Which matrices have an inverse. higher mathematics

Consider the problem of defining the operation inverse to matrix multiplication.

Let A be a square matrix of order n. Matrix A^(-1) , which together with the given matrix A satisfies the following equalities:

A^(-1)\cdot A=A\cdot A^(-1)=E,

called reverse. The matrix A is called reversible, if there is an inverse for it, otherwise - irreversible.

It follows from the definition that if inverse matrix A^(-1) exists, then it is square of the same order as A . However, not every square matrix has an inverse. If the determinant of matrix A zero(\det(A)=0) , then there is no inverse for it. Indeed, applying the theorem on the determinant of the product of matrices for the identity matrix E=A^(-1)A, we obtain a contradiction

\det(E)=\det(A^(-1)\cdot A)=\det(A^(-1))\det(A)=\det(A^(-1))\cdot0=0

since the determinant of the identity matrix is equal to 1. It turns out that the difference from zero of the determinant of a square matrix is the only condition for the existence of an inverse matrix. Recall that a square matrix whose determinant is equal to zero is called degenerate (singular), otherwise - non-singular (non-singular).

Theorem 4.1 on the existence and uniqueness of the inverse matrix. square matrix A=\begin(pmatrix)a_(11)&\cdots&a_(1n)\\ \vdots&\ddots&\vdots\\ a_(n1)&\cdots&a_(nn) \end(pmatrix), whose determinant is non-zero, has an inverse matrix and, moreover, only one:

A^(-1)=\frac(1)(\det(A))\cdot\! \begin(pmatrix)A_(11)&A_(21)&\cdots&A_(1n)\\ A_(12)&A_(22)&\cdots&A_(n2)\\ \vdots&\vdots&\ddots&\vdots\\ A_(1n )&A_(2n)&\cdots&A_(nn) \end(pmatrix)= \frac(1)(\det(A))\cdot A^(+),

where A^(+) is the matrix transposed for the matrix composed of the algebraic complements of the elements of the matrix A .

The matrix A^(+) is called attached matrix with respect to the matrix A .

Indeed, the matrix \frac(1)(\det(A))\,A^(+) exists under the condition \det(A)\ne0 . We must show that it is inverse to A , i.e. satisfies two conditions:

\begin(aligned)\mathsf(1))&~A\cdot\!\left(\frac(1)(\det(A))\cdot A^(+)\right)=E;\\ \mathsf (2))&~ \!\left(\frac(1)(\det(A))\cdot A^(+)\right)\!\cdot A=E.\end(aligned)

Let's prove the first equality. According to item 4 of Remarks 2.3, it follows from the properties of the determinant that AA^(+)=\det(A)\cdot E. So

A\cdot\!\left(\frac(1)(\det(A))\cdot A^(+)\right)= \frac(1)(\det(A))\cdot AA^(+) = \frac(1)(\det(A))\cdot \det(A)\cdot E=E,

which was to be shown. The second equality is proved similarly. Therefore, under the condition \det(A)\ne0, the matrix A has an inverse

A^(-1)=\frac(1)(\det(A))\cdot A^(+).

We prove the uniqueness of the inverse matrix by contradiction. Let besides the matrix A^(-1) there exists one more inverse matrix B\,(B\ne A^(-1)) such that AB=E . Multiplying both sides of this equality on the left by the matrix A^(-1) , we get \underbrace(A^(-1)AB)_(E)=A^(-1)E. Hence B=A^(-1) , which contradicts the assumption B\ne A^(-1) . Therefore, the inverse matrix is unique.

Remarks 4.1

1. It follows from the definition that the matrices A and A^(-1) are permutable.

2. The matrix inverse to a nondegenerate diagonal one is also diagonal:

\Bigl[\operatorname(diag)(a_(11),a_(22),\ldots,a_(nn))\Bigr]^(-1)= \operatorname(diag)\!\left(\frac(1 )(a_(11)),\,\frac(1)(a_(22)),\,\ldots,\,\frac(1)(a_(nn))\right)\!.

3. The matrix inverse to a nondegenerate lower (upper) triangular matrix is lower (upper) triangular.

4. Elementary matrices have inverses, which are also elementary (see item 1 of Remarks 1.11).

Inverse Matrix Properties

The matrix inversion operation has the following properties:

\begin(aligned)\bold(1.)&~~ (A^(-1))^(-1)=A\,;\\ \bold(2.)&~~ (AB)^(-1 )=B^(-1)A^(-1)\,;\\ \bold(3.)&~~ (A^T)^(-1)=(A^(-1))^T\ ,;\\ \bold(4.)&~~ \det(A^(-1))=\frac(1)(\det(A))\,;\\ \bold(5.)&~~ E^(-1)=E\,. \end(aligned)

if the operations indicated in equalities 1-4 make sense.

