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InterviewSolution
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.
| 501. |
If `A=[2-1 3 1]`and `B=[1 4 7 2]`, find `3A-2Bdot` |
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Answer» `3A-2B=3 [(2,-1),(3,1)]-2[(1,4),(7,2)]` `=[(6,-3),(9,3)]-[(2,8),(14,4)]` `=[(6-2, -3-8),(9-14,3-4)]` `=[(4,-11),(-5,-1)]` |
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| 502. |
Let `A+2B=[(1,2,0),(6,-1,3),(-5,3,1)] and 2A-B=[(2,-1,5),(2,-1,6),(0,1,2)],` then find `tr(A)-tr(B).` |
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Answer» Here to find the value of tr (A) - tr(B), we need not to find the matrices A and B. We can find tr(A)-tr (B) using the properties of trace of matrix, i.e., `A+2B=[(1,2,0),(6,-3,3),(-5,3,1)]` `implies tr (A+2B)=-1` or `tr (A)+2tr (B)=-1` (1) `2A-B=[(2,-1,5),(2,-1,6),(0,1,2)]` `implies tr (2A-B)=3` or `2tr (A)-tr (B) =3` (2) Solving (1) and (2), we get tr(A)=1 and tr (B)=-1 `implies` tr(A)- tr (B)=2 |
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| 503. |
Let A be a `nxxn` matrix such that`A ^(n) = alpha A,` where `alpha ` is a real number different from 1 and - 1. The matrix `A + I_(n)` isA. singularB. invertibleC. scalar matrixD. None of these |
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Answer» Correct Answer - B Let `B=A + I_(n)` `therefore A= B-I_(n)` Given, `A^(n) =alpha A ` `rArr (B-I_(n) ) ^(n) = alpha (B-I_(n))` `rArr B^(n) -""^(n)C_(1) B^(n-1) + ""^(n) C_(2) B^(n-2) +...+ (-1) ^(n) I_(n)` `= alpha B - alphaI_(n)` `rArr B( B^(n-1) -""^(n)C_(1) B^(n-2) + ""^(n) C_(2) B^(n-3) +...+ (-1) ^(n-1) I_(n)-alphaI_(n))` `= [(-1)^(n+1) - alpha ]I_(n ) ne 0 [ because alpha ne pm 1]` Hence, B is invertible. |
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| 504. |
If `A=[[(-1+isqrt(3))/(2i),(-1-isqrt(3))/(2i)],[(1+isqrt(3))/(2i),(1-isqrt(3))/(2i)]]`, `I = sqrt(-1) and f (x) = x^(2) + 2, ` then `f(A)` equals toA. `[[1,0],[0,1]]`B. `((3-isqrt(3))/2)[[1,0],[0,1]]`C. `((5-isqrt(3))/2)[[1,0],[0,1]]`D. `(2+isqrt(3))[[1,0],[0,1]]` |
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Answer» Correct Answer - D `because omega = (-1+isqrt(3))/2 and omega ^(2) =(-1-isqrt(3))/2` ltvrgt Also, `omega ^(3) = 1 and omega + omega ^(2) = -1 ` Thus, ` A = [[-iomega,-iomega^(2)],[iomega^(2),iomega ]]` ` A^(2) = [[-iomega,-iomega^(2)],[iomega^(2),iomega ]][[-iomega,-iomega^(2)],[iomega^(2),iomega ]]= [[-omega^(2)+omega,0],[0,-omega^(2)+omega]]` Now, `f (A) =A^(2) + 2I= [[-omega^(2)+omega,0],[0,-omega^(2)+omega]]+ [[2,0],[0,2]]` `= [[-omega^(2)+omega+2,0],[0,-omega^(2)+omega+2]] ` `=(-omega^(2)+omega+2)[[1,0],[0,1]]= (2 + isqrt(3)) [[1,0],[0,1]]` |
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| 505. |
if `A` and `B`are squares matrices such that `A^(2006)=O` and `A B=A+B,` then `det(B)`equalsA. `-1`B. 0C. 1D. None of these |
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Answer» Correct Answer - B `because AB = A+B` `rArr B= AB - A = (B-I)` `rArr det (B) = det (A) cdot det (B-I) = 0" "[because A^(2006 )=0rArr det A^(2006) = 0 ] [therefore det A=0]` |
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| 506. |
The number of `2x2`matrices `X`satisfying the matrix equation `X^2=I(Ii s2x2u n i tm a t r i x)`is1 (b) 2(c) 3 (d) infinite |
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Answer» Correct Answer - D `because X^(2) = I rArr (X^(-1) X) X=X^(-1)I` `rArr Ix = X^(-1)` `rArr X = X^(-1)` which is self invertible involutory matrix. There are many such matrices which are inverse of their own. |
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| 507. |
Find non-zero values of `x`satisfying the matrix equation:`x[2x2 3x]+2[8 5x4 4x]=2[x^2+8 24 10 6x]` |
