How to get the original matrix from the inverse matrix. Finding the inverse matrix online

Let there be a square matrix of nth order

Matrix A -1 is called inverse matrix in relation to 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 the 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 in which the number of rows and columns coincide.

Theorem for the existence condition of an inverse matrix

In order for a matrix to have an inverse matrix, it is necessary and sufficient that it be non-singular.

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 matrix A into the table for solving systems of equations using the Gaussian method and assign matrix E to it on the right (in place of the right-hand sides of the equations).
Using Jordan transformations, reduce matrix A to a matrix consisting of unit 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 under the matrix A of the original table you get the identity matrix E.
Write down the inverse matrix A -1, which is located 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 matrix A and assign the identity matrix E to the right. Using Jordan transformations, we reduce matrix A to the identity matrix E. The calculations are given 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 was obtained. Therefore, the calculations were performed correctly.

Answer:

Solving matrix equations

Matrix equations can look like:

AX = B, HA = B, AXB = C,

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

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

For example, to find the matrix from the 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 matrix is equal to (see example 1)

Matrix method in economic analysis

Along with others, they are also used 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 necessary comparative assessment functioning of organizations and their structural divisions.

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

At the first stage the system is being formed economic indicators and on its basis, a source data matrix is compiled, which is a table in which system numbers are shown in its individual rows (i = 1,2,....,n), and in vertical columns - numbers of indicators (j = 1,2,....,m).

At the second stage For each vertical column, the largest of the available indicator values is identified, which is taken as one.

After this, all amounts reflected in this column are divided by highest 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 matrix indicator is assigned a certain weight coefficient k. The value of the latter is determined by expert opinion.

On the last one, fourth stage found rating values Rj are grouped in order of their increase or decrease.

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

In order to find the inverse matrix online, you will need to indicate the size of the matrix itself. To do this, click on the “+” or “-” icons until you are satisfied with the number of columns and rows. Next, enter the required elements in the fields. Below is the “Calculate” button - by clicking it, you will receive an answer on the screen with a detailed solution.

In linear algebra, quite often one has to deal with the process of calculating the inverse matrix. It exists only for unexpressed matrices and for square matrices provided that the determinant is nonzero. In principle, calculating it is not particularly difficult, especially if you are dealing with a small matrix. But if you need more complex calculations or a thorough double-check of your decision, it is better to use this online calculator. With its help, you can quickly and accurately solve an inverse matrix.

Using this online calculator you can make your calculations much easier. In addition, it helps to consolidate the material obtained in theory - it is a kind of simulator for the brain. It should not be considered as a replacement for manual calculations; it can give you much more, making it easier to understand the algorithm itself. Besides, it never hurts to double-check yourself.

The matrix $A^(-1)$ is called the inverse of the square matrix $A$ if the condition $A^(-1)\cdot A=A\cdot A^(-1)=E$ is satisfied, where $E $ is the identity matrix, the order of which is equal to the order of the matrix $A$.

A non-singular matrix is a matrix whose determinant is not equal to zero. Accordingly, a singular matrix is one whose determinant is equal to zero.

The inverse matrix $A^(-1)$ exists if and only if the matrix $A$ is non-singular. If the inverse matrix $A^(-1)$ exists, then it is unique.

There are several ways to find the inverse of a matrix, and we will look at two of them. This page will discuss the adjoint matrix method, which is considered standard in most higher mathematics courses. The second method of finding the inverse matrix (the method of elementary transformations), which involves using the Gauss method or the Gauss-Jordan method, is discussed in the second part.

Adjoint matrix method

Let the matrix $A_(n\times n)$ be given. In order to find the inverse matrix $A^(-1)$, three steps are required:

Find the determinant of the matrix $A$ and make sure that $\Delta A\neq 0$, i.e. that matrix A is non-singular.
Compose algebraic complements $A_(ij)$ of each element of the matrix $A$ and write the matrix $A_(n\times n)^(*)=\left(A_(ij) \right)$ from the found algebraic complements.
Write the inverse matrix taking into account the formula $A^(-1)=\frac(1)(\Delta A)\cdot (A^(*))^T$.

The matrix $(A^(*))^T$ is often called adjoint (reciprocal, allied) to the matrix $A$.

If the solution is done manually, then the first method is good only for matrices of relatively small orders: second (), third (), fourth (). To find the inverse of a matrix higher order, other methods are used. For example, the Gaussian method, which is discussed in the second part.

