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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.
| 401. |
If `A+B^T=[(1,3),(4,5)] and A^T-B=[(7,8),(-1,3)]` , then find matrices A and B. |
| Answer» Correct Answer - `A= [{:(4, 1), (6, 4):}]; B = [{:(-2, -2),(2, 1):}]` | |
| 402. |
Find matrix X, if `[[3,5,-9],[-1,4,-7]]+X=[[6,2,3],[4,8,6]]` |
| Answer» `[{:(3,-3,12),(5,4,13):}]` | |
| 403. |
If `({:((1)/(2), -(3)/(5)),((4)/(6), -(1)/(7)):}) = ({:(-a, b), (c, -d):})({:(1, 0), (0, 1):})` then find a, b,c and d. |
| Answer» Correct Answer - `a = -(1)/(2), b = -(3)/(5), c = (4)/(6) and d = (1)/(7)` | |
| 404. |
If `5A=[{:(5,10,-15),(2,3," "4),(1,0,-5):}]`, find A. |
| Answer» `[{:(1,2,-3),((1)/(2),(3)/(5)," "(4)/(5)),((1)/(5),0,-1):}]` | |
| 405. |
Find matrices A and B, if `2A-B=[{:(6,-6,0),(-4," "2,1):}]" and "2B+A=[{:(" "3,2," "5),(-2,1,-7):}].` |
| Answer» `A=[{:(3,-2," "1),(-2," "1,-1):}]" and "B=[{:(0,4," "4),(0,0,-6):}]` | |
| 406. |
If `B = [{:(-1, 0), (2, 4):}] " and " f(x) = x^(2) - 4x + 5,` then find f(B). |
| Answer» Correct Answer - `[{:(10, 0), (-2, 5):}]` | |
| 407. |
Find matrices A and B, if `A+B=[{:(1,0," "2),(5,4,-6),(7,3," "8):}]" and "A-B=[{:(-5,-4,8),(11," "2,0),(-1," "7,4):}].` |
| Answer» `A=[{:(-2,-2,5),(8,3,-3),(3,5,6):}]" and "B=[{:(3," "2,-3),(-3," "1,-3),(4,-2," "2):}]` | |
| 408. |
Find matrices A and B, if A + B = \(\begin{bmatrix}1&0 & 2 \\[0.3em]5& 4 & -6\\[0.3em]7 & 3& 8\end{bmatrix}\) and A - B = \(\begin{bmatrix}-5&-4 & 8 \\[0.3em]11& 2 & 0\\[0.3em]-1 & 7& 4\end{bmatrix}.\) A + B = [(1,0,2)(5,4,-6)(7,3,8)]A - B = [(-5,-4,8)(11,2,0)(-1,7,4)]. |
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Answer» Add (A + B) and (A - B) We get (A + B) + (A - B) = \(\begin{bmatrix}1& 0& 2 \\[0.3em]5 & 4 &-6 \\[0.3em]7 & 3 & 8\end{bmatrix}\) + \(\begin{bmatrix}-5& -4& 8 \\[0.3em]11 & 2 &0 \\[0.3em]-1 & 7 & 4\end{bmatrix}\) 2A = \(\begin{bmatrix}-4& -4& 10 \\[0.3em]16 & 6 &-6 \\[0.3em]6 & 10 & 12\end{bmatrix}\) A = \(\begin{bmatrix}-2& -2& 5 \\[0.3em]8 & 3 &-3 \\[0.3em]3 & 5 & 6\end{bmatrix}\) Now Subtract (A - B) from (A + B) (A + B) - (A - B) = \(\begin{bmatrix}1& 0& 2 \\[0.3em]5 & 4 &-6\\[0.3em]7 & 3& 8\end{bmatrix}\) - \(\begin{bmatrix}-5& -4& 8 \\[0.3em]11 & 2 &0\\[0.3em]-1 & 7& 4\end{bmatrix}\) (2B) = \(\begin{bmatrix}6& 4& -6 \\[0.3em]-6 & 2 &-6\\[0.3em]8 & -4& 4\end{bmatrix}\) B = \(\begin{bmatrix}3& 2& -3 \\[0.3em]-3 & 1 &-3\\[0.3em]4 & -2&2\end{bmatrix}\) Conclusion: A = \(\begin{bmatrix}-2& -2&5 \\[0.3em]8 & 3 &-3\\[0.3em]3 & 5&6\end{bmatrix},\) B = \(\begin{bmatrix}3& 2&-3 \\[0.3em]-3 & 1 &-3\\[0.3em]4 & -2&2\end{bmatrix}\) |
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| 409. |
Compare the product , `({:(-5, 1),(6, -1), (-2, 0):}) ({:(4, -5, -1), (5, 6, 1):})((5), (-4), (1))` |
| Answer» Correct Answer - `[(-193), (232), (-78)]` | |
| 410. |
If `A = [{:(2, -3), (4, 1):}], B = [{:(2, 3), (5, 0):}]" and " C = [{:(-1, 2),(0, 5):}]` then find A(B + C). |
| Answer» Correct Answer - `[{:(-13, -5), (9, 25):}]` | |
| 411. |
If A and B are two matrices such that A + B = `[{:(3, 8),(11, 6):}] " and " A-B = `[{:(5, 2),(-3, -6):}]`, then find the matrices A and B. |