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Answer» Given that `x[(2x, 2),(3,x)]+2[(8, 5x),(4,4x)]=2 [(x^(2)+8,24),(10, 6x)]` `implies [(2x^(2), 2x),(3x,x^(2))]+[(16,10x),(8,8x)]=[(2x^(2)+16, 48),(20,12x)]` `implies [(2x^(2)+16, 2x+10x),(3x+8, x^(2)+8x)]=[(2x^(2)+16,48),(20,12x)]` Comparing the elements, we get `2x+10x=48` `implies 12x=48` `implies x=4` This value of x also satisfies the equations `3x+8=20` and `x^(2)+8x=12x` |
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| 508. |
Let M be a column vector (not null vector) and `A=(MM^T)/(M^TM)` the matrix A is : (where `M^T` is transpose matrix of M) |
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Answer» `A=(m*m^T)/(m^Tm)` `mmT=[[a],[b],[c]]_(3*1)*[a,b,c]_(1*3)` `[[a^2,ab,ac],[ab,b^2,bc],[ac,bc,c^2]]_(3*3)` `m^T*m=[a,b,c]_(1*3)[[a],[b],[c]]=[a^2+b^2+c^2]` `A=1/(a^2+b^2+c^2)[[a^2,ab,ac],[ab,b^2,bc],[ac,bc,c^2]]` `=(a^2b^2c^2+a^2b^2c^2+a^2b^2c^2)-(a^2b^2c^2+a^2b^2c^2+a^2b^2c^2)` `A^m=0`. |
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| 509. |
Addition of matrices is defined if order of the matrices is ________ |
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Answer» Answer is Same. |
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| 510. |
State whether the statements is true or false:If two matrices A and B are of the same order, then 2A + B = B + 2A. |
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Answer» Answer is True |
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| 511. |
`[[2,3,-1,-1],[1,-1,-2,-4],[3,1,3,-2],[6,3,0,-7]]` Find rank of this matrix. |
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Answer» Given matrix is, `[[2,3,-1,-1],[1,-1,-2,-4],[3,1,3,-2],[6,3,0,-7]]` Applying `R_4=>R_4-R_1-R_2-R_3` `=[[2,3,-1,-1],[1,-1,-2,-4],[3,1,3,-2],[0,0,0,0]]` As complete `R_4` is `0`, so rank can not be `4`. Now, we will take matrix, `[[2,3,-1],[1,-1,-2],[3,1,3]]` Determinant of the above matrix, `D = [2(-1)-3(9)-1(4)] = -33` As, `D !=0`, so we have a minor of `3 xx 3`. So, rank of the given matrix is `3.` |
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| 512. |
25 0n using alementary row op ration R tie-32, in the following matrix equation we hate 3 0 3 1 |
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Answer» `[[4,2],[3,3]]=[[1,2],[0,3]][[2,0],[1,1]]` `[[-5,-7],[3,3]]=[[1,-7],[0,3]][[2,0],[1,1]]` `A`. |
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| 513. |
If `A` and `B` are two square matrices such that `B=-A^(-1)BA`, then `(A+B)^(2)` is equal toA. OB. `A^(2) + B^(2)`C. `A^(2) + 2 AB + B^(2)`D. `A + B` |
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Answer» Correct Answer - B `because B=- A^(-1) BA` ` rArr AB=- BA ` `rArr AB + BA = 0 ` Now , `(A+B)^(2) =(A+B) (A+B) ` `= A^(2) + AB + BA + B^(2)` `= A^(2) + 0 + B^(2)` `= A^(2) + B^(2)` |
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| 514. |
If `I_n` is the identity matrix of order n, then rank of `I_n` isA. 1B. nC. 0D. none of these |
| Answer» Correct Answer - B | |
| 515. |
Let `A`be an nth-order square matrix and `B`be its adjoint, then `|A B+K I_n|`is (where `K`is a scalar quantity)`(|A|+K)^(n-2)`b. `(|A|+)K^n`c. `(|A|+K)^(n-1)`d. none of theseA. `(abs(A) +k)^(n-2) `B. `(abs(A) +k)^(n)`C. `(abs(A) +k)^(n-1)`D. `(abs(A) +k)^(n+1)` |
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Answer» Correct Answer - B ` because ` B = adj A `rArr AB = A("ajd " A) = abs(A) I_(n)` `therefore AB + KI_(n )= abs(A) I_(n) + kI_(n) = (abs(A) + k ) I_(n)` `rArr abs( AB + KI_(n ))= abs((abs(A) + k))I_(n) = (abs(A) + k )^(n)` |
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| 516. |