Example No. 1

Find the inverse of matrix $A=\left(\begin(array) (cccc) 5 & -4 &1 & 0 \\ 12 &-11 &4 & 0 \\ -5 & 58 &4 & 0 \\ 3 & - 1 & -9 & 0 \end(array) \right)$.

Since all elements of the fourth column are equal to zero, then $\Delta A=0$ (i.e. the matrix $A$ is singular). Since $\Delta A=0$, there is no inverse matrix to matrix $A$.

Answer: matrix $A^(-1)$ does not exist.

Example No. 2

Find the inverse of matrix $A=\left(\begin(array) (cc) -5 & 7 \\ 9 & 8 \end(array)\right)$. Perform check.

We use the adjoint matrix method. First, let's find the determinant of the given matrix $A$:

$$ \Delta A=\left| \begin(array) (cc) -5 & 7\\ 9 & 8 \end(array)\right|=-5\cdot 8-7\cdot 9=-103. $$

Since $\Delta A \neq 0$, then the inverse matrix exists, therefore we will continue the solution. Finding algebraic complements

\begin(aligned) & A_(11)=(-1)^2\cdot 8=8; \; A_(12)=(-1)^3\cdot 9=-9;\\ & A_(21)=(-1)^3\cdot 7=-7; \; A_(22)=(-1)^4\cdot (-5)=-5.\\ \end(aligned)

We compose a matrix of algebraic additions: $A^(*)=\left(\begin(array) (cc) 8 & -9\\ -7 & -5 \end(array)\right)$.

We transpose the resulting matrix: $(A^(*))^T=\left(\begin(array) (cc) 8 & -7\\ -9 & -5 \end(array)\right)$ (the resulting matrix is often is called the adjoint or allied matrix to the matrix $A$). Using the formula $A^(-1)=\frac(1)(\Delta A)\cdot (A^(*))^T$, we have:

$$ A^(-1)=\frac(1)(-103)\cdot \left(\begin(array) (cc) 8 & -7\\ -9 & -5 \end(array)\right) =\left(\begin(array) (cc) -8/103 & 7/103\\ 9/103 & 5/103 \end(array)\right) $$

So, the inverse matrix is found: $A^(-1)=\left(\begin(array) (cc) -8/103 & 7/103\\ 9/103 & 5/103 \end(array)\right) $. To check the truth of the result, it is enough to check the truth of one of the equalities: $A^(-1)\cdot A=E$ or $A\cdot A^(-1)=E$. Let's check the equality $A^(-1)\cdot A=E$. In order to work less with fractions, we will substitute the matrix $A^(-1)$ not in the form $\left(\begin(array) (cc) -8/103 & 7/103\\ 9/103 & 5/103 \ end(array)\right)$, and in the form $-\frac(1)(103)\cdot \left(\begin(array) (cc) 8 & -7\\ -9 & -5 \end(array )\right)$:

$$ A^(-1)\cdot(A) =-\frac(1)(103)\cdot \left(\begin(array) (cc) 8 & -7\\ -9 & -5 \end( array)\right)\cdot\left(\begin(array) (cc) -5 & 7 \\ 9 & 8 \end(array)\right) =-\frac(1)(103)\cdot\left( \begin(array) (cc) -103 & 0 \\ 0 & -103 \end(array)\right) =\left(\begin(array) (cc) 1 & 0 \\ 0 & 1 \end(array )\right) =E $$

Answer: $A^(-1)=\left(\begin(array) (cc) -8/103 & 7/103\\ 9/103 & 5/103 \end(array)\right)$.

Example No. 3

Find the inverse matrix for the matrix $A=\left(\begin(array) (ccc) 1 & 7 & 3 \\ -4 & 9 & 4 \\ 0 & 3 & 2\end(array) \right)$. Perform check.

Let's start by calculating the determinant of the matrix $A$. So, the determinant of the matrix $A$ is:

$$ \Delta A=\left| \begin(array) (ccc) 1 & 7 & 3 \\ -4 & 9 & 4 \\ 0 & 3 & 2\end(array) \right| = 18-36+56-12=26. $$

Since $\Delta A\neq 0$, then the inverse matrix exists, therefore we will continue the solution. We find the algebraic complements of each element of a given matrix:

We compose a matrix of algebraic additions and transpose it:

$$ A^*=\left(\begin(array) (ccc) 6 & 8 & -12 \\ -5 & 2 & -3 \\ 1 & -16 & 37\end(array) \right); \; (A^*)^T=\left(\begin(array) (ccc) 6 & -5 & 1 \\ 8 & 2 & -16 \\ -12 & -3 & 37\end(array) \right) . $$

