Linear Algebra is an essential branch of A-level mathematics that covers many topics considered o set the foundation for more complex mathematics. As the name implies, Linear Algebra considers linear equations whose order or power is 1. That is, all equations are set up as;

From the above equations, it can be seen that all the x’s are linear as they do not have an order of two and above. Linear Algebra comprises topics such as Matrices, Eigenvalues and Eigenvectors, Vectors, Linear Transformations, Linear dependence and Independence, and Singular Value Decomposition.

All of the above topics are very important topics for further mathematical computations. For example, the matrix is essential for calculating Gaussian Elimination. This article will consider the essentials of Linear Algebra which ae the Matrices, Eigenvalues and Eigenvectors.

**The Matrix**

A matrix is a rectangular set of real or complex values enclosed in a bracket.

**Knowing the Basics of a Matrix**

**Matrix Formation**

This matrix

The ‘x’ does not mean a multiplication operation. It is used to differentiate the number of rows in a matrix from the number of columns. For a matrix is usually number of rows (m) X number of columns (n)

**Column and Row Matrix**

A matrix that is given as

is a column matrix and is expressed as m X 1 matrix.

A matrix given as

is a row matrix and is given as 1 X n matrix

**Square Matrix**

A matrix whose number of columns is the same as its number of rows. i.e. n = m is a square matrix. 2 X 2, 3 X 3, and 4 X 4 matrixes are all square matrices.

**Square Matrix**

A single element matrix is a matrix that has just one element, i.e. [a]. It is a 1 X 1 matrix.

**Identifying Elements in a Matrix**

When a matrix value (elements) is not expressed in terms of values. They are usually given variables in the form below

From the above, the very first element has 11 has its subscript because it is the first element when considering from the row aspect (first 1 of the subscript) and the first element when considering from the column angle (second 1 of the subscript). Now, a_{23} is termed so because it is the element in the second row and third column.

**Identity Matrix**

An Identity matrix is a matrix whose diagonals are 1, and the reset elements are all zeros. The type of matrix does not matter. As long as the diagonals are all 1s and other elements zero, then it is a diagonal matrix

Any matrix multiplied by an Identity matrix will give the same matrix as an answer.

**Additional and Subtraction of Matrices**

The Addition and subtraction of a matrix are only possible if the two matrices has the same rows and columns. That is, the two matrices must be equal in terms of their n x m. To add or subtract, the elements in the same position of the two matrices are what will be added or subtracted.

**Example 1**

Find the respective A + B and A – B

**Solution**

Clearly, the row of the two matrices is the same and can be added or subtracted together.

**Multiplication of Matrices**

When multiplying a matrix, You can multiply a matrix by a scalar quantity (one digit) or multiply two matrices together. For the multiplication operation to work on two matrices, the first matric number of columns must be the same as the number of rows of the second matrix.

**Example 2**

**Solution**

Clearly, the number of columns of the first matrix is 3 while the row of the second matrix is 3

The next step will be to multiply each element in the first column of the second matrix with all the elements in the row of the first matrix. Add up the values together. That will be the first element in the first row of the new matrix

After that, multiply all the elements in the second column of the second matrix with the elements in the first row and add up together. That will be the second element in the first row of the new matrix

Finally, multiply all the elements in the third column of the second matrix with the elements in the first row and add them up together. That will be the third element in the first row of the new matrix

That is;

Now to get the bottom of the new matrix we are trying to form, multiply all of the columns in the same order followed in getting in the first row. The only difference is that they will be multiplied against the second row of the first matrix.

That will give;

The above steps are how to compute matrices Multiplication.

**Transpose of a Matrix**

The transpose of a matrix is the exchange of its rows for its columns and vice versa. For example, the transpose of a Matrix A

**The Determinant of the Matrix**

The determinant of a matrix is generally important for computing problems like the eigenvectors. Finding it is very easy for a 2 x 2 and 3 x 3 matrices, which are basically the main matrices that students will need to deal with.

**Example**

Find the determinant of the 2 x 2 Matrix

**Solution**

**Step 1:** To Find the 3 x 3 matrix determinant, we have to acknowledge that there are certain signs that exist in matrices and they are given as follows

We can see that we cancelled everything elements in the row and columns that connect with 3, leaving 7, 4, 4 2 in the form of 2 x 2 matrix that we can put in the form

Now we will need to find the determinant of the 2 x 2 matrix, which will be (7 x 2) – (4 x 4)

**Now we will arrange it in the form**

**3 [7 x 2) – (4 x 4)] …….. eqn (1)**

We can see that the 3 which caused the whole debacle is multiplying the mini determinant formed. Now we will term this as eqn (1). Now recall we also have to do 6 and 1 in the first row as outlined in step 2. So let’s do it for 6

We have to add or subtract each of them together, but the signs depend on the matrix signs we talked about

Looking at the first row,

3 will be +,

6 will be –

and

1 will be +

So that will leave us with

**+ 3[7 x 2) – (4 x 4)] – 6[2 x 2) – (4 x 5)] + 1[2 x 4) – (7 x 5)]**

3[14 – 16] – 6[4 – 20] + [8 – 35]

3(-2) – 6(-16) + (-27)

-6 + 96 – 27

= 63

It is important to take note of the above steps as they are consistent in finding the 3 x 3 matrix.

