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Matrix Determinant

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Have you ever considered how to find out if a system of simultaneous equations contains a solution? You may set up a system of equations to compare deals on something you're looking to buy to compare several factors across the different options, but how do you check if you've set up the system correctly and if there are solutions to compare?

You can store this system in a matrix and then find the determinant of this matrix to dictate if you have a solution available.

Read further to find out more about how this works.

A matrix is an array used to store, display and compute data. The internals are called elements and the matrix will have \(m\) columns and \(n\) rows.

To understand what a determinant is and how to apply it, we must first understand what a matrix is.

**A matrix is a way of displaying information**- for example, a system of simultaneous equations can be written in matrix form where your columns are for a variable and rows are for your equations. The solutions would then form a column vector. The matrix notation makes it easier to carry out transformations and solve data sets- especially when there are more than 2 equations to solve!

But how do we go about solving a matrix? That's where determinants come in - we use them to solve matrices.

The general matrix notation is that \(m\) denotes the number of columns and \(n\) denotes the number of rows. The internals of the matrix can then be written as:\[A_{m,n} = \begin{bmatrix}a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\\vdots & \vdots & \ddots & \vdots \\a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix}\]

For more information and examples see our article on Core Matrices.

**\(2\times 2\) matrix example**\[A_{2,2}=\begin{bmatrix}3&7\\12&-3\end{bmatrix}\]

**\(2\times 3\) matrix example**\[A_{2,3}=\begin{bmatrix}2&-4&19\\11&23&5\end{bmatrix}\]

**\(4\times 3\) matrix example**\[A_{4,3}=\begin{bmatrix}2&8&4\\-2&-5&-3\\13&9&7\\-7&3&-2\end{bmatrix}\]

Matrices are a really useful way of displaying and storing a lot of information and they are widely used throughout mathematics, physics and engineering at the higher levels of those disciplines.

So now we know the general basics of matrices but what is a determinant and why is it relevant?

The determinant is a value we can figure out for any square matrix that we can then use to calculate the inverse matrix.

A square matrix is a matrix that has an equal number of rows and columns, \(m=n.\)

As you can see below, square matrices have an equal number of rows and columns to form a square shape

**\(2\times 2\) matrix example **\[A_{2,2}=\begin{bmatrix}3&7\\12&-3\end{bmatrix}\]

**\(3\times 3\) matrix example **\[A_{3,3}=\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}\]

**An invertible matrix** is a matrix** for which we can find another matrix such that their product is the identity matrix \((I)\)**.

Our initial matrix can be denoted as \(A\) and the second matrix is the inverse of this matrix so is denoted as \(A^{-1}\). This gives the identity \[AA^{-1}=I.\] You can think of inverse matrices as the reciprocals of the matrix world.

The determinant also tells us **if** a matrix is invertible. Let the determinant of matrix A be denoted as \(\det{A}.\)

- If the determinant of the matrix A is given as \(\det{A} = 0\) the matrix is
**s****ingular**and therefore does not have an inverse. There is no solution to this matrix. - If the determinant of the matrix A is given as \(\det{A} \neq 0\) the matrix is
**non-s****ingular**and therefore does have an inverse. There is a solution to this matrix.

For more information and examples on inverting a matrix see our article Inverting Matrices.

So now we know what a determinant is and what it's used for but we still need to find out how they work.

Let's start with the most basic form - a \(2\times 2\) matrix determinant. The method to calculate the determinant of a \(2\times 2\) matrix is most basically explained by cross multiplying and then subtraction of these multiplied values.

Let's consider the below matrix,\[A_{2,2} = \begin{bmatrix}a_{1,1} & a_{1,2} \\a_{2,1} & a_{2,2}\end{bmatrix}\]This is the notation we used before but lets write it with distinct elements so the methodology is easier to follow. Therefore,\[A_{2,2} = \begin{bmatrix}a&b\\c&d\end{bmatrix}\]Our first step is to cross multiply- we multiply top left, bottom right and then top right, bottom left- and then subtract the second multiplication from the first. Therefore,\[\det{A}=ad-cb\]In our original notation this would be,\[\det{A}=a_{1,1}a_{2,2}-a_{1,2}a_{2,1}\]Let's now apply this to an example.

Find the determinant of matrix \(A\) below and then identify if the matrix is invertible.\[A=\begin{bmatrix}4&9\\-2&8\end{bmatrix}\]

**Solution**

**Step 1. Find the determinant**

\[\begin{align} \det{A}&=ad-cb\\&=(4\cdot 8)-(9\cdot -2)\\&=32-(-18)\\&=50\end{align}\]

**Step 2. Identify if matrix \(A\) is invertible**

\(\det{A} \neq 0\) so the matrix \(A\) is non-singular and as such invertible.

we move next to learn how to find the determinant of a \(3\times 3\) matrix.

Now we've seen how to find the determinant of a \(2\times 2\) matrix but we can also encounter \(3\times 3\) matrices in Further Maths so let's look at how to find the determinants of these now.

The process is slightly more complex than a \(2\times 2\) matrix determinant but follows the same principles. Let's consider the matrix below,\[A_{3,3}=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\]The way we calculate the determinant of this is to break it down into a series of \(2\times 2\) matrices.

To do this we go along the top row and multiply each of the elements there by the determinant of its minor.

The minor in a \(3\times 3\) matrix are the elements left if you cross out the row and column stemming from your root element.

Note the sign convention here with the determinant formula - it goes \(+,-,+\).

