Inverse matrix examples with 3x3 solution. Method of elementary transformations (Gauss and Gauss-Jordan methods for finding inverse matrices)

This topic is one of the most hated among students. Worse, probably, are the qualifiers.

The trick is that the very concept of an inverse element (and I’m not just talking about matrices) refers us to the operation of multiplication. Even in the school curriculum, multiplication is considered a complex operation, and multiplication of matrices is generally a separate topic, to which I have an entire paragraph and video lesson dedicated.

Today we will not go into the details of matrix calculations. Let’s just remember: how matrices are designated, how they are multiplied, and what follows from this.

Review: Matrix Multiplication

First of all, let's agree on notation. A matrix $A$ of size $\left[ m\times n \right]$ is simply a table of numbers with exactly $m$ rows and $n$ columns:

\=\underbrace(\left[ \begin(matrix) ((a)_(11)) & ((a)_(12)) & ... & ((a)_(1n)) \\ (( a)_(21)) & ((a)_(22)) & ... & ((a)_(2n)) \\ ... & ... & ... & ... \\ ((a)_(m1)) & ((a)_(m2)) & ... & ((a)_(mn)) \\\end(matrix) \right])_(n)\]

To avoid accidentally mixing up rows and columns (believe me, in an exam you can confuse a one with a two, let alone some rows), just look at the picture:

What's happening? If you place the standard coordinate system $OXY$ in the upper left corner and direct the axes so that they cover the entire matrix, then each cell of this matrix can be uniquely associated with coordinates $\left(x;y \right)$ - this will be the row number and column number.

Why is the coordinate system placed in the upper left corner? Yes, because it is from there that we begin to read any texts. It's very easy to remember.

Why is the $x$ axis directed downwards and not to the right? Again, it's simple: take a standard coordinate system (the $x$ axis goes to the right, the $y$ axis goes up) and rotate it so that it covers the matrix. This is a 90 degree clockwise rotation - we see the result in the picture.

In general, we have figured out how to determine the indices of matrix elements. Now let's look at multiplication.

Definition. Matrices $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$, when the number of columns in the first coincides with the number of rows in the second, are called consistent.

Exactly in that order. One can be confused and say that the matrices $A$ and $B$ form an ordered pair $\left(A;B \right)$: if they are consistent in this order, then it is not at all necessary that $B$ and $A$ those. the pair $\left(B;A \right)$ is also consistent.

Only matched matrices can be multiplied.

Definition. The product of matched matrices $A=\left[ m\times n \right]$ and $B=\left[ n\times k \right]$ is the new matrix $C=\left[ m\times k \right]$ , the elements of which $((c)_(ij))$ are calculated according to the formula:

\[((c)_(ij))=\sum\limits_(k=1)^(n)(((a)_(ik)))\cdot ((b)_(kj))\]

In other words: to get the element $((c)_(ij))$ of the matrix $C=A\cdot B$, you need to take the $i$-row of the first matrix, the $j$-th column of the second matrix, and then multiply in pairs elements from this row and column. Add up the results.

Yes, that’s such a harsh definition. Several facts immediately follow from it:

1. Matrix multiplication, generally speaking, is non-commutative: $A\cdot B\ne B\cdot A$;
2. However, multiplication is associative: $\left(A\cdot B \right)\cdot C=A\cdot \left(B\cdot C \right)$;
3. And even distributively: $\left(A+B \right)\cdot C=A\cdot C+B\cdot C$;
4. And once again distributively: $A\cdot \left(B+C \right)=A\cdot B+A\cdot C$.

The distributivity of multiplication had to be described separately for the left and right sum factor precisely because of the non-commutativity of the multiplication operation.

If it turns out that $A\cdot B=B\cdot A$, such matrices are called commutative.

Among all the matrices that are multiplied by something there, there are special ones - those that, when multiplied by any matrix $A$, again give $A$:

Definition. A matrix $E$ is called identity if $A\cdot E=A$ or $E\cdot A=A$. In the case of a square matrix $A$ we can write:

The identity matrix is ​​a frequent guest when solving matrix equations. And in general, a frequent guest in the world of matrices. :)

And because of this $E$, someone came up with all the nonsense that will be written next.

What is an inverse matrix

Since matrix multiplication is a very labor-intensive operation (you have to multiply a bunch of rows and columns), the concept of an inverse matrix also turns out to be not the most trivial. And requiring some explanation.