**Cofactors and Adjoints**

Now in our above discussion for finding the 3 x 3 matrices, we talked about the mini determinants for the first three rows. Do you know that we can replace the answers of those determinants in place of where the first three initial elements are?

Also, we can find the mini determinant for every single element.

Now we can arrange all the determinant answers in place of the elements that trigger them.

For example, element 3 of the first row and first column has its mini determinant as – 2. So we can replace 3 with – 2

Now all of these mini determinants are what we would have called the cofactors. This means that – 2 in the first row and first column would have been the cofactor of 3 in the first row and first column

However, we have to remember the placement signs that come with the 3 x 3 matrices which are

As such, for us to get the Co factors, we have to impact the mini determinants with the cofactors. That means that the signs will be alongside the cofactors.

As such, our cofactors matrix is

**Adjoint**

Now the Adjoint of a Matrix is simply the transpose of the cofactors of that matrix

**Inverse of A Matrix**

Now to find the inverse of a matrix A, will need to divide all the elements of the Matrix adjoints by the determinant value of the matrix. That is the steps of finding an inverse is as follows

Find the determinant of the 3 x 3 matrix

Find the Cofactors of the matrix

Get the Adjoint from the cofactors

Divide the element of the cofactors by the determinant.

**Eigenvalues ****and Eigenvectors**

Eigen Values and Eigenvectors employ matrices to an extent and are used in Engineering and other science concepts. To find the Eigenvectors, students must first find the eigenvalues

The formula for Eigen Vectors and Eigen Values are given as

**AX **=** LX**

Now the Eigenvalues are standard fixed values and are straightforward to compute. On the other hand, the Eigenvectors are varied vectors, so the aim is to only find the simplest vectors based on the Eigen values.

Now if we can also make eqn (a) as

AX – LX = 0

(A – LI)x = 0, where I is an identity matrix and is only there because that is the only way L can be subtracted from A

So in Matrix Form, all that we have is

The above is a rule for Eigenvalues.

**How to Calculate the Eigenvalues?**

Step 1: Find the determinant of Eqn C and find the Ls

Step 2: Solve for the Ls

**Example**

Find the Eigen values and Eigen vectors of the Matrix A

**Solution**

**Step 1: Find the determinant of Eqn C and find the Ls**

Using eqn (c), we will have the determinant of matrix A in respect to the Eigenvalues as

**Step 2: Solve for the Ls**

Following the process of finding the 3 x 3 matrix determinant, we will have something like this below

Now since it is a 3 x 3 matrix, we must have 3 Ls as answers. This is normal. For every n x n by matrices, we have n Ls and n vectors.

So we will get 3 Ls from eqn (d)

**First L**

**How to Find the Eigenvalues?**

Now to find the first eigenvectors, we will use the first eigenvalue.

Now, remember that the formula is actually

AX = LX

Since the first eigenvalue is – 5.47

This will lead to

(3 – 5.47)X_{1} + 6X_{2 } + X_{3} = 5.47X_{1}

2X_{1} + (7 – 5.47)X_{2 } + 4X_{3} = 5.47X_{2}

5X_{1} + 4X_{2 } + 5.47X_{3} = 5.47X_{3}

Following the pattern, we will have

X_{1 }= (6X_{2 } + X_{3})/7.94 ……… (e)

X_{1 }= (3.94X_{2 } – 4X_{3})/2 ………. (f)

X_{1 }= – 4X_{2}/5 ………… (g)

WE Will find the eigenvector based on its smallest value which is 1.

Now since X_{1 }= – 4/5X_{2}

Let X_{2 } = 1,

Now since X_{1 }= – 4/5(1) = 4/5

Now substituting the values of X_{2 } = 1, and X_{1 }= – 4/5 in eqn (e) we will have

4/5= (6(1) + X_{3})/7.94 ……… (e)

4/5 x 7.94 – 6 = X_{3}

X_{3} = 0.352

Therefore, the first possible Eigenvectors for the scalar 5.47

X = 1, 4/5 and 0.352

Now iof students use the answer of X_{1 }and X_{2} On eqn (f) they will have different vectors. This goes to prove that eigenvectors are no in any way limited to just unique answers.

Students can attempt to find the other Eigenvectors with respect to the two other scalers -5 – 7.2i, -5 + 7.2i. The eigen vectors will be in the form of complex numbers based on the two scalrs

**Conclusion**

The matrices, Vector and Linear Transformations are the foundation of linear algebra and many concepts of mathematics. All three fields are extremely vast and are standalone topics in A-Level mathematics. What this article has shown, however, are the essentials of how all three topics can be handled. The philosophy of the topics was discussed; as such, students can use all the information here to build momentum on applying them to other areas of mathematics.

For all three topics, many examples were considered to show students how they work. The examples were strategically collected so students could use their knowledge to handle a problem they encountered. Irrespective of where students may have to deal with any of the three concepts, these examples ensured that they would be able to scale through. Nevertheless, it is recommended that students practice more on the topic themselves.