Calculation of the inverse matrix. Inverse matrix. Finding the inverse matrix using the method of elementary transformations

The inverse matrix for a given matrix is such a matrix, multiplying the original one by which gives the identity matrix: A mandatory and sufficient condition for the presence of an inverse matrix is that the determinant of the original matrix is not equal to zero (which in turn implies that the matrix must be square). If the determinant of a matrix is equal to zero, then it is called singular and such a matrix does not have an inverse. In higher mathematics, inverse matrices are important and are used to solve a number of problems. For example, on finding the inverse matrix a matrix method for solving systems of equations was constructed. Our service site allows calculate inverse matrix online two methods: the Gauss-Jordan method and using the matrix of algebraic additions. The first one involves a large number of elementary transformations inside the matrix, the second one involves the calculation of the determinant and algebraic additions to all elements. To calculate the determinant of a matrix online, you can use our other service - Calculation of the determinant of a matrix online

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For any non-singular matrix A there is a unique matrix A -1 such that

A*A -1 =A -1 *A = E,

where E is the identity matrix of the same orders as A. The matrix A -1 is called the inverse of matrix A.

In case someone forgot, in the identity matrix, except for the diagonal filled with ones, all other positions are filled with zeros, an example of an identity matrix:

Finding the inverse matrix using the adjoint matrix method

The inverse matrix is defined by the formula:

where A ij - elements a ij.

Those. To calculate the inverse matrix, you need to calculate the determinant of this matrix. Then find the algebraic complements for all its elements and compose a new matrix from them. Next you need to transport this matrix. And divide each element of the new matrix by the determinant of the original matrix.

Let's look at a few examples.

Find A -1 for a matrix

Solution. Let's find A -1 using the adjoint matrix method. We have det A = 2. Let us find the algebraic complements of the elements of matrix A. In this case, the algebraic complements of the matrix elements will be the corresponding elements of the matrix itself, taken with a sign in accordance with the formula

We have A 11 = 3, A 12 = -4, A 21 = -1, A 22 = 2. We form the adjoint matrix

We transport the matrix A*:

We find the inverse matrix using the formula:

We get:

Using the adjoint matrix method, find A -1 if

Solution. First of all, we calculate the definition of this matrix to verify the existence of the inverse matrix. We have

Here we added to the elements of the second row the elements of the third row, previously multiplied by (-1), and then expanded the determinant for the second row. Since the definition of this matrix is different from zero, then its inverse matrix exists. To construct the adjoint matrix, we find the algebraic complements of the elements of this matrix. We have

According to the formula

transport matrix A*:

Then according to the formula

Finding the inverse matrix using the method of elementary transformations

In addition to the method of finding the inverse matrix, which follows from the formula (the adjoint matrix method), there is a method for finding the inverse matrix, called the method of elementary transformations.

Elementary matrix transformations

The following transformations are called elementary matrix transformations:

1) rearrangement of rows (columns);

2) multiplying a row (column) by a number other than zero;

3) adding to the elements of a row (column) the corresponding elements of another row (column), previously multiplied by a certain number.

To find the matrix A -1, we construct a rectangular matrix B = (A|E) of orders (n; 2n), assigning to matrix A on the right the identity matrix E through a dividing line:

Let's look at an example.

Using the method of elementary transformations, find A -1 if

Solution. We form matrix B:

Let us denote the rows of matrix B by α 1, α 2, α 3. Let us perform the following transformations on the rows of matrix B.

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 higher order matrix, 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 do occur in test papers.

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 to expand 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.

Let's continue the conversation about actions with matrices. Namely, during the study of this lecture you will learn how to find the inverse matrix. Learn. Even if math is difficult.

What is an inverse matrix? Here we can draw an analogy with inverse numbers: consider, for example, the optimistic number 5 and its inverse number. The product of these numbers is equal to one: . Everything is similar with matrices! The product of a matrix and its inverse matrix is equal to – identity matrix, which is the matrix analogue of the numerical unit. However, first things first – let’s first solve an important practical issue, namely, learn how to find this very inverse matrix.

What do you need to know and be able to do to find the inverse matrix? You must be able to decide qualifiers. You must understand what it is matrix and be able to perform some actions with them.

There are two main methods for finding the inverse matrix:
by using algebraic additions And using elementary transformations.

Today we will study the first, simpler method.

Let's start with the most terrible and incomprehensible. Let's consider square matrix The inverse matrix can be found using the following formula:

Where is the determinant of the matrix, is the transposed matrix of algebraic complements of the corresponding elements of the matrix.

The concept of an inverse matrix exists only for square matrices, matrices “two by two”, “three by three”, etc.

Designations: As you may have already noticed, the inverse matrix is denoted by a superscript

Let's start with the simplest case - a two-by-two matrix. Most often, of course, “three by three” is required, but, nevertheless, I strongly recommend studying a simpler task in order to understand the general principle of the solution.

Example:

Find the inverse of a matrix

Let's decide. It is convenient to break down the sequence of actions point by point.

1) First we find the determinant of the matrix.

If your understanding of this action is not good, read the material How to calculate the determinant?

Important! If the determinant of the matrix is equal to ZERO– inverse matrix DOESN'T EXIST.

In the example under consideration, as it turned out, , which means everything is in order.

2) Find the matrix of minors.

To solve our problem, it is not necessary to know what a minor is, however, it is advisable to read the article How to calculate the determinant.

