 # Week 11

Example 1. There are three cans. One of them holds exactly 10 litres of milk and is full. The other two cans can hold 7 litres and 3 litres respectively. There is no graduation mark on the cans. A customer

asks for 5 litres of milk. How would you give him the amount he ask? He would not be satisfied by eye estimates.

Example 2. Let n be a positive integer. Find the number of common factors of n2 + 3n + 1 and n2 + 4n + 3.

Example 3. Find the matrices A and B if 2A + 3B = I2 and A + B = 2At.

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says “the only way to learn mathematics is to do mathematics”.

Solution 1. The man takes an empty vessel other than these.

With the help of 3 litres can he takes out 9 litres of milk from the 10 litres can and pours it in the extra can. So, 1 litre milk remains in the 10 litres can. With the help of 7 litres can he takes out 7 litres of milk from the extra can and pours it in the 10 litres can. The 10 litres can now has 1 + 7 = 8 litres of milk.

With the help of 3 litres can he takes out 3 litres milk from the 10 litres can. The 10 litres can now has 8 – 3 = 5 litres of milk, which he gives to the customer.

Solution 2. Recall that, two positive integers p and q will have a common factor other than one, only when |pq| is a divisor of both p and q. We have,

n2 + 3n + 1 = n2 + 2n + 1 + n = (n + 1)2 + 1; and

n2 + 4n + 3 = n2 + 3n + n + 3 = (n + 3)(n + 1).

Let p = n2 + 4n + 3 and q = n2 + 3n + 1.

Then pq = (n2 + 4n + 3) – (n2 + 3n + 1) = n + 2.

Certainly, neither p nor q is divisible by n + 2.

Hence, the common factor of n2 + 3n + 1 and n2 + 4n + 3 is only 1.

Solution 3. Given equations are

• 2A + 3B = I2    (1)
• A + B = 2At      (2)

Since the sum of two matrices is I2, from equation (1), both of them must be of order two (since the addition of two matrices is possible only when they are of the same order).

Transposing equation (1), we get

• (2A + 3B)t = (I2)t
• Or, (2A)t + (3B)t = I2
• Or, 2At + 3Bt = I2        (3)

Transposing equation (2), we get

• (A + B)t = (2At)t
• Or, (A)t + (B)t = 2(At)t
• Or, At + Bt = 2A                       (4)

From equations (3) and (4), we can either eliminate At or Bt. If we eliminate At, we will get a relation involving Bt and A. But, if we eliminate Bt, we will get a relation involving At and A. Therefore, we will eliminate Bt.

Multiplying equation (4) by 3 and then subtracting equation (3) from it, we get

• 3(At + Bt) – (2At + 3Bt) = 6AI2
• Or, 3At + 3Bt – 2At – 3Bt = 6AI2
• Or, At = 6AI2            (5)

Substituting the value of At from equation (5) in equation (2), we find that

• A + B = 2(6AI2)
• Or, A + B = 12A – 2I2
• Or, 11AB = 2I2        (6)

Using (1) and (6), we now find the value of A and B.

Multiplying (3) by 3 and adding with (1), we find that

• 3(11AB) + (2A + 3B) = 3(2I2) + I2
• Or, 33A – 3B + 2A + 3B = 6I2 + I2
• Or, 35A = 7I2
• Or, A = I2 / 5

Similarly, substituting the value of A in (1), we find that B = I2 / 5.

Therefore, required solutions are