Key Definition

Well, it's time to know the truth.

Definition. A matrix $B$ is called the inverse of a matrix $A$ if

The inverse matrix is ​​denoted by $((A)^(-1))$ (not to be confused with the degree!), so the definition can be rewritten as follows:

It would seem that everything is extremely simple and clear. But when analyzing this definition, several questions immediately arise:

1. Does an inverse matrix always exist? And if not always, then how to determine: when it exists and when it does not?
2. And who said that there is exactly one such matrix? What if for some initial matrix $A$ there is a whole crowd of inverses?
3. What do all these “reverses” look like? And how, exactly, should we count them?

As for calculation algorithms, we will talk about this a little later. But we will answer the remaining questions right now. Let us formulate them in the form of separate statements-lemmas.

Basic properties

Let's start with how the matrix $A$ should, in principle, look in order for $((A)^(-1))$ to exist for it. Now we will make sure that both of these matrices must be square, and of the same size: $\left[ n\times n \right]$.

Lemma 1. Given a matrix $A$ and its inverse $((A)^(-1))$. Then both of these matrices are square, and of the same order $n$.

Proof. It's simple. Let the matrix $A=\left[ m\times n \right]$, $((A)^(-1))=\left[ a\times b \right]$. Since the product $A\cdot ((A)^(-1))=E$ exists by definition, the matrices $A$ and $((A)^(-1))$ are consistent in the order shown:

\[\begin(align) & \left[ m\times n \right]\cdot \left[ a\times b \right]=\left[ m\times b \right] \\ & n=a \end( align)\]

This is a direct consequence of the matrix multiplication algorithm: the coefficients $n$ and $a$ are “transit” and must be equal.

At the same time, the inverse multiplication is also defined: $((A)^(-1))\cdot A=E$, therefore the matrices $((A)^(-1))$ and $A$ are also consistent in the specified order:

\[\begin(align) & \left[ a\times b \right]\cdot \left[ m\times n \right]=\left[ a\times n \right] \\ & b=m \end( align)\]

Thus, without loss of generality, we can assume that $A=\left[ m\times n \right]$, $((A)^(-1))=\left[ n\times m \right]$. However, according to the definition of $A\cdot ((A)^(-1))=((A)^(-1))\cdot A$, therefore the sizes of the matrices strictly coincide:

\[\begin(align) & \left[ m\times n \right]=\left[ n\times m \right] \\ & m=n \end(align)\]

So it turns out that all three matrices - $A$, $((A)^(-1))$ and $E$ - are square matrices of size $\left[ n\times n \right]$. The lemma is proven.

Well, that's already good. We see that only square matrices are invertible. Now let's make sure that the inverse matrix is ​​always the same.

Lemma 2. Given a matrix $A$ and its inverse $((A)^(-1))$. Then this inverse matrix is ​​the only one.

Proof. Let's go by contradiction: let the matrix $A$ have at least two inverses - $B$ and $C$. Then, according to definition, the following equalities are true:

\[\begin(align) & A\cdot B=B\cdot A=E; \\ & A\cdot C=C\cdot A=E. \\ \end(align)\]

From Lemma 1 we conclude that all four matrices - $A$, $B$, $C$ and $E$ - are squares of the same order: $\left[ n\times n \right]$. Therefore, the product is defined:

Since matrix multiplication is associative (but not commutative!), we can write:

\[\begin(align) & B\cdot A\cdot C=\left(B\cdot A \right)\cdot C=E\cdot C=C; \\ & B\cdot A\cdot C=B\cdot \left(A\cdot C \right)=B\cdot E=B; \\ & B\cdot A\cdot C=C=B\Rightarrow B=C. \\ \end(align)\]

We got the only possible option: two copies of the inverse matrix are equal. The lemma is proven.

The above arguments repeat almost verbatim the proof of the uniqueness of the inverse element for all real numbers $b\ne 0$. The only significant addition is taking into account the dimension of matrices.

However, we still do not know anything about whether every square matrix is ​​invertible. Here the determinant comes to our aid - this is a key characteristic for all square matrices.

