Lecture 3. Inverse matrix

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At this lesson, we will consider the application of determinant theory to solving algebraic problems. First, let’s introduce two concepts. A square matrix with a determinant equal to zero is called degenerate, but if the determinant is different from zero, the matrix is called non-degenerate. We formulate a number of conditions that are equivalent to non-degeneracy of the matrix. So, the following theorem holds. A matrix A is non-degenerate if and only if any of the following conditions are met: first, the rows or columns of the matrix are linearly independent, the second, equivalent condition, the matrix A has the inverse, i.e. it is invertible, and the third condition is that a system of linear equations with the main matrix A has only one solution. Each of these conditions is equivalent to the fact that the matrix is non-degenerate, i.e. it has a non-zero element. Let’s solve the following problem. Let’s determine at what value of the parameter k, the three specified geometric vectors are coplanar. Vectors are defined by their coordinates. You should remember that vectors are called coplanar if they are parallel to a single plane. We consider a geometric space denoted by E3. It is known that the coplanarity of vectors is equivalent to their linear dependence, and this in turn is equivalent to the fact that the rows of the matrix made up of these vectors are linearly dependent. By our theorem, this means that its determinant is zero. So, we need to calculate the determinant and determine when it turns to zero. Let’s set up this determinant by writing vectors in rows, and at the same time apply one of the schemes for calculating the third-order determinant. Let’s consider the so-called rule of triangles, i.e. we multiply the elements that are the vertices of two triangles plus the element standing on the main diagonal, we take them with a plus sign. Next, the elements that we take with a minus sign, the secondary diagonal, and the elements located at the vertices of two triangles. We get a number 2k -1. Obviously, the determinant turns to zero only when k equals 1/2. Thus, the original vectors are coplanar only when the parameter k is 1/2. If we lay off these vectors from one point, they will lie in the same plane. For all the other values of the parameter k, the vectors will no longer be coplanar. For example, if k is set to zero or one, we get a non-planar vector. Let’s look at another example. Let’s take the system of linear equations written on the slide and try to determine how many solutions it has. And we will answer the question without solving this system. Note that this system is homogeneous, which means that the zero vector is its solution. So, the system is compatible, which means that there are two cases, either this system has only one zero solution, or along with the zero solution there are others, but, therefore, there are infinitely many of them. So, we need to choose which option holds for our system, either the zero solution is the only one, or there are infinitely many solutions. You should remember that a system has one solution if and only if the determinant of its main matrix is not zero, i.e. the matrix is non-degenerate. Let’s make up a determinant and calculate it, and make a conclusion depending on the values. However, we will not use the formula to calculate the determinant. Note that the third column of the matrix can be obtained by subtracting 2 from the first column, i.e. 1 minus-1 is 2, 2 minus 1 is 1, 3 minus 4 is -1, which means that the columns are linearly dependent. This means that the determinant of this matrix is zero. So, if the matrix is degenerate, the system has infinitely many solutions. Now, let’s remember what the inverse matrix is. We introduced this concept at the previous lecture. By definition, a matrix is called the inverse of matrix A if its product with matrix A gives the unit matrix, and, in any order. According to the above theorem, a matrix has an inverse if and only if it is non-degenerate. It turns out that there is a formula that allows us to construct a non-degenerate matrix in the case when it exists. This formula looks like this (see the video). It contains algebraic complements of the elements of the original matrix. Let me remind you that to create an algebraic complement, it is necessary to cross out the row and column that contains the corresponding element from the original matrix. And do not forget to multiply the resulting determinant by a coefficient equal to minus one to the power of i+j, where i and j are the indices of this element. Note that if we consider the first row in a matrix and find the algebraic complement of each of its elements for it, we write these complements to the column. So, we do the same with each row. If we further multiply the original matrix by the one that we’ve made up, for example, the first element in the first row will give us nothing more than the expansion of the determinant of the matrix A along the first row. After performing similar calculations for all the elements, we will get that the main diagonal will have determinants, and in all other places there will be 0, and therefore, we take the determinant for the sign of the matrix, reduce it with the written coefficient and get the unit matrix. Thus, the matrix constructed is the inverse of the original one. Let’s take the following example. Let’s take a second-order matrix and try to find the inverse of it. First, let’s check whether the inverse exists, and then use the specified formula. So, we calculate the determinant according to a well-known rule, we get 10. Since the matrix is non-degenerate, it means that the determinant exists. We will find algebraic complements for each element, and I will note again that first we consider the first row, and we will write the resulting complements in the column. So, we immediately put the necessary minus signs before those complements that have an odd sum of indices. Next, we calculate. Note that after crossing out the row and column, we get the matrix consisting of one element. In this case, its determinant is equal to this element. We do this with each element of the original matrix. We get the following numbers (see the video). Thus, the inverse matrix takes the following form (see the video). The coefficient 1/10 is multiplied by the matrix of the obtained complements. Let’s add the coefficient to the matrix by multiplying all its elements by one-tenth. We get the answer. Now, let’s solve the following system of equations. Let’s see that the main matrix for this system is just the same matrix from our example. In B, we write a column of