Let A and B be square matrices of the same order such that `A^(2)=I` and `B^(2)=I`, then which of the following is CORRECT ?A. IF A and B are inverse to each other, then `A=B`.B. If `AB=BA`, then there exists matrix `C=(AB+BA)/2` such that `C^(2)=C`.C. If `AB=BA`, then there exists matrix `D=AB-BA` such that `D^(n)=O` for some `n in N`.D. If `AB=BA` then `(A+B)^(5)=16 (A+B)`. |
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Answer» We have `A^(2)=I` and `B^(2)=I` `:. A=A^(-1)` and `B=B^(-1)` If A and B are inverse to each other, then `A=B^(-1)=B`. `C=(AB+BA)/2` `implies C^(2)=((AB)^(2)+(BA)^(2)+AB.BA+BA.AB)/4` Now, `ABBA=AIA=A A=I` Similarly, `BA AB=I` Also `(AB)^(2)=ABAB` `=A AB B" "("if "AB=BA)` `=A^(2)B^(2)` `=I` Similarly `(BA)^(2)=I` `:. C^(2)=I` Thus, if `AB=BA`, then `C^(2)=I` `D=AB-BA` `implies D^(2)=(AB)^(2)+(BA)^(2)-ABxxBA-BAxxAB` `=I+I-I-I" "("if "AB=BA)` `=O` Thus, if `AB=BA`, then `D^(n)=O`. `(A+B)^(5)=A^(5)+5A^(4)B+10A^(3)B^(2)+10A^(2)B^(3)+5AB^(4)+B^(5)` `=A+5B+10A+10B+5A+B` `=16(A+B)` |
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| 517. |
Let B is an invertible square matrix and B is the adjoint of matrix A such that `AB=B^(T)`. ThenA. A is an identity matrixB. B is symmetric matrixC. A is a skew-symmetric matrixD. B is skew symmetic matrix |
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Answer» Given that `AB=B^(T)` and `B=` adj. A `implies A=B^(T)B^(-1)` `implies |A|=|B^(T)|xx|B^(-1)|=1` ...(i) Now, adj. `A=` adj `(B^(T) B^(-1))` `="adj"(B^(-1))xx"adj"(B^(T))` `=("adj B")^(-1)xx"adj"(B^(T))` `implies ("adj. B")^(T)=("adj. B")xx("adj. A")=("adj. B")B=|B|I` `implies ("adj. B")^(T)=|B|. I` `implies ("adj. (adj. A)")^(T)=|"adj. A"|I` `:. |A|^(n-2) A^(T)=|A|^(n-1)I` `implies A^(T)=|A|I` `implies A^(T)=I` `implies A=I` `:.` B=adj. A=adj. `I=I` |
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| 518. |
The following are the steps in finding the matrix B, if `B + ({:(2, 3), (4, 5):}) = ({:(5, 4), (3, 2):})`. Arrange them in sequential order. (A) `therefore ({:(p, q), (r, s):}) + ({:(2, 3), (4, 5):}) = ({:(5, 4), (3, 2):})` (B) Let ` B = ({:(p, q), (r, s):})` (C) `({:(p+2, q+3), (r+4, s+5):}) = ({:(5, 4), (3, 2):})` (D) ` P + 2 = 5, q + 3 = 4, r+ 4 = 3, s + 5 = 2` `rArr p = 3, q = 1, r = -1, s = -3` (E) `therefore B = ({:(3, 1), (-1, -3):})`A. BACDEB. BADCEC. BDCAED. BADEC |
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Answer» Correct Answer - A The required sequential order is BACDE. |
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| 519. |
`(AB)^(-1)` =____.A. `A^(-1) B^(-1)`B. `B^(-1) A`C. `AB^(-1)`D. `B^(-1)A^(-1)` |
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Answer» Correct Answer - D `(AB)^(-1) = B^(-1)A^(-1).` |
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| 520. |
If `|{:(2, -4), (9, d-3):}|`=4, then d = _____.A. 13B. 26C. -13D. -26 |
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Answer» Correct Answer - C Given `|{:(2, -4), (9, d-3):}| = 2(d-3) - (-4 xx 9) = 4` 2d- 6 +36 = 4 2d = -26, d = -13. |
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| 521. |
What is the total number of 2 × 2 matrices with each entry 0 or 1? |
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Answer» We are given with the information that, Each element of the 2 × 2 matrix can be filled in 2 ways, either 0 or 1. We need to find the number of total 2 × 2 matrices with each entry 0 or 1. Let A be 2 × 2 matrix such that, A = \( \begin{bmatrix}a_{11} & a_{12} \\[0.3em]a_{21} & a_{22}\\[0.3em]\end{bmatrix}\) Note that, There are 4 elements in the matrix. So, If 1 element can be filled in 2 ways, either 0 or 1. That is, Number of ways in which 1 element can be filled = 21 Then, Number of ways in which 4 elements can be filled = 24 ⇒ Number of ways in which 4 elements can be filled = 16 Thus, Total number of 2 × 2 matrices with each entry 0 or 1 is 16. |
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| 522. |
