If , then
is equal to
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)}
If , then (A2)` is equal to
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 is
If and
then
is equal to
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)}
The number of all possible matrices of order 3×3 with each entry 0 or 1 is
23x3=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
, then (AxB) is equal to
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
If A is a symmetric matrix, then B'AB is a symmetric metrix. So, B'AB is a symmetric matrix.
The transpose of is
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
What is the maximum number of different elements needed to write a skew symmetric matrix of order n?
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
If A, B are symmetric matrices of same order then the matrix AB-BA is a
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.
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