Nature of solutions of a set of equations using matrices in linear algebra. Today it’s all about the nature of solutions of a set of equations using matrices.

Want to know about the related topics? Do check out the following:

**The rank of a rectangular matrix**

**How to get the rank of any square matrix**

**DOWNLOAD READ & PRINT – Nature of solutions of a set of equations using matrices (pdf)**

**Nature of solutions of a set of equations using matrices | Linear algebra**

Suppose I have a set of equations like

Now I have to find out the nature of the solutions of this set of equations without solving them.

That means whether there will be no solution or many solutions or exactly one solution.

So for that, the best way is to use matrices.

First of all, I will write down two matrices.

One of them is the coefficient matrix, say .

Another one is the augmented matrix, say, .

Therefore, in this case, the matrix is

Also the matrix is

Here the matrix is a square matrix. But the matrix is a rectangular matrix.

Next, I’ll find the ranks of these two matrices.

Now I have three possible nature of solutions of this set of equations. If

- the ranks of and are equal to the number of variables, the solutions will be unique.
- the ranks of and are equal but are less than the number of variables, the number of solutions will be infinite.
- the rank of is less than the rank of , there will be no solution.

Here I’ll give now two examples of that.

I hope that helps.

**Examples of nature of solutions of a set of equations using matrices**

Disclaimer: None of these examples is mine. I have chosen these from some book. I have also given the due reference at the end of the post.

So here are the examples.

**Example 1**

According to Stroud and Booth (2011)* “Determine the rank of and of for the following set of equations and hence determine the nature of the solutions. Do *not* solve the equations.

”

**Solution**

Now here the given set of equations is

Therefore the coefficient matrix is

Also, the augmented matrix is

So, I’ll start with the rank of the matrix .

**Step 1**

First of all, I’ll try to reduce the matrix to an upper triangular matrix. At first, I’ll subtract 4 times row 1 from row 2. Simultaneously, I’ll also subtract row 1 from row 3.

So in the mathematical term, it will be

Row 2 – 4Row 1, Row 3 – Row 2.

Therefore the equivalent matrix will be

Now here I have to stop. And this is because I can’t reduce it further.

Next, I’ll find out the determinant of the matrix .

Thus it will be

Therefore the rank of the matrix is 3.

Now I’ll find out the rank of the matrix .

**Step 2**

So I’ll reduce the matrix to an upper triangular matrix. At first, I’ll subtract 4 times row 1 from row 2. Simultaneously, I’ll also subtract row 1 from row 3.

Thus in the mathematical term, it will be

Row 2 – 4Row 1, Row 3 – Row 2.

Therefore the equivalent matrix will be

Now here I have to stop. And this is because I can’t reduce it further.

Next, I’ll find out the determinant of any submatrix of the matrix .

So I’ll start with a 3 × 3 submatrix of .

First of all, I’ll choose a 3 × 3 submatrix, say

Now, this is the matrix .

Also, I already know the rank of the matrix . And that is 3.

Therefore the rank of the matrix is also 3.

Thus I can say that the rank of the matrix = the rank of the matrix = 3.

Now the number of the unknown in this set of equations is also 3.

Therefore I can conclude that this set of equations has unique solutions.

Hence I can conclude that this is the answer to the given example.

Now comes the second example.

**Example 2**

According to Stroud and Booth (2011)* “Determine the rank of and of for the following set of equations and hence determine the nature of the solutions. Do *not* solve the equations.

”

**Solution**

Now here the given set of equations is

Therefore the coefficient matrix is

Also, the augmented matrix is

So, I’ll start with the rank of the matrix .

**Step 1**

First of all, I’ll try to reduce the matrix to an upper triangular matrix. And for that, I’ll subtract row 1 from row 2. Simultaneously, I’ll also subtract twice row 1 from row 3.

Thus in the mathematical term, it will be

Row 2 – Row 1, Row 3 – 2 Row 2.

Therefore the equivalent matrix will be

Now here I have to stop. And this is because I can’t reduce it further.

Next, I’ll find out the determinant of the matrix .

Thus it will be

Therefore the rank of the matrix is *not* 3.

Now I’ll choose a 2 × 2 submatrix of , say

So the determinant of this submatrix will be

Thus I can say that the rank of the matrix is *not *2.

So I can say that the rank of the matrix is 1.

Now I’ll find out the rank of the matrix .

**Step 2**

As you can see, I’ll try to reduce the matrix to an upper triangular matrix.

And for that, at first, I’ll subtract row 1 from row 2.

Simultaneously, I’ll also subtract twice row 1 from row 3.

So in the mathematical term, it will be

Row 2 – Row 1, Row 3 – 2 Row 2.

Therefore the equivalent matrix will be

Next, I’ll subtract row 3 from row 2.

Thus Row 2 – Row 3 gives

Then, I’ll subtract twice row 2 from row 3.

Thus Row 3 – 2 Row 2 gives

In the end, I’ll subtract twice row 3 from row 2.

Thus Row 2 – 2 Row 3 gives

Now I interchange row 2 and 3 to get

And here I have to stop.

This is because I can’t reduce it further.

Now I’ll find out the determinant of any submatrix of the matrix .

**Step 3**

So I’ll start with a 3 × 3 submatrix of .

First, I’ll choose a 3 × 3 submatrix of , say

So the determinant of this submatrix will be

Therefore the rank of the matrix is *not* .

Now I’ll choose a 2 × 2 submatrix of , say

So the determinant of this submatrix will be

Thus I can say that the rank of the matrix is not .

So I can say that the rank of the matrix is 1.

Thus I can say that the rank of the matrix = the rank of the matrix = 1.

Now the number of the unknown in this set of equations is also 3.

Therefore I can conclude that this set of equations has infinitely many solutions.

Hence I can conclude that this is the answer to the given example.

Dear friends, this is the end of today’s post on the nature of solutions using matrices in linear algebra. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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