In the tip above you can see the sign convention goes \(+,-,+\). These are the cofactors for the first row of a \(3\times 3\) matrix.

While slightly beyond the scope you are expected to deal with here, there are cofactors for every element in the matrix.

This means we could also find the determinant of a matrix using rows 2 or 3 as our root elements and take the minors from there- we would just need to apply the correct cofactors to do so.

However, all you need to concern yourself with for now is the top row and \(+,-,+\).

Let's now have a look at how to apply this to an example.

Find the determinant of the below matrix.\[A_{3,3}=\begin{bmatrix}4&8&12\\7&19&2\\0&5&2\end{bmatrix}\]

**Solution**

We apply our formula for the determinant.\[\begin{align}\det{A}&=a(ei-fh)-b(di-fg)+c(dh-eg)\\&=4[(19\cdot 2)-(2\cdot 5)]-8[(7\cdot 2)-(2\cdot 0)]+12[(7\cdot 5)-(19\cdot 0)]\\&=4[(38)-(10)]-8[(14)-(0)]+12[(35)-(0)]\\&=4(28)-8(14)+12(35)\\&=112-112+420\\&=420 \end{align}\]

We move now to know more the determinant of a diagonal matrix, after defining it.

To calculate the determinate of a diagonal matrix, we must first understand what one is.

A **diagonal matrix** is a matrix that has **all non-diagonal elements as 0**. This doesn't mean that the diagonal elements themselves can't hold the value of 0 but it does mean any non-diagonal element is 0.

It takes the form of,\[A=\begin{bmatrix}a_{1,1} & 0 &0& \cdots & 0 \\0 & a_{2,2} & 0&\cdots & 0 \\0&0&a_{3,3}&\cdots &0\\\vdots & \vdots &\vdots & \ddots & \vdots \\0 & 0 &0& \cdots & a_{m,n} \end{bmatrix}\]

The determinant of a diagonal matrix can be found while multiplying the diagonal elements.

The **determinant of a diagonal matrix is the product of the diagonal elements**. Therefore,\[\det{A}=a_{1,1}\cdot a_{2,2}\cdot a_{3,3} \cdot \quad \cdots \quad \cdot a_{m,n}\]

If the diagonal elements are not all non-zero values then the matrix cannot be non-singular as a product with a \(0\) in will always return a solution of \(0\) and as we've seen previously this makes the matrix singular and uninvertible.

Let's have a look at this in an example.

Find \(\det{A}\), where,\[A_{5,5}=\begin{bmatrix}13&0&0&0&0\\0&-6&0&0&0\\0&0&7&0&0\\0&0&0&-1&0\\0&0&0&0&3\end{bmatrix}\]

**Solution**

We know the determinant of a diagonal matrix is the product of the diagonal elements. \[\begin{align} \det{A}&=a_{1,1}\cdot a_{2,2}\cdot a_{3,3}\cdot a_{4,4}\cdot a_{5,5}\\&=(13)\cdot (-6)\cdot (7)\cdot (-1)\cdot (3)\\&=1638.\end{align}\]

Can we calculate the determinant of an inverse matrix? The answer is, YES!

Our final matrix to consider when looking at determinants is that of the inverse matrix.

For the inverse matrix to exist, we know the original matrix must have had a determinant that was a non-zero value. We also earlier likened the inverse matrix to a reciprocal of the original matrix- that is going to come into play again here.

**The determinant of an inverse matrix** is equal to the inverse or reciprocal of the original matrix. In mathematical terms this means that the determinant of an inverse matrix takes the following form,\[\det{A^{-1}}=\frac{1}{\det{A}}.\]

Let's take the follwoing example.

Take matrix \(A\) below and find if it is invertible. If matrix \(A\) is invertible, find the determinant of that inverse matrix.

\[A=\begin{bmatrix}6&2\\12&9\end{bmatrix}\]

**Solution**

**Step 1. Find the determinant of \(A\)**

\[\begin{align} \det{A}&=ad-cb\\&=(6\cdot 9)-(12\cdot 2)\\&=54-24\\&=30\end{align}\]

**Step 2. Identify if matrix \(A\) is invertible**

\(\det{A} \neq 0\) so the matrix \(A\) is non-singular and as such invertible.

**Step 3. Find the determinant of the inverse matrix**

\[\begin{align}\det{A^{-1}}&=\frac{1}{\det{A}}\\&=\frac{1}{30}. \end{align}\]

- The determinant of a \(2\times 2\) matrix is given by,\[\det{A}=ad-bc\]
- The determinant of a \(3\times 3\) matrix uses the top row elements and the determinate of their respective minors to calculate the determinant. This is given by,\[\begin{align}\det{A}&=a\cdot \det{\begin{bmatrix}e&f\\h&i\end{bmatrix}}-b\cdot \det{\begin{bmatrix}d&f\\g&i\end{bmatrix}}+c\cdot \det{\begin{bmatrix}d&e\\g&h\end{bmatrix}} \\ & =a(ei-fh)-b(di-fg)+c(dh-eg)\end{align} \]
- The determinant of a diagonal matrix is the product of all the diagonal elements- which are the only elements that can be non-zero. This is given by, \[\det{A}=a_{1,1}\cdot a_{2,2}\cdot a_{3,3} \cdot \quad \cdots \quad \cdot a_{m,n}\]
- The determinant of an inverse matrix is the inverse of the determinant of the original matrix. This is given by, \[\det{A^{-1}}=\frac{1}{\det{A}}\]

Yes- they will both have the same determinant

The determinant of the matrix will be a non-zero value

More about Matrix Determinant

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