(1,-2,3)]`. If B is the inverse of A, then `alpha` is : |
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Answer» Correct Answer - b |
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| 489. |
If `A=[{:(1,-5),(-3," "2),(4,-2):}]" and "B=[{:(3," "1),(2,-1),(-2," "3):}]`, find the matrix C such that `A+B+C` is a zero matrix. |
| Answer» `C=[{:(-4," "4),(" "1,-1),(-2,-1):}]` | |
| 490. |
If A and B are two square matices such that AB= A and BA = B, then find `(A^(2006) B^(2006))^(-1)`.A. `A^(-1) B^(-1)`B. `B^(-1) A^(-1)`C. ABD. Cannot be determined |
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Answer» Correct Answer - B (i) If `AB = A, BA = B " then " A^(2) =A, B^(2) = B. ` (ii) `(AB)^(-1) = B^(-1)A^(-1). ` |
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| 491. |
If `A = [{:(3, 4), (-1, 2):}] " and B"= [{:(2, -3), (4, -5):}]` , then find the determinant of AB.A. 10B. 20C. 12D. 15 |
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Answer» Correct Answer - B `"det" (AB) = "det" A * "det"B`. |
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| 492. |
The rank of the matrix `{:[(1,2,3,0),(2,4,3,2),(3,2,1,3),(6,8,7,5)]:}`, isA. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - C | |
| 493. |
Find the values of x and y for which `[{:(x,y),(3y,x):}][{:(1),(2):}]=[{:(3),(5):}].` |
| Answer» Correct Answer - `x=1,y=1` | |
| 494. |
If the trace of the matrix A is 4 and the trace of matrix B is 7, the find trace of the matrix AB.A. 4B. 7C. 28D. Cannot be determined |
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Answer» Correct Answer - D Trace `(AB) ne "Trace"(A) * "Trace" (B).` |
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| 495. |
If `A = [{:(3, 0), (0, 3):}]`, then find `A^(n),` (where `n in N)`A. `[{:(3n, 0), (0, 3n):}]`B. `[{:(3, 0), (0, 3):}]`C. `[{:(1, 0), (0, 1):}]`D. `I_(2 xx 2)` |
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Answer» Correct Answer - C `(i) A = 3I. ` (ii) ` A^(n) = 3^(n) I^(n).` |
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| 496. |
If A is an invertible matrix then `det(A^-1)` is equal to (A) 1 (B) `1/|A|` (C) `|A|` (D) none of theseA. det (A)B. `1/(det (A))`C. 1D. none of these |
| Answer» Correct Answer - B | |
| 497. |
Let A be any `3xx2` matrix. Then prove that det. `(A A^(T))=0`. |
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Answer» Let `A=[(a,p),(b,q),(c,r)]` `:. A^(T)=[(a,b,c),(p,q,r)]` `:. A A^(T)=[(a^(2)+p^(2),ab+pq,ac+pr),(ab+pq,b^(2)+q^(2),bc+qr),(ac+pr,bc+qr,c^(2)+r^(2))]` det. `(A A^(T))=|(a,p,0),(b,q,0),(c,r,0)||(a,b,c),(p,q,r),(0,0,0)|=0` |
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| 498. |
find the real value of x for which the matrix `=[(x+1,3,5),(1,x+3,5),(1,3,x+5)]` is non-singular. |
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Answer» Let `A=[(x+1,3,5),(1,x+3,5),(1,3,x+5)]` `therefore|A|=[(x+1,3,5),(1,x+3,5),(1,3,x+5)]` Applying `C_(1)rarr C_(1)+C_(2)+C_(3),` then `therefore|A|=[(x+9,3,5),(x+9,x+3,5),(x+9,3,x+5)]` Applying `R_(2)rarrR_(2)-R_(1) and R_(3)rarrR_(3)-R_(1),` then `|A|=|(x+9,3,5),(0,x,0),(0,0,x)|=x^(2)(x+9)` `therefore` A is non-singular. `therefore" " |A|!=0 rArr x^(2)(x+9)!=0 ` `therefore" " x!=0,-9` Hence,` x in R - {0,-9}.` |
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| 499. |
Which of the following is (are) NOT the square of a `3xx3`matrix with real entries?`[1 0 0 0 1 0 0 0-1]`(b) `[-1 0 0 0-1 0 0 0-1]``[1 0 0 0 1 0 0 0 1]`(d) `[1 0 0 0-1 0 0 0-1]`A. `{:[(1,0,0),(0,1,0),(0,0,-1)]:}`B. `{:[(-1,0,0),(0,-1,0),(0,0,-1)]:}`C. `{:[(1,0,0),(0,1,0),(0,0,1)]:}`D. `{:[(-1,0,0),(0,-1,0),(0,0,-1)]:}` |
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Answer» Correct Answer - A::B Let `{:A=[(1,0,0),(0,1,0),(0,0,-1)]:}`.Then, `absA=-1`. If A is a prefect square of matrix `A_1`. Then, `A_(1)^2=A` `rArr abs(A_1)^2=absArArr abs(A_1)^2=-1rArr A_1` cannot be a real matrix So, option(a) correct. Let `{:R=[(-1,0,0),(0,-1,0),(0,0,-1)]:}` be the sqare of a `3xx3` matrix `B_1`, Then, `B=B_1^2rArrabs(B_1)^2=absBrArrabs(B_1)^(2) = {:abs((-1,0,0),(0,-1,0),(0,0,-1))=-1:}` So, B cannot be the perfect square of a real matrix. In option (c), the given matrix is the identity matrix `I_3` such that `I_3=I_3^2` Conisder the matrix `{:[(1,0,0),(0,-1,0),(0,0,-1)]:}` given in option (d). Clearly, `{:[(1,0,0),(0,-1,0),(0,0,-1)]=[(1,0,0),(0,0,1),(0,-1,0)][(1,0,0),(0,0,1),(0,-1,0)]:}` Thus, the matrix given in option(d) is the perfect square of a real matrix. |
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| 500. |
If `A=((1,2),(0,1))` `P=((cos theta , sin theta),( -sin theta, cos theta))`, `Q=P^T AP` find `PQ^(2014)P^T` |
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Answer» `Q=P^TAP` `QP^T=P^TAPP^T` `QP^T=P^TA` `PQ^2014 P^T=PQ^2013QP^T=PQ^2013P^TA` `=PQ^2012QP^TA=PQ^2012P^TA^2` `=PW^2011QP^TA^2=PQ^2011P^TA^3` `=PP^TA^2014=IA^2014=A^2014`. `A=[[1,2],[0,1]]` `A^2=[[1,2],[0,1]]*[[1,2],[0,1]]` `=[[1,4],[0,1]]` `A^3=[[1,2],[0,1]]*[[1,2],[0,1]]*[[1,2],[0,1]]` `A^3=[[1,6],[0,1]]` `A^2014=[[1,4028],[0,1]]`. |
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