Lemma 3. Given a matrix $A$. If its inverse matrix $((A)^(-1))$ exists, then the determinant of the original matrix is ​​nonzero:

\[\left| A\right|\ne 0\]

Proof. We already know that $A$ and $((A)^(-1))$ are square matrices of size $\left[ n\times n \right]$. Therefore, for each of them we can calculate the determinant: $\left| A\right|$ and $\left| ((A)^(-1)) \right|$. However, the determinant of a product is equal to the product of the determinants:

\[\left| A\cdot B \right|=\left| A \right|\cdot \left| B \right|\Rightarrow \left| A\cdot ((A)^(-1)) \right|=\left| A \right|\cdot \left| ((A)^(-1)) \right|\]

But according to the definition, $A\cdot ((A)^(-1))=E$, and the determinant of $E$ is always equal to 1, so

\[\begin(align) & A\cdot ((A)^(-1))=E; \\ & \left| A\cdot ((A)^(-1)) \right|=\left| E\right|; \\ & \left| A \right|\cdot \left| ((A)^(-1)) \right|=1. \\ \end(align)\]

The product of two numbers is equal to one only if each of these numbers is non-zero:

\[\left| A \right|\ne 0;\quad \left| ((A)^(-1)) \right|\ne 0.\]

So it turns out that $\left| A \right|\ne 0$. The lemma is proven.

In fact, this requirement is quite logical. Now we will analyze the algorithm for finding the inverse matrix - and it will become completely clear why, with a zero determinant, no inverse matrix in principle can exist.

But first, let’s formulate an “auxiliary” definition:

Definition. A singular matrix is ​​a square matrix of size $\left[ n\times n \right]$ whose determinant is zero.

Thus, we can claim that every invertible matrix is ​​non-singular.

How to find the inverse of a matrix

Now we will consider a universal algorithm for finding inverse matrices. In general, there are two generally accepted algorithms, and we will also consider the second one today.

The one that will be discussed now is very effective for matrices of size $\left[ 2\times 2 \right]$ and - partially - size $\left[ 3\times 3 \right]$. But starting from the size $\left[ 4\times 4 \right]$ it is better not to use it. Why - now you will understand everything yourself.

Algebraic additions

Get ready. Now there will be pain. No, don’t worry: a beautiful nurse in a skirt, stockings with lace will not come to you and give you an injection in the buttock. Everything is much more prosaic: algebraic additions and Her Majesty the “Union Matrix” come to you.

Let's start with the main thing. Let there be a square matrix of size $A=\left[ n\times n \right]$, whose elements are called $((a)_(ij))$. Then for each such element we can define an algebraic complement:

Definition. Algebraic complement $((A)_(ij))$ to the element $((a)_(ij))$ located in the $i$th row and $j$th column of the matrix $A=\left[ n \times n \right]$ is a construction of the form

\[((A)_(ij))=((\left(-1 \right))^(i+j))\cdot M_(ij)^(*)\]

Where $M_(ij)^(*)$ is the determinant of the matrix obtained from the original $A$ by deleting the same $i$th row and $j$th column.

Again. The algebraic complement to a matrix element with coordinates $\left(i;j \right)$ is denoted as $((A)_(ij))$ and is calculated according to the scheme:

1. First, we delete the $i$-row and $j$-th column from the original matrix. We obtain a new square matrix, and we denote its determinant as $M_(ij)^(*)$.
2. Then we multiply this determinant by $((\left(-1 \right))^(i+j))$ - at first this expression may seem mind-blowing, but in essence we are simply figuring out the sign in front of $M_(ij)^(*) $.
3. We count and get a specific number. Those. the algebraic addition is precisely a number, and not some new matrix, etc.

The matrix $M_(ij)^(*)$ itself is called an additional minor to the element $((a)_(ij))$. And in this sense, the above definition of an algebraic complement is a special case of a more complex definition - what we looked at in the lesson about the determinant.

Important note. Actually, in “adult” mathematics, algebraic additions are defined as follows:

1. We take $k$ rows and $k$ columns in a square matrix. At their intersection we get a matrix of size $\left[ k\times k \right]$ - its determinant is called a minor of order $k$ and is denoted $((M)_(k))$.
2. Then we cross out these “selected” $k$ rows and $k$ columns. Once again you get a square matrix - its determinant is called an additional minor and is denoted $M_(k)^(*)$.
3. Multiply $M_(k)^(*)$ by $((\left(-1 \right))^(t))$, where $t$ is (attention now!) the sum of the numbers of all selected rows and columns . This will be the algebraic addition.

Look at the third step: there is actually a sum of $2k$ terms! Another thing is that for $k=1$ we will get only 2 terms - these will be the same $i+j$ - the “coordinates” of the element $((a)_(ij))$ for which we are looking for an algebraic complement.