Let's prove property 2: if the product AB is non-degenerate square matrices of the same order has an inverse matrix, then (AB)^(-1)=B^(-1)A^(-1).

Indeed, the determinant of the product of matrices AB is not equal to zero, since

\det(A\cdot B)=\det(A)\cdot\det(B), where \det(A)\ne0,~\det(B)\ne0

Therefore, the inverse matrix (AB)^(-1) exists and is unique. Let us show by definition that the matrix B^(-1)A^(-1) is inverse with respect to the matrix AB . Really.

Finding the inverse matrix.

In this article, we will deal with the concept of an inverse matrix, its properties and ways of finding it. Let us dwell in detail on solving examples in which it is required to construct an inverse matrix for a given one.

Page navigation.

Inverse matrix - definition.

Finding the inverse matrix using a matrix of algebraic additions.

Properties of the inverse matrix.

Finding the inverse matrix by the Gauss-Jordan method.

Finding the elements of the inverse matrix by solving the corresponding systems of linear algebraic equations.

Inverse matrix - definition.

The concept of an inverse matrix is introduced only for square matrices whose determinant is different from zero, that is, for non-singular square matrices.

Definition.

Matrixis called the inverse of the matrix, whose determinant is different from zero, if equalities are true , where E is the identity matrix of order n on the n.

Finding the inverse matrix using a matrix of algebraic additions.

How to find the inverse matrix for a given one?

First, we need the concepts transposed matrix, the matrix minor, and the algebraic complement of the matrix element.

Definition.

Minork-th order matrices A order m on the n is the determinant of the order matrix k on the k, which is obtained from the elements of the matrix A located in the selected k lines and k columns. ( k does not exceed the smallest number m or n).

Minor (n-1)th order, which is made up of the elements of all rows, except i-th, and all columns except j-th, square matrix A order n on the n let's denote it as .

In other words, the minor is obtained from the square matrix A order n on the n crossing out elements i-th lines and j-th column.

For example, let's write, minor 2nd order, which is obtained from the matrix selection of elements of its second, third rows and first, third columns . We also show the minor, which is obtained from the matrix deleting the second row and third column . Let us illustrate the construction of these minors: and .

Definition.

Algebraic addition element of a square matrix is called the minor (n-1)th order, which is obtained from the matrix A, deleting elements of its i-th lines and j-th column multiplied by .

The algebraic complement of an element is denoted as . Thus, .

For example, for a matrix the algebraic complement of the element is .

Secondly, we will need two properties of the determinant, which we discussed in the section matrix determinant calculation:

Based on these properties of the determinant, the definitions operations of multiplying a matrix by a number and the concept of an inverse matrix, we have the equality , where is a transposed matrix whose elements are algebraic complements .

Matrix is indeed the inverse of the matrix A, since the equalities . Let's show it

Let's compose inverse matrix algorithm using equality .

Let's analyze the algorithm for finding the inverse matrix using an example.

Example.

Given a matrix . Find the inverse matrix.

Solution.

Calculate the matrix determinant A, expanding it by the elements of the third column:

The determinant is non-zero, so the matrix A reversible.

Let's find a matrix from algebraic additions:

Let's perform the transposition of the matrix from algebraic additions:

Now we find the inverse matrix as :

Let's check the result:

Equality are executed, therefore, the inverse matrix is found correctly.

Properties of the inverse matrix.

Concept of inverse matrix, equality , the definitions of operations on matrices, and the properties of the determinant of a matrix make it possible to substantiate the following inverse matrix properties:

Finding elements of the inverse matrix by solving the corresponding systems of linear algebraic equations.

Consider another way to find the inverse matrix for a square matrix A order n on the n.

This method is based on the solution n systems of linear inhomogeneous algebraic equations with n unknown. The unknown variables in these systems of equations are the elements of the inverse matrix.

The idea is very simple. Denote the inverse matrix as X, that is, . Since by definition of the inverse matrix , then

Equating the corresponding elements by columns, we get n systems of linear equations

We solve them in any way and form an inverse matrix from the found values.

