Examples of infinite matrix in the following topics:
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- It is possible to solve this system using the elimination or substitution method, but it is also possible to do it with a matrix operation.
- Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: $X$ is the matrix representing the variables of the system, and $B$ is the matrix representing the constants.
- Using matrix multiplication, we may define a system of equations with the same number of equations as variables as:
- To solve a system of linear equations using an inverse matrix, let $A$ be the coefficient matrix, let $X$ be the variable matrix, and let $B$ be the constant matrix.
- If the coefficient matrix is not invertible, the system could be inconsistent and have no solution, or be dependent and have infinitely many solutions.
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- $\displaystyle{\begin{matrix} \lim\limits_{x \to p} & (f(x) + g(x)) & = & \lim\limits_{x \to p} f(x) + \lim\limits_{x \to p} g(x) \\ \lim\limits_{x \to p} & (f(x) - g(x)) & = & \lim\limits_{x \to p} f(x) - \lim\limits_{x \to p} g(x) \\ \lim\limits_{x \to p} & (f(x)\cdot g(x)) & = & \lim\limits_{x \to p} f(x) \cdot \lim\limits_{x \to p} g(x) \\ \lim\limits_{x \to p} & \left ( \frac{f(x)}{g(x)} \right ) & = & \frac{\lim\limits_{x \to p} f(x)} {\lim\limits_{x \to p} g(x)} \end{matrix}}$
- These rules are also valid for one-sided limits, for the case $p = \pm$, and also for infinite limits using the following rules:
- $\displaystyle{\begin{matrix} &q + \infty &=& \infty \text{ for } q \neq - \infty \\ &q \cdot \infty &=& \infty \text{ if } q > 0 \\ &q \cdot \infty &=& -\infty \text{ if } q < 0 \\& \frac{q}{\infty} &=& 0 \text{ if } q \neq \pm \infty \end{matrix}}$
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- Solve a system of equations in three variables, differentiating between systems that have no solutions and ones that have infinitely many solutions
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- Dependent systems have an infinite number of solutions.
- An infinite number of
solutions can result from several situations.
- $\left\{\begin{matrix}
\begin {aligned} 2x + y - 3z &= 0 \\
4x + 2y - 6z &= 0 \\
x - y + z &= 0
\end {aligned}
\end{matrix} \right.$
- $\left\{\begin{matrix}
\begin {aligned} x - 3y + z &= 4\\
-x + 2y - 5z &= 3 \\
5x - 13y + 13z &= 8
\end {aligned}
\end{matrix} \right.$
- $\left\{\begin{matrix}
\begin {aligned} -y - 4z &= 7 \\
2y + 8z &= -12
\end {aligned}
\end {matrix} \right.$
-
- $egin{matrix} x-2y &= &-1\ 3x+5y &= &8\ 4x+3y &=& 7 nd{matrix} $
- $egin{matrix} x+y &= &1\ 2x+y &= &1\ 3x+2y &=& 3 nd{matrix} $
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- A left inverse of a matrix $A\in \mathbf{R}^{n \times m}$ is defined to be a matrix $B$ such that
- You can readily verify that any matrix of the form
- But there are infinitely many other left inverses.
- Here you can readily verify that any matrix of the form
- So there is an infinite set of solutions $x_3$ .
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- $\left[\begin{matrix} 7x & =&7z\\ 8x+y & =&5z+2w\\ y & = & 3z\\ 3y &= & 6z+w \end{matrix} \right].$
- $\left[ {\begin{matrix} 8z+y & =&5z+2w\\ y & = & 3z\\ 3y &= & 6z+w \end{matrix} } \right]$
- Since these two equations are equivalent, we have an infinite number of solutions.
- So we'll simply choose one solution, and know that there are infinitely many multiples of this one solution.
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- where $X$ is the matrix whose columns are the $\mathbf{x}_i$$\mathbf{c}$ vectors and $\mathbf{c}$ is the vector of unknown expansion coefficients.
- As you well know, matrix equation has a unique solution $\mathbf{c}$ if and only if the $\mathbf{x}_i$ are linearly independent.
- We have emphasized throughout this course that functions are vectors too, they just happen to live in an infinite dimensional vector space (for instance, the space of square integrable functions).
- In general, the sum will require an infinite number of coefficients $c_i$ , since a function has an infinite amount of information.
- So, let us define a dot (or inner) product for functions on an interval $[-l,l]$ (this could be an infinite interval)
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- First let's look at the $2\times 2$ matrix formulation of the problem:
- (For a 2 by 2 matrix $\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$ the determinant is . ) In our case the matrix whose determinant must be zero is which equals
- Remember, $\omega_0 ^2$ is fixed, as are the elements of the matrix $T$ .
- This is a rather unusual thing for a matrix to do.
- The only way to make the high frequency modes be purely sinusoidal is to let there be a continuously infinite number of springs and masses.
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- Infinite sequences and series can either converge or diverge.
- If the limit of is infinite or does not exist, the series is said to diverge.
- An easy way that an infinite series can converge is if all the $a_{n}$ are zero for sufficiently large $n$s.
- Such a series can be identified with a finite sum, so it is only infinite in a trivial sense.
- An infinite sequence of real numbers shown in blue dots.