If \( \begin{bmatrix}9&-1 &4 \\[0.3em]-2 & 1 &3\\[0.3em]\end{bmatrix}\) = A +\( \begin{bmatrix}1&2 &-1 \\[0.3em]0 & 4 &9\\[0.3em]\end{bmatrix}\) , then find matrix A. |
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Answer» We are given that, \( \begin{bmatrix}9&-1 &4 \\[0.3em]-2 & 1 &3\\[0.3em]\end{bmatrix}\) = A +\( \begin{bmatrix}1&2 &-1 \\[0.3em]0 & 4 &9\\[0.3em]\end{bmatrix}\) We need to find the matrix A. In order to find A, Shift the matrix in addition with A to left hand side of the equation. Just like in algebraic property, X = A + Y ⇒ A = X – Y Similarly, \( \begin{bmatrix}9&-1 &4 \\[0.3em]-2 & 1 &3\\[0.3em]\end{bmatrix}\) = A +\( \begin{bmatrix}1&2 &-1 \\[0.3em]0 & 4 &9\\[0.3em]\end{bmatrix}\) ⇒ A = \( \begin{bmatrix}9&-1 &4 \\[0.3em]-2 & 1 &3\\[0.3em]\end{bmatrix}\) - \( \begin{bmatrix}1&2 &-1 \\[0.3em]0 & 4 &9\\[0.3em]\end{bmatrix}\) Subtraction in matrices is done by subtraction of corresponding elements in the matrices. ⇒ A = \( \begin{bmatrix}9-1&-1-2 &4-(-1) \\[0.3em]-2-0 & 1-4 &3-9\\[0.3em]\end{bmatrix}\) ⇒ A = \( \begin{bmatrix}8&-3 &5 \\[0.3em]-2 & -3 &-6\\[0.3em]\end{bmatrix}\) Thus, We get A = \( \begin{bmatrix}8&-3 &5 \\[0.3em]-2 & -3 &-6\\[0.3em]\end{bmatrix}\). |
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| 523. |
If\( \begin{bmatrix}x &x-y \\[0.3em]2x+y & 7\\[0.3em]\end{bmatrix}\)= \( \begin{bmatrix}3 &1 \\[0.3em]8 & 7\\[0.3em]\end{bmatrix}\), then find the value of y. |
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Answer» We are given that, \( \begin{bmatrix}x &x-y \\[0.3em]2x+y & 7\\[0.3em]\end{bmatrix}\)= \( \begin{bmatrix}3 &1 \\[0.3em]8 & 7\\[0.3em]\end{bmatrix}\) We need to find the value of y. We know by the property of matrices, \(\begin{bmatrix} a_{11}& a_{12} \\[0.3em] a_{21} & a_{22} \\[0.3em] \end{bmatrix}\)= \(\begin{bmatrix} b_{11}& b_{12} \\[0.3em] b_{21} & b_{22} \\[0.3em] \end{bmatrix}\) This implies, a11 = b11, a12 = b12, a21 = b21 and a22 = b22 So, if we have \( \begin{bmatrix}x &x-y \\[0.3em]2x+y & 7\\[0.3em]\end{bmatrix}\)= \( \begin{bmatrix}3 &1 \\[0.3em]8 & 7\\[0.3em]\end{bmatrix}\) Corresponding elements of two elements are equal. That is, x = 3 …(i) x – y = 1 …(ii) 2x + y = 8 …(iii) 7 = 7 To solve for y, We have equations (i), (ii) and (iii). From equation (i), x = 3 Substituting the value of x = 3 in equation (ii), x – y = 1 ⇒ 3 – y = 1 ⇒ y = 3 – 1 ⇒ y = 2 Thus, We get y = 2. |
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| 524. |
For a 2x2 matrix, A = [aij], whose elements are given by aij = i /j , write the value of a12. |
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Answer» aij=i/j => a12=1/2 [Here i = 1 and j = 2] |
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| 525. |
If a matrix has 5 elements, write all possible orders it can have. |
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Answer» We are given that, A matrix has 5 elements. We need to find all the possible orders. We know that, If there is a matrix A, of order m × n. Then, There are mn elements. Or, If a matrix has mn elements, then The order of the matrix = m × n or n × m For example, If a matrix is of order 1 × 2, then There are 2 elements in the matrix. [ a11 , a12]1x2 = 2 elements Or, If a matrix is of order 2 × 1, then There are 2 elements in the matrix. \( \begin{bmatrix}a_{11} \\[0.3em]a_{21}\\[0.3em]\end{bmatrix}_{2\times1}\) = 2 elements Similarly, If a matrix has 5 elements, then The order of this matrix are 1 × 5 or 5 × 1. Thus, Possible orders of a matrix having 5 elements are 1 × 5 and 5 × 1 |
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| 526. |
For a 2 × 2 matrix A = [aij] whose elements are given by aij =\(\frac{i}{j}\), write the value of a12. |