| Answer» Correct Answer - `A = [{:(4, 5), (4, 0):}]; B = [{:(-1, 3), (7, 6):}]` | |
| 412. |
If `A xx [{:(-3, 4),(5, 10):}] = [{:(13, 6):}]`, then find A. |
| Answer» Correct Answer - `[-2 (7)/(5)]` | |
| 413. |
Two friends Jack and Jill attend IIT entrance test which has three sections, Mathematics, Physics and Chemistry.Each question in Mathematics, Physics and Chemistry carry 5 marks, 8 marks and 3 marks respectively. Jack attempted 10 questions in Mathematics, 12 in Physics and 6 in Chemistry while Jill attemped 18, 5 and 9 questions in Mathematics, Physics and Chemistry respectively Assuming that all the questions attempted were correct, find the individual marks obtained by the boys by showing the above information as a matrix product. |
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Answer» Correct Answer - Marks obtained by Jack = 164 Marks obtained by Jill = 157 |
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| 414. |
If the rank of the matrix `[[-1,2,5],[2,-4,a-4],[1,-2,a+1]]` is `1` then the value of `a` is (A) `-1` (B) 2 (C) `-6` (D) 4A. `2, if a=-6`B. `2, if a=1`C. `1, if a=2`D. `1, if a=-6` |
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Answer» Correct Answer - B::D Let `A= [[-1, 2, 5],[2, -4, a-4],[1, -2, a+1]]` Applying `R_(2) rarr R_(2) 2 R_(1) and R_(3) rarr R_(3) + R_(1)`, then `A = [[-1, 2, 5],[0,0,a+6],[0,0,a+6]]` Applying `R_(3) rarr R_(3) - R_(2)`, then `A = [[-1, 2, 5],[0,0,a+6],[0,0,0]]` For `a = - 6 , p (A) = 1` For `a = 1,2,p (A) = 2` |
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| 415. |
If `A=[[3,-3,4],[2,-3,4],[0,-1,1]]` , thenA. `adj(adjA)=A`B. `abs(adj(adj(A)))=1`C. `abs(adj(A))=1`D. None of these |
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Answer» Correct Answer - A::B::C Here, `abs(A) = abs((3, -3, 4),(2, -3, 4),(0, -1, 1))` `= 3 (-3+4) + 3 (2-0) + 4 (-2+0)= 1 ne 0` `because adj (adjA) = abs(A)^(3-2) A = A ` ...(i) and `abs(adj (A) ) = abs(A)^(3-1) = abs(A)^(2) = 1^(2) = 1 ` Also, `abs(adj(adj(A))) = abs(A ) = 1 ` [ from Eq. (i) ] |
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| 416. |
If matrix a satisfies the equation `A^(2)=A^(-1)`, then prove that `A^(2^(n))=A^(2^((n-1))), n in N`. |
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Answer» `A^(2^(n))=A^(2.2^(n-1))=(A^(2))^(2^(n-1))` `=(A^(-1))^(2^(n-1))=(A^(2^(n-1)))^(-1)=(A^(2.2^(n-1)))^(-1)` `=((A^(2))^(2^(n-1)))^(-1)=((A^(-1))^(-1))^(2^((n-2)))=A^(2^((n-2)))` |
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| 417. |
Fiven the matrix `A= [[-1,3,5],[1,-3,-5],[-1,3,5]]` and X be the solution set of the equation `A^(x)=A,` where `x in N -{1}`. Evaluate `prod((x^(3)+1)/(x^(3)-1))` where the contincued extends for all `x in X`. |
