Matrix Linear Systems Equation

• zack7
In summary, the conversation discusses a homework problem where the goal is to find the solution to the equation $\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$, where c is a scalar. The solution involves calculating and reducing the matrix A - I, which is the 2 x 2 identity matrix. The conversation includes some mistakes in calculation, but ultimately arrives at the solution $\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$ or $\vec {x} = x1\begin{pmatrix} 1 \\ 0\end{pmatrix}$, where c or x1 is an arbitrary number.
zack7

Homework Statement

Question in the image.

Homework Equations

I got a but can seem to get the answer for b which is $$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$

where for any scalar c

The Attempt at a Solution

I managed to get it to $$\vec{x}(A-1)=0$$

$$= \begin{pmatrix} 0 3 \\ -1 0\end{pmatrix}=\begin{pmatrix} 1 0 \\ 0 1\end{pmatrix}$$
Then I reduced the matrix to reduced row echolon form
but after that I seem not to get the answer.

Thank you

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zack7 said:

Homework Statement

Question in the image.

Homework Equations

I got a but can seem to get the answer for b which is $$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$

where for any scalar c

The Attempt at a Solution

I managed to get it to $$\vec{x}(A-1)=0 You made a mistake in how you calculated A - I. (It's not A - 1. IOW, you don't subtract 1 from each entry of A.) zack7 said: = \begin{pmatrix} 0 3 \\ -1 0\end{pmatrix}$$=\begin{pmatrix} 1 0 \\ 0 1\end{pmatrix}$$Then I reduced the matrix to reduced row echolon form but after that I seem not to get the answer. Thank you Mark44 said: You made a mistake in how you calculated A - I. (It's not A - 1. IOW, you don't subtract 1 from each entry of A.) I get (A-I), which would then be$$\begin{pmatrix} 0 3 \\ 0 0\end{pmatrix}$$but this would mean x1 and x2 = 0 Last edited by a moderator: zack7 said: but if I factorize the x out I get (A-1), how would I simplfy that since it is not possible to subtract? No, you don't get A - 1. Here's the way it goes: Ax = x => Ax - x = 0 => (A - I)x = 0 In the last equation, that is A - I, the 2 x 2 identity matrix, NOT the matrix whose entries are all 1's. Calculate A - I, and row reduce it to find the solution set. Mark44 said: No, you don't get A - 1. Here's the way it goes: Ax = x => Ax - x = 0 => (A - I)x = 0 In the last equation, that is A - I, the 2 x 2 identity matrix, NOT the matrix whose entries are all 1's. Calculate A - I, and row reduce it to find the solution set. Thanks, I understand that part,careless mistake but when I reduce it I get x1 and x2 = 0 zack7 said: I get (A-I), which would then be \begin{pmatrix} 0 3 \\ 0 0\end{pmatrix}$$
but this would mean x1 and x2 = 0
No. The top row says 0x1 + 3x2 = 0, so x1 is arbitrary.

Mark44 said:
No. The top row says 0x1 + 3x2 = 0, so x1 is arbitrary.

then why is the answer $$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$
$$\vec {x} = x1\begin{pmatrix} 1 \\ 0\end{pmatrix}$$

Last edited by a moderator:
Fixed your LaTeX. You need to put the [noparse]$$[/noparse] thing at the end of your expression, too. zack7 said: then why is the answer$$\vec{x} = c\begin{pmatrix} 1 \\ 0\end{pmatrix}$$isn't it then$$\vec{x} = x1\begin{pmatrix} 1 \\ 0\end{pmatrix}
Those are pretty much the same thing. In the first, c is an arbitrary number; in the second, x1 is an arbitrary number.

1. What is a matrix linear systems equation?

A matrix linear systems equation is a system of equations where the unknown variables are represented in a matrix form. It can be solved using various methods, such as Gaussian elimination or matrix inversion.

2. How do you solve a matrix linear systems equation?

To solve a matrix linear systems equation, you can use methods such as Gaussian elimination, where you manipulate the equations to reduce the system to a triangular form, or matrix inversion, where you use the inverse of the coefficient matrix to solve for the unknown variables.

3. What is the difference between a matrix linear systems equation and a regular linear equation?

The main difference between a matrix linear systems equation and a regular linear equation is that in a matrix equation, the unknown variables are represented in a matrix form, while in a regular linear equation, the unknown variables are represented as constants or variables.

4. What are some real-world applications of matrix linear systems equations?

Matrix linear systems equations have many real-world applications, such as in engineering for solving systems of equations in circuit analysis, in economics for solving optimization problems, and in computer graphics for transformations and rotations.

5. Can a matrix linear systems equation have more than one solution?

Yes, a matrix linear systems equation can have multiple solutions. This occurs when the system of equations is consistent and has either an infinite number of solutions or a unique solution. If the system is inconsistent, there will be no solutions.

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