Using the formula $A^(-1)=\frac(1)(\Delta A)\cdot (A^(*))^T$, we get:

$$ A^(-1)=\frac(1)(26)\cdot \left(\begin(array) (ccc) 6 & -5 & 1 \\ 8 & 2 & -16 \\ -12 & - 3 & 37\end(array) \right)= \left(\begin(array) (ccc) 3/13 & -5/26 & 1/26 \\ 4/13 & 1/13 & -8/13 \ \ -6/13 & -3/26 & 37/26 \end(array) \right) $$

So $A^(-1)=\left(\begin(array) (ccc) 3/13 & -5/26 & 1/26 \\ 4/13 & 1/13 & -8/13 \\ - 6/13 & -3/26 & 37/26 \end(array) \right)$. To check the truth of the result, it is enough to check the truth of one of the equalities: $A^(-1)\cdot A=E$ or $A\cdot A^(-1)=E$. Let's check the equality $A\cdot A^(-1)=E$. In order to work less with fractions, we will substitute the matrix $A^(-1)$ not in the form $\left(\begin(array) (ccc) 3/13 & -5/26 & 1/26 \\ 4/13 & 1/13 & -8/13 \\ -6/13 & -3/26 & 37/26 \end(array) \right)$, and in the form $\frac(1)(26)\cdot \left( \begin(array) (ccc) 6 & -5 & 1 \\ 8 & 2 & -16 \\ -12 & -3 & 37\end(array) \right)$:

$$ A\cdot(A^(-1)) =\left(\begin(array)(ccc) 1 & 7 & 3 \\ -4 & 9 & 4\\ 0 & 3 & 2\end(array) \right)\cdot \frac(1)(26)\cdot \left(\begin(array) (ccc) 6 & -5 & 1 \\ 8 & 2 & -16 \\ -12 & -3 & 37\ end(array) \right) =\frac(1)(26)\cdot\left(\begin(array) (ccc) 26 & 0 & 0 \\ 0 & 26 & 0 \\ 0 & 0 & 26\end (array) \right) =\left(\begin(array) (ccc) 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end(array) \right) =E $$

The check was successful, the inverse matrix $A^(-1)$ was found correctly.

Answer: $A^(-1)=\left(\begin(array) (ccc) 3/13 & -5/26 & 1/26 \\ 4/13 & 1/13 & -8/13 \\ -6 /13 & -3/26 & 37/26 \end(array) \right)$.

Example No. 4

Find the matrix inverse of matrix $A=\left(\begin(array) (cccc) 6 & -5 & 8 & 4\\ 9 & 7 & 5 & 2 \\ 7 & 5 & 3 & 7\\ -4 & 8 & -8 & -3 \end(array) \right)$.

For a fourth-order matrix, finding the inverse matrix using algebraic additions is somewhat difficult. However, such examples in tests meet.

To find the inverse of a matrix, you first need to calculate the determinant of the matrix $A$. The best way to do this in this situation is by decomposing the determinant along a row (column). We select any row or column and find the algebraic complements of each element of the selected row or column.

For example, for the first line we get:

$$ A_(11)=\left|\begin(array)(ccc) 7 & 5 & 2\\ 5 & 3 & 7\\ 8 & -8 & -3 \end(array)\right|=556; \; A_(12)=-\left|\begin(array)(ccc) 9 & 5 & 2\\ 7 & 3 & 7 \\ -4 & -8 & -3 \end(array)\right|=-300 ; $$ $$ A_(13)=\left|\begin(array)(ccc) 9 & 7 & 2\\ 7 & 5 & 7\\ -4 & 8 & -3 \end(array)\right|= -536;\; A_(14)=-\left|\begin(array)(ccc) 9 & 7 & 5\\ 7 & 5 & 3\\ -4 & 8 & -8 \end(array)\right|=-112. $$

The determinant of the matrix $A$ is calculated using the following formula:

$$ \Delta(A)=a_(11)\cdot A_(11)+a_(12)\cdot A_(12)+a_(13)\cdot A_(13)+a_(14)\cdot A_(14 )=6\cdot 556+(-5)\cdot(-300)+8\cdot(-536)+4\cdot(-112)=100. $$

$$ \begin(aligned) & A_(21)=-77;\;A_(22)=50;\;A_(23)=87;\;A_(24)=4;\\ & A_(31) =-93;\;A_(32)=50;\;A_(33)=83;\;A_(34)=36;\\ & A_(41)=473;\;A_(42)=-250 ;\;A_(43)=-463;\;A_(44)=-96. \end(aligned) $$

Matrix of algebraic complements: $A^*=\left(\begin(array)(cccc) 556 & -300 & -536 & -112\\ -77 & 50 & 87 & 4 \\ -93 & 50 & 83 & 36\\ 473 & -250 & -463 & -96\end(array)\right)$.