So today we're using a slightly simplified definition. But as we will see later, it will be more than enough. The following thing is much more important:

Definition. The allied matrix $S$ to the square matrix $A=\left[ n\times n \right]$ is a new matrix of size $\left[ n\times n \right]$, which is obtained from $A$ by replacing $(( a)_(ij))$ by algebraic additions $((A)_(ij))$:

\\Rightarrow S=\left[ \begin(matrix) ((A)_(11)) & ((A)_(12)) & ... & ((A)_(1n)) \\ (( A)_(21)) & ((A)_(22)) & ... & ((A)_(2n)) \\ ... & ... & ... & ... \\ ((A)_(n1)) & ((A)_(n2)) & ... & ((A)_(nn)) \\\end(matrix) \right]\]

The first thought that arises at the moment of realizing this definition is “how much will have to be counted!” Relax: you will have to count, but not that much. :)

Well, all this is very nice, but why is it necessary? But why.

Main theorem

Let's go back a little. Remember, in Lemma 3 it was stated that the invertible matrix $A$ is always non-singular (that is, its determinant is non-zero: $\left| A \right|\ne 0$).

So, the opposite is also true: if the matrix $A$ is not singular, then it is always invertible. And there is even a search scheme for $((A)^(-1))$. Check it out:

Inverse matrix theorem. Let a square matrix $A=\left[ n\times n \right]$ be given, and its determinant is nonzero: $\left| A \right|\ne 0$. Then the inverse matrix $((A)^(-1))$ exists and is calculated by the formula:

\[((A)^(-1))=\frac(1)(\left| A \right|)\cdot ((S)^(T))\]

And now - everything is the same, but in legible handwriting. To find the inverse matrix, you need:

1. Calculate the determinant $\left| A \right|$ and make sure it is non-zero.
2. Construct the union matrix $S$, i.e. count 100500 algebraic additions $((A)_(ij))$ and place them in place $((a)_(ij))$.
3. Transpose this matrix $S$, and then multiply it by some number $q=(1)/(\left| A \right|)\;$.

That's all! The inverse matrix $((A)^(-1))$ has been found. Let's look at examples:

\[\left[ \begin(matrix) 3 & 1 \\ 5 & 2 \\\end(matrix) \right]\]

Solution. Let's check the reversibility. Let's calculate the determinant:

\[\left| A\right|=\left| \begin(matrix) 3 & 1 \\ 5 & 2 \\\end(matrix) \right|=3\cdot 2-1\cdot 5=6-5=1\]

The determinant is different from zero. This means the matrix is ​​invertible. Let's create a union matrix:

Let's calculate the algebraic additions:

\[\begin(align) & ((A)_(11))=((\left(-1 \right))^(1+1))\cdot \left| 2 \right|=2; \\ & ((A)_(12))=((\left(-1 \right))^(1+2))\cdot \left| 5 \right|=-5; \\ & ((A)_(21))=((\left(-1 \right))^(2+1))\cdot \left| 1 \right|=-1; \\ & ((A)_(22))=((\left(-1 \right))^(2+2))\cdot \left| 3\right|=3. \\ \end(align)\]

Please note: the determinants |2|, |5|, |1| and |3| are determinants of matrices of size $\left[ 1\times 1 \right]$, and not modules. Those. If there were negative numbers in the determinants, there is no need to remove the “minus”.

In total, our union matrix looks like this:

\[((A)^(-1))=\frac(1)(\left| A \right|)\cdot ((S)^(T))=\frac(1)(1)\cdot ( (\left[ \begin(array)(*(35)(r)) 2 & -5 \\ -1 & 3 \\\end(array) \right])^(T))=\left[ \begin (array)(*(35)(r)) 2 & -1 \\ -5 & 3 \\\end(array) \right]\]

OK it's all over Now. The problem is solved.

