Ok so if a system has no solutions, that means that no value will make it true

infinite solutions means that all numbers wil make the system true

example of no solution

a simplification that gives you
3=0
that is false so no solution

example of infinite
if after simplify you get
x=x
you can put all numer for x and get right answer

the simplified form of an equation with no solution will be in the form a=b where a is not equal to b, basically it gives you an obviously false statement like 3=4

simplified form on an equation with infinity solutions would end up with variable=variable or x=x or y=y

4 0

3 years ago

Which would be be the right answer??????????????????

Serjik [45]

Answer:

so many question marks i like trains?

4 0

3 years ago

Consider the following two ordered bases of R3:

grigory [225]

Answer:

Let $A = (a_1, ..., a_n)$ and $B = (b_1, ..., b_n)$ bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors $b_1, ..., b_n$ at base A.

The, we case correspond to find the coordinates of vectors of C,

$\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] \}$

at base B.

1. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

$\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&-1\\0&-1&1&-1\end{array}\right]$

Using rows operation we obtain the echelon form of the matrix

$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right]$

now we use backward substitution

$c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\2\\1\end{array}\right]$

2. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

$\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&0\\0&-1&1&-1\end{array}\right]$

Using rows operation we obtain the echelon form of the matrix

$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right]$

now we use backward substitution $c=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\3\\2\end{array}\right]$

3. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$