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)

2 Answers 2

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:


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.


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:


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.



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.


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.


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.

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.
  4. Write adj 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

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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:

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Matrices are rectangular arrays of numbers or other mathematical objects. We define matrices and how to add and multiply them, discuss some special matrices such as the identity and zero matrix, learn about transposes and inverses, and define orthogonal and permutation matrices.


Jeffrey R. Chasnov

Текст видео

[MUSIC] We're going to learn about what an inverse matrix is now. Inverse matrices are very important in matrix algebra. And it's important to know that not all matrices are invertible. If a matrix is invertible, we say is has an inverse matrix, but many matrices are not invertible. So we're going to touch on that topic in this video and then, we're going to go into more in depth later on. So what is an inverse matrix? So if a matrix is invertible, then we can write that A times its inverse, which we denote as A to the -1, is equal to the identity matrix, okay. So it's like the multiplication inverse for numbers. And it doesn't matter the order that you write this, so this is the same as A inverse A, okay? A has to be a square matrix. A has to be n by n, a square matrix. And as I said, not all matrices are invertible. So there are some certain relationships. If you have the product to two invertible square matrices, and you want to find the inverse of that, this works similar to the way the transpose works. This is B inverse times A inverse. It's a relatively simple proof, and I'm going to leave it to you guys for an exercise. The other one is what if you were looking for the inverse of the transpose matrix, okay? This is the inverse matrix transposed. So that's another fact or theorem that'll I'll leave it to you to do in the exercises, okay. So what I want to do in this video is really compute the inverse of a two by two matrix. So we can set up some equations right, so let's look at the two by two matrix. So we're going to have a general two by two, a, b, c, d, and I want to multiply this matrix by its inverse to get the identity matrix. So the inverse is the matrix that we don't know, so that's going to be an unknown matrix. So let me write that as the first column, I'll write as x1 y1, and the second column, I'll write as x2 y2 and that's supposed to be the identity matrix so let me write that out. That's supposed to equal 1 0 0 1. Okay, so the goal here is to determine what is the inverse of this two by two matrix, right? So I should make this clear, this one is our matrix A. This one is our matrix A inverse, and this one is the identity matrix. Okay, so we want to find A inverse. Okay, how do we do that? Well, we have to write down equations. Okay, if we write down the equations that this matrix product represents, then maybe we can solve for the unknowns. So let's see, what is this? So we use matrix multiplication. So we have ax1 + by1. Ax1 + by1 is equal to 1, right? And then we have ax2 + by2. And what is that equal to? That's equal to 0, right? Then in the second row, we have cx1 + dy1. And that one is equal to 0. And finally, we have a cx2 + dy2. And that's this last element here which is 1, okay. We have four unknowns, x1, y1, and x2, y2. And we have four equations. So you think we can solve for the unknowns. Okay, so how would we do that? I think the easiest way is to use these equations with a 0 on the right-hand side. Those are called homogeneous equations. And we can solve those equations for the y's. So we can solve for y1, we can solve for y2, so let's do that. I can put that over here, now for y1 first. So dy1 = -cx1. So then we solve for y1. So we get y1, right, minus cx1 divided by d, so minus c over d x1, right. And the second equation we can solve for y2. So we can subtract, -ax2 and divided by b. So -a over b, so this is y2 = -a over b x2, okay? So we can eliminate y1 and y2 in favor of x1 and x2. So now, we go to the first equation, so let's say we eliminate y1. So we get ax1, + by1, so by1. So instead of plus, we can use minus here, -bc over d. X1 = 1, right? We can, Multiply by d and divide by d, right? And then solve this equation for x1. So you can factor out an x1, and then you have times a d minus b c, divided by d, and then you can bring that to the other side. So if you let me skip some algebra steps and hopefully, don't make a mistake. We can have x1 =, so we have ad- bc over d, and we flip that so that's d over ad- bc. Okay, so we've got x1. If we've got x1, then we also have y1, right? So y1 is in terms of x1, so it's minus c over d x1. The d cancels, so it becomes minus c, Over ad- bc. Okay, we're good. Now we need to find x2 and y2. So we have y2 here in terms of x2. So now, we need to use the fourth equation, right? We just use the first equation, so now we need the fourth equation. So we substitute in y2, let me write it down here. So we have cx2 + dy2, so that becomes -ad over b x2 = 1. Again, we do the same thing, so we have a b here in the denominator. So we have bc and then b. And then we can solve this for x2, so we get x2, let me do it down here first. So we got x2 = b over bc- ad. So bc- ad is just the negative of ad- bc. So we can change the sign here so we can write x2 = -b over ad- bc. Okay, we're almost done. We have x2, so now we just need y2. So y2 is = -a over bx2, so the minus signs cancel, that becomes then a over ad- bc. Okay, so we've got the result. So let me put it here, I'll use a different color so you see that. So sorry, I'm going to write on my face and maybe a little bit lower. So we have A inverse then, okay. So what is A inverse? So x1, x2, so there's always a 1 over ad- bc, so 1 over ad- bc. And then we have the matrix, the matrix is x1 x2, so d- b and then we have y1 y2. So we have -ca, okay? That's the inverse matrix. So interesting, a b c d, the inverse of that is 1 over ad- bc. And then the matrix itself, the diagonal elements switch, so we have da. And then we just negate the off-diagonal elements, -b and -c. This piece here ad- bc, is what we call the determinant of A. Is ad- bc, and that's for a two by two matrix. So that is this two by two matrix, the determinant is ad- bc. And interesting thing here is that, if ad- bc = 0, you're dividing by 0, so there is no inverse matrix. So the determinant of a = 0, then the inverse does not exist, okay? The matrix is not invertible. That's a very important result, and we'll go into it more for the general n by n matrix. Okay, this formula for the inverse of a two by two matrix is something I typically don't remember. But if you're a student in a matrix algebra course, it's probably a good idea to memorize this formula, okay. So what have we done? We've introduced the idea of the inverse matrix. A, A inverse is equal to inverse A is the identity matrix. We have some identities. A B inverse is B inverse A inverse. A transpose inverse equals A inverse transpose. And then we've derived using solving a system of four equations and four unknowns. We derive the formula for the inverse of a two by two matrix. I'm Jeff Chasnov, thanks for watching, and I'll see you in the next video.

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 $A^<-1>$.

Watch the video: Math: Inverse matrix (December 2021).