The matrix of minors has the same dimensions as the matrix, that is, in this case.
The only thing left to do is find four numbers and put them instead of asterisks.

Let's return to our matrix
Let's look at the top left element first:

How to find it minor?
And this is done like this: MENTALLY cross out the row and column in which this element is located:

The remaining number is minor of this element, which we write in our matrix of minors:

Consider the following matrix element:

Mentally cross out the row and column in which this element appears:

What remains is the minor of this element, which we write in our matrix:

Similarly, we consider the elements of the second row and find their minors:

Ready.

It's simple. In the matrix of minors you need CHANGE SIGNS two numbers:

These are the numbers that I circled!

– matrix of algebraic additions of the corresponding elements of the matrix.

And just...

4) Find the transposed matrix of algebraic additions.

– transposed matrix of algebraic complements of the corresponding elements of the matrix.

5) Answer.

Let's remember our formula
Everything has been found!

So the inverse matrix is:

It is better to leave the answer as is. NO NEED divide each element of the matrix by 2, since the result is fractional numbers. This nuance is discussed in more detail in the same article. Actions with matrices.

How to check the solution?

You need to perform matrix multiplication or

Examination:

Received already mentioned identity matrix is a matrix with ones by main diagonal and zeros in other places.

Thus, the inverse matrix is found correctly.

If you carry out the action, the result will also be an identity matrix. This is one of the few cases where matrix multiplication is commutative, more details can be found in the article Properties of operations on matrices. Matrix Expressions. Also note that during the check, the constant (fraction) is brought forward and processed at the very end - after the matrix multiplication. This is a standard technique.

Let's move on to a more common case in practice - the three-by-three matrix:

Example:

Find the inverse of a matrix

The algorithm is exactly the same as for the “two by two” case.

We find the inverse matrix using the formula: , where is the transposed matrix of algebraic complements of the corresponding elements of the matrix.

1) Find the determinant of the matrix.

Here the determinant is revealed on the first line.

Also, don’t forget that, which means everything is fine - inverse matrix exists.

2) Find the matrix of minors.

The matrix of minors has a dimension of “three by three” , and we need to find nine numbers.

I'll take a closer look at a couple of minors:

Consider the following matrix element:

MENTALLY cross out the row and column in which this element is located:

We write the remaining four numbers in the “two by two” determinant.

This two-by-two determinant and is the minor of this element. It needs to be calculated:

That’s it, the minor has been found, we write it in our matrix of minors:

As you probably guessed, you need to calculate nine two-by-two determinants. The process, of course, is tedious, but the case is not the most severe, it can be worse.

Well, to consolidate – finding another minor in the pictures:

Try to calculate the remaining minors yourself.

Final result:
– matrix of minors of the corresponding elements of the matrix.

The fact that all the minors turned out to be negative is purely an accident.

3) Find the matrix of algebraic additions.

In the matrix of minors it is necessary CHANGE SIGNS strictly for the following elements:

In this case:

We do not consider finding the inverse matrix for a “four by four” matrix, since such a task can only be given by a sadistic teacher (for the student to calculate one “four by four” determinant and 16 “three by three” determinants). In my practice, there was only one such case, and the customer of the test paid quite dearly for my torment =).

In a number of textbooks and manuals you can find a slightly different approach to finding the inverse matrix, but I recommend using the solution algorithm outlined above. Why? Because the likelihood of getting confused in calculations and signs is much less.

Methods for finding the inverse matrix. Consider a square matrix

Let us denote Δ = det A.

The square matrix A is called non-degenerate, or not special, if its determinant is nonzero, and degenerate, or special, IfΔ = 0.

A square matrix B is for a square matrix A of the same order if their product is A B = B A = E, where E is the identity matrix of the same order as the matrices A and B.

Theorem . In order for matrix A to have an inverse matrix, it is necessary and sufficient that its determinant be different from zero.

The inverse matrix of matrix A, denoted by A- 1, so B = A - 1 and is calculated by the formula

, (1)

where A i j are algebraic complements of elements a i j of matrix A..

Calculating A -1 using formula (1) for high-order matrices is very labor-intensive, so in practice it is convenient to find A -1 using the method of elementary transformations (ET). Any non-singular matrix A can be reduced to the identity matrix E by means of EDs of only columns (or only rows). If the EDs perfected over matrix A are applied in the same order to the identity matrix E, then the result is an inverse matrix. It is convenient to perform EP on matrices A and E simultaneously, writing both matrices side by side through a line. Let us note once again that when searching for the canonical form of a matrix, in order to find it, you can use transformations of rows and columns. If you need to find the inverse of a matrix, you should use only rows or only columns during the transformation process.

Example 1. For matrix find A -1 .

Solution.First we find the determinant of matrix A
This means that the inverse matrix exists and we can find it using the formula: , where A i j (i,j=1,2,3) are algebraic additions of elements a i j of the original matrix.

Where .

Example 2. Using the method of elementary transformations, find A -1 for the matrix: A = .

Solution.We assign to the original matrix on the right an identity matrix of the same order: . Using elementary transformations of the columns, we will reduce the left “half” to the identity one, simultaneously performing exactly the same transformations on the right matrix.
To do this, swap the first and second columns: ~ . To the third column we add the first, and to the second - the first, multiplied by -2: . From the first column we subtract the second doubled, and from the third - the second multiplied by 6; . Let's add the third column to the first and second: . Multiply the last column by -1: . The square matrix obtained to the right of the vertical bar is the inverse matrix of the given matrix A. So,
.