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Answer» `A= [[-1,3,5],[1,-3,-5],[-1,3,5]][[-1,3,5],[1,-3,-5],[-1,3,5]]=[[-1,3,5],[1,-3,-5],[-1,3,5]]` = A `therefore A^(2) = A^(3) = A^(4) = A^(5) = …= A` but given `A^(x) = A` `rArr x = 2, ,3, 4, 5, … [because x ne 1,` given] `therefore prod ((x^(3)+1)/(x^(3)-1)) = prod ((x+1)/(x-1)) prod (x^(2)-x+1)/(x^(2)+x+1))` On putting `x = 2, 3, 4, 5...` `prod ((x^(3)+1)/(x^(3)-1)) =lim_(nrarr infty) prod_(x=2)^(n) ((x+1)/(x-1)) prod_(x=2)^(n) (x^(2)-x+1)/(x^(2)+x+1))` `lim _(n rarr infty)((3cdot 4cdot 5...(n-1) n (n+1) )/(1cdot 2cdot 3 ... (n-3) (n-2)(n-1)))` `xxlim _(n rarr infty)((3cdot7cdot... (n^(2)-m+1) )/(7cdot 13 ... (n^(2)-n+1) (n^(2) + n+1)))` `=lim _(n rarr infty) (n(n+1))/(2)xx 3/(n^(2)+n+1)` `= 3/2lim _(n rarr infty) ((1+1/n))/((1+1/n+1/n^(2)))= 3/2cdot ((1+0))/((1+0+0))=3/2` |
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| 418. |
Let A and B be matrices of order n. Provce that if (I - AB) is invertible, (I - BA) is also invertible and `(I-BA)^(-1) = I + B (I- AB)^(-1)A, ` where I be the dientity matrix of order n. |
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Answer» Here, `I -BA = BIB^(-1)-BABB^(-1) = B(I-AB)B^(-1)` ...(i) Hence, `abs(I-BA) = abs(B)abs(I-AB) abs(B^(-1)) = abs(B) abs(I-AB) 1/abs(B)` `= abs(I-AB)` If `abs(I-AB)ne 0, "then" abs(I-BA) ne 0` i.e. if `(I -AB)` is invertible, then `(I - BA)` is also invertible. Now, `(I-BA) [ I + B (I-AB)^(-1)A]` `= (I- BA) + (I-BA) B (I-AB)^(-1)A ` [ using Eq. (i)] `= (I- BA) +B (I-AB) B^(-1) B (I-AB)^(-1)A ` `=(I-BA)B(I-AB)(I-AB^(-1))A` `= (I - BA) + BA= I` Hence, `(I- BA)^(-1)=I + B(I-AB)^(-1) A.` |
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| 419. |
Find the value of x and y that satisfy the equations `[(3,-2),(3,0),(2,4)] [(y,y),(x,x)]=[(3,3),(3y,3y),(10,10)]` |
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Answer» Given `[(3,-2),(3,0),(2,4)] [(y,y),(x,x)]=[(3,3),(3y,3y),(10,10)]` or `[(3y-2x,3y-2x),(3y,3y),(2y+4x,2y+4x)]=[(3,3),(3y,3y),(10,10)]` Comparing elements we have `3y-2x=3` `2y+4x+10` Solving (1) and (2), we get `x=3//2, y=2` |
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| 420. |
If `A=[{:(2,3,-1),(1,4,2):}]` and `B=[{:(2,3),(4,5),(2,1):}]` then AB and BA are defined and equal. |
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Answer» False, Since AB is defined `therefore AB=[{:(2,3,-1),(1,4,2):}][{:(2,3),(4,5),(2,1):}]=[{:(14,20),(22,25):}]` Also, BA is defined `therefore BA=[{:(2,3),(4,5),(2,1):}][{:(2,3,-1),(1,4,2):}]` `=[{:(7,18,4),(13,32, 6),(5,10,0):}]` `ABneBA` |
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| 421. |
If A is skew-symmetric matrix then `A^(2)` is a symmetric matrix. |
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Answer» True `because [A^(2)]=[A]^(2)` `=[-A]^(2)` `[because A=-A]` `=A^(2)` Hence, `A^(2)` is symmetric matrix |
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| 422. |
If `A={:((a-a,b-b)):}" then "((1)/(ab))A=`A. `((1)/(b)+(1)/(b) (1)/(a)-(1)/(a))`B. `((1)/(a)-(1)/(a)(1)/(b)-(1)/(b))`C. `((1)/(b)-(1)/(b)(1)/(a)-(1)/(a))`D. `(a^(2)b" "-a^(2)b" "b^(2)a" "-b^(2)a)` |
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Answer» Correct Answer - C Apply scalar product of a matrix. |
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| 423. |
If `A={:[(0,-1),(1,0)]:}andB={:[(-1,0),(3,2)]:}`, then `AB^(2)` =A. `{:[(-3,-4),(-1,0)]:}`B. `{:[(-3,-4),(1,0)]:}`C. `{:[(-3,-4),(0,1)]:}`D. `{:[(-3,4),(1,0)]:}` |
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Answer» Correct Answer - B (i) `AB^(2)=(AB)B`. (ii) Find `B^(2)`, then find `AB^(2)`. |