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Answer» We are given with, A matrix of order 2 × 2, A = [aij]. aij = \(\frac{i}{j}\) We need to find the value of a12. Here, If A is of the order 2 × 2 then, Number of rows of A = 2 Number of columns of A = 2 We can easily find the elements using the representation of element, . Compare aij with a12. We get, i = 1 j = 2 Putting these values in aij = \(\frac{i}{j}\), a12 = \(\frac{1}{2}\) Thus, The value of a12 = \(\frac{1}{2}\). |
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| 527. |
If the matrix AB s zero, thenA. It is not necessary that either A=O or B=OB. A=O or B=OC. A=O and B=OD. all the above statements are wrong |
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Answer» Correct Answer - A If `{:A=[(1,0),(0,0)],B=[(0,0),(0,1)],"then " AB=[(0,0),(0,0)]:}` However, `A ne O, B ne O`. So, option (a) is correct. |
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| 528. |
If `A=[(1,2,-1),(-1,1,2),(2,-1,1)]`, then det (adj(adj A) isA. `14^4`B. `14^3`C. `14^2`D. 14 |
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Answer» Correct Answer - A We know that for a square matrix of order n adj (adjA)`=abs(A)^(n-2)A," if " absA ne 0`. `rArr "det (adj(adj A))" abs(abs(A)^(n-2)A)` `rArr "det (adj(adj A)) "(abs(A)^(n-2))^nabsA` `rArr " det(adj)(adj A))" =abs(A)^(n^2-2n+1)` Here, n=3 and and `absA=14`. `:. " det(adj(adj A))"=14^4` |
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| 529. |
For any `2xx2`matrix, if `A (a d j A)=[10 0 0 10]`, then `|A|`is equal to(a) 20 (b) 100 (c) 10 (d) 0A. 20B. 100C. 10D. 0 |
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Answer» Correct Answer - C We know that for any square matrix A. `A(adjA) =absAI` `:. A (adjA)={:[(10,0),(0,10)]:}=10I_2rArrabsAI_2=10I_2rArrabsA=10` |
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| 530. |
If `A^5=O` such that `A^n != I` for `1 |
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Answer» Correct Answer - D We have, `A^4(I-A)=A^4-A^5=A^4-O=A^4 ne I` `A^3(l-A)=A^3-A^4 ne I` `and , (I+A)(I-A)=I-A^2 ne I` Hence, `(I-A)^(-1)ne A^4,A^3,I+A`. |
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| 531. |
If A, B are square matrices of order 3,A is non0singular and AB = O, then B is aA. null matrixB. singular matrixC. unit matrixD. non-singular matrix. |
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Answer» Correct Answer - A It is given that `absAne0`. So, `A^(-1)`exists. Now, `AB=O` `rArr A^(-1) (AB) =A^(-1) O` `rArr (A^(-1)A) B=OrArr IB =OrArr B=O` |
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| 532. |
If `A=[1a0 1]`, then `A^n`(where `n in N)`equals`[1n a0 1]`(b) `[1n^2a0 1]`(c) `[1n a0 0]`(d) `[nn a0n]`A. `{:A=[(1,na),(0,1)]:}`B. `{:A=[(1,n^2a),(0,1)]:}`C. `{:A=[(1,na),(0,0)]:}`D. `{:A=[(1,2a),(0,n)]:}` |
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Answer» Correct Answer - B We have, `{:A=[(1,a),(0,1)]:}` `:.A^2={:[(1,a),(0,1)][(1,a),(0,1)]=[(1,2a),(0,1)]:}` It can proved by the principle of mathematical induction that `A^n={:[(1,na),(0,1)]:}` for all `n in N. |
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| 533. |
If `{:A=[(n,0,0),(0,n,0),(0,0,n)]and B=[(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3)]:}` , then AB is equal toA. BA=IB. nBC. `B^n`D. A + B |
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Answer» Correct Answer - B We have, `A=nl_3`. `:. AB=(nl_3)B=n(I_3B)=nB`. |
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| 534. |
Let for `A=[(1,0,0),(2,1,0),(3,2,1)]`, there be three row matrices `R_(1), R_(2)` and `R_(3)`, satifying the relations, `R_(1)A=[(1,0,0)], R_(2)A=[(2,3,0)]` and `R_(3)A=[(2,3,1)]`. If B is square matrix of order 3 with rows `R_(1), R_(2)` and `R_(3)` in order, then The value of det. `(2A^(100) B^(3)-A^(99) B^(4))` isA. `-27`B. `-9`C. `-3`D. 9 |
