# 2.6: The Matrix Inverse - Mathematics

2.6: The Matrix Inverse - Mathematics

## Find the inverse of a matrix in $GL(2,,, Bbb Z_)$.

What are the necessary steps and reasoning for calculating the following matrix in GL(2,$Bbb Z_<11>$):

I found the answer to be $egin 9&9 10&8 end$, however I lost 2 points (out of 10) to the question--which is the reason for my confusion.

I found the determinant to be -8, which is 3 in $ℤ_<11>$ and therefore, in calculating the inverse, $left(frac13 ight)$ = 4 in $ℤ_<11>$. While I was computing the inverse, I had this intermediary matrix $M^ <-1>= egin 20& <-24><-12>&8 end$ which I then converted to mod 11 to obtain my final answer. My professor circled the -12 and -24 in the matrix and asked "What is '24' inverse?" This was where I lost points, and I'm completely unsure why. I talked to my professor about this, and I came away thinking I understood but upon further thought I'm still unsure. His reasoning was along the lines of -12 and -24 are additive inverses. but then I get lost. I don't understand why my method of computation is "incorrect." If anything is unclear, or if anything is wrong with the formatting (this is my first post on this website), please let me know and I'll do my best to elaborate on/fix anything! Thanks.

(Also, if it helps I can post my entire computation with every step in detail)

Question: "Why the inverse of a matrix involves division by the determinant?"

Answer: We use the adjugate matrix and the determinant to prove existence of an inverse of a matrix as follows:

The "adjugate matrix" $ad(A)$ has the property that $ad(A)A=Aad(A)=det(A)I$ where $det(-): Mat(n,k) ightarrow k$ is a map with $det(AB)=det(A)det(B)$ . Here $Mat(n,k)$ is the set of $n imes n$ -matrices with coefficients in $k$ . $det(A)$ is the "determinant" of the matrix $A$ as defined in your linear algebra course.

Lemma: A square matrix $A$ has an inverse iff $det(A) eq 0$ .

Proof: If $det(A) eq 0$ it follows $A^<-1>:=frac<1>ad(A)$ is an inverse. Conversely assume there is a matrix $B$ with $AB=BA=I$ . It follows $det(AB)=det(A)det(B)=1$ and hence $det(A) eq 0$ .

Hence the adjugate matrix and the determinant map implies the existence of an inverse of $A$ : The matrix $A$ has a unique inverse $A^<-1>$ iff $det(A) eq 0$ .

Example: Let egin A= egin a & b c & d end end

and define the adjunct matrix $ad(A)$ by

There is in general for any $n imes n$ -matrix $A$ a unique matrix $ad(A)$ with $ad(A)A=Aad(A)=det(A)I$ . This result is proved in any serious linear algebra course. Hence the above proves the Lemma explicitly for any $2 imes 2$ -matrix.

## A Real Life Example: Bus and Train A group took a trip on a bus, at $3 per child and$3.20 per adult for a total of $118.40. They took the train back at$3.50 per child and $3.60 per adult for a total of$135.20.

How many children, and how many adults?

First, let us set up the matrices (be careful to get the rows and columns correct!):

This is just like the example above:

So to solve it we need the inverse of "A":

Now we have the inverse we can solve using:

There were 16 children and 22 adults!

The answer almost appears like magic. But it is based on good mathematics.

Calculations like that (but using much larger matrices) help Engineers design buildings, are used in video games and computer animations to make things look 3-dimensional, and many other places.

The calculations are done by computer, but the people must understand the formulas.

As stated in the comment of saulspatz, a row operation can be considered as a left multiplication by some other matrix. Consider $A = left( egin 1 & 2 4 & 5 end ight).$

Subtracting first row from the second 4 times is the same as multiplying by $R_1$ : $R_1A = left( egin 1 & 0 -4 & 1 end ight)left( egin 1 & 2 4 & 5 end ight) = left( egin 1 & 2 0 & -3 end ight)$

Multiplying the second row by $-1/3$ is the same as multiplying by $R_2$ : $R_2R_1A=left( egin 1 & 0 0 & -frac<1> <3> end ight)left( egin 1 & 2 0 & -3 end ight) = left( egin 1 & 2 0 & 1 end ight)$

Subtracting second row from the first row 2 times is the same as multiplying by $R_3$ : $R_3R_2R_1A = egin 1 & -2 0 & 1 end egin 1 & 2 0 & 1 end = egin 1 & 0 0 & 1 end = I.$

Now we see that $(R_3R_2R_1)A = I$ , so by the definition of inverse matrix $R_3R_2R_1=A^<-1>$ . However, when we apply the same operation to the matrix $B$ , we find out that: $R_3R_2R_1B = (R_3R_2R_1)B=A^<-1>B$

## How to find the inverse of a 3ࡩ matrix

The formula to calculate the inverse of a 3ࡩ matrix is as follows:

Next we will see how to calculate the inverse of a 3ࡩ matrix by solving an exercise step by step:

### Example

Find the inverse of the following 3ࡩ matrix:

To determine the inverse of the matrix, we have to apply the following formula:

Therefore, we have to calculate the determinant of the matrix and verify that it is different from 0. So we evaluate the determinant of the 3ࡩ matrix using cofactor expansion:

The determinant of the matrix is not 0, so the matrix is invertible.

