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**Simplifying Matrices – Explanation & Examples**

Matrix notation is one of the significant branches of numerical analysis used to represent linear equations and transformations. The term matrix was first introduced by the mathematician James Joseph Sylvester in 1850. Before that, they were known as arrays. Matrices form a very fundamental portion of mathematics. They are used to simplify tons of data, consisting of several rows, into one factor. In this article, we will discuss **Simplifying Matrices. **

*Simplifying matrices refer to the solution of matrices by applying different operations.*

In this article, we will be covering the following topics:

- What is a matrix?
- How to simplify matrices?
- Scalar multiplication.
- Multiplication of matrices.
- Addition or subtraction of matrices.
- Steps for simplifying matrices.
- We are simplifying matrices involving inverse.
- Simplifying determinants.
- Practice problems
- Examples

**What is a Matrix?**

Before we dive into the realm of simplifying matrices, the first question arises what a matrix is?

*The matrix is defined as a rectangular array of numbers or expressions arranged in rows and columns in mathematics. Matrix operations are used to deal with linear equations that are referred to as the system of linear equations.*

For instance, consider an equation (1) which is given below:

x + 3 (y) = 0,

It has infinitely many solutions, and it forms a straight line. But if we consider another equation (2) which is given as:

2 (x) + y = 5

Now, solving both the equations simultaneously gives only one solution for both x and y that is the point:

x = -3

y = 1

It can also be regarded as the point where both the equations are intersecting with each other.

The two equations can be depicted using a matrix notation. The matrix notation for the above system of linear equation will be a 2 x 2 matrix given as

[ 1 3 2 1 ]

where row 1 of the matrix mentioned above represents the equation (1) and row, 2 represents the equation (2). Therefore, matrix operations can be used to simplify matrices of complex equations.

**How To Simplify Matrices?**

Solving matrices using matrix operation is often considered a difficult job as one has to keep track of large numbers or expressions. Still, this problem can be solved by simplifying the matrices. Simplification of the matrices reduces the computational complexity and makes it much more accurate, and makes the operations much more effortless.

Following is the description of some matrix operations that can be used to simplify matrices and matrix equations.

**Scalar Multiplication:**

*Scalars are the numbers written along with the matrices, not on the inside of the matrices but the outside. *

So, firstly while simplifying the matrices, look for the scalar numbers being multiplied with the matrix. If a scalar is being multiplied in matrix algebra, it will be multiplied with all the entries inside the matrix. For instance, consider the following equation:

2A + 3B = 3

If A and B are the 2-by-2 matrices, then scalar two will be multiplied with all the matrix A entries. Similarly, scalar three will be multiplied with all the elements of matrix B.

Let’s understand this concept better with the help of an example.

*Example 1*

Simplify.

5A + B + C

Where

A = [ 2 1 ; -1 3 ]

B = [ 3 4 ; 2 1 ]

C = [ 1 5 ; 4 2 ]

Solution

After analyzing the question, it can be seen there is only one scalar multiplication which is 5A.

Let’s first solve for 5A:

5A = 5 [ 2 1 ; -1 3 ]

5A = [ 10 5 ; -5 15 ]

Now, simplifying the equation and performing the addition operation:

5A + B + C

5A + B + C = [ 10 5 ; -5 15 ] + [ 3 4 ; 2 1 ] + [ 1 5 ; 4 2 ]

5A + B + C = [ 10 + 3 + 1 5 + 4 + 5 ; -5 + 2 + 4 15 + 1 + 2 ]

5A + B + C = [ 14 14 ; 1 18 ]

So, the above expression of matrices has been simplified into a single matrix.

**Multiplication of the Matrices:**

*Matrices can be multiplied if the number of columns in the first matrix being multiplied is equal to the number of rows of the second matrix. *

This condition must be satisfied to find the product of any two matrices. For example, consider two matrices A and B; the product of two matrices can be found out by lining up rows and columns. Multiply across the lines and then add the values. This will give you the first entry of the new matrix and so on.

If matrix A is given as:

A = [ 1 2 6 1 ]

Similarly, matrix B is given as:

B = [ 2 ; 5 ; 2 ; 1]

As we can see that the number of columns of matrix A is four, and similarly, the number of rows of matrix B is 4, so these matrices are capable of multiplication. The multiplication of the two matrices gives a new matrix C which is shown as:

C = [ 6 7 14 31 ]

This matric C is obtained by obtaining the dot product of each row of matrix A with each column of matrix B and then adding it. In these matrices, the dot product of the row of matrix A with the column of matrix B is obtained.

Remember, matrix multiplication is non-commutative.

*Example 2*

Find the product of the following matrices:

A = [ 1 2 ; 3 4 ]

B = [ 0 2 ; 1 3 ]

Solution

To find these two matrices’ product, we first have to evaluate if these matrices are capable of multiplication. For that, we need to check their order.

Order of matrix A = 2 x 2

Order of matrix B = 2 x 2

Since the number of matrix A columns is equal to the number of rows of B, these matrices can be multiplied.

Now for the multiplication:

A x B = [ 1 2 ; 3 4 ] x [ 0 2 ; 1 3 ]

Multiplication will be performed by first multiplying row 1 of matrix A with column 1 of matrix B and then column 2 of matrix B. Similarly, next up, we will multiply row 2 of matrix A with both matrix B columns.

A x B = [ 1×0 + 2×1 1×2 + 2×3 ; 3×0 + 4×1 3×2 + 4×3 ]

A x B = [ 0+2 2+6 ; 0+4 6+12 ]

A x B = [ 2 8 ; 4 18 ]

This indicates the product of the two matrices, A and B.

