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QUESTION: 1

If , then is equal to

Solution:

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)}

QUESTION: 2

If , then (A^{2})` is equal to

Solution:

A = {(2,1) (1,2)}

A^{2} = {(2,1) (1,2)} * {(2,1) (1,2)}

A^{2} = {(4 + 1) (2 + 2)} {(2 + 2) (1 + 4)}

A^{2} = {5,4} {4,5}

QUESTION: 3

The transpose of is

Solution:

QUESTION: 4

If and then is equal to

Solution:

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)}

QUESTION: 5

The number of all possible matrices of order 3×3 with each entry 0 or 1 is

Solution:

2^{3x3}=2^{9}=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 2^{9}=512

QUESTION: 6

, then (AxB) is equal to

Solution:

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)}

QUESTION: 7

If A is a symmetric matrix, then B’ AB is

Solution:

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

QUESTION: 8

The transpose of is

Solution:

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

QUESTION: 9

What is the maximum number of different elements needed to write a skew symmetric matrix of order n?

Solution:

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

= n^{2} - 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** = (n^{2 }- n) /2

QUESTION: 10

If A, B are symmetric matrices of same order then the matrix AB-BA is a

Solution:

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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