Therefore, substituting the value of the determinant in the formula, the inverse of the matrix will be:

Now we have to calculate the adjoint of matrix A. To do this, we have to replace each element of matrix A by its cofactor in order to obtain the cofactor matrix, and then transpose the cofactor matrix.

Remember that the formula to compute the i, j cofactor of a matrix is as follows:

Where Mij is the i, j minor of the matrix, that is, the determinant that results from deleting the i-th row and the j-th column of the matrix.

So the cofactors of the elements of matrix A are:

Once we have calculated all the cofactors, we simply have to replace each entry of A by its cofactor:

And we transpose the cofactor matrix to find the adjugate matrix:

Thus, we substitute the adjugate matrix in the formula of the inverse matrix:

And finally, we multiply each term in the matrix by the fraction:

Once we have seen how to compute the inverse of a 3ࡩ matrix, you can practice with the following exercises solved step by step to fully understand the concept.

### Practice

#### Problem 4

Invert the following 3ࡩ dimension matrix by the adjoint matrix method:

The inverse matrix formula is:

First we solve the determinant of the matrix using cofactor expansion:

The determinant is nonzero, therefore, the matrix can be inverted.

Once we have solved the determinant, we find the cofactor of each element of matrix A:

Now we calculate the cofactor matrix by replacing each element by its cofactor:

And we transpose the cofactor matrix to obtain the adjoint of matrix A:

So the inverted matrix is:

#### Problem 5

Find the inverse of the following matrix of order 3 by the adjoint matrix algorithm:

The formula to find the inverse of a matrix of order 3 is as follows:

So, first we have to calculate the determinant of the matrix to check the invertibility of the matrix:

The determinant of matrix A is equal to zero so the matrix is non-invertible, in other words, the inverse of matrix A does not exist.

#### Problem 6

Perform the matrix inversion of the following 3ࡩ dimension matrix by the adjugate matrix method:

To invert the 3ࡩ matrix we have to apply the following formula:

First we evaluate the determinant of the 3ࡩ matrix:

The determinant is nonzero, therefore, the matrix can be inverted.

Once we have solved the determinant, we find the cofactors of all the elements of matrix A:

Now we calculate the cofactor matrix by replacing each element of matrix A by its cofactor:

We transpose the cofactor matrix to find the adjugate matrix:

And we apply the formula of the inverse of a 3ࡩ matrix:

## MULTIPLICATIVE INVERSES OF MATRICES

In this section multiplicative identity elements and multiplicative inverses are introduced and used to solve matrix equations. This leads to another method for solving systems of equations.

IDENTITY MATRICES The identity property for real numbers says that a * I = a and I * a = a for any real number a. If there is to be a multiplicative identity matrix I, such that:

for any matrix A, then A and I must be square matrices of the same size. Otherwise it would not be possible to find both products. For example, let A be the 2 X 2 matrix and let represent the 2 X 2 identity matrix. To find I, use the fact that IA = A, or Multiplying the two matrices on the left side of this equation and setting the elements of the product matrix equal to the corresponding elements of A gives the following system of equations with variables x11, x12, x21, and x22 Notice that this is really two systems of equations in two variables. Use one of the methods of the previous chapter to find the solution of this system: x11 = 1, x12 = x21 = 0, and X22 = 1. From the solution of the system, the 2 X 2 identity matrix is Check that with this definition of I, both Al = A and IA = A.

Example 1 VERIFYING THE IDENTITY PROPERTY

Let Verify that MI = M and IM = M The 2 X 2 identity matrix found above suggests the following generalization:

For any value of n there is an n X n identity matrix having l's down the diagonal and 0's elsewhere. The n x n identity matrix is given by l where: Here aij = 1 when i = j (the diagonal elements) and aaj = 0 otherwise.

Example 2 - STATING AND VERIFYING THE 3 X 3 IDENTITY MATRIX

Let K = Given the 3 X 3 identity matrix I and show that KI = K.

