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Determinants

- Calculus
- Absolute Maxima and Minima
- Absolute and Conditional Convergence
- Accumulation Function
- Accumulation Problems
- Algebraic Functions
- Alternating Series
- Antiderivatives
- Application of Derivatives
- Approximating Areas
- Arc Length of a Curve
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- Derivative Functions
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- Derivatives
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- Determining Volumes by Slicing
- Direction Fields
- Disk Method
- Divergence Test
- Eliminating the Parameter
- Euler's Method
- Evaluating a Definite Integral
- Evaluation Theorem
- Exponential Functions
- Finding Limits
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- First Derivative Test
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- General Solution of Differential Equation
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- Implicit Differentiation Tangent Line
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- Determinants
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- Expression Math
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- Faces Edges and Vertices
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- Finding Rational Zeros
- Finding the Area
- Forms of Quadratic Functions
- Fractional Powers
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- Fractions
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- Fractions and Factors
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- Function Basics
- Functional Analysis
- Functions
- Fundamental Counting Principle
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- Generating Terms of a Sequence
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- Greatest Common Divisor
- Growth and Decay
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- Hyperbolas
- Imaginary Unit and Polar Bijection
- Implicit differentiation
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- Inequalities Maths
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- Integers
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- Iterative Methods
- L'Hopital's Rule
- Law of Cosines in Algebra
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- Limits of Accuracy
- Linear Expressions
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- Math formula
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- Notation
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- Surds
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- The Quadratic Formula and the Discriminant
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Jetzt kostenlos anmeldenDid you know that the area of a triangle, the volume of a parallelepiped, and solving a linear system of equations have something in common? Well, yes. They are all calculated via the use of a mathematical quantity, the **determinant**.

The determinant of a matrix is a unique number that is associated with a square matrix.

This means that the number can be used to represent that matrix. It has great use in solving linear equations, showing how changes in area and volume occur during linear transformations and explain how variables in integrals change. The determinant of a matrix M is represented as$det\left[M\right]or\left|M\right|$.

Determinants are of three basic types based on the kind of square matrix. They are the first-order determinant - for a 1 by 1 matrix wherein the determinant of the matrix and the elements of the matrix are the same. $\left[A\right]=\left|A\right|$. Second-order determinants and third-order determinants which would both be explained in detail hereafter.

Although, there are higher-order determinants such as those of 4 by 4, 5 by 5 matrices, and so on, but we shall not be discussing them. However, we shall begin with the matrix of order 2 afterwards.

The determinant of a 2 by 2 matrix is often referred to as a second-order determinant. This is achieved by subtracting the product of the right diagonal elements from the product of the left diagonal elements.

So, for a matrix $M=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$, we have,

$\left|M\right|=ad-bc$

Find the second-order determinant of the matrix $C=\left[\begin{array}{cc}5& -6\\ 4& 3\end{array}\right]$.

**Solution:**

$\left|\begin{array}{c}C\end{array}\right|=\left|\begin{array}{cc}5& -6\\ 4& 3\end{array}\right|\phantom{\rule{0ex}{0ex}}=(5\times 3)-(-6\times 4)\phantom{\rule{0ex}{0ex}}=15-(-24)\phantom{\rule{0ex}{0ex}}=15+24\phantom{\rule{0ex}{0ex}}=39.\phantom{\rule{0ex}{0ex}}$

The determinant of a matrix of order 3 is also called the third-order determinant. This does not follow the simple approach of the determinant of order 2. In fact, it requires the production of a **minor**.

A minor is the determinant of the obtained matrix derived by deleting the respective row and column associated with an element.

When this occurs, a matrix of order 3 becomes of order 2. For instance, if a matrix$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$, the determinant is obtained by producing 3 minors obtained by the elimination of the rows and columns associated with the elements of the first row separately. This means that **a**, **b**, and **c** would produce each minor as follows.

When the rows and columns associated with the element a is eliminated you have:

$={M}_{11}\mathrm{minor}=\left|\begin{array}{cc}e& f\\ h& i\end{array}\right|$

The minor ${M}_{11}=ei-fh$

When the rows and columns associated with the element b is eliminated you have:

$={M}_{12}\mathrm{minor}=\left|\begin{array}{cc}d& f\\ g& i\end{array}\right|$

The minor${M}_{12}=di-fg$

When the rows and columns associated with the element c is eliminated you have:

$={M}_{13}\mathrm{minor}=\left|\begin{array}{cc}d& e\\ g& h\end{array}\right|$

The minor ${M}_{13}=dh-eg$

To note clearly what the minors stand for,

${M}_{11}$ is the minor associated with the element in the first row and first column,

${M}_{12}$ is the minor associated with the element in the first row and second column,

and

${M}_{13}$ is the minor of the element in the first row and third column.

