Week 25

Example 1. Suppose ABC is a triangle with angles A, B, C and the corresponding sides a, b, c satisfies the equation a cos Bb cos A = 3c/5. Find the value of tan A/ tan B.

Example 2. 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. Using the projection theorem, we have a cos B + b cos A = c. Combining it with the given information a cos Bb cos A = 3c/5, we get a cos B = 4c/5 and b cos A = c/5. Therefore,

Alternatively, applying the law of cosines we have

With the help of this, we get

As another alternative, consider the triangle ABC as shown in the figure. Through C draw CD perpendicular to AB with foot point D. We have a cos B = DB and b cos A = AD. Since AB = c, we can write AD + DB = c. By the given

condition, we have DBAD = 3c/5. From these two relations, we get AD = c/5 and DB = 4c/5. Hence, DB/AD = 4. Therefore,

as before.

Solution 2. 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