Matrices Test - 5

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.