Monthly Archives: May 2021

Inverting the Matrix

– Chandrahas M. Halai

Whenever any non-zero real number is multiplied by its inverse we get 1. Using a similar analogy we define an inverse of a matrix such that when the matrix is multiplied with its inverse we get an Identity matrix as the result.

Let A^-1 denote the inverse matrix for the n x n square matrix A. Then, we have

But the question that arises now is how does one go about finding the inverse of a matrix?

If we transform the square matrix A into an Identity matrix by performing elementary row operations on it, then by performing the same sequence of elementary row operations on the identity matrix on the right it will be transformed into the inverse matrix of A.

We have

This will be transformed into

Let,

Let us do the following elementary row operations to reduce the elements in the second and third rows of the first column to zeroes:

Thus, we get

This same result can be obtained by multiplying matrix A with another matrix which we can call the elimination matrix.

Let us consider the following matrix equation:

BC = D

If we observe the matrix multiplication we find that the first row of matrix D is the combination of the rows of matrix C given by the elements of the first row of matrix B. Similarly, the second row of matrix D is the combination of the rows of matrix C given by the elements of the second row of matrix B. And so on.

Thus to perform the above two elementary row operations on Matrix A we can multiply it with the following elimination matrix E₁:

Now, we have to reduce the element in the third row, second column to zero. For this we need to do the following elementary row operation:

To achieve this we now multiply by elimination matrix E₂, as shown below:

Now, we have to convert the element in third row, third column to 1. We do this with the following row operation:

To achieve this we now multiply by elimination matrix E₃, as shown below:

Now, we have to reduce the elements in the first and second rows of the third column to zeroes. For this we need to do the following elementary row operations:

To achieve this we now multiply by elimination matrix E₄, as shown below:

Now, we have to convert the element in second row, second column to 1. We do this with the following row operation:

To achieve this we now multiply by elimination matrix E₅, as shown below:

Now, we have to reduce the element in the first row, second column to zero. For this we need to do the following elementary row operation:

To achieve this we now multiply by elimination matrix E₆, as shown below:

Let us summarise what we have done in the following matrix equation:

E₆ E₅ E₄ E₃ E₂ E₁ A = I

Let us multiply all the elimination matrices into a single elimination matrix E. Hence, now we have

EA = I

When the product of two matrices is an Identity matrix, the two matrices are inverses of each other. This means that the elimination matrix E is the inverse of matrix A.

Let us verify this. We have

Hence, it is now verified that the elimination matrix E is the inverse of matrix A.

Going around in a Circle

5 Replies

– Chandrahas M. Halai

Two friends go for a morning walk on a circular road. They start their walk at the same time with different speeds. The speed of one friend is 5 kms/hr and of the other is 4 kms/hr. If the circumference of the road is 500 metres then what is the time of their meeting?

The method to solve such a problem is given in the book Ganita Kaumudi (गणित कौमुदी) composed by Narayan Pandit in the year 1356 CE. Narayan Pandit gives a sutra to solve the above kinds of problems.

सङ्गमकालः परिधौ गत्यन्तरभाजिते भवति ॥४०॥

Meaning:

The time of the meeting is given by the circumference divided by the difference in the speeds.

Let us see how we get this result:

Let c be the circumference of the circular road. Let v₁ and v₂ be the speeds of the two friends and let v₁ > v₂. Also, let t be the time of their meeting.

Let x be the distance walked by the second friend till the time of their meeting.

The first friend who is faster will complete one more round of the road than the slower one. Therefore the distance covered by the first friend at the time of their meeting is = c + x.

We have,

time = distance / speed

Therefore for the first friend, we have

And for the second friend, we have

Hence, we have

Substituting this in (1), we have

This proves the sutra.

Let us apply the sutra to solve the given problem.

Hence, the two friends will meet after half hour of starting their walk.

When do we meet next?

4 Replies

– Chandrahas M. Halai

योजनत्रिशती पन्थाः पुरयोरन्तरं तयोः ।

एकादशगतिस्त्वेको नवयोजनगः परः ॥

युगपन्निर्गतौ स्वस्वपुरतो लिपिवाहकौ ।

समागमद्वयं ब्रूहि गच्छतोश्च निवृत्तयोः ॥ ४४ ॥

Meaning:

The distance between two towns is 300 yojanas. Two letter carriers start their journey from their respective towns (simultaneously) to the other town, one with a speed of 11 yojanas per day and the other with 9 yojanas per day. Tell us quickly the times of their two meetings, the first during their onward journey and second while returning back.

1 yojana = approximately 12kms as per Vishnu Purana.

The above problem is from the book Ganita Kaumudi (गणित कौमुदी) composed by Narayan Pandit in the year 1356 CE as mentioned in the final verses of the book.

How does one solve such a problem?

Narayan Pandit has given a sutra to solve just such kind of problems. The sutra is as given below:

अध्वनि गतियोगहृते प्रजायते प्रथमसङ्गमे कालः ।

तस्मिन् योगे द्विगुणे योगात् तस्मात् पुनर्योगः ॥३९॥

Meaning:

The time of the first meeting is given by the distance between the two towns divided by the sum of the speeds. The time interval between the first meeting and the second meeting during the return journey is twice the time taken for the first meeting.

Let us understand how we get these results.

Let A and B be the two towns and d be the distance between them. Let v_A and v_B be the speeds of the letter carriers starting from towns A and B respectively. Let the M₁ be the point of their first meeting and M₂ the point of their second meeting. Let the time of their first meeting from the start of their journey be t₁ days and the time of their second meeting be t₂ days.

Let distance AM₁ = x

and distance BM₂ = y

Therefore,

BM₁ = d – x

AM₂ = d – y

We have, time = distance / speed

Hence, we have