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Answer» Correct Answer - A `overset(B)([(-,R_(1),-),(-,R_(2),-),(-,R_(3),-)])overset(A)([(1,0,0),(2,1,0),(3,2,1)])=overset(C)([(1,0,0),(2,3,0),(2,3,1)])` (1) `:.` (det. B) (det. A)=3 `:.` (det. B)=3 [as det. A=1] det. `(2A^(100)B^(3)-A^(99)B^(4))` = det. `(A^(99) (2A-B)B^(3))` `=("det. A")^(99)xxdet. (2A-B)xx("det B")^(3)` Now from (1), we get `B=A^(-1) C=[(1,0,0),(-2,1,0),(1,-2,1)][(1,0,0),(2,3,0),(2,3,1)]` `=[(1,0,0),(0,3,0),(-1,-3,1)]` `:. 2A-B=[(1,0,0),(4,-1,0),(7,7,1)]` `:.` det. `(2A-B)=-1` `:.` det. `(2A^(100) B^(3)-A^(99) B^(4))=(1)^(99) (-1) (3)^(3)=-27` |
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| 535. |
Let for `A=[(1,0,0),(2,1,0),(3,2,1)]`, there be three row matrices `R_(1), R_(2)` and `R_(3)`, satifying the relations, `R_(1)A=[(1,0,0)], R_(2)A=[(2,3,0)]` and `R_(3)A=[(2,3,1)]`. If B is square matrix of order 3 with rows `R_(1), R_(2)` and `R_(3)` in order, then The value of det. `(2A^(100) B^(3)-A^(99) B^(4))` isA. `-2`B. `-1`C. 2D. 3 |
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Answer» Correct Answer - D `overset(B)([(-,R_(1),-),(-,R_(2),-),(-,R_(3),-)])overset(A)([(1,0,0),(2,1,0),(3,2,1)])=overset(C)([(1,0,0),(2,3,0),(2,3,1)])` (1) `:.` (det. B) (det. A)=3 `:.` (det. B)=3 [as det. A=1] det. `(2A^(100)B^(3)-A^(99)B^(4))` = det. `(A^(99) (2A-B)B^(3))` `=("det. A")^(99)xxdet. (2A-B)xx("det B")^(3)` Now from (1), we get `B=A^(-1) C=[(1,0,0),(-2,1,0),(1,-2,1)][(1,0,0),(2,3,0),(2,3,1)]` `=[(1,0,0),(0,3,0),(-1,-3,1)]` `:. 2A-B=[(1,0,0),(4,-1,0),(7,7,1)]` `:.` det. `(2A-B)=-1` `:.` det. `(2A^(100) B^(3)-A^(99) B^(4))=(1)^(99) (-1) (3)^(3)=-27` |
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| 536. |
Let `A=[(1,0,0),(1,0,1),(0,1,0)]` satisfies `A^(n)=A^(n-2)+A^(2)-I` for `n ge 3`. And trace of a square matrix X is equal to the sum of elements in its proncipal diagonal. Further consider a matrix `underset(3xx3)(uu)` with its column as `uu_(1), uu_(2), uu_(3)` such that `A^(50) uu_(1)=[(1),(25),(25)], A^(50) uu_(2)=[(0),(1),(0)], A^(50) uu_(3)=[(0),(0),(1)]` Then answer the following question : The value of `|uu|` equals |
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Answer» Correct Answer - B `A^(n)-A^(n-2)=A^(2)-I implies A^(50)=A^(48)+A^(2)-I` Further, `A^(48)=A^(46)+A^(2)-I` `A^(46)=A^(44)+A^(2)-I` `{:(vdots,vdots,vdots,vdots):}` `(A^(4)=A^(2)+A^(2)-I)/(A^(50)=25 A^(2)-24I)` Here, `A^(2)=[(1,0,0),(1,0,1),(0,1,0)][(1,0,0),(1,0,1),(0,1,0)]=[(1,0,0),(1,1,0),(1,0,1)]` `implies A^(50)=[(25,0,0),(25,25,0),(25,0,25)]-24 [(1,0,0),(0,1,0),(0,0,1)]` `=[(1,0,0),(25,1,0),(25,0,1)]` `:. |A^(50)|=1` Also, `tr(A^(50))=1+1+1=3`. Further, `[(1,0,0),(25,1,0),(25,0,1)][(x),(y),(z)]=[(1),(25),(25)]implies [(x),(y),(z)]= uu_(1)=[(1),(0),(0)]` Similarly, `uu_(2)=[(0),(1),(0)]` and `uu_(3)=[(0),(0),(1)]implies uu=[(1,0,0),(0,1,0),(0,0,1)]`, i.e., `|uu|=1` |
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| 537. |
Let `A=[(1,0,0),(1,0,1),(0,1,0)]` satisfies `A^(n)=A^(n-1)+A^(2)-I` for `n ge 3`. And trace of a square matrix X is equal to the sum of elements in its proncipal diagonal. Further consider a matrix `underset(3xx3)(uu)` with its column as `uu_(1), uu_(2), uu_(3)` such that `A^(50) uu_(1)=[(1),(25),(25)], A^(50) uu_(2)=[(0),(1),(0)], A^(50) uu_(3)=[(0),(0),(1)]` Then answer the following question : Trace of `A^(50)` equals |
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Answer» Correct Answer - D `A^(n)-A^(n-2)=A^(2)-I implies A^(50)=A^(48)+A^(2)-I` Further, `A^(48)=A^(46)+A^(2)-I` `A^(46)=A^(44)+A^(2)-I` `{:(vdots,vdots,vdots,vdots):}` `(A^(4)=A^(2)+A^(2)-I)/(A^(50)=25 A^(2)-24I)` Here, `A^(2)=[(1,0,0),(1,0,1),(0,1,0)][(1,0,0),(1,0,1),(0,1,0)]=[(1,0,0),(1,1,0),(1,0,1)]` `implies A^(50)=[(25,0,0),(25,25,0),(25,0,25)]-24 [(1,0,0),(0,1,0),(0,0,1)]` `=[(1,0,0),(25,1,0),(25,0,1)]` `:. |A^(50)|=1` Also, `tr(A^(50))=1+1+1=3`. Further, `[(1,0,0),(25,1,0),(25,0,1)][(x),(y),(z)]=[(1),(25),(25)]implies [(x),(y),(z)]= uu_(1)=[(1),(0),(0)]` Similarly, `uu_(2)=[(0),(1),(0)]` and `uu_(3)=[(0),(0),(1)]implies uu=[(1,0,0),(0,1,0),(0,0,1)]`, i.e., `|uu|=1` |