Let's analyze this method with an example.

Example.

Given a matrix . Find the inverse matrix.

Solution.

Accept . Equality gives us three systems of linear nonhomogeneous algebraic equations:

We will not describe the solution of these systems; if necessary, refer to the section solution of systems of linear algebraic equations.

From the first system of equations we have , from the second - , from the third - . Therefore, the desired inverse matrix has the form . We recommend checking to make sure the result is correct.

Summarize.

We considered the concept of an inverse matrix, its properties and three methods for finding it.

Example of Inverse Matrix Solutions

Exercise 1. Solve SLAE using the inverse matrix method. 2 x 1 + 3x 2 + 3x 3 + x 4 = 1 3 x 1 + 5x 2 + 3x 3 + 2x 4 = 2 5 x 1 + 7x 2 + 6x 3 + 2x 4 = 3 4 x 1 + 4x 2 + 3x 3 + x4 = 4

Form start

End of form

Solution. Let's write the matrix as: Vector B: BT = (1,2,3,4) Principal Determinant Minor for (1,1): = 5 (6 1-3 2)-7 (3 1-3 2)+4 ( 3 2-6 2) = -3 Minor for (2,1): = 3 (6 1-3 2) -7 (3 1-3 1)+4 (3 2-6 1) = 0 Minor for (3 ,1): = 3 (3 1-3 2)-5 (3 1-3 1)+4 (3 2-3 1) = 3 Minor for (4,1): = 3 (3 2-6 2) -5 (3 2-6 1)+7 (3 2-3 1) = 3 Minor determinant ∆ = 2 (-3)-3 0+5 3-4 3 = -3

Transposed matrix Algebraic complements ∆ 1.1 = 5 (6 1-2 3)-3 (7 1-2 4)+2 (7 3-6 4) = -3 ∆ 1.2 = -3 (6 1-2 3) -3 (7 1-2 4)+1 (7 3-6 4) = 0 ∆ 1.3 = 3 (3 1-2 3)-3 (5 1-2 4)+1 (5 3-3 4 ) = 3 ∆ 1.4 = -3 (3 2-2 6) -3 (5 2-2 7)+1 (5 6-3 7) = -3 ∆ 2.1 = -3 (6 1-2 3)-3 (5 1-2 4)+2 (5 3-6 4) = 9 ∆ 2.2 = 2 (6 1-2 3)-3 (5 1-2 4)+1 (5 3- 6 4) = 0 ∆ 2.3 = -2 (3 1-2 3)-3 (3 1-2 4)+1 (3 3-3 4) = -6 ∆ 2.4 = 2 (3 2- 2 6)-3 (3 2-2 5)+1 (3 6-3 5) = 3 ∆ 3.1 = 3 (7 1-2 4)-5 (5 1-2 4)+2 (5 4 -7 4) = -4 ∆ 3.2 = -2 (7 1-2 4)-3 (5 1-2 4)+1 (5 4-7 4) = 1 ∆ 3.3 = 2 (5 1 -2 4)-3 (3 1-2 4)+1 (3 4-5 4) = 1 ∆ 3.4 = -2 (5 2-2 7)-3 (3 2-2 5)+1 ( 3 7-5 5) = 0 ∆ 4.1 = -3 (7 3-6 4) -5 (5 3-6 4)+3 (5 4-7 4) = -12 ∆ 4.2 = 2 ( 7 3-6 4) -3 (5 3-6 4) +3 (5 4-7 4) \u003d -3 ∆ 4.3 \u003d -2 (5 3-3 4) -3 (3 3-3 4) +3 (3 4-5 4) = 9 ∆ 4.4 = 2 (5 6-3 7)-3 (3 6-3 5)+3 (3 7-5 5) = -3 Inverse Matrix Result Vector X X = A -1 ∙ B X T = (2,-1,-0.33.1) x 1 = 2 x 2 = -1 x 3 = -0.33 x 4 = 1

see also SLAE solutions by the inverse matrix method online. To do this, enter your data and get a decision with detailed comments.