free terms, given that the matrix A is not degenerate, we can say that this system has exactly one solution. To find it, let’s remember that such a system can be written in matrix form. In the form AX equals B, where A and B are input matrices, and X is a column of unknowns. Since the matrix A has an inverse, we multiply this equation by the inverse matrix on the left. We get that the solution X is written as the product of the inverse matrix by column B. We have found the inverse matrix. Let’s use this and multiply the specified matrices. Let me remind you that when multiplying, we multiply a row by a column, and then we get an unknown matrix X, which gives us the solution to this system. Let’s solve a more complex matrix equation. First, we denote the matrix data with the letters A, B, and E. There is the unit matrix in the right part, so we have denoted it with the letter E. The inverse matrix of A is known to us, now we will find the inverse matrix of B. By the formula, you must write down the coefficient. To do this, we need the determinant of the matrix. We calculate it and then find the required algebraic complements. The calculation is duplicated in the first row where the matrix B is entered. So, crossing out rows and columns, do not forget to take into account the signs of the complements, after that we get this answer (see the video). Let’s perform some transformation. The coefficient 1/0.01 can be converted to the number 100. In this case, I suggest adding the multiplier of 10 to the matrix in order to get rid of fractions. After that, we get the following (see the video). Next, we perform the specified transformation with the matrix equation, multiplying the inverse of A on the left and the inverse of B on the right. So that, after we place the parentheses on the left side of the equality, there are unit matrices to the left and right of the matrix X. So, the matrix X is expressed like this (see the video). We all know the matrices. We substitute the found matrices into the resulting formula and perform the calculations. Our coefficients are decreasing. We remember that the product of the matrix by the unit does not change the original matrix, so you need to multiply only two matrices. Afterwards, you get this answer (see the video). So, the matrix found is the solution to the original equation. Now, let’s talk about the so-called Cramer’s formulas, which in some cases allow us to solve a system of linear equations. Let me remind you that at the previous lecture we considered these formulas on a particular example, when we considered a system consisting of two equations with two variables. We will take an arbitrary system with n equations and n variables. In this case, the main matrix is square. We assume that this matrix is non-degenerate. In this case, you can get the following formulas. To write them down, let’s consider the vectors and columns of coefficients for variables. So, A1 is a column of coefficients for the first variable, A2 – for the second variable, and so on. An is a column of coefficients corresponding to n variables. And, of course, the column of free terms. Using these notations, our system can be written in this form (see the video). Ai are matrices, and xi are numerical coefficients. The determinant of the matrix A can be understood as the determinant from columns Ai. Since we assumed that our matrix A is non-degenerate, i.e. it has a non-zero determinant, it means that this system has exactly one solution. To find a solution, we can use the following theorem. The theorem states that if a system of equations has only one solution, each coordinate of this solution can be found by the written formula. Let’s consider how the fraction is formed, through which ci is calculated. First, the denominator is the determinant of the matrix, and in the numerator we make up the determinant, in which we substitute columns Ai, but instead of the i-th column, we write column B. For example, when calculating the first coordinate c1, we replace the first column with B, leave all other columns Ai, divide by the determinant of the matrix A. When we find c2, we change the second column, substitute B instead, and so on. Briefly, the determinant that is obtained by replacing the i-th column with column B is denoted by Delta i . In this case, each coordinate ci is equal to the quotient of Delta i by Delta. Delta is the determinant of the matrix A. These are the formulas that are obtained. Let’s consider the following example. Let’s take a system of linear equations, where the determinant is a third-order matrix, and we apply the above formulas. Let’s calculate the Delta first. To do this, I suggest using elementary transformations, given that adding another row multiplied by an arbitrary number does not change the original determinant. So, we add 1 to the second row, and subtract the first two from 3. After that, we get the following matrix, which as you can see is triangular. By the property formulated at the previous lecture, its value is equal to the product of the elements standing on the main diagonal, that is, 2. Now, we count the three determinants Delta 1, Delta 2, and Delta 3. So, in order to find Delta 1, we replace the first column of the main matrix with the column of free terms. To calculate it, we are supposed to use the property again. Note that the first and third columns are proportional, namely, the third column is equal to the first two, which means that this determinant is zero. Let’s make up Delta 2. In this case, we replace the second column of the main matrix with column B, get the same columns 406 and 203 only in a different order, which means that we again have a zero value for this determinant. We find Delta 3. There are no proportional rows and columns here. I suggest using elementary transformations. After that we again get a triangular matrix, from which we easily find its determinant, which is equal to 4. So, substituting the obtained values into the formula, we get the answer. The first two values are zero because Delta 1 and Delta 2 are zero, but Delta 3 is two. So, we get the solution to the system. The first two coordinates are zero, and the third coordinate is 2. Thus, the only solution to this system is the triple 002. In conclusion, note that the two ways of solving systems discussed in the lecture assume that the system taken has a square basic matrix, the determinant of which is not equal to zero. In this case, we can solve such a system either using the inverse matrix or using Cramer’s formulas. If the main matrix is degenerate, we cannot apply these methods. To do this, we have a general method, the Gauss method.

 

 


Last modified: Пятница, 18 декабря 2020, 3:52