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| 538. |
Let `A=[(1,0,0),(1,0,1),(0,1,0)]` satisfies `A^(n)=A^(n-1)+A^(2)-I` for `n ge 3`. And trace of a square matrix X is equal to the sum of elements in its proncipal diagonal. Further consider a matrix `underset(3xx3)(uu)` with its column as `uu_(1), uu_(2), uu_(3)` such that `A^(50) uu_(1)=[(1),(25),(25)], A^(50) uu_(2)=[(0),(1),(0)], A^(50) uu_(3)=[(0),(0),(1)]` Then answer the following question : The values of `|A^(50|` equals |
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Answer» Correct Answer - B `A^(n)-A^(n-2)=A^(2)-I implies A^(50)=A^(48)+A^(2)-I` Further, `A^(48)=A^(46)+A^(2)-I` `A^(46)=A^(44)+A^(2)-I` `{:(vdots,vdots,vdots,vdots):}` `(A^(4)=A^(2)+A^(2)-I)/(A^(50)=25 A^(2)-24I)` Here, `A^(2)=[(1,0,0),(1,0,1),(0,1,0)][(1,0,0),(1,0,1),(0,1,0)]=[(1,0,0),(1,1,0),(1,0,1)]` `implies A^(50)=[(25,0,0),(25,25,0),(25,0,25)]-24 [(1,0,0),(0,1,0),(0,0,1)]` `=[(1,0,0),(25,1,0),(25,0,1)]` `:. |A^(50)|=1` Also, `tr(A^(50))=1+1+1=3`. Further, `[(1,0,0),(25,1,0),(25,0,1)][(x),(y),(z)]=[(1),(25),(25)]implies [(x),(y),(z)]= uu_(1)=[(1),(0),(0)]` Similarly, `uu_(2)=[(0),(1),(0)]` and `uu_(3)=[(0),(0),(1)]implies uu=[(1,0,0),(0,1,0),(0,0,1)]`, i.e., `|uu|=1` |
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| 539. |
If `A = [{:("sin" theta, "tan" theta), ("tan" theta, "sin" theta):}]` has no multiplicative inverse, then_____.A. `q = 0^(@)`B. `q = 45^(@)`C. `q = 60^(@)`D. q = 30 |
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Answer» Correct Answer - A Since A has no multiplicative inverse, |A| = 0, `rArr "sin"^(2) theta - "tan"^(2)theta = 0` `rArr "sin"^(2)theta = "tan"^(2)theta` `rArr "sin"theta = "tan" theta rArr = theta =0^(@)`. |
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| 540. |
If A and B are two matrices such that A + B and AB are both defined, then …(a) A and B are two matrices not necessarily of same order. (b) A and B are square matrices of same order. (c) Number of columns of a is equal to the number of rows of B.(d) A = B. |
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Answer» (b) A and B are square matrices of same order |
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| 541. |
If `A=[{:(omega,0),(0,omega):}]`, where `omega` is cube root of unity, then what is `A^(100)` equal to ?A. AB. `-A`C. OD. none of these |
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Answer» Correct Answer - A We have, `A={:omega[(1,0),(0,1)]=omegaI_2:}` `:. A^100=omega^100(I_2)^100=omega^100I_2=omegaI_2=A`. |
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| 542. |
If `A = [[a,b,c],[x,y,z],[p,q,r]], B= [[q , -b,y],[-p,a,-x],[r,-c,z]]` and if A is invertible, then which of the following is not true?A. `abs(A) = abs(B)`B. `abs(A) = -abs(B)`C. `abs(adjA) = abs(adjB)`D. A is invertible `hArr` B is invertble |
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Answer» Correct Answer - A `because abs(B) = abs((q, -b, y ),(-p, a ,-x),(r, -c, z))` Applying `R-(2) rarr (-1) R_(2)`, then `abs(B) = abs((q, -b, y ),(p, -a ,x),(r, -c, z))` Appluing `C_(2) rarr (-1) C_(2),` then `abs(B) = abs((q, b, y ),(p, a ,x),(r, c, z)) = abs(B^(T)) = abs((q,p,r),(b,a,c),(y,x,z))` `= -abs((b,a,c),(q,p,r),(y,x,z)) [ R_(1) harr R_(2)]` `= abs((b,a,c),(y,x,z),(q,p,r)) [ R_(1) harr R_(3)0]` `= -abs((a,b,c),(x,y,z),(p,q,r)) =-abs(A)` `rArr abs(B) = -abs(A)` Also, `abs(adj B) = abs(B)^(2)` ` = abs(A)^(2) = abs(adjA) [because abs(A) ne 0 , "then"abs(B) ne 0]` |