Task 2. Write the system of equations in matrix form and solve it using the inverse matrix. Check the obtained solution. Solution:xml:xls

Example 2. Write the system of equations in matrix form and solve using the inverse matrix. Solution:xml:xls

Example. A system of three linear equations with three unknowns is given. Required: 1) find its solution using Cramer's formulas; 2) write the system in matrix form and solve it using matrix calculus. Guidelines. After solving by Cramer's method, find the button "Inverse matrix solution for initial data". You will receive an appropriate decision. Thus, the data will not have to be filled in again. Solution. Denote by A - the matrix of coefficients for unknowns; X - column matrix of unknowns; B - matrix-column of free members:

Vector B: BT =(4,-3,-3) Given these notations, this system of equations takes the following matrix form: A*X = B. If the matrix A is nondegenerate (its determinant is nonzero, then it has an inverse matrix A -1. Multiplying both sides of the equation by A -1, we get: A -1 * A * X \u003d A -1 * B, A -1 * A \u003d E. This equality is called matrix notation of the solution of the system of linear equations. To find a solution to the system of equations, it is necessary to calculate the inverse matrix A -1 . The system will have a solution if the determinant of the matrix A is non-zero. Let's find the main determinant. ∆=-1 (-2 (-1)-1 1)-3 (3 (-1)-1 0)+2 (3 1-(-2 0))=14 So, the determinant is 14 ≠ 0, so we continue solution. To do this, we find the inverse matrix through algebraic additions. Let we have a non-singular matrix A:

We calculate algebraic additions.

∆ 1,1 =(-2 (-1)-1 1)=1

∆ 1,2 =-(3 (-1)-0 1)=3

∆ 1,3 =(3 1-0 (-2))=3

∆ 2,1 =-(3 (-1)-1 2)=5

∆ 2,2 =(-1 (-1)-0 2)=1

∆ 2,3 =-(-1 1-0 3)=1

∆ 3,1 =(3 1-(-2 2))=7

∆ 3,2 =-(-1 1-3 2)=7

X T =(-1,1,2) x 1 = -14 / 14 = -1 x 2 = 14 / 14 =1 x 3 = 28 / 14 =2 Examination. -1 -1+3 1+0 2=4 3 -1+-2 1+1 2=-3 2 -1+1 1+-1 2=-3 doc:xml:xls Answer: -1,1,2.

Definition 1: A matrix is called degenerate if its determinant is zero.

Definition 2: A matrix is called non-singular if its determinant is not equal to zero.

Matrix "A" is called inverse matrix, if the condition A*A-1 = A-1 *A = E (identity matrix) is satisfied.

A square matrix is invertible only if it is nonsingular.

Scheme for calculating the inverse matrix:

1) Calculate the determinant of the matrix "A" if ∆ A = 0, then the inverse matrix does not exist.

2) Find all algebraic complements of the matrix "A".

3) Compose a matrix of algebraic additions (Aij )

4) Transpose the matrix of algebraic complements (Aij )T

5) Multiply the transposed matrix by the reciprocal of the determinant of this matrix.

6) Run a check:

At first glance it may seem that it is difficult, but in fact everything is very simple. All solutions are based on simple arithmetic operations, the main thing when solving is not to get confused with the "-" and "+" signs, and not to lose them.

Now let's decide together practical task, calculating the inverse matrix.

Task: find the inverse matrix "A", shown in the picture below:

We solve everything exactly as indicated in the plan for calculating the inverse matrix.

1. The first thing to do is to find the determinant of the matrix "A":

Explanation:

We have simplified our determinant by using its main functions. First, we added to the 2nd and 3rd row the elements of the first row, multiplied by one number.

Secondly, we changed the 2nd and 3rd columns of the determinant, and according to its properties, we changed the sign in front of it.

Thirdly, we took out the common factor (-1) of the second row, thereby changing the sign again, and it became positive. We also simplified line 3 the same way as at the very beginning of the example.

We have a triangular determinant, in which the elements below the diagonal are equal to zero, and by property 7 it is equal to the product diagonal elements. As a result, we got ∆ A = 26, hence the inverse matrix exists.

A11 = 1*(3+1) = 4

A12 \u003d -1 * (9 + 2) \u003d -11

A13 = 1*1 = 1

A21 = -1*(-6) = 6

A22 = 1*(3-0) = 3

A23 = -1*(1+4) = -5

A31 = 1*2 = 2

A32 = -1*(-1) = -1

A33 = 1+(1+6) = 7

3. The next step is to compile a matrix from the resulting additions:

5. We multiply this matrix by the reciprocal of the determinant, that is, by 1/26:

6. Well, now we just need to check:

During the verification, we received an identity matrix, therefore, the decision was made absolutely correctly.

2 way to calculate the inverse matrix.

