4x-3y=1 -8x+6y=-2 elimination method

Suppose A is n x n matrix and the equation Ax = 0 has only the trivial solution. Explain why A has n pivot columns and A is row

vovangra [49]

Answer:

Remember, a homogeneous system always is consistent. Then we can reason with the rank of the matrix.

If the system Ax=0 has only the trivial solution that's mean that the echelon form of A hasn't free variables, therefore each column of the matrix has a pivot.

Since each column has a pivot then we can form the reduced echelon form of the A, and leave each pivot as 1 and the others components of the column will be zero. This means that the reduced echelon form of A is the identity matrix and so on A is row equivalent to identity matrix.

4 0

3 years ago

Which is the graph of the inequality y ≥ 2x + 3

Charra [1.4K]

It's B, remember when you have _ ( referring to =) you have straight line, but when it's just < or > you have dotted line

7 0

2 years ago

A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha

8_murik_8 [283]

To model this situation, we are going to use the decay formula: $A=Pe^{rt}$
where
$A$ is the final pupolation
$P$ is the initial population
$e$ is the Euler's constant
$r$ is the decay rate
$t$ is the time in years

A. We know for our problem that the initial population is 1,250, so $P=1250$ ; we also know that after a year the population is 1000, so $A=1000$ and $t=1$ . Lets replace those values in our formula to find $r$ :
$A=Pe^{rt}$
$1000=1250e^{r}$
$e^{r}= \frac{1000}{1250}$
$e^{r}= \frac{4}{5}$
$ln(e^{r})=ln( \frac{4}{5} )$
$r=ln( \frac{4}{5} )$
$r=-02231$

Now that we have $r$ , we can write a function to model this scenario:
$A(t)=1250e^{-0.2231t}$ .

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C.
- The function is decreasing
- The function doe snot have a x-intercept
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0:
$A(0)=1250e^{(-0.2231)(0)$
$A_{0}=1250e^{0}$
$A_{0}=1250$
- Over the interval [0,10], the function will have a minimum at t=10:
$A(10)=1250e^{(-0.2231)(10)$
$A_{10}=134.28$

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: $m= \frac{A(0)-A(10)}{10-0}$
where
$m$ is the rate of change
$A(10)$ is the function evaluated at 10
$A(0)$ is the function evaluated at 0
We know from previous calculations that $A(10)=134.28$ and $A(0)=1250$ , so lets replace those values in our formula to find $m$ :
$m= \frac{134.28-1250}{10-0}$
$m= \frac{-1115.72}{10}$
$m=-111.572$
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.