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| 543. |
If `A^(3)=O`, then prove that `(I-A)^(-1) =I+A+A^(2)`.A. `I-A`B. `(I-A)^(-1)`C. `(I+A)^(-1)`D. none of these |
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Answer» Correct Answer - B We have, `(I-A)(I+A+A^2)=I+A+A^2-A-A^2-A^3=I-O=I` `:. (I-A)^(-1)=I+A+A^2` |
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| 544. |
If A and B are square matrices such that B = –A–1BA, then (a) AB + BA = 0 (b) (A + B)2 = A2 + B2 (c) (A + B)2 = A2 + 2AB + B2 (d) (A + B)2 = A + B |
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Answer» Correct option (a, b) Explanation: B = –A–1 BA ⇒ AB = –(AA–1) (BA) ⇒ AB = –IBA ⇒ AB = –BA ⇒ AB + BA = 0 Now (A + B)2 = (A + B) (A + B) = A2 + AB + BA + B2 = A2 + B2 |
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| 545. |
Let three matrices `A = [[2,1],[4,1]],B=[[3,4],[2,3]]and C= [[3,-4],[-2,3]],` then tr `(A) + tr ((ABC)/2)+tr((A(BC)^(2))/4) + tr ((A(BC)^(3))/8) +...+infty` equals toA. 4B. 9C. 12D. 6 |
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Answer» Correct Answer - D `because BC = [[3,4],[2,3]] [[3,-4],[-2,3]]=[[1,0],[0,1]]=I` `therefore tr (A) + tr ((ABC)/2) + tr ((A(BC)^(2))/4) +tr((A(BC)^(3))/8) +...` `= tr (A) + tr (A/2) + tr(A/2^(2))+tr(A/2^(3))+..." upto " infty` `=tr (A) + 1/2 tr (A) + 1/2^(2) tr (A) + ... " upt0 " infty ` `= (tr(A))/(1-(1/2)) = 2 tr (A) = 2(2+1) = 6` |
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| 546. |
If A is non-singular and `(A-2I) (A-4I) = O`, then `1/6 A + 4/3 A^(-1)` is equal toA. `O`B. `I`C. `2I`D. `6I` |
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Answer» Correct Answer - B we have, `(A-2I) (A-4I) = 0` `rArr A^(2) - 4A - 2 A + 8 I^(2) = 0` `rArr A^(2) - 6 A+ 8 I =0` `rArr A^(-1) (A^(2)-6A=8I) = A^(-1)0` ` rArr A- 6 I + 8 A^(-1) = 0` `rArr 1/6 A + 4/3 A^(-1) = I` |
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| 547. |
If `{:A=[(0,1,-1),(2,1,3),(3,2,1)]:}` then `(A(adj A)A^(-1))A=`A. `2{:[(3,0,0),(0,3,0),(0,0,3)]:}`B. `{:[(-6,0,0),(0,-6,0),(0,0,-6)]:}`C. `{:[(0,1//6,-1//6),(2//6,1//6,3//6),(3//6,2//6,1//6)]:}`D. none of these |
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Answer» Correct Answer - A We have, `{:absA=[(0,1,-1),(2,1,3),(3,2,1)]:}=0+7-1=6` `:. (A(adj A)A^(-1))A=(A(adjA))(A^(-1)A)` `rArr(A(adj A)A^(-1))A=(absAI)I` `rArr(A(adj A)A^(-1))A=absAI={:[(6,0,0),(0,6,0),(0,0,6)]=2[(3,0,0),(0,3,0),(0,0,3)]:}` Hence, option (a) is correct. |
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| 548. |
If `A=[[0, 1,2],[1,2,3],[3,a,1]]and A^(-1)[[1//2,-1//2,1//2],[-4,3,b],[5//2,-3//2,1//2]]` thenA. `a = 1, b = -1`B. `a = 2, b = -1/2`C. `a = -1, b=1`D. `a=1/2, b=1/2` |
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Answer» Correct Answer - A We have, `A A^(-1) = I` `rArr [[0,1,2],[1,2,3],[3,a,1]] [[1/2,-1/2,1/2],[-4,3,6],[5/2,-3/2,1/2]]= [[1,0,0],[0,1,0],[0,0,1]]` `rArr [[1,0,b+1],[0,1,2(b+1)],[4(1-a),3(a-1),ab+2]]= [[1,0,0],[0,1,0],[0,0,1]]` On comparing, we get `b+ 1 + 0 , ab + 2 =1, a- 1 = 0` `therefore a = 1, b= -1` |
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| 549. |
α, β ,γ are roots of x3 + ax2 + b = 0, then the value of||(α, β ,γ), (β, γ, α), (γ, α,β)| is(a) –a3 (b) a3– 3b(c) a3 (d) None of these |
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Answer» Correct option (c) a3 Explanation: α + β + γ = –a, α β + β γ + γ α = 0, α β γ = -b It can be shown that |(α, β ,γ), (β, γ, α), (γ, α,β)| = (α + β + γ)(α2 + β2 + γ2 - α β - β γ - γ α) = (α + β + γ)(α + β + γ)2 -3(α β + β γ + γ α)) = –(–a)((–a)2–3 x 0) = a3 |
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| 550. |
Find the inverse of the matrix `[{:(-1,1,2),(1,2,3),(3,1,1):}]` |
| Answer» `[{:(1,-1," "1),(-8," "7,-5),(5,-4," "3):}]` | |