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3 years ago

Exercise 2.1.47 Prove that ifA^-1 exists and AX=0 then X =0.

Mathematics

1 answer:

stepan [7]3 years ago

5 0

Answer:

Step-by-step explanation:

If $A^{-1}$ exist, then we have that $A^{-1}\cdot A=1$ .

Therefore, in the case that $AX=0$ , if we multiply both

terms in the equation by $A^{-1}$ , then we have

$A \cdot A^{-1}X=A^{-1}0$

and so the equation tells us that

$X=A\cdot A^{-1}X=A^{-1}0=0$ .

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6 soccer balls for $150

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Which of the following square root of -80

Marta_Voda [28]

You can factor -80 as

$-80 = (-1)\cdot 16 \cdot 5$

So, we have

$\sqrt{-80} = \sqrt{(-1)\cdot 16 \cdot 5}$

The square root of a product is the product of the square roots:

$\sqrt{-80} = \sqrt{(-1)}\sqrt{16}\sqrt{5}$

Since $i^2=-1$ and $4^2=16$ , we have

$\sqrt{-80} = 4i\sqrt{5}$

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3 years ago

How many 6 digit different locker combinations are possible if no repeat numbers are allowed?

Alex_Xolod [135]

Since we are trying to find the number of sequences can be made <em>without repetition</em>, we are going to use a combination.

The formula for combinations is:

$_n C _k = \dfrac{n!}{k! (n - k)!}$

$n$ is the total number of elements in the set
$k$ is the number of those elements you are desiring

Since there are 10 total digits, $n = 10$ in this scenario. Since we are choosing 6 digits of the 10 for our sequence, $k = 6$ in this scenario. Thus, we are trying to find $_{10} C _6$ . This can be found as shown:

$_{10} C _6 = \dfrac{10!}{6! \cdot 4!} = \dfrac{10 \cdot 9 \cdot 8 \cdot 7}{4!} = \dfrac{5040}{24} = 210$

There are 210 total combinations.

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3 years ago

How does technology affect the economy

lys-0071 [83]

Answer:

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2 years ago

Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.

Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

$\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]$

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

$\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

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

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

$\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

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

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

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3 years ago