Furthermore, a 3 by 3 matrix has 9 minors, namely b${M}_{11},{M}_{12},{M}_{13},{M}_{21},{M}_{22},{M}_{23},{M}_{31},{M}_{32}$and ${M}_{33}.$ In addition minors are important because they are used to define cofactors. These cofactors are in turn used in calculating determinants.

A cofactor is used in calculating determinants, and it has the same numeric value as the minor but may have a different sign. The cofactor is expressed as:

${C}_{ij}={(-1)}^{i+j}{M}_{ij}$

where

i is the row index,

j is the column index,

${C}_{ij}$ is the cofactor corresponding to the ${i}^{th}$ row and ${j}^{th}$ column,

and

${M}_{ij}$ is the minor corresponding to the ${i}^{th}$row and${j}^{th}$ column.

Remember that only the sign may change when comparing a cofactor to its corresponding minor. Thus you should observe that what i+j is an even number, then C_{ij} = M_{ij}. Meanwhile, instances where i+j is odd, then C_{ij} = -M_{ij}. Therefore, M_{11} = C_{11}, M_{12} = -C_{12}, M_{13} = C_{13} etc.

After the application of the formula, an easier sign formula was introduced as follows:

$\left[\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right]$

Now we know how to break the 3 by 3 matrix into 2 by 2 minors to find their determinants. After this, we find the cofactor by multiplying the determinant of the minors with the corresponding signs. Then each cofactor is multiplied by its corresponding element. A summation of all products gives the determinant of a 3 by 3 matrix. Therefore, for a matrix M,

$M=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$,

$\left|\begin{array}{c}M\end{array}\right|=a\left({C}_{11}\right)+b\left({C}_{12}\right)+c\left({C}_{13}\right)$

You should note that the determinants can be calculated by using any row or column, not just the first row. For instance, the determinant of matrix M can also be calculated (using the second row in this case) as:

$\left|M\right|=d\left({C}_{21}\right)+e\left({C}_{22}\right)+f\left({C}_{23}\right)$

The examples herein shall use only the first row so as to enable you to try out the examples using different rows or columns.

Find the determinant of the matrix$D=\left[\begin{array}{ccc}1& 4& 2\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$.

**Solution:**

Step 1.

Bring out your minors.

${M}_{11}=\left|\begin{array}{cc}5& 6\\ 8& 9\end{array}\right|=(5\times 9)-(6\times 8)\phantom{\rule{0ex}{0ex}}{M}_{11}=45-48=-3\phantom{\rule{0ex}{0ex}}{M}_{12}=\left|\begin{array}{cc}4& 6\\ 7& 9\end{array}\right|=(4\times 9)-(6\times 7)\phantom{\rule{0ex}{0ex}}{M}_{12}=36-42=-6\phantom{\rule{0ex}{0ex}}{M}_{13}=\left|\begin{array}{cc}4& 5\\ 7& 8\end{array}\right|=(4\times 8)-(5\times 7)\phantom{\rule{0ex}{0ex}}{M}_{13}=32-35=-3$Step 2.

Since you have the determinants of the minors, find the cofactors

${C}_{11}=+1\left({M}_{11}\right)=+1(-3)=-3\phantom{\rule{0ex}{0ex}}{C}_{12}=-1\left({M}_{12}\right)=-1(-6)=6\phantom{\rule{0ex}{0ex}}{C}_{13}=+1\left({M}_{13}\right)=+1(-3)=-3$

Step 3

Now you calculated the cofactors, find the sum of all products between the elements and their corresponding cofactors.

$\left|\begin{array}{c}M\end{array}\right|=a\left(C11\right)+b\left(C12\right)+c\left(C13\right)\phantom{\rule{0ex}{0ex}}\left|\begin{array}{c}D\end{array}\right|=1(-3)+4\left(6\right)+2(-3)\phantom{\rule{0ex}{0ex}}\left|\begin{array}{c}D\end{array}\right|=-3+24-6\phantom{\rule{0ex}{0ex}}\left|\begin{array}{c}D\end{array}\right|=15$

To calculate the determinant of a matrix$M=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$ using a diagonal method, you need to follow the following steps.

Step 1

Write out the elements in the first and second column of the matrix to the right-hand side of the matrix as seen below.

Step 2

Draw three diagonal lines rightwards, each cutting through three elements, and calculate the product of these elements as seen below.

Step 3

Sum all products of these elements. Thus,

$\mathrm{Rightwards}\mathrm{diagonal}\mathrm{product}=aei+bfg+cdh$

Step 4

Repeat step 1 and draw three diagonal lines leftwards, each cutting through three elements and calculating the product of these elements as seen below.

