Explore topic-wise InterviewSolutions in .

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.

1.

Show that all positive integral powers of a symmetric matrix aresymmetric.

Answer» Let `A` is a symmetric matrix, then, we have to prove,
`(A^n)^T = A^n`
We know, `(AB)^T = B^TA^T`
`:. (A^n)^T = (A*A*...*A)^T = A^T*A^T*A^T...A^T = (A^T)^n = A^n`
`=> (A^n)^T = A^n`.
It shows all positive integral powers of a symmetric matrix are symmetric.
Discussion
2.

If A is a skew-symmetric matrix and `n`is an odd natural numbr, write whether `A^n`is symmetric or skew-symmetric or neither of the two.

Answer» Here, `A` is a skew-symmetric matrix.
`:. A^T = -A.`
Now, We know, `(A^n)^T = (A^T)^n`
`=>(A^n)^T = (-A)^n`
As, `n` is an odd natural number,
`:. (A^n)^T = -A^n`
It means, `A^n` will be a skew-symmetric matrix for a odd number `n`, if `A` is a skew-symmetric matrix.
Discussion
3.

Show that the matrix `B^TA B`is symmetric or skew-symmetric according as A is symmetric orskew-symmetric.

Answer» A matrix `A` is symmetric if `A = A^T`.
A matrix `A` is skew-symmetric if `A = -A^T`.
(i) When `A` is symmetric matrix, then `A = A^T`
Then, `(B^TAB)^T = B^TA^T(B^T)^T = B^TAB`...[As `A^T = A and (B^T)^T = B`]
`:. (B^TAB)^T = B^TAB`
`:. B^TAB` is a symmetric matrix when `A` is symmetric

(ii)When `A` is skew-symmetric matrix, then `A = -A^T`
Then, `(B^TAB)^T = B^TA^T(B^T)^T = B^T(-A)B`...[As `A^T =- A and (B^T)^T = B`]
`:. (B^TAB)^T =- B^TAB`
`:. B^TAB` is a skew-symmetric matrix when `A` is skew-symmetric
Discussion
4.

A matrix which is both symmetric as well as skew-symmetric is a nullmatrix.

Answer» A matrix `A` is symmetric if `A = A^T`.
`=> a_(ij) = a_(ji),` for all `i,j->(1)`
Here `a_(ij)` are elements of matrix with `i` and `j` representing rows and columns of matrix.
A matrix `A` is skew-symmetric if `A = -A^T`.
`=> a_(ij) = -a_(ji),` for all `i,j`
`=>-a_(ij) = a_(ji)->(2)`
From (1) and (2),
`a_(ij) = -a_(ij)`
`=>2a_(ij) = 0`
`=>a_(ij) = 0`
So, each element of the matrix will be `0`. thus, it will be a null matrix.
Discussion
5.

Express the matrix `[3-2-4 3-2-5-1 1 2]`as the sum of a symmetric and skew-symmetric matrix.

Answer» A matrix `A` can be expressed as the sum of symmetric and skew-symmetric matrix in following way:`A = 1/2(A+A^T)+1/2(A-A^T)`
Here,
`A = [[3,-2,-4],[3,-2,-5],[-1,1,2]]`
`:. A^T = [[3,3,-1],[-2,-2,1],[-4,-5,2]]`
`=>A+A^T = [[3,-2,-4],[3,-2,-5],[-1,1,2]]+[[3,3,-1],[-2,-2,1],[-4,-5,2]] = [[6,1,-5],[1,-4,-4],[-5,-4,4]]`
`=>A-A^T = [[3,-2,-4],[3,-2,-5],[-1,1,2]]-[[3,3,-1],[-2,-2,1],[-4,-5,2]] = [[0,-5,-3],[5,0,-6],[3,6,0]]`
`:. A = 1/2 [[6,1,-5],[1,-4,-4],[-5,-4,4]]+ [[0,-5,-3],[5,0,-6],[3,6,0]]`
`=>A = [[3,1/2,-5/2],[1/2,-2,-2],[-5/2,-2,2]]+[[0,-5/2,-3/2],[5/2,0,-3],[3/2,3,0]]`
Discussion
6.

The sales figure of two car dealers during January 2013 showed thatdealer A sold 5 deluxe, 3 premium and 4 standard cars, while dealer B sold 7deluxe, 2 premium and 3 standard cars. Total sales over the 2 month period ofJanuary-February revealed that dealer A sold 8 deluxe 7 premium and 6standard cars. In the same 2 month period, dealer B sold 10 deluxe, 5 premiumand 7 standard cars. Write 2 x 3 matrices summarizing sales data for Januaryand 2 month period for each dealer.

Answer» Here, we will create two matrices `A` and `B` for the given details.
Sales data for Dealer `A` in matrix form,
`A = [[5,3,4],[8,7,6]].`
Here, first row represents car sold in January month and
second row represents car sold over the two months period of January-February.
First column represents number of deluxe cars , second column represents number of premium cars
and third colum represents the number of standard cars.

Sales data for Dealer `B` in matrix form,
`B = [[7,2,3],[10,5,7]].`
Here, first row represents car sold in January month and
second row represents car sold over the two months period of January-February.
First column represents number of deluxe cars , second column represents number of premium cars
and third colum represents the number of standard cars.
Discussion
7.