1. Elementary transformation of matrices

2. Inverse matrix through an elementary converter.

Elementary matrix transformation includes:

1. Multiplying a string by a non-zero number.

2. Adding to any line of another line, multiplied by a number.

3. Swapping the rows of the matrix.

4. Applying a chain of elementary transformations, we obtain another matrix.

A -1 = ?

1. (A|E) ~ (E|A -1 )

2. A -1*A=E

Consider it on practical example with real numbers.

Exercise: Find the inverse matrix.

Solution:

Let's check:

A little clarification on the solution:

We first swapped rows 1 and 2 of the matrix, then we multiplied the first row by (-1).

After that, the first row was multiplied by (-2) and added to the second row of the matrix. Then we multiplied the 2nd row by 1/4.

The final stage of the transformation was the multiplication of the second row by 2 and the addition from the first. As a result, we have an identity matrix on the left, therefore, the inverse matrix is the matrix on the right.

After checking, we were convinced of the correctness of the solution.

As you can see, calculating the inverse matrix is very simple.

In concluding this lecture, I would also like to devote some time to the properties of such a matrix.

Let there be a square matrix of the nth order

Matrix A -1 is called inverse matrix with respect to the matrix A, if A * A -1 = E, where E is the identity matrix of the nth order.

Identity matrix- such a square matrix, in which all elements along the main diagonal, passing from the upper left corner to the lower right corner, are ones, and the rest are zeros, for example:

inverse matrix may exist only for square matrices those. for those matrices that have the same number of rows and columns.

Inverse Matrix Existence Condition Theorem

For a matrix to have an inverse matrix, it is necessary and sufficient that it be nondegenerate.

The matrix A = (A1, A2,...A n) is called non-degenerate if the column vectors are linearly independent. The number of linearly independent column vectors of a matrix is called the rank of the matrix. Therefore, we can say that in order for an inverse matrix to exist, it is necessary and sufficient that the rank of the matrix is equal to its dimension, i.e. r = n.

Algorithm for finding the inverse matrix

Write the matrix A in the table for solving systems of equations by the Gauss method and on the right (in place of the right parts of the equations) assign matrix E to it.
Using Jordan transformations, bring matrix A to a matrix consisting of single columns; in this case, it is necessary to simultaneously transform the matrix E.
If necessary, rearrange the rows (equations) of the last table so that the identity matrix E is obtained under the matrix A of the original table.
Write the inverse matrix A -1, which is in the last table under the matrix E of the original table.

Example 1

For matrix A, find the inverse matrix A -1

Solution: We write down the matrix A and on the right we assign the identity matrix E. Using Jordan transformations, we reduce the matrix A to the identity matrix E. The calculations are shown in Table 31.1.

Let's check the correctness of the calculations by multiplying the original matrix A and the inverse matrix A -1.

As a result of matrix multiplication, the identity matrix is obtained. Therefore, the calculations are correct.

Answer:

Solution of matrix equations

Matrix equations can look like:

AX = B, XA = B, AXB = C,

where A, B, C are given matrices, X is the desired matrix.

Matrix equations are solved by multiplying the equation by inverse matrices.

For example, to find the matrix from an equation, you need to multiply this equation by on the left.

Therefore, to find a solution to the equation, you need to find the inverse matrix and multiply it by the matrix on the right side of the equation.

Other equations are solved similarly.

Example 2

Solve the equation AX = B if

Solution: Since the inverse of the matrix equals (see example 1)

Matrix method in economic analysis

Along with others, they also find application matrix methods . These methods are based on linear and vector-matrix algebra. Such methods are used for the purposes of analyzing complex and multidimensional economic phenomena. Most often, these methods are used when it is necessary to compare the functioning of organizations and their structural divisions.

In the process of applying matrix methods of analysis, several stages can be distinguished.

At the first stage the formation of a system of economic indicators is carried out and on its basis a matrix of initial data is compiled, which is a table in which system numbers are shown in its individual lines (i = 1,2,....,n), and along the vertical graphs - numbers of indicators (j = 1,2,....,m).

At the second stage for each vertical column, the largest of the available values of the indicators is revealed, which is taken as a unit.

After that, all the amounts reflected in this column are divided by the largest value and a matrix of standardized coefficients is formed.

At the third stage all components of the matrix are squared. If they have different significance, then each indicator of the matrix is assigned a certain weighting coefficient k. The value of the latter is determined by an expert.

On the last fourth stage found values of ratings Rj grouped in order of increasing or decreasing.