**Addition or Subtraction:**

*Matrices of the same order can be added or subtracted. *

For matrix simplification, put all the values of individual matrices into equations within one big matrix and then perform the required operation, i.e. addition or subtraction. Therefore, solving the matrices gives a single simplified matrix.

**Steps For Simplifying Matrices:**

Following steps must be followed while simplifying the matrices:

- Solve for any scalar multiplication in the equation.
- Multiply matrices and simplify.
- Perform the operations of addition or subtraction and simplify.

Let us have a better understanding with the help of some examples.

*Example 3*

Simplify.

2A + BC + 3B

Where

A = [ 1 3 ; 1 1 ]

B = [ 3 1 ; 2 1 ]

C = [ 1 3 ; 4 4 ]

Solution

Since as it can be seen there are two scalar multiplications 2A and 3B.

So let’s first solve for 2A:

2A = 2 [ 1 3 ; 1 1 ]

2A = [ 2 6 ; 2 2 ]

Now, for 3B:

3B = 3 [ 3 1 ; 2 1 ]

3B = [ 9 3 ; 6 3]

Now let’s solve for the matrix multiplication:

BC = [ 3 1 ; 2 1 ] x [ 1 3 ; 4 4 ]

BC = [ 3×1 + 1×4 3×3 + 1×4 ; 2×1 + 1×4 2×3 + 1×4 ]

BC = [ 7 13 ; 6 10 ]

Now, simplifying the equation and performing the addition operation:

2A + BC + 3B = [ 2 6 ; 2 2 ] + [ 9 3 ; 6 3 ] + [ 7 13 ; 6 10 ]

2A + BC + 3B = [ 2 + 9 + 7 6 + 2 + 13 ; 2 + 6 + 6 2 + 3 + 10 ]

2A + BC + 3B = [ 18 22 ; 14 15 ]

So, the above expression of matrices has been simplified into a single matrix.

**Simplifying Matrices Involving Inverse of a Matrix:**

Another important concept when it comes to matrices is simplifying the inverse of a matrix. So, let’s take a look at that concept.

Let A and B be two matrices of the same order, then A and B are said to be multiplicative inverses of each other if and only if:

AB = BA = I

Where I am the identity matrix of the same order.

The inverse of the matrix A is written as A-1 then,

A x A-1 = A-1 x A = I

The inverse of the matrix can be computed as

A-1 = Adj A / | A |

Adjoint A is obtained by interchanging the 1st entry with its diagonal entry and reversing all the other entries’ signs. And | A | stands for the determinant of A, which is obtained by subtracting the diagonal entries’ product.

Let us discuss an example of matrix simplification involving multiplicative inverse.

*Example 4*

Solve the following equation:

( AB ) -1 + 3A

Where

A = [ 4 -1 ; 0 2 ]

B = [ -4 1 ; 1 -1 ]

C = [ 3 -2 ; 1 1 ]

Solution

First of all, solving for 3A:

3A = 3 [ 4 -1 ; 0 2 ]

3A = [ 12 -3 ; 0 6 ]

Now multiplying the two matrices A and B:

AB = [ 4 -1 ; 0 2 ] x [ -4 1 ; 1 -1 ]

AB = [ 4x-4 + -1×1 -1×1 + -1x-1 ; 0x-4 + 0x1 0x1 + 2x-1 ]

AB = [ -17 5 ; 2 -2 ]

Now finding the inverse of AB:

( AB ) -1 = Adj AB / | AB |

Adj A = [ -2 -5 ; -2 -17 ]

| AB | = {(-2) x (-17) – (2) x (5)}

| AB | = 34 – 10

| AB | = 24

So the inverse is given by:

( AB ) -1 = [ -0.08 -0.2 ; -0.08 -0.7 ]

Now, putting the values into the original expression

( AB ) -1 + 3A = [ -0.08 -0.2 ; -0.08 -0.7 ] + [ 12 -3 ; 0 6 ]

( AB ) -1 + 3A = [ 11.92 -3.2 ; -0.08 5.3 ]

So, this is the simplified matrix.

**Simplifying Determinants:**

A determinant can be simplified by using the following three techniques:

- Factor a constant from a row or column.

- Interchange two rows or two columns.

- Add a multiple of a row or column to the other row or column.

The matrices of order 3 x 3 or higher-order matrices involve a lot of difficult computation so, the above-mentioned techniques of determinant simplification can be used to reduce the computational complexity.

To further strengthen the concept of simplifying matrices, consider the following practice problems.

### Practice Problems

- Simplify the following matrix equation

a. A + B + C

b. ( BC ) -1 + 2A

c. 3A + AC + B

d. A + ( B – C )

e. 3A – 2B

Where,

A = [ 1 -2 ; 3 4 ]

B = [ 0 -3 ; 1 1 ]

C = [ 1 2 ; 2 3 ]

2. Show that A + C = C + A. Use the matrices given in question no 1.

3. Let A be n x n matrix, then A2 = I where I is an identity matrix. Determine whether A = I or not?

4. Evaluate the determinant of the matrix A = [ 1 0 ; 0 1 ].

5. Evaluate the inverse of the above matrix.

### Answers

a. [ 2 -3 ; 6 8 ]

b. [ 0.33 -7 ; 7 10 ]

c. [ 0 -12 ; 21 31 ]

d. [ 0 -7 ; 2 2 ]

e. [ 3 0 ; 7 10 ]

2. They both side of the equation are equal.

3. No. A is not necessarily the identity matrix.

4. Determinant = 1

5. Inverse = Matrix A.