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| 424. |
If `A={:((3,1),(-4,5)):},B={:((-3,1),(4,-5)):}andC={:((2,1),(-1,3)):}`, thenA. AB=ACB. AC=BCC. BC=CBD. A(B+C)=AB+AC |
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Answer» Correct Answer - D (i) Try from products AB, BC AC and verify from options. |
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| 425. |
If `A={:[(1,7),(5,6)]:},B={:[(-3,2),(-1,5)]:}andC={:[(2,-3),(4,1)]:}`, then find A(B+C) and AB+AC. |
| Answer» Correct Answer - `{:[(20,41),(13,31)]:},{:[(-10,37),(-21,40)]:}{:[(20,41),(13,31)]:}` | |
| 426. |
If the matrix `{:[(a,a+b),(a+b+c,a+b+c+d)]:}` is symmetric, then which of the following holds ?A. a=0B. b=0C. c=0D. d=0 |
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Answer» Correct Answer - C If A is symmetric matrix, then `A^(T)=A`. |
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| 427. |
Let A and B be matrices of orders 3 × 2 and 2 × 4 respectively. Write the order of matrix AB. |
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Answer» We are given that, Order of matrix A = 3 × 2 Order of matrix B = 2 × 4 We need to find the order of matrix AB. We know that, Matrix A × Matrix B = Matrix AB If order of matrix A is (m × n) and order of matrix B is (r × s), Then, Matrices A and B can be multiplied if and only if n = r. That is, Number of columns in A = Number of rows in B Also, The order of resulting matrix AB comes out to be m × s. Applying it, Number of columns in A = 2 Number of rows in B = 2 This means, Matrices A and B can be multiplied, And its order will be given as : Order of matrix AB = 3 × 4 Thus, Order of matrix AB = 3 × 4 |
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| 428. |
If `A={:[(4,-2),(3,-1),(2,5)]:}andB={:[(-7,2),(5,9),(3,4)]:}`, then 2A+7B=A. `{:[(41,10),(40,60),(26,38)]:}`B. `{:[(-41,10),(41,61),(26,38)]:}`C. `{:[(14,5),(30,-60),(24,26)]:}`D. `{:[(-41,10),(41,61),(25,38)]:}` |
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Answer» Correct Answer - D First find the scalar product and add the two matrices. |
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| 429. |
Which of the following is a `1xx3` matrix ?A. `[{:(1),(2),(3):}]`B. `[1,2,3,4]`C. [4,5,6]D. All the above. |
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Answer» Correct Answer - C The matrix with one row and 3 columns has an order of `1xx3` |
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| 430. |
If `A={:[(2,3),(4,1)]:},B={:[(9,5),(-6,1)]:}andC={:[(2,-6),(-3,1)]:}` then find (A+B)+C. |
| Answer» Correct Answer - `{:[(13,2),(-5,3)]:}` | |
| 431. |
The product of `{:[(a,b),(c,d)]:}and{:[(x),(y)]:}` is _________ . |
| Answer» Correct Answer - `{:[(ax,by),(cx,dy)]:}` | |
| 432. |
If `A={:[(p,0),(0,p)]:}" then "A^(n+1)` is _________ .A. `{:[(p,0),(0,p)]:}`B. `{:[(p^(n+1),0),(0,p^(n+1))]:}`C. `{:[(np,0),(0,np)]:}`D. None of these |
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Answer» Correct Answer - B Given `A={:[(p,0),(0,P)]:}` `A=P={:[(1,0),(0,1)]:}` `A=PI` `A^(n+1)=P^(n+1)xxI^(n+1)=P^(n+1)I(becauseI^(n)=I)` `=P^(n+1){:[(1,0),(0,1)]:}={:[(p^(n+),0),(0,P^(n+1))]:}`. |