The above matrix methods should be used, for example, when comparative analysis various investment projects, as well as in assessing other economic performance indicators of organizations.

Typically, inverse operations are used to simplify complex algebraic expressions. For example, if the problem contains the operation of division by a fraction, you can replace it with the operation of multiplying by a reciprocal, which is the inverse operation. Moreover, matrices cannot be divided, so you need to multiply by the inverse matrix. Calculating the inverse of a 3x3 matrix is quite tedious, but you need to be able to do it manually. You can also find the reciprocal with a good graphing calculator.

Steps

Using the attached matrix

Transpose the original matrix. Transposition is the replacement of rows with columns relative to the main diagonal of the matrix, that is, you need to swap the elements (i, j) and (j, i). In this case, the elements of the main diagonal (starts in the upper left corner and ends in the lower right corner) do not change.

To swap rows for columns, write the elements of the first row in the first column, the elements of the second row in the second column, and the elements of the third row in the third column. The order of changing the position of the elements is shown in the figure, in which the corresponding elements are circled with colored circles.

Find the definition of each 2x2 matrix. Each element of any matrix, including the transposed one, is associated with a corresponding 2x2 matrix. To find a 2x2 matrix that corresponds to a certain element, cross out the row and column in which this element is located, that is, you need to cross out five elements of the original 3x3 matrix. Four elements that are elements of the corresponding 2x2 matrix will remain uncrossed out.

For example, to find the 2x2 matrix for the element that is located at the intersection of the second row and the first column, cross out the five elements that are in the second row and first column. The remaining four elements are elements of the corresponding 2x2 matrix.
Find the determinant of each 2x2 matrix. To do this, subtract the product of the elements of the secondary diagonal from the product of the elements of the main diagonal (see figure).
Detailed information about 2x2 matrices corresponding to certain elements of a 3x3 matrix can be found on the Internet.

Create a matrix of cofactors. Record the results obtained earlier in the form of a new matrix of cofactors. To do this, write the found determinant of each 2x2 matrix where the corresponding element of the 3x3 matrix was located. For example, if a 2x2 matrix is considered for the element (1,1), write down its determinant in position (1,1). Then change the signs of the corresponding elements according to a certain pattern, which is shown in the figure.

Sign change scheme: the sign of the first element of the first line does not change; the sign of the second element of the first line is reversed; the sign of the third element of the first line does not change, and so on line by line. Please note that the signs "+" and "-", which are shown in the diagram (see figure), do not indicate that the corresponding element will be positive or negative. In this case, the “+” sign indicates that the sign of the element does not change, and the “-” sign indicates that the sign of the element has changed.
Detailed information about cofactor matrices can be found on the Internet.
This is how you find the associated matrix of the original matrix. It is sometimes called the complex conjugate matrix. Such a matrix is denoted as adj(M).

Divide each element of the adjoint matrix by the determinant. The determinant of the matrix M was calculated at the very beginning to check that the inverse matrix exists. Now divide each element of the adjoint matrix by this determinant. Record the result of each division operation where the corresponding element is located. So you will find the matrix, the inverse of the original.

The determinant of the matrix shown in the figure is 1. Thus, the associated matrix here is the inverse matrix (because dividing any number by 1 does not change it).
In some sources, the division operation is replaced by the multiplication operation by 1/det(M). In this case, the end result does not change.

Write down the inverse matrix. Write the elements located on the right half of the large matrix as a separate matrix, which is an inverse matrix.

Enter the original matrix into the calculator's memory. To do this, click the Matrix button, if available. For a Texas Instruments calculator, you may need to press the 2 nd and Matrix buttons.

Select the Edit menu. Do this using the arrow buttons or the corresponding function button located at the top of the calculator's keyboard (the location of the button depends on the calculator model).

Enter the matrix designation. Most graphing calculators can work with 3-10 matrices, which can be denoted letters A-J. As a general rule, just select [A] to denote the original matrix. Then press the Enter button.

Enter the matrix size. This article talks about 3x3 matrices. But graphical calculators can work with large matrices. Enter the number of rows, press the Enter button, then enter the number of columns and press the Enter button again.

Enter each element of the matrix. A matrix will be displayed on the calculator screen. If a matrix has already been entered into the calculator before, it will appear on the screen. The cursor will highlight the first element of the matrix. Enter the value of the first element and press Enter. The cursor will automatically move to next element matrices.