Step 5

Sum all products of these elements. Thus,

$\mathrm{Leftwards}\mathrm{diagonal}\mathrm{product}=ceg+afh+bdi$

Step 6

Subtract the leftward diagonal product from the rightward diagonal product.

$\left|M\right|=Rightwarddiagonalproduct-leftwarddiagonalproduct\phantom{\rule{0ex}{0ex}}\left|\begin{array}{c}M\end{array}\right|=(aei+bfg+cdh)-(ceg+afh+bdi)\phantom{\rule{0ex}{0ex}}\left|\begin{array}{c}M\end{array}\right|=aei+bfg+cdh-ceg-afh-bdi$

If the coding system is to be used, then the determinant would be:

$\left|\begin{array}{c}M\end{array}\right|=({r}_{1}{c}_{1}\times {r}_{2}{c}_{2}\times {r}_{3}{c}_{3})+({r}_{1}{c}_{2}\times {r}_{2}{c}_{3}\times {r}_{3}{c}_{1})+({r}_{1}{c}_{3}\times {r}_{2}{c}_{1}\times {r}_{3}{c}_{2})-\phantom{\rule{0ex}{0ex}}\left(\right({r}_{1}{c}_{3}\times {r}_{2}{c}_{2}\times {r}_{3}{c}_{1})+({r}_{1}{c}_{1}\times {r}_{2}{c}_{3}\times {r}_{3}{c}_{2})+({r}_{1}{c}_{2}\times {r}_{2}{c}_{1}\times {r}_{3}{c}_{3}\left)\right)$

Find the determinant of the matrix

$D=\left[\begin{array}{ccc}1& 4& 2\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$.**Solution:**

Using the diagonal method:

$\mathrm{The}\mathrm{rightwards}\mathrm{diagonal}\mathrm{product}=(1\times 5\times 9)+(4\times 6\times 7)+(2\times 4\times 8)\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{rightwards}\mathrm{diagonal}\mathrm{product}=45+168+64\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{rightwards}\mathrm{diagonal}\mathrm{product}=277\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{leftwards}\mathrm{diagonal}\mathrm{product}=(2\times 5\times 7)+(1\times 6\times 8)+(4\times 4\times 9)\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{leftwards}\mathrm{diagonal}\mathrm{product}=70+48+144\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{leftwards}\mathrm{diagonal}\mathrm{product}=262\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left|D\right|=rightwardsdiagonalproduct-leftwardsdiagonalprodct\phantom{\rule{0ex}{0ex}}\left|D\right|=277-262\phantom{\rule{0ex}{0ex}}\left|D\right|=15$There are several properties of determinants in matrices that would ease both understandings as well as solving matrix operations. They are:

The determinants of a matrix are the same across any row or column.

The determinant is equal to 0 when all elements of a row or column are 0.

The determinant of an identity matrix is 1.

When a matrix A is multiplied by a scalar c, the determinant of the new matrix cA is equal to the product of the determinant A and c to the power of the number of rows n of the square matrix. $\left|cA\right|={c}^{n}\left|A\right|$.

When two rows or columns of a matrix are interchanged, the determinant of the new matrix is the product of -1 and the determinant of the old matrix.

When the rows and columns of a matrix are interchanged (transposed), the determinant of the transposed matrix is equal to the determinant of the original matrix. $\left|A\right|=\left|{A}^{T}\right|$.

The determinant of an inverse matrix is expressed as: $\left|{A}^{-1}\right|=\frac{1}{\left|A\right|}$.

When two rows or columns of a matrix are identical, the determinant of the matrix is 0.

The determinant of matrix n times is its determinant n times. $\left|{A}^{n}\right|={\left(\left|A\right|\right)}^{n}$

The determinant of ABC equals to the product of determinants A, B and C. $\left|ABC\right|=\left|A\right|\times \left|B\right|\times \left|C\right|$.

When a matrix P is derived from adding n-times a row of Q to a different role. Then, $\left|P\right|=\left|Q\right|$.

Where $adj\left(A\right)$ is the adjoint of a matrix A, then, $\left|adj\left(A\right)\right|={\left(\left|A\right|\right)}^{n-1}$ and $\left|adj\left(adj\right(A\left)\right)\right|={\left(\left|A\right|\right)}^{(n-1)\times (n-1)}$.

- The determinant of a matrix is a number that is associated with a square matrix.
- The determinant of a 2 by 2 matrix is often referred to as a second-order determinant.
- The determinant of a 3 by 3 matrix is also called the third-order determinant.
- A minor is the determinant of the obtained matrix derived by deleting the respective row and column associated with an element.
- Cofactor is used in calculating determinants, and it has the same numeric value as the minor but may have a different sign.

More about Determinants

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