Solving Matrix equstions of the type: AX = C (FIND The X) - (NOT HW)

In summary: Similarly, if you have two equations with like terms, you can combine them by adding and subtracting like terms.
  • #1
rugh
2
0
Hi,
Last day I learned about matrix equations like this:
AX = C
A–1AX = A–1C
IX = A–1C
X = A–1C

The objective of this is to Find X.

Here they done it by right multiplication of the Inverse matrix (A^-1).

My Questions are:

1) Where can I get more of these exercises and in higher level? (This exercise was fairly simple, but what about something like p^Tx^-1P^-1PX=APX)
Do you have some book that also shows in understandable way how to solve this?
Because when it's get harder I don't know how to approach it.
Do you have any tips? (Or good book)

2) The same question, but for the system of linear equations like this:

X - ky + z = 2 + kz
kx - 4y - 6z = k^2 + 2k - 9
2x + 4y + 8z = kx+2k(y+z-1) + 13 - k^2

Again, How to approach , and maybe some "Rule of thumbs" in general in solving equations.
For example I don't multiply with "-" and all the solution goes wrong…

Thanks a lot :)
 
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  • #2
rugh said:
Hi,
Last day I learned about matrix equations like this:
AX = C
A–1AX = A–1C
IX = A–1C
X = A–1C

The objective of this is to Find X.

Here they done it by right multiplication of the Inverse matrix (A^-1).

My Questions are:

1) Where can I get more of these exercises and in higher level? (This exercise was fairly simple, but what about something like p^Tx^-1P^-1PX=APX)
Do you have some book that also shows in understandable way how to solve this?
Because when it's get harder I don't know how to approach it.
Do you have any tips? (Or good book)

2) The same question, but for the system of linear equations like this:

X - ky + z = 2 + kz
kx - 4y - 6z = k^2 + 2k - 9
2x + 4y + 8z = kx+2k(y+z-1) + 13 - k^2

Again, How to approach , and maybe some "Rule of thumbs" in general in solving equations.
For example I don't multiply with "-" and all the solution goes wrong…

Thanks a lot :)

Hey rugh and welcome to the forums.

What environment are you learning this in? Is it self-study or are you taking a course in this?

In terms of finding the inverse, there is actually more than one method. The first way to learn this is by studying Gaussian Elimination. From that you can go on to other methods, but I would suggest learning that first since that gives you an intuitive idea of how we calculate the inverse from first principles and will help you in other ways like with LU decomposition.

In terms of rearranging matrix equations, you can do either add/subtract a matrix from each side, scalar multiply both matrices on both sides, premultiply a matrix or postmultiply a matrix. Note that if you multiply one side on the left, then you have to do it on the same side on the right. For example if AXA^-1 = C then A^-1 x AXA^-1 = A^-1 x C and not C x A^-1: these are not the same thing.

To approach the 2nd question it will help if you understand Gaussian elimination. Once you understand this, you will be able to understand all these kinds of problems.
 
  • #3
rugh said:
2) The same question, but for the system of linear equations like this:

X - ky + z = 2 + kz
kx - 4y - 6z = k^2 + 2k - 9
2x + 4y + 8z = kx+2k(y+z-1) + 13 - k^2

Again, How to approach , and maybe some "Rule of thumbs" in general in solving equations.
For example I don't multiply with "-" and all the solution goes wrong…

Thanks a lot :)
If you are attempting problems like this, then presumably you know how to do things like "combining like terms". For example, if, in the first equation, x- ky+ z= 2+ kz, subtracting kz from both sides gives x- ky + (1- k)z= 2 so that we have combined the z terms. The second equation has no x, y, or z on the right so we can leave that. The third equation is the same as
2x+ 4y+ 8z= kx+ 2ky+ 2kz- 2k+ 13- k^2 (distributive law). If we subtract kx, 2ky, and 2kz from both sides, then we have (2- k)x+ (4- 2k)y+ (8- 2k)z= -2k+ 13- k^2.

That is, the three equations are
x- ky+ (1- k)z= 2
kx- 4y- 6z= k^2+ 2k- 9
(2- k)x+ (4- 2k)y+ (8- 2k)z= -2k+ 13- k^2.

We can write that as the matrix equation
[tex]\begin{bmatrix}1 & -k & 1- k \\ k & -4 & -6 \\ 2- k & 4- 2k & 4- 2k\end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix}2 \\ k^2+ 2k- 9\\ -2k+ 13- k^2\end{bmatrix}[/tex]

Now you can think of that as "Ax= b" where A is the matrix
[tex]\begin{bmatrix}1 & -k & 1- k \\ k & -4 & -6 \\ 2- k & 4- 2k & 4- 2k\end{bmatrix}[/tex]
and the solution is, as you say, [itex]x= A^{-1}b[/itex].

Of course, whether the matrix A has an inverse and what it is depends upon k.
 
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  • #4
Thank all of you for your replay,
I know the Gauss Elimination, I just searching for more exercises and ways of solving the equations I wrote in the 2 Sections.
I just want more exercises, are there any recommended books or web pages?

10X
 
  • #5


Hi there,

I can understand your interest in learning more about matrix equations and how to solve them. Matrix equations are used in many areas of science, such as physics, engineering, and computer science, so it's important to have a good understanding of them.

To answer your first question, there are many resources available for practicing and learning about matrix equations. Some good books that explain the concepts and provide exercises at different levels include "Linear Algebra and Its Applications" by David C. Lay and "Matrix Computations" by Gene H. Golub and Charles F. Van Loan. You can also find online resources, such as Khan Academy, that offer interactive lessons and practice problems.

When it comes to approaching harder matrix equations, it's important to have a good grasp of the fundamentals and to practice regularly. One tip is to break the problem down into smaller steps and use the properties of matrix operations to simplify the equations. Also, make sure to review the properties of inverse matrices, as they are crucial in solving these types of equations.

For your second question, the system of linear equations you provided can also be solved using matrix methods. One approach is to represent the equations in matrix form, where the coefficients of each variable are arranged in a matrix. Then, you can use techniques such as Gaussian elimination or Cramer's rule to solve for the variables. As for the "rule of thumbs" in solving equations, it's important to always double check your work and make sure you are applying the correct operations to both sides of the equation.

I hope this helps and good luck in your studies!
 

FAQ: Solving Matrix equstions of the type: AX = C (FIND The X) - (NOT HW)

What is a matrix equation?

A matrix equation is a mathematical equation in which matrices are used to represent the relationships between variables. It is represented in the form AX = C, where A and C are matrices and X is the unknown variable.

How do you solve a matrix equation?

To solve a matrix equation, you need to use the properties and operations of matrices. First, you need to isolate the variable X by using inverse operations. Then, you can use the inverse of A to solve for X.

What is the inverse of a matrix?

The inverse of a matrix is a matrix that when multiplied by the original matrix, results in an identity matrix. It is denoted by A-1 and is used to solve matrix equations.

Can all matrix equations be solved?

No, not all matrix equations can be solved. A matrix equation is only solvable if the matrix A is invertible, which means it has a non-zero determinant. If the determinant of A is zero, then the matrix equation cannot be solved.

What is the importance of solving matrix equations?

Solving matrix equations is important in many areas of science, such as physics, engineering, and economics. It allows us to model and solve complex systems and make predictions about the behavior of these systems. It is also used in data analysis and machine learning.

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