Transpose Of A Matrix - Free MCQ Practice Test with solutions, JEE Maths

A’ = {(1,-1,5) (0,0,2)}   B = {(-2,0) (0,2) (3,4)}
B’ = {(-2,0,3) (0,2,4)}
2B’ = 2{(-2,0,3) (0,2,4)}
2B’ = {(-4,0,6) (0,4,8)}
(A’ - 2B’) = {(1,-1,5) (0,0,2)}  - {(-4,0,6) (0,4,8)}
= {(5,-1,-1) (0,-4,-6)}
 (A’ - 2B’)’ = {(5,0) (-1,-4) (-1,-6)}

A = {(2,1) (1,2)}
A² =  {(2,1) (1,2)} * {(2,1) (1,2)}
A² = {(4 + 1) (2 + 2)} {(2 + 2) (1 + 4)}
A² = {5,4} {4,5}

The transpose of the given column matrix is a row matrix with the same elements in the same order. So, the transpose of the given matrix is ([4, -1, 2]), which matches option B.

A = {1, 2, 3}   B = {(-1), (-2), (0), (-3)}
A’ = {(1), (2), (3)}  B’ = {-1, -2, 0 , -3}
A’B’ = {(-1, -2, 0, 3) (-2, -4 , 0, -6) (-3, -6, 0, -9)}

2^3x3=2⁹=512.
 

The number of elements in a 3 X 3 matrix is the product 3 X 3=9.

Each element can either be a 0 or a 1.

Given this, the total possible matrices that can be selected is 2⁹=512

 A ={(1,2) (4,3)} B = {(3,2) (-1,1)}
AB= {[(1*3)+(2*(-1)) (1*2)+(2*1)] [(4*3)+(3 *(-1)) (4*2)+(3*1)]} 
= {(1,4) (9,11)}

If A is a symmetric matrix, then B'AB is a symmetric metrix. So, B'AB is a symmetric matrix.

The transpose of a matrix is a new matrix whose rows are the columns of the original. (This makes the columns of the new matrix the rows of the original). Here is a matrix and its transpose:        is matrix and its transpose is  

For a skew symmetric matrix , as we know all the diagonal elements are zero and the upper triangular elements are the same as that of lower triangular elements in such a fashion that the matrix A = -(transpose A) satisfies.
 
therefore , for a matrix A of dimension n *n , the diagonal elements are zero i.e there would be n zeros in the diagonal
no. of elements remaining to be distinct = total no. of elements -
diagonal elements = (n * n ) - n 
                                   = n² - n

Now as we already said that the the upper traingular half elements are same as that of lower triangular half.

therefore the maximum number of distinct elements are                             = (n² - n) /2

A and B are symmetric matrices, therefore, we have:
A′=A and B′=B..........(i)
 
Consider
(AB−BA)′=(AB)′ − (BA)′,[∵(A−B)′=A′B′]
 
=B′A′− A′B',[∵(AB)′= B′A]
 
=BA−AB [by (i) ]
 
=−(AB−BA)
 
∴(AB−BA) ′=−(AB−BA)
 
Thus, (AB−BA) is a skew-symmetric matrix.