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2 years ago

What is the determinant of the coefficient matrix of the system –11 –2 0 55

Mathematics

2 answers:

Alina [70]2 years ago

7 0

The first two rows of coefficients are identical, so by inspection, the determinant is 0.

Mademuasel [1]2 years ago

7 0

Answer:

The determinant of coefficient matrix of the given system is 0.

Step-by-step explanation:

Given system of equation

$-x-y-z=-3$

$-x-y-z=8$

$3x+2y+z=0$

The coefficient matrix of the system

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

Let A= $\left[\begin{array}{ccc}-1&-1&-1\\-1&-1&-1\\3&2&1\end{array}\right]$

X= $\left[\begin{array}{}x&y&z\end{array}\right]$

B= $\left[\begin{array}-3 &8&0\end{array}\right]$

Therefore , we can AX=B

$\left[\begin{array}{ccc}-1&-1&-1\\1&-1&-1\\3&2&1\end{array}\right]$ $\left[\begin{array}{}x&y&z \end{array}\right]$ = $\left[\begin{array}{}-3&8&0\end{array}\right]$

The determinant of coefficient matrix of the given system is given by

$\begin{vmatrix}-1&-1&-1\\-1&-1&-1\\3&2&1\end{vmatrix}$

By using determinant property : when two rowsor two columns are identical then the value of determinant is equal to zero .

$\therefore \begin{vmatrix}A\end{vmatrix}=0$

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4 0

1 year ago

Read 2 more answers

It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the

JulijaS [17]

Answer:

1) n=48

2) n=298

3) n=426

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63$

And rounded up we have that n=48

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7$

And rounded up we have that n=298

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: