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snow_lady [41]
3 years ago
7

13. Determine whether B = {(-1, 1,-1), (1, 0, 2), (1, 1, 0)} is a basis of R3.

Mathematics
1 answer:
Sholpan [36]3 years ago
5 0

Answer:  Yes, the given set of vectors is a basis of R³.

Step-by-step explanation:  We are given to determine whether the following set of three vectors in R³ is a basis of R³ or not :

B = {(-1, 1,-1), (1, 0, 2), (1, 1, 0)} .

For a set to be a basis of R³, the following two conditions must be fulfilled :

(i) The set should contain three vectors, equal to the dimension of R³

and

(ii) the three vectors must be linearly independent.

The first condition is already fulfilled since we have three vectors in set B.

Now, to check the independence, we will find the determinant formed by theses three vectors as rows.

If the value of the determinant is non zero, then the vectors are linearly independent.

The value of the determinant can be found as follows :

D\\\\\\=\begin{vmatrix} -1& 1 & -1\\ 1 & 0 & 2\\ 1 & 1 & 0\end{vmatrix}\\\\\\=-1(0\times0-2\times1)+1(2\times1-1\times0)-1(1\times1-0\times1)\\\\=(-1)\times(-2)+1\times2-1\times1\\\\=2+2-1\\\\=3\neq 0.

Therefore, the determinant is not equal to 0 and so the given set of vectors is linearly independent.

Thus, the given set is a basis of R³.

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Answer: 51/8 inches

Step-by-step explanation:

Since you must cut down the length of copper tubbing from 15 1/8 inches to 8 3/4 inches, the leftover piece will be:

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The LCM will be 8

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Step-by-step explanation:

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Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

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4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

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Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

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4(b)

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