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| 433. |
If the orders of matrices `A^(T),BandC^(T)" are "3xx4,2xx3and1xx2` respectively, then the order of the matrix `(AB^(T))C` isA. `3xx2`B. `2xx3`C. `4xx2`D. `4xx1` |
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Answer» Correct Answer - D If A is `mxxn` matrix, B is `nxxp` matrix, then the order of AB is `mxxp`. |
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| 434. |
The number of elements that a square matrix of order n has below its leading diagonal, isA. `(n(n+1))/2`B. `(n(n-1))/2`C. `((n-1)(n-1))/2`D. `((n+1)(n+1))/2` |
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Answer» Correct Answer - B There `aren^2` elements in a square matrix of order n out of which n elements other than diagonal elemenets is `(n^2-n)`. Half of these elements are above the diagonal and half are below the diagonal. So, required number of elements is `(n^2-n)/2=1/2n(n-1)`. |
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| 435. |
If a matrix has 7 elements, then the order of the matrix can beA. `4xx3`B. `3xx4`C. `4xx1`D. None of these |
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Answer» Correct Answer - D Number of rows `xx` Number of columns=7 |
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| 436. |
The product of `{:[(2,-4,4)]:},and{:[(1),(3),(5)]:}` is ________ . |
| Answer» Correct Answer - [10] | |
| 437. |
If `{:[(2,-1,0),(9,2,4),(6,3,-9)]:}`, find (k+l) A. |
| Answer» Correct Answer - `{:[(2k+2l,-k-l,0),(9k+9l,2k+2l,4k+4l),(6k+6l,3k+3l,-9k-9l)]:}` | |
| 438. |
The product of `{:[(1),(2)]:}` and `{:[(3,4)]:}` is ________ . |
| Answer» Correct Answer - `{:[(3,4),(6,8)]:}` | |
| 439. |
If `{:[(p),(q),(r)]:}andB={:[(3,4,5)]:}`, then AB isA. `{:[(3p,4p,5p),(3q,4q,5q),(3p,4q,5r)]:}`B. `{:[(3p,3q,3p),(4p,4q,4q),(5q,5p,5r)]:}`C. `{:[(3p,4p,5p),(3q,4q,5q),(3r,4r,5r)]:}`D. None of these |
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Answer» Correct Answer - C Multiply the matrices A and B. |
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| 440. |
If `A={:[(2,-3,1)]:}andB={:[(4),(2),(-2)]:}`, then find `2A^(T)+B`A. `{:[(4),(8),(5)]:}`B. `{:[(8),(-4),(0)]:}`C. `{:[(8),(4),(0)]:}`D. `{:[(8),(-8),(0)]:}` |
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Answer» Correct Answer - B Find `A^(T)` and find the sum. |
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| 441. |
If the matrix `A={:[(5,-3),(5,-3)]:}," then "A^(n+1)` = ________ .A. `{:[(5,-3),(5,-3)]:}`B. `{:[(5,3),(5,3)]:}`C. `2^(n){:[(5,-3),(5,-3)]:}`D. `2^(n+1){:[(5,-3),(5,-3)]:}` |
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Answer» Correct Answer - C `A={:[(5,-3),(5,-3)]:}` `A^(2)={:[(5,-3),(5,-3)]:}{:[(5,-3),(5,-3)]:}` `{:[(25-15,-15+9),(25-15,-15+9)]:}` `{:[(10,-6),(10,-6)]:}=2{:[(5,-3),(5,-3)]:}` `A^(3)=A^(2)*A` `{:[(10,-6),(10,-6)]:}{:[(5,-3),(5,-3)]:}` `{:[(50-30,-30+18),(50-30,-30+18)]:}={:[(20,-12),(20,-12)]:}` `=4{:[(5,-3),(5,-3)]:}=2^(2){:[(5,-3),(5,-3)]:}` `A^(n+1)=2^(n){:[(5,-3),(5,-3)]:}`. |
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| 442. |