If `A ,Ba n dC`three matrices of the same order, then prove that`A=B rArr A+C=B+C`

Answer» Let `A, B and C` are `m xxn` order matrices such that,
`A = [a_(ij)]_(mxxn)`
`B = [b_(ij)]_(mxxn)`
`C = [c_(ij)]_(mxxn)`
It is given that ,
`A = B`
`=> a_(ij) = b_(ij)` , `1 le i le m and 1 le j le n`
`=> a_(ij) + c_(ij) = b_(ij) + c_(ij)` , ` 1 le i le m and 1 le j le n`
`=>(A+C)_(ij) = (B+C)_(ij)`
`=>A+C = B+C`
Discussion
8.

Find the value of `lambda,`a non-zero scalar, if `lambda[1 0 2 3 4 5]+2[1 2 3-1-3 2]=[4 4 10 4 2 14]`

Answer» `lambda[[1,0,2],[3,4,5]] +2[[1,2,3],[-1,-3,2]] = [[4,4,10],[4,2,14]]`
So, `lambda+2 = 4 => lambda = 2`
`2lambda+6 = 10=> lambda = 2`
`3lambda - 2 = 4=> lambda = 2`
`4lambda - 6 = 2=>lambda = 2`
`5lambda +2 = 12=> lambda = 2`
So, solving all the equations, we get the same solution i.e. `lambda = 2`.
Discussion
9.

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

Answer» We will prove it using mathematical induction.
For, `n = 1`,
`L.H.S = (A^T)^1 = A^T`
`R.H.S. = (A^1)^T = A^T`
As, `L.H.S.= R.H.S`, our equation is true for `n = 1`.
Let, our equation is true for `n = k` where `k` is a natural number.
Then, `(A^T)^k = (A^K)^T->(1)`.
Now, we have to prove, for `n = k+1`, given equation is true.
For, `n = k+1`,
`L.H.S. = (A^T)^k+1`
`=(A^T)^k*(A^T)^1`
From (1),
`=(A^k)^T*A^T`
`=(A^k*A)^T`
`=(A^(k+1))^T`
`R.H.S = (A^(k+1))^T`.
`:. L.H.S. = R.H.S.`, for `n = k+1`.
Thus, our equation is true for `n = k+1`.
`:. (A^T)^n = (A^n)^T`.
Discussion
10.

In a certain city there are 30 colleges. Each college has 15 peons, 6clerks, 1 typist and 1 section officer. Express the given information as acolumn matrix. Using scalar multiplication, find the total number of posts ofeach kind in all the colleges.

Answer» A college has 15 peons, 6 clerks, 1 typist and 1 section officer.
So, in matrix representation,
`A = [[15],[6],[1],[1]]`
Here, first row represents number of peons, second row represents number of clerks.
Third row represents number of typist and fourth row represents number of section officer.
As there are `30` colleges , so total number of posts,
`X = 30A = 30[[15],[6],[1],[1]] = [[450],[180],[30],[30]]`
`:.` Total number of peons `=450`
Total number of clerks`=180`
Total number of typists`=30`
Total number of section officers `=30`
Discussion
11.

If `A=[[costheta,isintheta],[isintheta,costheta]],`then prove by principal of mathematical induction that`A^n=[[cosntheta, i sinn theta],[isin n theta, cos n theta]]`for all n`in` `NN`.

Answer» Here, it is given that,
`A = [[cos theta, isintheta],[isintheta,cos theta]]`.
We have to prove,
`A^n = [[cos ntheta, isin ntheta],[isin ntheta,cos ntheta]]`.
Now, for `n = 1`.
`A^1 = [[cos 1theta, isin 1theta],[isin 1theta,cos 1theta]]`.
`=>A = [[cos theta, isintheta],[isintheta,cos theta]]`, which is true.
So, given equation is true for `n = 1`.
Now, let given equation is true for `n = k`.
Then, `A^k = [[cos ktheta, isin ktheta],[isin ktheta,cos ktheta]]`->(1).Now, we have to prove given equation is true for `n = k+1`.
We have to prove,
`A^(k+1) = [[cos (k+1)theta, isin (k+1)theta],[isin (k+1)theta,cos (k+1)theta]]`
Now, `L.H.S. = A^(k+1) = A^k*A = [[cos ktheta, isin ktheta],[isin ktheta,cos ktheta]] [[cos theta, isintheta],[isintheta,cos theta]]`
`= [[cos kthetacostheta + i^2sin kthetasin theta, icoskthetasin theta + isinkthetacos theta],[isin ktheta cos theta+ i cosktheta sin theta,i^2sin ktheta sin theta+cos k theta cos theta]] `
`= [[cos kthetacostheta - sin kthetasin theta, i(coskthetasin theta + sinkthetacos theta)],[i(sin ktheta cos theta+ cosktheta sin theta),-sin kthetasin theta+cos k theta cos theta]] `
`=[[cos (k+1)theta, isin (k+1)theta],[isin (k+1)theta,cos (k+1)theta]] = R.H.S.`
Thus, our equation is true for `n = k+1`.
`:.A^n = [[cos ntheta, isin ntheta],[isin ntheta,cos ntheta]]`.
Discussion