1. Given the set of vectors. < b m a t r i x

1. Given the set of vectors. \begin{Bmatrix} \begin{bmatrix} 1 \\ -5 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ -7 \\ 6 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -4 \\ 3 \\ 0 \end{bmatrix} \end{Bmatrix} a. Is the set of vectors linearly independent? Justify your answer. b. Is the set of vectors a basis for $\mathbb{R}^3$? Justify your answer. c. Does the set span $\mathbb{R}^3$? Justify your answer. Hint: think about rows and pivots.

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00:01 In this question we are given a set of vectors u and w and we need to check if this set is a linearly independent set orthogonal set or and also if it spans the vector space r3 so first of all we we are going to check if this set is linearly independent or not so clearly it can be seen that this this vector u is equal to one two three minus three and if we multiply this whole vector by minus 2 we will get minus 2 minus 4 and 6 which is nothing but equal to v that is minus 2 times of u vectors is equal to v vector so this is clearly seen that this u and v are dependent so this the the set which is given to us of u v and w is not linearly independent is it is linearly dependent now to check if this set is orthogonal to check orthogonal just take the take the dot product of these vectors and if it comes equal to zero then it is orthogonal so let us check the first vector is one two minus three the second vector is minus two minus four and six we need to check if their dot product is equal to zero so let us check the broad product of these two vectors will be minus 2 plus 2 multiplied by minus 4 will give minus 4 and minus 3 multiplied by 6 will give minus 18 so this is clearly not equal to 0 so these 2 vectors are not orthogonal so we can say that this set is not orthogonal vectors and now we need to check if this set span r 3 so for this we need to check if this if the linear combination of them is equal to any vector of 3 and then we will solve the linear system of equations and check if the solution is consistent that is if the solution exists then this can span the vector space r3 so let us take if we multiply this by x and this multiplied by y and this multiplied by z and put it equal to a vector of r 3 a bc that is x multiplied by 1 2 minus 3 plus y minus 2 minus 4 6 plus z 3 6 9 equal to a b c so we will get the linear system and if we put it into the matrix form we will get 1 2 minus 3 minus 2 minus 4 1 2 minus 3 1 2 minus 3 and this is 6 and this is 3 6 9 and the augmented matrix for this will be a b b now we will check if this is if this can be put in a colon form or not so for this let us make a transformation r2 goes to r2 minus 2 r1 and r3 goes to r3 plus 3 times of r1 so what we will get 1 2 3 a and here we will get 0 minus 4 it is minus 2 minus 4 plus 4 is equal to 0 and 6 minus 6 will be 0 again and here minus 3 plus 3 will give 0 and here is 6 minus 6 will be give 0 and 9 minus 9 plus 3 times of 3 is 9 it will give 18...

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