Answer. $\left[ \begin(array)(*(35)(r)) 2 & -1 \\ -5 & 3 \\\end(array) \right]$

Task. Find the inverse matrix:

\[\left[ \begin(array)(*(35)(r)) 1 & -1 & 2 \\ 0 & 2 & -1 \\ 1 & 0 & 1 \\\end(array) \right] \]

Solution. We calculate the determinant again:

\[\begin(align) & \left| \begin(array)(*(35)(r)) 1 & -1 & 2 \\ 0 & 2 & -1 \\ 1 & 0 & 1 \\\end(array) \right|=\begin(matrix ) \left(1\cdot 2\cdot 1+\left(-1 \right)\cdot \left(-1 \right)\cdot 1+2\cdot 0\cdot 0 \right)- \\ -\left (2\cdot 2\cdot 1+\left(-1 \right)\cdot 0\cdot 1+1\cdot \left(-1 \right)\cdot 0 \right) \\\end(matrix)= \ \ & =\left(2+1+0 \right)-\left(4+0+0 \right)=-1\ne 0. \\ \end(align)\]

The determinant is nonzero—the matrix is ​​invertible. But now it’s going to be really tough: we need to count as many as 9 (nine, motherfucker!) algebraic additions. And each of them will contain the determinant $\left[ 2\times 2 \right]$. Flew:

\[\begin(matrix) ((A)_(11))=((\left(-1 \right))^(1+1))\cdot \left| \begin(matrix) 2 & -1 \\ 0 & 1 \\\end(matrix) \right|=2; \\ ((A)_(12))=((\left(-1 \right))^(1+2))\cdot \left| \begin(matrix) 0 & -1 \\ 1 & 1 \\\end(matrix) \right|=-1; \\ ((A)_(13))=((\left(-1 \right))^(1+3))\cdot \left| \begin(matrix) 0 & 2 \\ 1 & 0 \\\end(matrix) \right|=-2; \\ ... \\ ((A)_(33))=((\left(-1 \right))^(3+3))\cdot \left| \begin(matrix) 1 & -1 \\ 0 & 2 \\\end(matrix) \right|=2; \\ \end(matrix)\]

In short, the union matrix will look like this:

Therefore, the inverse matrix will be:

\[((A)^(-1))=\frac(1)(-1)\cdot \left[ \begin(matrix) 2 & -1 & -2 \\ 1 & -1 & -1 \\ -3 & 1 & 2 \\\end(matrix) \right]=\left[ \begin(array)(*(35)(r))-2 & -1 & 3 \\ 1 & 1 & -1 \ \2 & 1 & -2 \\\end(array) \right]\]

That's it. Here is the answer.

Answer. $\left[ \begin(array)(*(35)(r)) -2 & -1 & 3 \\ 1 & 1 & -1 \\ 2 & 1 & -2 \\\end(array) \right ]$

As you can see, at the end of each example we performed a check. In this regard, an important note:

Don't be lazy to check. Multiply the original matrix by the found inverse matrix - you should get $E$.

Performing this check is much easier and faster than looking for an error in further calculations when, for example, you are solving a matrix equation.

Alternative way

As I said, the inverse matrix theorem works great for sizes $\left[ 2\times 2 \right]$ and $\left[ 3\times 3 \right]$ (in the latter case, it’s not so “great” "), but for larger matrices the sadness begins.

But don’t worry: there is an alternative algorithm with which you can calmly find the inverse even for the matrix $\left[ 10\times 10 \right]$. But, as often happens, to consider this algorithm we need a little theoretical background.

Elementary transformations

Among all possible matrix transformations, there are several special ones - they are called elementary. There are exactly three such transformations:

1. Multiplication. You can take the $i$th row (column) and multiply it by any number $k\ne 0$;
2. Addition. Add to the $i$-th row (column) any other $j$-th row (column), multiplied by any number $k\ne 0$ (you can, of course, do $k=0$, but what's the point? ? Nothing will change).
3. Rearrangement. Take the $i$th and $j$th rows (columns) and swap places.

Why these transformations are called elementary (for large matrices they do not look so elementary) and why there are only three of them - these questions are beyond the scope of today's lesson. Therefore, we will not go into details.

Another thing is important: we have to perform all these perversions on the adjoint matrix. Yes, yes: you heard right. Now there will be one more definition - the last one in today's lesson.

Adjoint matrix

Surely at school you solved systems of equations using the addition method. Well, there, subtract another from one line, multiply some line by a number - that’s all.

So: now everything will be the same, but in an “adult” way. Ready?

