No. of notes of Rs. 10 = 12

No. of notes of Rs. 50 = 4

No. of notes of Rs. 100 = 7

(Take the variable representing the no. of coins to be equal to n, where n can range from 1 to 4. You'll see that it will be 4, because for the other cases, it results in a fractional number of notes, which is absurd....aint it??)

2. Give a, b, c, d, e coefficients starting from left side of the equation (reactants) to the right side (products):

For P: 2a + b = 4c + e

For H: 3b = 2d + 4e

For I: 4a = e

For O: 4b = d

Now, take every term on L.H.S., except 'e' terms for all the equations. Set e=1 and form a 4x4 matrix on L.H.S. with 4 equations. Solve the system using Gaussian Elimination and get the values of all the co-efficients (for e=1 certainly). If negative values are there for some co-efficients, it means that those (products or reactants) should be on the other side (R.H.S). Verify your answer by checking the no. of atoms of each element to be equal on both sides of the equation. Multiply both sides by a suitable factor for a nicer look. The final balanced equation will look like this (multiples may be different as per choice):

3. Use the same strategy as in the last question (for three variables here and c=1). On solving the system, you'll get two values for the same variable ('b'). It means that the reaction described by the equation is invalid.The chemist is just blabing!

4. Write down two matrices A & B with general elements. Then first, write down the general element of A.B (inner product), then after taking the transpose, note down the change in the subscripts of the product elements, which when observed shows the interchanging of i and j. Similarly, do the same for B(T) and A(T), and write down the general element of the product B(T)A(T). It will be the same as that of (A.B)(T).If 'a' belongs to A and 'b' to B, then,

General element of (AB)T is: C ij = summation of [a(jk).b(ki)] where k varies from 1 to n.

General element of BT.AT is: Dij (Believe me it's just the same)

So the question comes out to be proved!

5. Take a general permutation matrix P(ab) of order nxn (by observing the general behavior of a permutation matrix), which exchanges the rows 'a' and 'b'. It is a fact to be noted that when P(ab) is raised to the power n, the result is the original permutation matrix( i.e P(ab) ). So, first verify the base case (for k=1), then set k=n and use the above described fact to prove that it equals the original permutation matrix, and lastly multiply both sides of the (k=n) equation by P(ab), which results in P(ab) raised to the power 2, which can be easily proved to be a permutation matrix.prove yourself by proving it!

6. For swapping of two rows a and b in a matrix, follow the steps in order:

i) Ra' <--- Ra – Rb

ii) Rb' <--- Rb + Ra

iii) Ra' <--- Ra – Rb

iv) Ra' <--- (-1)Ra

If you have found a better solution than those described above or you want to write you response, your comments are welcome!

## 2 comments:

the last one is brilliant!

i appreciate ur work it is interesting may allah bless u ........

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