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

**Example 2**. Find the matrices *A* and *B* if 2*A* + 3*B* = *I*_{2} and *A* + *B* = 2*A ^{t}*.

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 *B* – *b* cos *A* = 3*c*/5, we get *a* cos *B* = 4*c*/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 *DB* – *AD* = 3*c*/5. From these two relations, we get *AD* = *c*/5 and *DB* = 4*c*/5. Hence, *DB*/*AD* = 4. Therefore,

as before.

**Solution 2**. Given equations are

- 2
*A*+ 3*B*=*I*_{2}(**1**) *A*+*B*= 2*A*(^{t}**2**)

Since the sum of two matrices is *I*_{2}, 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

- (2
*A*+ 3*B*)= (^{t}*I*_{2})^{t} - Or, (2
*A*)+ (3^{t}*B*)=^{t}*I*_{2} - Or, 2
*A*+ 3^{t}*B*=^{t}*I*_{2}(**3**)

Transposing equation (2), we get

- (
*A*+*B*)= (2^{t}*A*)^{t}^{t} - Or, (
*A*)+ (^{t}*B*)= 2(^{t}*A*)^{t}^{t} - Or,
*A*+^{t}*B*= 2^{t}*A*(**4**)

From equations (3) and (4), we can either eliminate *A ^{t}* or

*B*. If we eliminate

^{t}*A*, we will get a relation involving

^{t}*B*and

^{t}*A*. But, if we eliminate

*B*, we will get a relation involving

^{t}*A*and

^{t}*A*. Therefore, we will eliminate

*B*.

^{t}Multiplying equation (4) by 3 and then subtracting equation (3) from it, we get

- 3(
*A*+^{t}*B*) – (2^{t}*A*+ 3^{t}*B*) = 6^{t}*A*–*I*_{2} - Or, 3
*A*+ 3^{t}*B*– 2^{t}*A*– 3^{t}*B*= 6^{t}*A*–*I*_{2} - Or,
*A*= 6^{t}*A*–*I*_{2}(**5**)

Substituting the value of *A ^{t}* from equation (5) in equation (2), we find that

*A*+*B*= 2(6*A*–*I*_{2})- Or,
*A*+*B*= 12*A*– 2*I*_{2} - Or, 11
*A*–*B*= 2*I*_{2}(**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(11
*A*–*B*) + (2*A*+ 3*B*) = 3(2*I*_{2}) +*I*_{2} - Or, 33
*A*– 3*B*+ 2*A*+ 3*B*= 6*I*_{2}+*I*_{2} - Or, 35
*A*= 7*I*_{2} - Or,
*A*=*I*_{2}

Similarly, substituting the value of *A* in (1), we find that ** B = I_{2} / 5**.

Therefore, required solutions are