Definition. Let a matrix $A=\left[ n\times n \right]$ and an identity matrix $E$ of the same size $n$ be given. Then the adjoint matrix $\left[ A\left| E\right. \right]$ is a new matrix of size $\left[ n\times 2n \right]$ that looks like this:

\[\left[ A\left| E\right. \right]=\left[ \begin(array)(rrrr|rrrr)((a)_(11)) & ((a)_(12)) & ... & ((a)_(1n)) & 1 & 0 & ... & 0 \\((a)_(21)) & ((a)_(22)) & ... & ((a)_(2n)) & 0 & 1 & ... & 0 \\... & ... & ... & ... & ... & ... & ... & ... \\((a)_(n1)) & ((a)_(n2)) & ... & ((a)_(nn)) & 0 & 0 & ... & 1 \\\end(array) \right]\]

In short, we take the matrix $A$, on the right we assign to it the identity matrix $E$ of the required size, we separate them with a vertical bar for beauty - here you have the adjoint. :)

What's the catch? Here's what:

Theorem. Let the matrix $A$ be invertible. Consider the adjoint matrix $\left[ A\left| E\right. \right]$. If using elementary string conversions bring it to the form $\left[ E\left| B\right. \right]$, i.e. by multiplying, subtracting and rearranging rows to obtain from $A$ the matrix $E$ on the right, then the matrix $B$ obtained on the left is the inverse of $A$:

\[\left[ A\left| E\right. \right]\to \left[ E\left| B\right. \right]\Rightarrow B=((A)^(-1))\]

It's that simple! In short, the algorithm for finding the inverse matrix looks like this:

1. Write the adjoint matrix $\left[ A\left| E\right. \right]$;
2. Perform elementary string conversions until $E$ appears instead of $A$;
3. Of course, something will also appear on the left - a certain matrix $B$. This will be the opposite;
4. PROFIT!:)

Of course, this is much easier said than done. So let's look at a couple of examples: for sizes $\left[ 3\times 3 \right]$ and $\left[ 4\times 4 \right]$.

Task. Find the inverse matrix:

\[\left[ \begin(array)(*(35)(r)) 1 & 5 & 1 \\ 3 & 2 & 1 \\ 6 & -2 & 1 \\\end(array) \right]\ ]

Solution. We create the adjoint matrix:

\[\left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 3 & 2 & 1 & 0 & 1 & 0 \\ 6 & -2 & 1 & 0 & 0 & 1 \\\end(array) \right]\]

Since the last column of the original matrix is ​​filled with ones, subtract the first row from the rest:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 3 & 2 & 1 & 0 & 1 & 0 \\ 6 & - 2 & 1 & 0 & 0 & 1 \\\end(array) \right]\begin(matrix) \downarrow \\ -1 \\ -1 \\\end(matrix)\to \\ & \to \left [ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 5 & -7 & 0 & -1 & 0 & 1 \\\end(array) \right] \\ \end(align)\]

There are no more units, except for the first line. But we don’t touch it, otherwise the newly removed units will begin to “multiply” in the third column.

But we can subtract the second line twice from the last - we get one in the lower left corner:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 5 & -7 & 0 & -1 & 0 & 1 \\\end(array) \right]\begin(matrix) \ \\ \downarrow \\ -2 \\\end(matrix)\to \\ & \left [ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right] \\ \end(align)\]

Now we can subtract the last row from the first and twice from the second - this way we “zero” the first column:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 1 & 5 & 1 & 1 & 0 & 0 \\ 2 & -3 & 0 & -1 & 1 & 0 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right]\begin(matrix) -1 \\ -2 \\ \uparrow \\\end(matrix)\to \\ & \ to \left[ \begin(array)(rrr|rrr) 0 & 6 & 1 & 0 & 2 & -1 \\ 0 & -1 & 0 & -3 & 5 & -2 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right] \\ \end(align)\]

Multiply the second line by −1, and then subtract it 6 times from the first and add 1 time to the last:

\[\begin(align) & \left[ \begin(array)(rrr|rrr) 0 & 6 & 1 & 0 & 2 & -1 \\ 0 & -1 & 0 & -3 & 5 & -2 \ \ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right]\begin(matrix) \ \\ \left| \cdot \left(-1 \right) \right. \\ \ \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrr|rrr) 0 & 6 & 1 & 0 & 2 & -1 \\ 0 & 1 & 0 & 3 & -5 & 2 \\ 1 & -1 & 0 & 1 & -2 & 1 \\\end(array) \right]\begin(matrix) -6 \\ \updownarrow \\ +1 \\\end (matrix)\to \\ & \to \left[ \begin(array)(rrr|rrr) 0 & 0 & 1 & -18 & 32 & -13 \\ 0 & 1 & 0 & 3 & -5 & 2 \\ 1 & 0 & 0 & 4 & -7 & 3 \\\end(array) \right] \\ \end(align)\]

All that remains is to swap lines 1 and 3:

\[\left[ \begin(array)(rrr|rrr) 1 & 0 & 0 & 4 & -7 & 3 \\ 0 & 1 & 0 & 3 & -5 & 2 \\ 0 & 0 & 1 & - 18 & 32 & -13 \\\end(array) \right]\]

Ready! On the right is the required inverse matrix.