Adjoint matrix: $(A^*)^T=\left(\begin(array) (cccc) 556 & -77 & -93 & 473\\ -300 & 50 & 50 & -250 \\ -536 & 87 & 83 & -463\\ -112 & 4 & 36 & -96\end(array)\right)$.

Inverse matrix:

$$ A^(-1)=\frac(1)(100)\cdot \left(\begin(array) (cccc) 556 & -77 & -93 & 473\\ -300 & 50 & 50 & -250 \\ -536 & 87 & 83 & -463\\ -112 & 4 & 36 & -96 \end(array) \right)= \left(\begin(array) (cccc) 139/25 & -77/100 & -93/100 & 473/100 \\ -3 & 1/2 & 1/2 & -5/2 \\ -134/25 & 87/100 & 83/100 & -463/100 \\ -28/ 25 & 1/25 & 9/25 & -24/25 \end(array) \right) $$

The check, if desired, can be done in the same way as in the previous examples.

Answer: $A^(-1)=\left(\begin(array) (cccc) 139/25 & -77/100 & -93/100 & 473/100 \\ -3 & 1/2 & 1/2 & -5/2 \\ -134/25 & 87/100 & 83/100 & -463/100 \\ -28/25 & 1/25 & 9/25 & -24/25 \end(array) \right) $.

In the second part, we will consider another way to find the inverse matrix, which involves the use of transformations of the Gaussian method or the Gauss-Jordan method.

Inverse matrix is a matrix A−1, when multiplied by which the given initial matrix A results in the identity matrix E:

AA −1 = A −1 A =E.

Inverse matrix method.

Inverse matrix method- this is one of the most common methods for solving matrices and is used to solve systems of linear algebraic equations (SLAEs) in cases where the number of unknowns corresponds to the number of equations.

Let there be a system n linear equations With n unknown:

Such a system can be written as a matrix equation A* X = B,

Where
- system matrix,

- column of unknowns,

- column of free odds.

From the derived matrix equation, we express X by multiplying both sides of the matrix equation on the left by A-1, resulting in:

A -1 * A * X = A -1 * B

Knowing that A -1 * A = E, Then E * X = A -1 * B or X = A -1 * B.

The next step is to determine the inverse matrix A-1 and multiplied by the column of free terms B.

Inverse matrix to matrix A exists only when det A≠ 0 . In view of this, when solving SLAEs using the inverse matrix method, the first step is to find det A. If det A≠ 0 , then the system has only one solution, which can be obtained using the inverse matrix method, but if det A = 0, then such a system inverse matrix method can't be solved.

Solving the inverse matrix.

Sequence of actions for inverse matrix solutions:

We obtain the determinant of the matrix A. If the determinant is greater than zero, we solve the inverse of the matrix further; if it is equal to zero, then we cannot find the inverse matrix here.
Finding the transposed matrix AT.
We look for algebraic complements, after which we replace all elements of the matrix with their algebraic complements.
We assemble the inverse matrix from algebraic additions: we divide all the elements of the resulting matrix by the determinant of the initially given matrix. The final matrix will be the required inverse matrix relative to the original one.

Below algorithm inverse matrix solutions essentially the same as the one above, the difference is only in a few steps: first of all we define the algebraic complements, and after that we calculate the allied matrix C.

Determine whether a given matrix is square. If the answer is negative, it becomes clear that there cannot be an inverse matrix for it.
Determine whether a given matrix is square. If the answer is negative, it becomes clear that there cannot be an inverse matrix for it.
We calculate algebraic complements.
We compose a union (mutual, adjoint) matrix C.
We compose the inverse matrix from algebraic additions: all elements of the adjoint matrix C divide by the determinant of the initial matrix. The final matrix will be the required inverse matrix relative to the given one.
We check the work done: multiply the initial and resulting matrices, the result should be an identity matrix.

This is best done using an attached matrix.

Theorem: If we assign an identity matrix of the same order to a square matrix on the right side and, using elementary transformations over the rows, transform the initial matrix on the left into the identity matrix, then the one obtained on the right side will be the inverse of the initial one.

An example of finding an inverse matrix.

Exercise. For matrix find the inverse using the adjoint matrix method.

Solution. Add to the given matrix A on the right is a 2nd order identity matrix:

From the 1st line we subtract the 2nd:

From the second line we subtract the first 2:

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