The 3 X 3 identity matrix is By the definition of matrix multiplication, MULTIPLICATIVE INVERSES For every nonzero real number a, there is a multiplicative inverse l/a such that Recall that l/a can also be written a^(-1). In the rest of this section, a method is developed for finding a multiplicative inverse for square matrices. The multiplicative inverse of a matrix A is written A^(-1). This matrix must satisfy the statements The multiplicative inverse of a matrix can be found using the matrix row transformations given in the previous tutorial and repeated here for convenience.

MATRIX ROW TRANSFORMATIONS

The matrix row transformations are:

interchanging any two rows of a matrix

multiplying the elements of any row of a matrix by the same nonzero scalar k and Let the unknown inverse matrix be By the definition of matrix inverse, AA^(-1) = 1, or  Setting corresponding elements equal gives the system of equations Since equations (1) and (3) involve only x and z, while equations (2) and (4) involve only y and w, these four equations lead to two systems of equations,

Writing the two systems as augmented matrices gives Each of these systems can be solved by the Gauss-Jordan method. However, since the elements to the left of the vertical bar are identical, the two systems can be combined into the one augmented matrix, and solved simultaneously as follows. Exchange the two rows to get a 1 in the upper left comer. Multiply the first row by -2 and add the results to the second row to get Now, to get a I in the second-row, second-column position, multiply the second row by 1/6. Finally, add the second row to the first row to get a 0 in the second column above the 1. The numbers in the first column to the right of the vertical bar give the values of x and z. The second column gives the values of y and w. That is,  To check, multiply A by A^(-1). The result should be I Verify that A - 1 A = I, also. Finally, The process for finding the multiplicative inverse A^(-1) n x n matrix A that has an inverse is summarized below.

FINDING AN INVERSE MATRIX
To obtain A^(-1) n x n matrix A for which A^(-1) exists, follow these steps.

1. Form the augmented matrix [A/I], where I is the n x n identity matrix.
2. Perform row transformations on [A|I] to get a matrix of the form [I|B].
3. Matrix B is A^(-1).
4. Verify by showing that BA = AB = I.

CAUTION Only square matrices have inverses, but not every square matrix has an inverse. If an inverse exists, it is unique. That is, any given square matrix has no more than one inverse. Note that the symbol A^(-1) does not mean 1/A the symbol A^(-1) is just the notation for the inverse of matrix A.

## Algorithm to find inverse of a matrix:

Suppose a square matrix A is given whose inverse is to be obtained.

1. Find |A|. If |A| = 0, write “Inverse does not exist”. If |A| 0 write “Inverse exists” and proceed to step 2.
2. Find cofactor of all elements of A.
3. Write matrix of the cofactor of A.
5. Whether the inverse is correct verify it by = I (Identitiy Matrix).

Suppose a 2*2 matrix A whose determinant is not equal to 0. where a,b,c,d are number, the inverse is

## 2.6: The Matrix Inverse - Mathematics  Количество зарегистрированных учащихся: 49 тыс.

This course is all about matrices, and concisely covers the linear algebra that an engineer should know. The mathematics in this course is presented at the level of an advanced high school student, but typically students should take this course after completing a university-level single variable calculus course. There are no derivatives or integrals in this course, but students are expected to have attained a sufficient level of mathematical maturity. Nevertheless, anyone who wants to learn the basics of matrix algebra is welcome to join. The course contains 38 short lecture videos, with a few problems to solve after each lecture. And after each substantial topic, there is a short practice quiz. Solutions to the problems and practice quizzes can be found in instructor-provided lecture notes. There are a total of four weeks in the course, and at the end of each week there is an assessed quiz. Download the lecture notes: http://www.math.ust.hk/

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Suppose you are given an equation in one variable such as $4x = 10$. Then you will find the value of $x$ that solves this equation by multiplying the equation by the inverse of 4: $color cdot 4x = color cdot 10$, so the solution will be $x = 2.5$.

Sometimes we can do something very similar to solve systems of linear equations in this case, we will use the inverse of the coefficient matrix. But first we must check that this inverse exists! The conditions for the existence of the inverse of the coefficient matrix are the same as those for using Cramer's rule, that is

1. The system must have the same number of equations as variables, that is, the coefficient matrix of the system must be square.

2. The determinant of the coefficient matrix must be non-zero. The reason, of course, is that the inverse of a matrix exists precisely when its determinant is non-zero.

3. To use this method follow the steps demonstrated on the following system:

$egin -x + 3y + z &= 1 2x + 5y &= 3 3x + y - 2z &= -2 end$

Step 1: Rewrite the system using matrix multiplication:

and writing the coefficient matrix as A, we have

Step 2: FInd the inverse of the coefficient matrix A. In this case the inverse is

Step 3: Multiply both sides of the equation (that you wrote in step #1) by the matrix $A^<-1>$.