Answer. $\left[ \begin(array)(*(35)(r))4 & -7 & 3 \\ 3 & -5 & 2 \\ -18 & 32 & -13 \\\end(array) \right ]$

Task. Find the inverse matrix:

\[\left[ \begin(matrix) 1 & 4 & 2 & 3 \\ 1 & -2 & 1 & -2 \\ 1 & -1 & 1 & 1 \\ 0 & -10 & -2 & -5 \\\end(matrix) \right]\]

Solution. We compose the adjoint again:

\[\left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & -2 & 0 & 1 & 0 & 0 \ \ 1 & -1 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\\end(array) \right]\]

Let's cry a little, be sad about how much we have to count now... and start counting. First, let’s “zero out” the first column by subtracting row 1 from rows 2 and 3:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & -2 & 0 & 1 & 0 & 0 \\ 1 & -1 & 1 & 1 & 0 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\\end(array) \right]\begin(matrix) \downarrow \\ -1 \\ -1 \\ \ \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 0 & -6 & -1 & -5 & -1 & 1 & 0 & 0 \\ 0 & -5 & -1 & -2 & -1 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\\end(array) \right] \\ \end(align)\]

We see too many “cons” in lines 2-4. Multiply all three rows by −1, and then burn out the third column by subtracting row 3 from the rest:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 0 & -6 & -1 & -5 & - 1 & 1 & 0 & 0 \\ 0 & -5 & -1 & -2 & -1 & 0 & 1 & 0 \\ 0 & -10 & -2 & -5 & 0 & 0 & 0 & 1 \\ \end(array) \right]\begin(matrix) \ \\ \left| \cdot \left(-1 \right) \right. \\ \left| \cdot \left(-1 \right) \right. \\ \left| \cdot \left(-1 \right) \right. \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & 4 & 2 & 3 & 1 & 0 & 0 & 0 \\ 0 & 6 & 1 & 5 & ​​1 & -1 & 0 & 0 \\ 0 & 5 & 1 & 2 & 1 & 0 & -1 & 0 \\ 0 & 10 & 2 & 5 & 0 & 0 & 0 & -1 \\\end (array) \right]\begin(matrix) -2 \\ -1 \\ \updownarrow \\ -2 \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr| rrrr) 1 & -6 & 0 & -1 & -1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 3 & 0 & -1 & 1 & 0 \\ 0 & 5 & 1 & 2 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array) \right] \\ \end(align)\]

Now is the time to “fry” the last column of the original matrix: subtract row 4 from the rest:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & -6 & 0 & -1 & -1 & 0 & 2 & 0 \\ 0 & 1 & 0 & 3 & 0 & -1 & 1 & 0 \\ 0 & 5 & 1 & 2 & 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array ) \right]\begin(matrix) +1 \\ -3 \\ -2 \\ \uparrow \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & -6 & 0 & 0 & -3 & 0 & 4 & -1 \\ 0 & 1 & 0 & 0 & 6 & -1 & -5 & 3 \\ 0 & 5 & 1 & 0 & 5 & 0 & -5 & 2 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array) \right] \\ \end(align)\]

Final throw: “burn out” the second column by subtracting line 2 from lines 1 and 3:

\[\begin(align) & \left[ \begin(array)(rrrr|rrrr) 1 & -6 & 0 & 0 & -3 & 0 & 4 & -1 \\ 0 & 1 & 0 & 0 & 6 & -1 & -5 & 3 \\ 0 & 5 & 1 & 0 & 5 & 0 & -5 & 2 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end( array) \right]\begin(matrix) 6 \\ \updownarrow \\ -5 \\ \ \\\end(matrix)\to \\ & \to \left[ \begin(array)(rrrr|rrrr) 1 & 0 & 0 & 0 & 33 & -6 & -26 & -17 \\ 0 & 1 & 0 & 0 & 6 & -1 & -5 & 3 \\ 0 & 0 & 1 & 0 & -25 & 5 & 20 & -13 \\ 0 & 0 & 0 & 1 & -2 & 0 & 2 & -1 \\\end(array) \right] \\ \end(align)\]

And again the identity matrix is ​​on the left, which means the inverse is on the right. :)

Answer. $\left[ \begin(matrix) 33 & -6 & -26 & 17 \\ 6 & -1 & -5 & 3 \\ -25 & 5 & 20 & -13 \\ -2 & 0 & 2 & - 1 \\\end(matrix) \right]$

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 DOES NOT 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.