If `A={:[(5,6),(7,8)]:}," then "(A+A^(T))/(2)` is _______ .A. symmetricB. skew symmetricC. diagonalD. unit matrix |
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Answer» Correct Answer - A `A={:[(5,6),(7,8)]:},A^(T)={:[(5,7),(6,8)]:}` `A+A^(T)={:[(5+5,6+7),(7+6,8+8)]:}` `A+A^(T)={:[(10,13),(13,16)]:}` `(1)/(2)(A+A^(T))={:[((10)/(2),(13)/(2)),((13)/(2),(16)/(2))]:}={:[(5,(13)/(2)),((13)/(2),8)]:}` `(1)/(2)[(A+A^(T))]^(T)=(1)/(2)[A+A^(T)]` `:." It is symmetric".` |
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| 443. |
If all the diagonal elements in a diagonal matrix is 0, then it is a ________ matrix. |
| Answer» Correct Answer - null | |
| 444. |
The number of column of the matrix `[{:(1,2,3),(4,5,6):}]` is ________A. 2B. 6C. 3D. 5 |
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Answer» Correct Answer - C Cound the number of vertical arrays of elements. |
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| 445. |
`{:[(0,-3),(3,0)]:}` is a ________ matrix. |
| Answer» Correct Answer - skew-symmetric | |
| 446. |
If A is any square matrix, then `(1)/(2) (A-A^(T))` is a ____matrix. |
| Answer» Correct Answer - skew-symmetric | |
| 447. |
If `(A + B^(T))^(T)` is a matrix of order `4 xx 3`, then the order of matrix B is____. |
| Answer» Correct Answer - ` 4 xx 3` | |
| 448. |
If A =\(\begin{vmatrix}α&2\\2&α\end{vmatrix}\)and |A3 | = 125, then α = ___________A = [(α, 2) (2, α)](a) ±3 (b) ±2 (c) ±5 (d) 0 |
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Answer» Correct option is : (a) ±3 |A3 | = 125 |A|3 = 53 …….[∵ |An | = |A|n , n ∈ N] ∴ |A| = 5 \(\begin{vmatrix}α&2\\2&α\end{vmatrix}\)=5 α2 – 4 = 5 α2 = 9 ∴ α = ± 3 |
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| 449. |
The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:(A) 27 (B) 18 (C) 81 (D) 512 |
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Answer» The correct answer is D. Now, each of the 9 elements can be filled in two possible ways. Therefore, by the multiplication principle, the required number of possible matrices is 29 = 512 |
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| 450. |
If `A=[{:(2,-1,3),(-4," "5,1):}]" and "B=[{:(2,3),(4,-2),(1,5):}]` then find AB and BA. Show that `ABneBA.` |
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Answer» Here, A is a `2xx3` matrix and B is a `3xx2` matrix. So, number of columns in A =number of rows in B. `:.` AB exists and it is a `2xx2` matrix. `AB=[{:(2,-1,3),(-4," "5,1):}][{:(2,3),(4,-2),(1,5):}]` `=[{:(2.2+(-1).4+3.1," "2.3+(-1).(-2)+3.5),(-4.2+5.4+1.1,-4.3+5.(-2)+1.5):}]` `=[{:(4-4+3,6+2+15),(-8+20+1,-12-10+5):}]=[{:(3," "23),(13,-17):}].` Again, B is a `3xx2` matrix and A is a `2xx3` matrix. So, number of columns in B=number of rows in A. `:.` BA exists and it is a `3xx3` matrix. `BA=[{:(2,3),(4,-2),(1,5):}][{:(2,-1,3),(-4," "5,1):}]` `=[{:(2.2+3.(-4),2.(-1)+3.5,2.3+3.1),(4.2+(-2).(-4),4.(-1)+(-2).5,4.3+(-2).1),(1.2+5.(-4),1.(-1)+5.5,1.3+5.1):}]` `=[{:(4-12,-2+15,6+3),(8+8,-4-10,12-2),(2-20,-1+25,3+5):}]=[{:(-8,13,9),(16,-14,10),(-18,24,8):}].` Clearly, `ABneBA.` |
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