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 ​​nonzero, 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.

Finding the inverse matrix.

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

Page navigation.

Inverse matrix - definition.

Finding the inverse matrix using a matrix from algebraic complements.

Properties of an inverse matrix.

Finding the inverse matrix using 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 nonzero, that is, for non-singular square matrices.

Definition.

Matrixcalled the inverse of a matrix, whose determinant is different from zero if the equalities are true , Where E– unit order matrix n on n.

Finding the inverse matrix using a matrix from algebraic complements.

How to find the inverse matrix for a given one?

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

Definition.

Minorkth order matrices A order m on n is the determinant of the order matrix k on k, which is obtained from the matrix elements 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 composed of elements of all rows except i-th, and all columns except jth, square matrix A order n on n let's denote it as .

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

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

Definition.

Algebraic complement element of a square matrix is ​​called minor (n-1)th order, which is obtained from the matrix A, crossing out elements of it i-th lines and jth column multiplied by .

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

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

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

Based on these properties of the determinant, the definition operations of multiplying a matrix by a number and the concept of an inverse matrix is ​​true: , where is a transposed matrix whose elements are algebraic complements.

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

Let's compose algorithm for finding the inverse matrix using equality .

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

Example.

Given a matrix . Find the inverse matrix.

Solution.

Let's calculate the determinant of the matrix A, decomposing it into the elements of the third column:

The determinant is nonzero, so the matrix A reversible.

Let's find a matrix of algebraic additions:

That's why

Let's transpose the matrix from algebraic additions:

Now we find the inverse matrix as :

Let's check the result:

Equalities are satisfied, therefore, the inverse matrix is ​​found correctly.

Properties of an inverse matrix.

The concept of an inverse matrix, equality , definitions of operations on matrices and properties of the determinant of a matrix make it possible to justify the following properties of inverse matrix:

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

Let's consider another way to find the inverse matrix for a square matrix A order n on 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. Let us 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 look at this method with an example.

Example.

Given a matrix . Find the inverse matrix.

Solution.

Let's accept . Equality gives us three systems of linear inhomogeneous algebraic equations:

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

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

Summarize.

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

Example of solutions using the inverse matrix method

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 + x 4 = 4

Beginning of the form

End of form

Solution. Let's write the matrix in the form: Vector B: B T = (1,2,3,4) Main 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 Determinant of minor ∆ = 2 (-3)-3 0+5 3-4 3 = -3

Transposed matrix Algebraic additions ∆ 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) = -3 ∆ 4.3 = -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 Results 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 solutions of SLAEs using the inverse matrix method online. To do this, enter your data and receive a solution with detailed comments.

Task 2. Write the system of equations in matrix form and solve it using the inverse matrix. Check the resulting 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 formulas; 2) write the system in matrix form and solve it using matrix calculus. Guidelines. After solving by Cramer's method, find the "Solving by inverse matrix method for source data" button. You will receive the appropriate solution. Thus, you will not have to fill in the data again. Solution. Let us denote by A the matrix of coefficients for unknowns; X - matrix-column of unknowns; B - matrix-column of free members:

Vector B: B T =(4,-3,-3) Taking into account these notations, this system of equations takes the following matrix form: A*X = B. If matrix A is non-singular (its determinant is non-zero, then it has an inverse matrix A -1... Multiplying both sides of the equation by A -1, we get: A -1 *A*X = A -1 *B, A -1 *A = E. This equality is called matrix notation of the solution to a 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 nonzero. Let's find the main determinant. ∆=-1 (-2 (-1)-1 1)-3 (3 (-1)-1 0)+2 (3 1-(-2 0))=14 So, determinant 14 ≠ 0, so we continue solution. To do this, we find the inverse matrix through algebraic additions. Let us have a non-singular matrix A:

We calculate algebraic complements.

∆ 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.

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.